### Math 152 diary, spring 2009: second section Later materialPrevious material In reverse order: the most recent material is first.

Monday, April 13 (Lecture #21)
We'll have an exam next Monday, April 20. Here is some review material. Here is an outline of (possibly!) useful approaches to sequence/series problems. On Thursday, April 16, students will work on the course coordinator's review problems. I will not be available this Thursday and Friday (out of town for a professional trip) but I will have a review session for the second exam on Sunday, April 19, from 5 to 7 PM. The location for this is SEC 117.

Why the Root Test?
The Root Test is another method for exploiting similarity with geometric series to diagnose absolute convergence (or divergence) of a series. We consider a series ∑an. Suppose we think that an "resembles" crn. Well, if we take the nth root of crn, we get (crn)1/n=c1/n(rn)1/n=c1/nrn·(1/n)=c1/nr. Now if n→∞, we have already seen that a sequence like {c1/n} has limit 1. So (crn)1/n→r as n→∞. So we can hope that the asymptotic behavior of (an)1/n as n→∞ can help analyze the convergence of ∑an.

Statement of the Root Test

The Root Test
Consider the series ∑n=0an, and the limit limn→∞|an|1/n. Suppose that this limit exists, and call its value, L. (L is what's used in our textbook.)
 If L<1, the series converges absolutely and therefore must converge. If L>1, the series diverges. If L=1, the Root Test supplies no information

Ludicrous example
Let's consider the series ∑n=1((5n+7)/(6n+8))n which I invented specifically to use the Root Test. I don't know any application where I've ever seen anything like this series which seems fairly silly to me. Well, anyway, the terms are all positive, so I can "forget" the absolute value signs. We take the nth root and remember that repeated exponentiations multiply:
[((5n+7)/(6n+8))n]1/n=((5n+7)/(6n+8))n·(1/n)=((5n+7)/(6n+8))1=((5n+7)/(6n+8)).
Now we need to discover the limit (if it exists!) of ((5n+7)/(6n+8)). But (L'Hôpital) this limit is 5/6. Since L=5/6<1, the series converges absolutely and must converge.
I don't know what the sum is. Oh well.

Less silly (maybe) example
This may look almost as ludicrous, but it turns out to be more significant. Again, though, this example is chosen to work well with the Root Test.
For which x's does the series ∑n=1nnxn converge?

The powers of n signal to me that probably I should try the Root Test. Here an is nnxn. We can't just discard the absolute value here, but we can push it through to the x because everything is multiplied. So:
|nnxn|1/n=(nn|x|n)1/n=nn·(1/n)|x|n·(1/n)=n|x|.
As I mentioned in class, as n→∞ jumping to the "conclusion" may be unwise. There are actually two cases. If x=0, the limit is 0, If x≠0, the limit does not exist (it is "∞"). So we can conclude that the series ∑n=1nnxn converges exactly when x=0.

Even less silly example
Let's try this: for which x's does ∑n=1xn/nn converge? I hope you will see the resemblance and contrast with the previous computation:
|xn/nn|1/n=(|x|n/nn)1/n=|x|n·(1/n)nn·(1/n)=|x|/n.
In this example, there aren't any special cases. For any x, limn→∞|x|/n=0=L. Since L<1 always, the series ∑n=1xn/nn converges absolutely for all x's and therefore converges for all x's.

Comment: Root vs. Ratio
As I mentioned in class, I have an emotional preference for the Ratio Test that I can't explain. But I will admit that analyzing the two previous examples with the Ratio Test would be very difficult. However, the Ratio Test works exceptionally well when series have factorials (you'll see why there are lots of series with factorials in the next lecture). So series with similar results to the two previous examples which I'd examine with the Ratio Test would be ∑n=0n!xn and ∑n=0xn/n!.

The next few examples were tedious to do in class, and I thank students for the patience they mostly displayed, since the reasons for doing them were not at all clear. I also made some dumb mistakes some of the time.

Example 76
For which x's does ∑n=1xn/n converge? We used the Ratio Test, and |an+1/an| simplified fairly easily to |x|[(n+1)/n]. Now L'H or simple algebraic manipulation shows that ρ=|x|. So we get guaranteed absolute convergence and therefore convergence when |x|<1 and divergence when |x|>1. For |x|=1, we don't get any information. I'll write the answer using interval notation now: if x is in (–1,1), the series converges. If x is in (1,∞), the series diverges. If x is in (–∞,–1), the series diverges. There's no information for x=1 or x=–1.

If you insist on knowing what happens at x=+/–1, let's "insert" these values of x into the series and investigate.
If x=+1, the series becomes ∑n=11n/n=∑n=11/n, the harmonic series. So the series diverges.
If x=–1, the series becomes ∑n=1(–1)n/n. This is (almost) the alternating harmonic series (it is off by a sign). So the series converges.
These results are reflected in the "improved" picture to the right.

Example 77
For which x's does ∑n=1xn/n2 converge? Again, the Ratio Test, and |an+1/an| simplified fairly easily to |x|[n/(n+1)]. And again manipulation shows that ρ=|x|. So we have absolute convergence and therefore convergence when |x|<1 and divergence when |x|>1. For |x|=1, we don't get any information. As intervals: if x is in (–1,1), the series converges. If x is in (1,∞), the series diverges. If x is in (–∞,–1), the series diverges. There's no information for x=1 or x=–1. (Looks a lot the same, huh?)

To see what happens at x=+/–1, put these values of x into the series and investigate the result directly (I don't know any other ways to do this).
If x=+1, the series becomes ∑n=11n/n2=∑n=11/n2. This is a p-series with p=2>1, so it converges.
If x=–1, the series becomes ∑n=1(–1)n/n2. But this series converges absolutely (it gives us the p-series just considered when the signs are stripped off) and therefore it must converve.
And again, look at the "improved" picture to the right.

Example 78
For which x's does ∑n=1n xn converge? The same limiting ratio is reported, and we get the same convergence/divergence/no information result, with the same initial picture.

When x=1, the series is ∑n=1n=1+2+3+4+... and this certainly diverges because the terms don't approach 0. The same reason shows that the series diverges when x=–1. So the result, as shown, is again slightly changed.

The reason for my going through all of these examples is that there basically aren't any others. Well, what I mean is that, qualitatively, there are no further types of behavior possible for this sort of series. So let me tell you the accepted vocabulary for what we are studying, and then describe the result.

What is a power series?
A power series centered at x0 (a fixed number) is an infinite series of the form ∑n=0cn(x–x0)n where the x is a variable and the cn are some collection of coefficients. It is sort of like an infinite degree polynomial. Usually I (and most people) like to take x0 to be 0 because this just makes thinking easier.

Convergence and divergence of power series
It turns out that the collection of examples we looked is a complete qualitative catalog of what can happen to the convergence and divergence of a power series. This isn't obvious, but it also isn't totally hard (it just involves comparing series to geometric series and needs no theoretical equipment beyond what we've already done). Here is the result: A power series centered at x0 always has an interval of convergence with the center of that interval equal to x0. Inside the interval of convergence, the power series converges absolutely and therefore converges. Outside the interval, the power series diverges. The power series may or may not converge on the two boundary points of the interval. The interval may have any length between 0 and ∞. Half the length of the interval is called the radius of convergence.

Going back to the examples
The seriesconverges
for x in
vergence equal to
Pictures: convergence in red and divergence in green
n=1nnxn[0]0
n=1xn/nn(–∞,+∞)
n=1xn/n[–1,1)1
n=1xn/n2[–1,1]1
n=1n xn(–1,1)1

These examples show that the interval of convergence of a power series may or may not include one or both of the endpoints of the interval. The reason for the number of examples was to show you, explicitly, that it is possible for the series to converge on neither or one or both of the boundary points. I wanted to show a "complete" collection of examples.
It turns out that behavior on the edge of the interval is probably only interesting (sigh) as an exam question in calc 2 (where it is frequently asked!) because of some results you'll be told about in a few lines. I mentioned that I could give a supporting argument for this result using nothing but geometric series and comparison (techniques on the level of this course) but we just don't have enough time!

A suspiciously simple question ... (the "IQ" test in class)
Suppose that you have a power series ∑n=0an (x–5)n centered at x0=5. You are told that the series converges when x=–1 and diverges when x=14. What can you say about the radius of convergence? For which x's must this series converge and for which x's must this series diverge? You are given no other information.

Answer The general theory, quoted above, states that a power series converges in an interval, and the center of the series, here x0=5, must be in the center of the interval. If the series converges at x=–1, then, since the distance from –1 to 5 is 6, the series must (by the general theory) converge at every number x whose distance to 5 is less than 6. I think to myself that "convergence spreads inward". What about divergence? Actually, "divergence spreads outward." The distance from 5 to 14, where we're told that the series diverges, is 9. Therefore any x whose distance to 5 is greater than 9 (left or right!) must be a place where the series diverges (because if it converged then the series would converge at 14, also, by the contagious (?) nature of convergence, and this isn't true).

What we can conclude from this information is the following:

• The series must converge at least in the interval [–1,11).
• The series must diverge at least in the interval (–∞,–4) and in the interval [14,∞).
• We can't conclude anything about the convergence of the series in the intervals [–4,–1) and [11,14).
• The radius of convergence of the series is a number R with 6≤R≤9. We don't know more than that.

I hope you note that if I had told you this information:
The series, centered at 5, diverged at –1 and converged at 14.
Then I would be lying ("I misspoke."). There is no such series. Convergence at 14 with center at 5 would immediately imply that the series converged at –1.

Calculus with power series
So I've said again and again in class that I'm never going to add up infinitely many numbers, and that the notion of infinite series is a short cut for the limit of the sequence of partial sums. All of this is true, but the real reason that people use infinite series with great energy and enthusiasm includes the following results about power series:

Hypothesis Suppose the power series ∑n=0an (x–x0)n has some positive radius of convergence, R, and suppose that f(x) is the sum of this series inside its radius of convergence.

Differentiation The series ∑n=0n an (x–x0)n–1 has radius of convergence R, and for the x's where that series converges, the function f(x) can be differentiated, and f´(x) is equal to the sum of that series.

Integration The series ∑n=0[an/(n+1)] (x–x0)n+1 has radius of convergence R, and for the x's where that series converges, the sum of that series is equal to an indefinite integral of f(x), that is ∫f(x)dx.

These results are not obvious at all, and they take some effort to verify, even in more advanced math courses. The results declare that for calculus purposes, a power series inside its radius of convergence can be treated just like a polynomial of infinite degree. You just differentiate and integrate the terms and the sums are the derivative and antiderivative of the original sum function.

Please note that other kinds of series many of you will likely see in applications later (such as Fourier series or wavelet series) do not behave as simply with calculus.

I wanted to show a use of this result. What follows is actually a standard useful application, although both the question and the solution technique may look silly to you now.

What is the value of ∑n=1n2/5n?
To compute ∑n=1n2/5n, look carefully at the individual terms. We're adding up n2/5n=(n2)(1/5n). In the context of sequences and series, the (1/5n) part might make me think of ∑n=1xn, a geometric series, with x=1/3. Generally, this is a geometric series with the first term, c, equal to x, and the ratio between successive terms, r, also equal to x. Well, if f(x)=∑n=1xn, then the series converges for |x|<1 and with f(x)=x/(1–x).

How can we get an n in the front of the series terms? Well, changing xn to n·something may push you (I hope!) to differentiate. So f´(x) is ∑n=1nxn–1. I also get the sum by differentiating x/(1–x). The Quotient Rule gives [1(1–x)–(–1)x]/(1–x)2=1/(1–x)2. So this must be ∑n=1nxn–1 when |x|<1.
Step 1 Differentiate.
But I need two n's (n2) in front and now the power is wrong. Let me fix up the power by multiplying by x, both the series and its sum:
x·∑n=1nxn–1=∑n=1nxn has sum x·(1/(1–x)2)=x/(1–x)2.
Step 2 Multiply by x.
Now I magically want another n to appear in front of the terms of ∑n=1nxn. This is just like the first step: we differentiate. The derivative of the series is ∑n=1n2xn–1. What about the sum of this series for |x|<1? We need to differentiate x/(1–x)2. (This is certainly the most irritating part of the whole exercise.) Again, the Quotient Rule:
The derivative of x/(1–x)2 is [1(1–x)2–x{2(1–x)1(–1)}]/(1–x)4 (parentheses are your friends!).
This is valid for |x|<1.
Step 3 Differentiate.
We need to fix up the power in the sum. It is n–1 and I want n. So let's multiply by x in both the series and the sum:
x·∑n=1n2xn–1=∑n=1n2xn and x·[1(1–x)2–x{2(1–x)1(–1)}]/(1–x)4 is the sum, for |x|<1.
Step 4 Multiply by x.
Now, clumsy as it looks, we have what's called a "closed form" (means just: a formula involving functions we know) for ∑n=1n2xn and we plug in x=1/5 into the formula. The answer is {1/5}·[1(1–{1/5})2–{1/5}{2(1–{1/5})1(–1)}]/(1–{1/5})4. That's the sum of the series.
Step 5 Plug in x=1/5.
This actually gives 15/32, just as I said last time. Of course, I could have asked a friend.

```> sum(n^2/5^n,n=1..infinity);
15
--
32```

QotD
Verify that 15/32 actually is the result of plugging in x=1/5 into the formula. This was the QotD for sections 1 through 3. Iwas a bit slower in the later lecture, and there I only asked (!) what the sum of nxn was, and how to use it to find ∑n=1n/5n.

I actually would have liked to give the following as the QotD but there wasn't time:
Discuss the convergence of &sumn=1(–1)nxn/sqrt(n) as thoroughly as possible. So: for which x's does this series converge? For which x's does it converge absolutely? For which x's does it converge conditionally? You can try this, and then look here for a solution.

Wednesday, April 8 (Lecture #20)
Scheduling ...
The scheduled subject for today's lecture is Ratio and Root Tests, and for the next lecture, the subject is Power Series. Today I will definitely motivate and state the Ratio Test, and probably won't get to the Root Test. I will, however, show an application to a power series. On Monday, I will continue with a statement of the Root test, and then formally define and state the principal properties of power series, and use both of the "tests" to analyze power series.

Two neat Tests for convergence
The last lecture discussed the relationship between absolute convergence and conditional convergence. Today we will begin to study the two standard "tests" which are used to diagnose absolute convergence. Both of these tests rely on some relationship with geometric series. Let me begin with an example.

n=0(50)n/n!
We met the sequence of individual terms {(50)n/n!} earlier. We showed that this specific sequence converges to 0. We did this by looking at what happens after the 100th term. Then each later sequence term is less than half the size of the term immediately before it. Eventually the terms squeeze down and their limit is 0.

But what about the series? Just knowing that the sequence of individual terms →0 is not enough information to guarantee that the series, the sum of the terms, is convergent. (That's what the harmonic series shows!) But look here:
∑n=0(50)n/n!=∑n=0100(50)n/n!+∑n=101(50)n/n!.
Let's ignore the first big lump -- I don't care how big it is. It actually doesn't influence whether the series converges or not. The convergence of the series depends on whether the infinite tail converges. Look at what can we say here:
50101/(101!)+50102/(102!)+50103/(103!)+...<50101/(101!)+50101/(101!)[1/2]+50101/(101!)[1/2]2+...
We can compare this infinite tail to a certain geometric series which is larger than it, and this geometric series must converge because it is a series with ratio 1/2 and 1/2<1.

Why -- what's happening?
The Ratio Test is a way of using the sort of logic that we considered above. It systematically checks if there is a valid comparison to a convergent or to a divergent geometric series. Here is the idea.
If we are studying a series ∑an, then we may hope that somehow an resembles crn, a term of a geometric series. We further hope that an+1 will resemble crn+1. But then if the "resemblance" is good enough, we might hope that the quotient, an+1/an, will be like (crn+1)/(crn), and this would be r. This is only a resemblance, so the textbook actually uses a different letter in its statement of the Ratio Test. And we put on absolute value signs, since whether or not a geometric series converges only depends on whether |r|, the absolute value of the ratio, is less than 1. Etc. So here is a formal statement:

The Ratio Test
Consider the series ∑n=0an, and the limit limn→∞|an+1/an|. Suppose that this limit exists, and call its value, ρ. (That's supposed to be the Greek letter rho.) Then:
 If ρ<1, the series converges absolutely and therefore must converge. If ρ>1, the series diverges. If ρ=1, the Ratio Test supplies no information

Applied to this problem
Let's see what happens when we try to apply this "Test" to the series ∑n=0(50)n/n!. Since an=(50)n/n!, the next term, an+1, will be (50)n+1/(n+1)!. Try to avoid possibly invalid shortcuts: just plug in n+1 everywhere that n appears. Then let's consider the absolute value of the ratio:

```|  50n+1|     50n+1
| ----- |    -----
| (n+1)!|    (n+1)!    (50n+1)n!     50(n!)    50(n!)    50
|-------| = ------- = ----------- = ------- = ------- = ----
|  50n  |      50n    (50n)(n+1)!    (n+1)!   (n+1)n!    n+1
| ----- |    -----
|  n!   |      n!

Step A     Step B     Step C       Step D   Step E   Step F
```
Let me try to describe sort of carefully the various steps. This is the first example, and I chose it not because it is especially difficult, but because the sequence of things to do is typical of many applications of the Ratio Test.

Step A Write |an+1/an|. I really try to avoid "premature simplification" here. That is, I try to just insert n+1 for n correctly, and then write things.
Step B In this case, the absolute value signs are not needed because everything involved is a positive number. This is not always going to be true!
Step C We have a compound fraction in Step B. I find them difficult to understand. Life is easier if we convert the compound fraction into a simple fraction, with one indicated division. So if you were in class you may have heard me mumbling, "The top of the bottom times the bottom of the bottom" which is the top of the simple fraction, and "The bottom of the top times the top of the bottom" which is the bottom of the simple fraction. O.k.: if you want, use numerator and denominator instead of top and bottom.
Step D Now I'll start the simplification. Since 50n+1=50n·50, we can cancel 50n from the top and bottom.
Step E Here is possibly the most novel part, algebraically, of this mess. We need to confront the factorials. (n+1)! is the product of all the positive integers from n+1 down to 1. Therefore it is the same as n+1 multiplied by the product of all the positive integers from n down to 1. Algebraically, this statement is the equation (n+1)!=(n+1)n!. I want to rewrite (n+1)! so that we can realize the cancellation of the n!'s.
Step F And here is the result which can be used to compute ρ.

The Ratio Test for n=0(50)n/n! leads us to consider limn→∞|an+1/an|=limn→∞50/(n+1)=0. So for this series, ρ=0. Since 0<1, the series converges absolutely and (using what we did last time) it converges.

I can identify the sum (and you will be able to also after a few more classes). It is e50. Partial sums of series of this type are exactly what your calculators use to compute values of the exponential function.

Another example
Let's consider ∑n=1n2/5n. Again, this is not a casual example. This sort of series occurs in the study of the statistical properties of certain types of component failures (it is involved with computing the standard deviation). Here an is n2/5n and an+1=(n+1)2/5n+1. So:

```| (n+1)2|    (n+1)2
| ----- |    -----
|  5n+1 |     5n+1     (n+1)25n    (n+1)2
|-------| = ------- = --------- = ------
|   n2  |      n2       n25n+1      n25
| ----- |    -----
|  5n   |     5n   ```

Well, again I just forget the absolute value signs because the terms are all positive. I rearrange from a compound fraction to a simple fraction. I cancel powers of 5. The result I need to consider is [(n+1)/n]2·(1/5). The core of this is what happens to (n+1)/n as n→∞. We can use L'Hôpital's Rule since this is an ∞/∞ case, and get 1/1, so the limit is 1. Or we can just rewrite, as some students suggested: (n+1)/n=1+(1/n), and this also →1 as n→∞. In any case, for the series ∑n=1n2/5n, we can compute limn→∞|an+1/an|= limn→∞[(n+1)/n]2·(1/5)=1/5=ρ. Since 1/5 is less than 1, the Ratio Test implies that the series converges absolutely and therefore converges.

The sum can actually be computed and it is 15/32 (really!). I will show you how to compute this in a few more classes.

 In all of these examples the terms are quotients, and essentially we are trying to compare the rates of growth of the top and the bottom. Exponentials (with a base>1) grow faster than any polynomial. For example, we could consider the infinite series ∑n=1∞n20/(1.01)n. The 20th term in this series is about 2.6·1019. That's B-I-G. Does this series converge? Well, the Ratio Test applies. If similar algebra is done, then |an+1/an| becomes [(n+1)/n]15/1.01 and, when n→∞, the limit is ρ=1/1.01 which is less than 1, so the series converges absolutely and therefore converges! I don't think this is obvious: {con|di}vergence all depends on the infinite tail -- you can't think about the "first few" terms. Here is a little more numerical information. If an=n20/(1.01)n, then a1,000=4.7·1040 (approximately and this is even bigger) and a10,000=6.1·1016 and a100,000=7.3·10–358. The last term is quite small, and the exponential growth has definitely surpassed the polynomial growth.

And another
We consider ∑n=072n/(n!)2. In this series we contrast the exponential growth on top with factorial growth on the bottom. Factorials increase faster (they are "superexponential"). In this case, some care is needed with the algebra using the Ratio Test. If an=72n/(n!)2 then an+1=72(n+1)/((n+1)!)2. Parentheses are your friends so use many of them in computations and you likely will make fewer errors!

```|  72(n+1)   |     72(n+1)
| --------  |    --------
| ((n+1)!)2 |    ((n+1)!)2    72(n+1)(n!)2
|-----------| = ---------- = ------------
|   72n     |       72n       72n((n+1)!)2
| -------   |    -------
|  (n!)2    |      (n!)2    ```
But 72(n+1)=72n+2=72n72 and so part of that cancels with 72n. Analysis of the factorials can be more confusing, but here it is:
((n+1)!)2=((n+1)n!)2=(n+1)2(n!)2
So part of that is canceled by the (n!)2. Therefore we need to compute limn→∞|an+1/an|=limn→∞72/(n+1)2=0=ρ. Since ρ<1, the series we are considering converges absolutely and therefore converges.

Again, this is not a "random" series. The sum of ∑n=072n/(n!)2 is close to a value of a Bessel function, J0(14). The series for J0(14) is actually ∑n=0(–1)n72n/(n!)2. It has an alternating sign, also. One simple place such functions occur is in the description of vibrations of circular drums (really!). The series with alternating signs must also converge, since we just verified that the series without signs converges since absolute convergence implies convergence.

Textbook example
The textbook analyzes (Example 4, page 591, section 10.5) what the Ratio Test tells about the series ∑n=11/n and the series ∑n=11/n2. Please see the textbook (I did this in detail in class). In both cases the resulting value of ρ is 1. Notice that one series (the harmonic series, a p-series with p=1) diverges and the other series (a p-series with p=2>1) converges. So truly if ρ=1 you can conclude nothing about the convergence or divergence of the original series.

It is certainly possible to have series where the limit of |an+1/an| doesn't even exist, so there isn't even any ρ to consider. I don't want to give such an example right now, but you should know that things can be very strange.

For which x's does ...
Here's the question. For which x's does the series ∑n=1xn/(3n+n) converge?

The Ratio Test does work here, but we need to be careful. First, the bottom is more complicated. And second, certainly the signs of the terms will vary because x can be negative.

|A·B|=|A|·|B| but |A|+|B| and |A+B| are not the same if the signs of A and B differ.
Look: |(–3)7|=|–21|=21 and |–3|·|7|=3·7=21 but |(–3)+7|=|4|=4 and |–3|+|7|=3+7=10. 10 and 4 are not the same.

If an=xn/(3n+n), then |an|=|x|n/(n+3n) because the bottom is always positive (so the signs agree) and the top is an absolute value of a product of x's, so it becomes a product of absolute values of x's. And |an+1| is similarly |x|n+1/(n+1+3n+1). Now we need to analyze the quotient. I am getting exhausted with all of this typing. I'll skip the compound fraction and just write the simple fraction which results:

``` |x|n+1(n+3n)         (n+3n)
-------------- = |x|·---------
|x|n(n+1+3n+1)       (n+1+3n+1)
```
We need to analyze the behavior of the somewhat complicated quotient (n+3n)/(n+1+3n+1) as n→∞. When we're done, we need to multiply by |x| in order to get ρ.

Informal analysis Well, as n increases, the polynomial growth doesn't matter at all. It is negligible compared to the exponential growth. So really we've got (approximately) just 3n/3n+1, and this is 1/3.

Formal analysis Look at (n+3n)/(n+1+3n+1) and divide the top and bottom by 3n. The result is ([n/3n]+1)/([n/3n]+[1/3n]+[3n+1/3n]) which is ([n/3n]+1)/([n/3n]+[1/3n]+3). What about [n/3n] as n→∞? We will use L'Hôpital's Rule since this is again ∞/∞. Remember that AB=eB ln(A), so that the quotient [n/3n]=[n/en ln(3)]. The derivative of the top (with respect to n) is 1, and the derivative of the bottom with respect to n is en ln(3)ln(3) (what's in the exponent is just a constant multiplying n, so the Chain Rule works easily. Therefore by L'H, limn→∞[n/3n]= limn→∞[1/en ln(3)ln(3)]= limn→∞[1/3nln(3)]=0. So (wow!) limn→∞(n+3n)/(n+1+3n+1)=limn→∞([n/3n]+1)/([n/3n]+[1/3n]+3)=1/3.

What about the Ratio Test limit? We need to multiply by |x| since we discarded it to get the fraction we just studied. So in this complicated case, ρ=|x|(1/3). We get convergence (actually absolute convergence) when ρ<1, which means that |x|<3. The x's which satisfy this are an interval from –3 to 3 (not including the endpoints). We get divergence when |x|>3. So for those x's satisfying either –∞<x<–3 or 3<x<∞ there is divergence. The Ratio Test doesn't work if x=3 or if x=–3. It turns out that this situation is typical, and we will look at more examples and more detail next time.

The Root Test is another result which relies on how a series resembles a geometric series. We'll discuss this next time.

 What happens at the "edges"? We saw that the Ratio Test doesn't give any information when x=3 or x=–3. So if we really needed to know what happens, we will need more work. Look at ∑n=1∞xn/(3n+n) when x=3. This is ∑n=1∞3n/(3n+n). The nth term is 3n/(3n+n) and if we divide the top and bottom by 3n we see that the nth term is 1/(1+[n/3n]). But we saw that as n→∞, n/3n→0 so that an→1. Any series which converges must have its nth term go to 0. Since this one doesn't, the series must diverges when x=3. Similarly, if you insert the value –3 for x in the series, you'll see that the terms do not→0, so the series must also diverge when x=3.

QotD
Use the Ratio Test to verify that ∑n=1[22n/{n!(n+1)!}] converges.
This is also a series which comes up in analyzing the vibrations of a circular membrane (a drum!). Really. Its approximate value is 3.87973.

Monday, April 6 (Lecture #19)
And now, where and what?
So we discussed series with positive (actually all we needed was non-negative) terms. The textbook has clean and coherent statements of the results: the dichotomy, the Comparison Test, the Limit Comparison Test, and the Integral Test. The technical meanings of these terms have been discussed: infinite series; sequence of partial sums; infinite tails; convergence of an infinite series.
I will repeat the fundamental dichotomy, one of this week's notable vocabulary words. (And dichotomy means, briefly, "a division into two, esp. a sharply defined one." As I mentioned in class, this will be very valuable to you when you repeat your whole life and must take SAT's again.).
• For a series with positive terms, bounded partial sums means the same thing as convergence.
• For a series with positive terms, unbounded partial sums means the same thing as divergence.

Series with terms of varying signs
So now we will complicate things a bit, and look at series whose signs vary. Let me start really easily but things will get more intricate rapidly. ( Varying stop signs) (These are varying signs, hah hah hah.)

1–1+1–1+1–...
This is just about the simplest example I could show. We got a formula for the nth term. We need the sign to alternate, and that will be given by (–1)something here. The sign will alternate if the "something here" is either n or n+1. The first term will be +1 and the second term will be –1 if we use n+1. So an explicit formula is an=(–1)n+1. Next I asked about convergences of the series ∑n=1(–1)n+1. For this we must consider the sequence of partial sums.
S1=1; S2=1–1=0; S3=1–1+1=1; S4=0, etc.
It isn't too difficult to see that Seven integer=0 and Sodd integer=1. The partial sums flip back and forth. This is exactly the kind of behavior we did not get when we considered series with all positive terms. There the partial sums just traveled "up", to the right. Well, this particular infinite series does not converge, since the partial sums do not approach a unique limit.
n=1(–1)n+1 diverges even though the sequence of partial sums is bounded.

2–(1{1/2})+(1{1/3})–(1{1/4})-(1{1/5})+...
This is a more complicated series. I suggested that we try to "guess" a formula by first getting a formula for the sign, and then a formula for the absolute value (the direction and magnitude, thinking about numbers as sort of one-dimensional vectors). In this case, the sign is surely given by (–1)n+1, just as before. The magnitude or absolute value is 1+{1/n}. The formula {n+1}/n was also suggested, another a good answer. So putting these together, an=(–1)n(1+{1/n}). And now we looked at the {con|di}vergence of ∑n=1(–1)n+1(1+{1/n}).

The partial sums are more complicated and more interesting.

S1=2; S2=2–(1{1/2})–{1/2}=.5; S3=2–(1{1/2})+{1{1/3})=11/6=1.8333; S4=2–(1{1/2})+(1{1/3})–(1{1/4})={7/12}=.58333
This is where I stopped in class, but, golly, I have a friend who could compute S17 either exactly ({4218925/2450448}) or approximately (1.72169). This is nearly silly. Richard Hamming, one of the twentieth century's greatest applied mathematicians, remarked that
The purpose of computing is insight, not numbers.

Let's try to get some insight. Look at the first four partial sums on the number line.

From S1 to S2, we move left since the second term in the series is negative. From S2 to S3 we move right, because the third term in the series is positive. But notice that we don't get to S1. because the jump right has magnitude 1{1/3} and this is less than 1{1/2}, the magnitude of the previous jump left.

I hope you are willing to believe that what's described persists in general.

• The even partial sums are increasing.
• The odd partial sums are decreasing.
• All of the even partial sums are less than all of the odd partial sums.

Does this series converge? Students had varied opinions about this, but the question was definitively settled by the observations of several clever students. The distance between any odd partial sum and any even partial sum will be at least 1, since the magnitude of the nth term is 1{1/n}, which is certainly >1. The successive partial sums can't get close to each other! But the collection of partial sums does not approach a unique limit.
n=1(–1)n+1(1+{1/n}) diverges.

1–1/2+1/3–1/4+1/5–...
Here an has sign (–1)n+1 again, and the absolute value or magnitude is 1/n. Does ∑n=1(–1)n+1(1/n) converge? The partial sums are more complicated and more interesting.

s1=1; s2=1–(1/2)–1/2=.5; s3=1–(1/2)+{1/3)=5/6=.8333; s4=1–(1/2)+(1/3)–(1/4)=7/12=.58333
Here's a picture of these partial sums. Things are a bit more crowded (that's good for convergence!) than in the previous picture.

The previous three qualitative properties still hold. Since the signs alternate, the partial sums wiggle left and right. Since the absolute values decrease, the odd sums are less than the even sums, and all of the even sums are less than all of the odd sums. But now the distance between the odd and even sums→0 since the magnitude of the terms is 1/n, and this→0. So here is a rather subtle phenomenon:
n=1(–1)n+1(1/n) converges.

The theorem on alternating series (Alternating Series Test)
The following is a major result of section 10.4 of the text, where it is called the Leibniz Test for Alternating Series.

Hypotheses Suppose that
• The terms of a series alternate in sign (+ to – to + etc.).
• The absolute value or magnitude of the terms decreases.
• The limit of the absolute values of these terms is 0.
Conclusion The series converges.
• This is a cute result and useful to analyze some special series. The most famous example is the alternating harmonic series, ∑j=1(–1)n+1(1/n), which we just saw. There are other examples in the textbook.
Notice that the alternating harmonic series converges but the original harmonic series with the signs stripped off, ∑n=1(1/n), diverges. To me this is somewhat subtle.

Some partial sums of the
alternating harmonic series
S10=0.6456349206
S100=0.6456349206
S1,000=0.6881721793
S10,000=0.6926474305
S100,000=0.6930971829
S1,000,000=0.6931466807
Finally, to the right is some "experimental evidence" which might help you believe that the alternating harmonic series converges.

The sum of the alternating harmonic series is ln(2). But the convergence is actually incredibly slow. The one millionth partial sum (which took almost 8 seconds for a moderately fast PC to compute) only has 5 accurate decimal digits. This is not the best and fastest way to compute things -- we will see much faster methods.

But what if ...
The sign distribution of terms in an infinite series could be more complicated. I suggested that we consider something like

```      7cos(36n7–2n2)+2sin(55n+88)
∑n=1∞ ---------------------------
2n```
Here the sign distribution of the top of the fraction defining an is quite complicated. These are the first 20 signs:
–1, 1, 1, –1, –1, –1, 1, –1, 1, –1, 1, 1, –1, 1, 1, –1, –1, 1, 1, –1
There's no nice pattern that I can see. The reason for the strange and truly unpredictable signs is because the positive integers and multiples of Pi do not nicely relate to one another (Pi is irrational). I believe that no one in the world can predict this sign sequence.

Does this series converge?

Please notice that with a few modifications, the corresponding question can be answered very easily. Look at:

```      7|cos(36n7–2n2)|+2|sin(55n+88)|
∑n=1∞ --------------------------------
2n```
Absolute values signs have been put around the cosine and sine functions. Now the series has all non-negative terms and we can use our comparison ideas. How big is the top? Since the values of both sine and cosince are in [–1,1], the top can't be any bigger than 9. The bottom is 2n. Therefore each term of this series is at most 9/2n. But this larger series is a geometric series with ratio 1/2<1 and so it must converge.

Proof via manipulative
One definition of manipulative (as a noun) is: "In teaching or learning arithmetic: a physical object which may be manipulated to help demonstrate a concept or practise an operation." There was a spectacular demonstration in class! It was inspired by thinking about old-fashioned folding carpenter's rulers. If we have an infinite series ∑n=1an, we could consider the associated series ∑n=1|an|, where we have stripped off the signs of the terms, and are just adding up the magnitudes. This is sort of like an unfolded carpenter's rule, stretched out as long as possible. It may happen that the series of absolute values, a series of positive terms, may converge. So when "the ruler" is stretched out as long as possible, it has finite length. Well, if we then fold up the ruler, so some segments point left (negative) and some point right (positive) then the resulting length will also be finite.

The picture here is an attempt to support this statement and to duplicate the physical effect of what I displayed in class. The top has the segments stretched out as far as possible. The next picture shows some of the segments rotating, aimed backwards (negatively). The last segment shows in red segments which are negative and in green the other segments, oriented postively. I hope this makes sense, and justifies the following:

The "folded" series compared to the "unfolded" series
If ∑n=1|an| converges, then ∑n=1an must converge also (and, actually, |n=1an|≤∑n=1|an|).

I really screwed up the last part of the statement of this result in the second lecture. Please read what is written here -- it is correct. I thank Ms. Zhou for calling this to my attention and certainly apologize for not writing the result correctly.

Proof via algebra
There is a verification of these statements in the textbook, using algebra, on p.584, Theorem 1, in section 10.4, if you would like to read it. Sigh.

And conversely?
Notice that the converse of the assertion about absolute values and series may not be correct. That is, a series may converge, and the series of absolute values of its terms may not. The simplest example, already verified, used the alternating harmonic series, covergent, and the harmonic series, divergent.

Vocabulary
A series ∑n=1an which has ∑n=1|an| converging is called absolutely convergent. Then the correct implication above is:

If a series is absolutely convergent, then it is a convergent series.
A series for which ∑n=1an converges and ∑n=1|an| diverges is called conditionally convergent. The alternating harmonic series is conditionally convergent.

Another example
Consider ∑n=1{sin(5n+8)}37/n5. I don't know very much about {sin(5n+8)}37 except that, for any n, this is a number inside the interval [–1,1]. Therefore ∑n=1|{sin(5n+8)}37/n5| has terms which are all smaller than ∑n=11/n5 (a p-series with p=5>1, so it must converge). The comparison test asserts that ∑n=1|(sin(5n+8)37/n5| converges, and therefore ∑n=1{sin(5n+8)}37/n5 itself must be a convergent series.

 How to use these ideas quantitatively What is I actually wanted to find ∑n=1∞{sin(5n+8)}37/n5, say to an accuracy of +/-.00001? Well, I could split this series up into SN+TN, a partial sum plus an infinite tail, and try to select N so that |TN|<.00001. Once I do that, well then I just (have a computer) compute the corresponding SN. So how can I get N with |TN|<.00001? Here is a way. I know that TN=∑n=N+1∞{sin(5n+8)}37/n5. This is an infinite series. I bet (using the result of a preceding paragraph that |TN|=|∑n=N+1∞{sin(5n+8)}37/n5|≤∑n=N+1∞1/n5 again since the biggest that sine can be absolute value is 1, and 137=1. We looked at an integral comparison technique for this series in the last lecture. There we learned that this infinite series was less than ∫x=N&infin[1/x5]dx and we can evaluate the improper integral. It is (see the link for computations!) exactly 1/[4N4]. If we want this to be less than .00001=1/105 I guess we could take N to be 13 (I cheated -- I used a computer). Then the value of the sum, to 5 decimal place accuracy, is S13=∑n=113{sin(5n+8)}37/n5 which is .00317. Maybe this specific example isn't too impressive, but the whole thing really does work in a large number of cases!

Given a series, take absolute values
The result just stated is a very powerful and easily used method. If you "give" me a series with random signs, the first thing I will do is strip off or discard the signs and try to decide if the series of positive (non-negative, really) terms converges.

QotD
The question was something like this:
The series ∑n=1((6n sin(5n–7)+8)/n17 converges. Explain why!

Wednesday, April 1 (Lecture #18)
Where are we? What have we done?
The technical meanings of these terms have been discussed: infinite series; sequence of partial sums; infinite tails; convergence of an infinite series.
We spent much of the last lecture discussing the specific example of the harmonic series, which is ∑n=11/n. This series diverges even though the sequence of terms, {1/n}, has limit 0.
We also discussed geometric series, in your text written as c+cr+cr2+cr3+...=∑n=0crn–1. This series converges if |r|<1, and its sum is c/(1–r) then. When |r|≥1 and c is not 0, the series will diverge.

I made an important comment: I can only cover what I hope is an overview of several useful examples in each lecture. I do not believe that the lectures alone give a complete view of the subject, and certainly the lectures alone are not enough for exam preparation. Please read the textbook and practice textbook problems.

The goal in today's lecture
I will show you some very useful (and usable!) techniques to decide if a series converges, and to approximate (as closely as you want) the sum of the series. So there will be lots of numbers today.

Series with positive terms
Today we will consider series whose terms are positive or, at worst, non-negative (≥0). (In the next lecture we'll discuss what happens if we allow different signs, but for now, only +'s.) What can we say in general about series with positive (or even just non-negative) terms? Well, the sequence of partial sums is increasing, since we're just adding more and more non-negative terms. What can happen? One thing is that the sequence of partial sums can tend to ∞ (hey, this is what happens to the infinite series ∑n=11: the sequence of partial sums is unbounded and the series diverges). Another thing that can happen to a positive series is that the sequence of partial sums can tend to a limit (a non-negative finite limit). This happens, for example, with positive geometric series with ratio less than 1. Then the sequence of partial sums is increasing and also bounded and the series converges. This is a consequence of the fact that "Bounded monotone sequences converge".

This theoretical alternative is everything.

• For a series with positive terms, bounded partial sums means the same thing as convergence.
• For a series with positive terms, unbounded partial sums means the same thing as divergence.
Today's vocabulary word
The principal meaning of the word dichotomy is (according to my online dictionary) "a division into two, especially a sharply defined one." So what's above is a dichotomy for series with positive terms.

A first use of comparison
I considered the series ∑n=11/(2n+n). This is a rather artificial example, and, as far as I know, does not occur in any interesting "real" application. But we can play with it a bit. So the first question I'd like to ask is: does this series converge?

One reason I'm starting with a series whose terms are 1/(2n+n) is that these terms are related to the terms of the harmonic series, which we investigated, and to the geometric series with ratio 1/2, which we also studied. What do we know?

Well, 1/(2n+n)<1/n always. So the partial sums for ∑n=11/(2n+n) are all less than the partial sums for ∑n=11/n. The second series is the harmonic series, and that series diverges and its partial sums are not bounded: they →∞ as n→∞. So what information do we get? The partial sums that we want to know about are less than something which goes to ∞. We get no information. From this alone, we cannot conclude that the smaller partial sums are either bounded or unbounded. We need something else.

We also know that 1/(2n+n)<1/2n always. Now we are comparing the series ∑n=11/(2n+n) with ∑n=11/2n. But this second, larger series does converge: it is a geometric series with c=1/2 and r=1/2<1. Its partial sums are bounded (all of them are less than 1, the sum of the series), and so the smaller partial sums of the series with terms 1/(2n+n) are also bounded by 1, and therefore must converge.

It converges!
The series ∑n=11/(2n+n) converges. We tried two comparisons and only one of these supplied enough information for a useful conclusion. Frequently several different approaches need to be tried to hope for useful information about an infinite series. Your tolerance for frustration should be high in order to increase your likely success.

The sum is (to 3 decimal places) ...
We know that ∑n=11/(2n+n) converges. Certainly, since all the terms are positive, I guess the sum will be positive. Also, since we compared this series to ∑n=11/2n, which has sum=1 (c=1/2 and r=1/2 and c/(1–r)=1), I also guess (no: I actually know!) that the sum is less than 1. So I know that
0<∑n=11/(2n+n)<1.

This is nice, but if this series occurs as the answer to some complicated question, we might want to know its sum more accurately. What if we wanted to know the sum to 3 decimal places (+/–.001)? This is a modest amount of accuracy. Let me show you a useful approach.

Well, ∑n=11/(2n+n)=SN+TN where SN=∑n=1N1/(2n+n) and TN=∑n=N+11/(2n+n). If we can find some nice specific value of N so that TN is guaranteed to be less than .001, then we will know that the corresponding finite sum, SN, will be within .001 of the true value of the sum of the whole infinite series. So what can we do? I will overestimate TN:
TN=∑n=N+11/(2n+n)<∑n=N+11/2n.

I choose the larger series to be one whose sum I can find easily. I just want some sort of answer -- I don't necessarily need the best answer, just some answer. Well, ∑n=N+11/2n is a geometric series. Be a bit careful in deducing c and r here, please. The first term, c, is 1/2N+1. I get this by looking at the "lower bound" in the ∑. The ratio between successive terms, r, is 1/2. So the sum of this overestimate is c/(1–r)=[1/2N+1]/(1–1/2)=[1/2N+1]/(1/2)=[2/2N+1]/(2/2)=1/2N. Do the algebra slowly and try not to make errors. This is how I would do this problem if I only had to make such estimates once in a while. If I needed to do this four or five times a day, well, heck, there are more systematic approaches. Well, what do we know? The infinite tail, TN, which we want to estimate, is positive and less than 1/2N. If we want TN<.001, then we can force it to be less by choosing N so that 1/2N<.001. Let's see: when N=10, I think we saw that 1/210=1/(1,024)<.001=1/(1,000).

Therefore S10=∑n=1101/(2n+n) will be within .001 of the "true value" of the sum of the whole series. It is easy for a computer or calculator to find this partial sum. It is .696.

Comparison Test
The Comparison Test applies to series with positive (or non-negative) terms. We have the following situation:

Suppose we know that 0<an≤bn for all n's.
If ∑n=1bn converges, then ∑n=1an converges.
If ∑n=1an diverges, then ∑n=1bn diverges.
No information is obtained if we know only either that the smaller series converges or the larger series diverges.
So convergence is "inherited" downward and divergence is "inherited" upward.

Another trick
There are two major tricks in the subject, and these two major tricks, in practice, handle about 99% of the examples that come up. One trick is comparison with geometric series, as we have just done. Here is a version of the other trick.
What can we say about the series 1+1/25+1/35+1/45+...=∑n=11/n5? This is a different kind of series. This is not a geometric series. The ratio connecting the first and the second terms is 1/32. The ratio between the second and third terms is 32/243. Since these numbers are not equal, this is not a geometric series. We need a different trick.

Comparison to a definite integral
Here is the trick. If an=1/n5, think of this quantity as an area of a rectangle whose width is 1 and whose height is 1/n5. Put this rectangle on the xy-plane so that its upper righthand corner is at the point (n,1/n5). The rectangles will all fit together as shown in the graph to the right. The corners are all on the curve y=1/x5. Everything is arranged so things work. Now look: the improper integral ∫x=1[1/x5]dx is larger than a2+a3+a4+... . I left out a1 because I don't want to integrate all the way to 0, since there is a different improperness there -- I just want to deal with the improperness at ∞. But look:
∫x=1[1/x5]dx=limA→∞x=1x=A[1/x5]dx=limA→–1/4x4|x=1x=A=limA→∞(–1/4A4)–(–1/4)=1/4. (Whew!)

Integrals are frequently easier to compute than sums. As far as I know, no one in the world knows either the true value of the sum of this infinite series or a good representation of its partial sums (and this series does occur in applications!). So what do we know? Don't forget, please, the initial term, a1=1. We know that
n=11/n5<1+∫x=1[1/x5]dx=5/4.

It converges!
The partial sums are all bounded above by 5/4. Since ∑n=11/n5 is a series of positive terms, bounded above is enough to imply that the series converges. The series converges and its sum is some positive number less than 5/4.

The sum is (to 3 decimal places) ...
Suppose I want the sum of this series to 3 decimal places. Then, just as before, I will write ∑n=11/n5=SN+TN, where SN=∑n=1N1/n5 and TN=∑n=N+11/n5. I will try to overestimate the infinite tail, TN, by something convenient, and then force it to be less than .001.

Here is how to overestimate this TN. Look at the picture to the right. The curve y=1/x5 is again there, and I am interested in what happens for x's bigger than N. I put the boxes representing TN (remember, this starts with 1/(N+1)5) under this portion of the curve. So (very very tricky!) I can make this estimate:
TN<∫x=N[1/x5]dx=limA→∞x=Nx=A[1/x5]dx=ETC.=1/[4N4].

I skipped some steps in the evaluation of the improper integral because it is about the same as the previous computation. Now I want to select N so that 1/[4N4]<.001. We decided if N=4, then 1/[4(44)]=1/(1,024)<1/1,000. So the fourth infinite tail will be less than one one-thousandth. And if we want the sum of the series to 3 decimal places, we just need to compute S4=∑n=141/n5 which is easy for a machine (or is even tolerable by hand, really). The value is 1.036 so that the sum of the series with error +/–.001 is 1.036.

By the way, every time I use this technique, I draw the pictures I've shown here and wrote in class. I don't do this often enough to have it "mechanized", so I need to remind myself how it works.

p-series in general
The other collection of examples you need to know are the p-series.
Suppose p is a positive number. Then the p-series is ∑n=11/np=1+1/2p+1/3p+1/4p+.... This series converges if p>1 and diverges if p<1. The reason that this is true is the Integral Test which I will write later, but PLEASE read the textbook about this -- there's not enough lecture time to discuss everything. Please note that the p-series is not a geometric series!
We just discussed the p-series for p=5. It converged. Last time we considered p=1, which is the harmonic series, and saw that it diverged. Let me investigate yet another example of a divergent p-series.

A divergent series
Take p=1/2. The p-series is 1+1/sqrt(2)+1/sqrt(3)+1/sqrt(4)+1/sqrt(5)+... and, according to what was written above, this series diverges. It diverges even though the terms get very small, because they don't get small enough fast enough. But if it diverges, there should be some partial sum which is bigger than 100. I would like to find a specific partial sum bigger than 100. Numerical questions like this actually arise in real applications, and an integral technique can be used to answer them without much difficulty.

A partial sum bigger than 100
The idea is the same and the idea is different. I'm sorry for writing such a silly sentence, but this sort of is the truth. Look at this picture. I have sketched y=1/sqrt(x). Since what I want is to underestimate SN I have placed the boxes over the curve. Then the upper left-hand corners of the boxes are on the curve. And to get all of the area representing SN over the curve, I will need to integrate from 1 to N+1. I don't think this is obvious or easy, but please look at the picture.

The estimate the picture implies is ∫x=1x=N+1[1/sqrt(x)]dx<SN. Now I can "easily" compute this integral (well, more easily than I can compute the partial sum!). Here:
x=1x=N+1[1/sqrt(x)]dx=2sqrt(x)|x=1x=N+1=2sqrt(N+1)–2sqrt(1)=2sqrt(N+1)–2.

This is an underestimate of SN. If I want to force SN to be at least 100, then this will be done if I know that 2sqrt(N+1)–2≥100 or sqrt(N+1)–1≥50 or sqrt(N+1)≥51 or (sigh!) N+1≥512=2601. So N should be at least 2600. Some computed partial sums are listed below.

 N SN 10 100 1,000 2,000 2,500 2,600 5.02 18.59 61.80 87.99 98.55 100.53

I was amused when I computed these numbers slightly before class because I didn't expect things to be so close. Usually the estimates gotten with these methods are fairly rough. What matters is that the method works -- it is effective, and usually easy to do.

 The Integral Test The textbook discusses the Integral Test in section 10.3. PLEASE read the textbook! Here is a version. Suppose f(x) is a positive decreasing function, defined for x≥1. Then the series ∑n=1∞f(n) converges exactly when the improper integral ∫1∞f(x) dx converges.

QotD
Does ∑n=11/(n2n) converge? (This is problem #19 in section 10.3 of the textbook.)
Solution Since 0≤1/(n2n)≤1/2n for all positive integer n, and since ∑n=11/2n converges (geometric series with absolute value of ratio equal to 1/2, less than 1) the series ∑n=11/(n2n) must converge by the comparison test.

 Here is a bit of a "dialog" from Maple. The first response shows that the program recognizes and can find the sums of (at least simple) geometric series. The second response, which just echoes the question, shows that the program can't automatically find a sum for the first series we investigated in this lecture.```> sum(1/2^n,n=1..infinity); 1 > sum(1/(2^n+n),n=1..infinity); infinity ----- \ 1 ) ------ / n ----- 2 + n n = 1```

Monday, March 30 (Lecture #17)
A recursive sequence
Many sequences that occur in applications are not defined by explicit formulas (Newton's method, for example). Here is a simple example of a sequence which is defined recursively -- that is, members of the sequence are defined in terms of previous elements of the sequence. (Again, my example is not random. I didn't have time to explain where it came from in class, but I will try to here.)
a1=1 and an+1=sqrt(an+5) for n>1.

So a2=sqrt(1+5)=sqrt(6). And a3=sqrt(sqrt(6)+5). Etc. Here I don't know if this sort of numerical computation helps much, but look at the first seven terms: 1, 2.449489743, 2.729375339, 2.780175415, 2.789296581, 2.790931132, 2.791223949. This is quite suggestive.

What can we see ...
A first observation Certainly I believe that the elements of this sequence are all positive. We are adding 5 and taking square root (remember that sqrt means non-negative square root here!). Also, I am fairly sure that the terms in this sequence are all less than 100 (here 100 is really just a random bound). How do I know this? Well, the first few are less than 100 (the numbers above). And if an<100, then an+5<100+5=105, so that sqrt(an+5)<sqrt(105). But the left-hand number is an+1 and that is less than sqrt(105), which is less than 100. So I have proved that all of the an's are between 0 and 100. That's not enough to conclude convergence, as we saw with earlier simpler examples. Sequences can wiggle. But this sequence does not wiggle.
A second observation Look a bit more closely at the terms I computed above. They seem to increase. Is that an accident? Well, if I know that an<an+1 then I can add 5 to both sides and get an+5<an+1+5. Also square rooting is increasing (remember the graph!) so that sqrt(an+5)<sqrt(an+1+5). But this inequality is exactly an+1<an+2. So increasingness is inherited by later terms of the sequence. Since I know the sequence in increasing for the beginning terms of the sequence because of the computations above, I know that the sequence will always be increasing.

What happens?
In this specific case we have an increasing sequence which is bounded. But increasing bounded sequences converge because they can't wiggle, and they can't (since they are bounded) jump out to "infinity". The sequence {an} defined above is an increasing bounded sequence, and it must converge. The sequence must "pile up" somewhere less than the bound. (A similar result is also true for decreasing bounded sequences -- please see the textbook. This fact is not supposed to be obvious!)

Its limit
Once I know that the sequence converges, I can use the equation an+1=sqrt(an+5) to find the limit quite nicely. So if I know that limn→∞an=L then certainly (since {an+1} is just about the same sequence, the numbers are all shoved along one place) I know that limn→∞an+1 is L also. So look:
If an+1=sqrt(an+5) take limn→∞. The result is L=sqrt(L+5). Square both sides, so L2=L+5. Then L2–L–5=0. And (quadratic formula) L=[1+/–sqrt((–1)2–4(1)(–5))]/2, and this is 1/2+/–sqrt(21)/2. Which root? As several students pointed out, the terms are positive, so take +, and the limit is 1/2+sqrt(21)/2. This is approximately 2.791287848 (close to the terms we computed above).

 Where the recursive sequence came from To the right is a computer-drawn graph (therefore we suppose it is quite accurate!) of y=x and y=sqrt(x+5) on the interval [–5,3]. Please notice that the graphs intersect at one point in the first quadrant. That point has coordinates (L,L) and is a point where sqrt(L+5)=L.

Now look at the picture below!
Start at 1 on the x-axis and go up until we hit the curve. That will be at the point (1,sqrt(1+5)) which is (1,sqrt(6)). Then move right until we hit the line. That will be at (sqrt(6),sqrt(6)). Then up and hit the curve at (sqrt(6),sqrt(sqrt(6)+5)), ETC. We move back and forth, sort of bouncing between the curves. If you look at the picture you will see that the points "accumulate" where the curves intersect, and (L,L) is that point. This "silly" picture is actually a simple version of a numerical method used to find roots.

Don't trust numbers always!
O.k., I will throw some numbers at you. If you consider the sequence {1+[1/n]} then 10-digit decimal approximations of the first 7 terms are:
2., 1.500000000, 1.333333333, 1.250000000, 1.200000000, 1.166666667, 1.142857143
This doesn't look bad. And I bet that this sequence converges, and its limit is 1: limn→∞(1+[1/n])=1. Fine.

I have a simple formula for another sequence. Here are the 10-digit decimal approximations of the first 7 terms:
2.000000000, 1.500000000, 1.333333333, 1.250000001, 1.200000003, 1.166666675, 1.142857160
This looks just about the same as the first sequence, just a little bit of fuzz in the bottom few digits. Well, the 1,000th term in this sequence, a1,000, is approximately 2·10435. Here is the formula: an=(1/n)+e[.000000000001]n5). This sequence grows really really quickly after a while (it is positive const multiplied by n5 inside the exponential function). So a few terms may not give very much information at all about asymptotics.

What are series?
Most of these phrases are quotes from the text. I'm also making a definite effort to use the notation in the text. So here we go:

• An infinite series "is an expression of the form ∑n=1an=a1+a2+a3+..."
Discussion This is a very strange concept, and you should realize just how strange it is. I don't know any person or computer or thing which could actually add up infinitely many numbers. I do know people and machines which could add, approximately or in certain cases, exactly, lots and lots (but finitely many!) numbers. So whatever you believe, the infinite series expression above is just some sort of symbolic "stuff" that needs to be correctly interpreted. It doesn't and can't literally mean, o.k., this is what you get when you add up all these numbers. That is just silly. Also, I should mention that in, say, Math 151, and even in this course, I have generally avoided using sigmas, that is, ∑'s, as an abbreviation for summation. I was trying to be gentle. Starting now I won't be so careful, because we'll be doing lots of things with summations, and the correct use of summation notation will just save a great deal of writing. Please ask questions if the notation becomes horribly intricate.
• There are actually two different sequences which people naturally associate with each infinite series, ∑n=1an. The situation can be confusing. One sequence is just the sequence of individual terms in the infinite series. This is {an}. This sequence by itself is rarely of huge interest in this context. The sequence which is of interest is the sequence of partial sums. So I need to define partial sums. Well, here are some of them:
S1=a1      S2=a1+a2      S3=a1+a2+a3      S4=a1+a2+a3+a4      S4=a1+a2+a3+a4+a5      ETC.
I don't think that "ETC." is very clear, so I'd better tell you precisely what SN is:
SN=∑n=1Nan (the sum of the first N terms of the infinite series).
This is the Nth partial sum. I hope you can see why this is called a partial sum. It is the sum of what some people call an initial segment of the infinite series. The sequence of partial sums is then the sequence {SN}. (The little n's are concealed inside the SN's as a summation index.)
• The sequence of partial sums may or may not converge. If the sequence of partial sums converges, then we say that the infinite series converges and the limit of the sequence of partial sums is called the sum of the infinite series.

 Metaphor? An infinite series ∑n=1∞an can be thought of as ∑n=1Nan+∑n=N+1∞an. So there is the Nth partial sum plus the "other" terms of the series, an infinite tail. These other terms I may sometimes call TN. I like to think of this maybe as some sort of animal. The partial sum is the body, and the infinite tail is ... well, the tail. The question of whether the series converges or not maybe is analogous to whether the whole weight of the animal is finite (this is a good analogy only for series whose terms are all positive -- we will deal later with series whose terms change sign). The weight will be finite exactly when the infinite tails→0 as n→∞. In fact, the first "few" terms of a series have nothing to do with convergence! You can change them, and the convergence of the series won't change at all. (If the series converges, the sum will change, but whether or not the series converges won't be changed.)

Maybe the simplest example
If all of the an's are 1, then the infinite series is 1+1+1+1+1+... and the partial sums are, well: S1=1 and S2=1+1=2 and S3=1+1+1=3 and ... well, I hope you are convinced that SN=N. The sequence {N} does not converge, so the infinite series 1+1+1+1+1+... diverges.

A simple and very deceptive example
As I mentioned in class, in this lecture my instructional strategy is the reverse of what I did last time. I will present as my first example something which is notorious and defies intuition. It is a very famous infinite series.
The harmonic series is the infinite series ∑n=11/n. Sometimes we might write this series as 1+1/2+1/3+1/4+... and the "..." is supposed to indicate, hey, you know the pattern, you understand the formula, etc. There are examples of infinite series where I certainly don't instantly "see" what ... means for those series. The harmonic series occurs in many physical applications, and also arises in analyzing lots of computer algorithms.

The sequence of the individual terms, {1/n}, isn't very complicated. It was one of our first sequence examples. The important question here is to understand the partial sums. Well, S1=1 and S2=1+1/2=3/2 and S3=1+1/2+1/3=11/6. This isn't helping. There is no known explicit formula for the partial sums of this series. I found decimal approximations for a bunch of partial sums. Look at the table to the right, please.

NSN
11.00000
21.50000
31.83333
102.92897
1005.18738
1,0007.48547
10,0009.78761
100,00012.09015
1,000,00014.37273
I emphasize that I know no simple shortcut for actually computing these darn numbers. The decimal approximations above were the result of lots of divisions and additions (an unreliable timing of the S1,000,000 computation is about 5 seconds). Just so you know what the heck these numbers mean, what would S1,000,001 be compared to the last table entry? Well, S1,000,001=∑n=11,000,0011/n=(n=11,000,0001/n)+(1/1,000,001)=S1,000,000+(1/1,000,001). Whew! Here are the actual (now 10-digit because the result wouldn't be observable in 5 digits!) approximations:
S1,000,000=14.39272672 and S1,000,001=14.39272772. They are different!

So what happens?
What can we conclude about the convergence of this series? The serious answer is nothing. We have weighed the body, Ssome numbers, but we have no idea what the size of Tthese numbers is: the "infinite tail" might be very thin (hey, less than 1/1,000,000) but it is very very long. Several ideas may occur.

 Primitive idea #1 It diverges because we're adding up infinitely many numbers, and therefore things get to be too darn large. Primitive idea #2 It converges because, although we're adding lots of numbers, the steps between the sums get smaller and smaller, so the sum can't get very large.

A simple argument
The numbers above are, ultimately, not very persuasive. But let me show you an elementary argument which will allow us to make a good decision. Please realize that this is a very clever argument! We first realized that the partial sums of the harmonic series were all positive and they were increasing: SN<SN+1. I will just look at some special partial sums. I wrote these lines on the board:
S1=1=2/2
S2=1+1/2=3/2
S4=1+1/2+1/3+1/4>1+1/2+(1/4+1/4)=1+1/2+2(1/4)=4/2
S8=1+1/2+1/3+1/4+1/5+1/6+1/7+1/8>1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)=1+1/2+2(1/4)+4(1/8)=5/2
Of course, in class (here is a reason to attend the lectures!) this was done sort of interactively. At this point I stopped and asked people what was the next line I should write. There was some conversation and this was produced as a result:
S16=1+1/2+1/3+1/4+1/5+1/6+1/7+1/8 +1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16 >1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16) >=1+1/2+2(1/4)+4(1/8)+8(1/16)=6/2

We can group the partial sums when N is a power of 2. Here is the general result: S2k>=(k+2)/2. For example, I asked if we could give one specific partial sum which would be larger than 100. Uhhhh ... if we took k=200, then (k+2)/2=101, larger than 100. So S2200 is bigger than 100.
How long would it take to add up this sort of partial sum? Well, 2200 is a large number. It is about 1.6·1060. And, let's see, I would hope I could do about 1011 additions and divisions in a second (that is an overestimate). And there are about 1/3 of 108 seconds in a year. So there would be maybe 1/3 of 1019 additions and divisions in a year. The age of the universe is sometimes estimated to be about 20 billion years, or 2·1010 years. So ... in one "universe age" we could compute maybe 2/3·1029 terms (this is a cruddy computational model, by the way). So all we would need is about ... uhhh ... 1031 universe ages. This sort of exercise is useful if it convinces you that computing some big partial sum is silly.

Therefore ...
Does the harmonic series converge? If it did converge, then eventually the partial sums would have to get close to a limit. But the estimate we have just seen shows that the partial sums get bigger and bigger and bigger. So the harmonic series diverges.
The discussion above is very special. We'll get a number of general tricks within a few lectures which can be used to show divergence rapidly. I just want to give you a reason for the divergence result. For the harmonic series, the infinite tail is very thin, but it is really really very long! Intuition?

Another kind of series
Let me turn to a series that behaves the way we'd like. A geometric series is one where successive terms are related by a constant ratio. In algebraic language, the text writes such a series as c+cr+cr2+cr3+...=∑n=1crn–1.
Actually, I just noticed that the text writes this as ∑n=0crn, so the exponent looks more normal but the index of summation begins with n=0. But I'll continue with the way I actually did it in class.

An (exceptional!) explicit formula of a partial sum
The partial sum of this series is SN=∑n=1Ncrn–1. Unlike almost any other series, it is possible to find an explicit formula for this SN. Here is the idea, which I think is usually shown in some high school (and maybe even some middle school!) math or physics courses:
SN=c+cr+cr2+cr3+...+crN
Multiply by r:
rSN=cr+cr2+cr3+cr4+...+crN+1
Subtract the second equation from the first:
SN–rSN=c+(all the inside terms cancel!)–crN+1
Factor the left-hand side:
(1–r)SN=c–crN+1.

Now solve for SN: if r is not equal to 1, SN=[(c–crN+1)/(1–r)].

Convergent geometric series
Now if |r|<1, powers of r, that is, the sequence {rn}, must approach 0. So in the formula for SN, when |r|<1, I know that rN+1→0 and SN→c/(1–r). When |r|>1, the series diverges (the powers of r grow).
If |r|<1, then the geometric series with first term c and ratio between successive terms r, ∑n=1crn–1, converges, and its sum is c/(1–r).

Example #1
Here is something which I hoped that many people saw before college. The infinite repeating decimal 0.731731731... represents a rational number (a quotient of integers). What rational number does it represent?

Here the most interesting problem is recognizing the implied geometric series. Decimal notation is very clever and conceals some true subtleties. So 0.731 itself means 731·(.001) which is 731/1000. What about 0.000731? This is 731·(.000001) which is 731·(.001)·(.001). That is, 10–3·10–3=10–6. Therefore 0.000731 is 731/(1000)2. And similarly 0.000000731 is 731/(1000)3. So we see (maybe not so "clearly"!) that:
0.731731731...=[731/1000]+[731/(1000)2]+731/(1000)3]+....
We therefore recognize that the repeating decimal indicates an infinite series whose first term, c=731/1000, and whose constant ratio between successive terms is r=1/1000 (this is certainly less than 1). The sum is then c/(1–r)=[731/1000]/(1–[1/1000])=[731/1000]/[999/1000]=731/1000.

Digression: how maybe this is done in earlier "grades"
A teacher might say the following:
Consider Q=0.731731731... and try to figure out another way of looking at Q. Well, 1,000Q=(1,000)·(0.731731731...)=731.731731731... so then:
1,000Q=731.731731731... and subtract
Q=0.731731731...
The result is 999Q=731, so that Q=731/999. Ain't that nice! My "excuse" for pointing out the geometric series approach is that I want to show you a use of geometric series, and also to maybe expose a bit of the structure of the decimal system, which is actually a very clever and intricate idea.

Example #2
A square of side length 5 has another square whose side length is half of that, placed outside but so that corners and an edge coincide. Another square whose side length is half of that, placed outside of both squares but so that corners and an edge coincide. And ...
My language is perhaps not too precise. A sort of picture of this object (just the first 6 squares) is shown to the right. The object is an example of a fractal. General information about fractals is here and a source which is very accessible is here.

The question What is the total area of all of the squares?

The first square has area 5·5=52. The second square has area (5/2)·(5/2)=52/4. The third square has area (5/2/2)·(5/2/2)=(5/22)·(5/22)=52/42

The pattern may convince you that the total area is the sum of
52+52/4+52/42+...
This is a geometric series whose first term is c=52, and the constant ratio between successive terms is r=1/4. The sum is then c/(1–r)=52/(1–[1/4])=100/3. This is the total area.

Questions about some other geometric quantities involving these squares can be asked. For example,
The question What is the total perimeter of all of the squares?

The first square has perimeter 4·5=20. The second square has perimeter 4·(5/2)=20/2. The third square has perimeter 4·(5/2/2)=4·(5/22)=20/22

The pattern may convince you that the total perimeter is the sum of
20+20/2+20/22+...
This is a geometric series whose first term is a=20, and the constant ratio between successive terms is r=1/2. The sum is then a/(1–r)=20/(1–[1/2])=40. This is the total perimeter.

Example #3
Bruno and Igor have a loaf of bread. Bruno eats half the loaf and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno. Bruno eats half of what he is given and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno...

The question How much bread (the total amount) does Bruno eat?

As I mentioned in class, I know people who can somehow "solve" these problems by inspection, that is, they read or listen to the problem, and ZAP!!! the answer is clear. (The same is true for the geometric problems mentioned just previously.) I am not one of these "zap" people -- some of the students seem to be! I would probably solve the problem by computing the amount of bread Bruno and Igor eat, for at least a few rounds. I would try to discover the pattern, and then I'd use this discovered pattern.
Round #Bruno eatsIgor eats
11/21/4
21/81/16
31/321/64

I filled out this table dynamically in class, with explanations being given as I did it. For example, I remarked that after Bruno ate half the loaf, Igor would receive the other half loaf. Igor would eat half of that, which is 1/4 loaf, and pass 1/4 loaf to Bruno. Bruno would eat half of a 1/4, which is 1/8 loaf, and pass the remaining 1/8 loaf to Igor, etc. It seems apparent ("clear") that Bruno eats 1/2+1/8+1/32+..., a quantity which we recognize as a geometric series. The first term, c, is 1/2, and the constant ratio between successive terms is 1/4. Therefore Bruno must eat c/(1–r)=(1/2)/(1–[1/4])=2/3 of the loaf. Poor Igor will eat 1–2/3=1/3 of the loaf. (Or you could compute what Igor eats directly as the sum of another geometric series.)

It is easy to change this problem. You could imagine the named people eating different quantities, or you could imagine there being more people, etc. Sums like this do arise in real applications, and I hope that you will be able to recognize them and cope with them.

QotD
Is the harmonic series 1+1/2+1/3+1/4+... a geometric series?
If your answer is YES, then you must supply me with the correct values of the parameters c and r which will give the harmonic series.
If your answer is NO, then you should give some specific fact about the harmonic series which explains why it is not a geometric series.

The answer is NO. If the harmonic series were c+cr+cr2+... then (using just the initial segment 1+1/2+1/3) we see that r=cr/c=(1/2)/1=1/2 and r=(cr2/cr)=(1/3)/(1/2)=2/3. Since 1/2 is not 2/3, the harmonic series is not a geometric series.

Many student answers did not give sufficient information. Just remarking that "The ratio is not constant" is not enough for me to conclude that students understand what is and what is not a geometric series. That bare sentence doesn't show what feature or fact about the harmonic series is used. So more is needed. Specific evidence must be given to show that the harmonic series is not a geometric series.

Students who answered "NO" and supported their assertion with the fact that the harmonic series diverges ignore the fact that some geometric series diverge also. For example, ∑n=110n=10+100+1,000+10,000+... is a geometric series, and it certainly does not converge.

Wednesday, March 25 (Lecture #16)
The previous (not given) QotD
The problem I would have given as the QotD is available, along with a solution, if you want to check whether you can solve a separable differential equation. Look here.

Exam warning
The second exam for these sections will be given on Monday, April 20 at the usual class time and place. This is certainly about a week later than I'd like, but we missed a day due to snow, there are conflicting exams, and some days have religious obstructions. I hope that this date will work for all of us. Further information about the exam will be available soon.

The workshop due tomorrow
To a certain extent, workshops are anti-exams. There is little real time pressure. You can reflect, do over, understand, etc. the solution as much as you want. So if you feel that exam performance doesn't necessarily show the level of your understanding and knowledge, your workshops can help. In this case, I bet that there will be a problem on the second exam closely resembling the workshop problem due tomorrow. So there would be several different "payoffs" for handing in a good writeup!

What are sequences and what are they used for?
A sequence is a sort of ordered list of numbers (precise definition below).
Much computation is done sequentially. For example better and better approximations to roots of equations are gotten by Newton's method. This scheme creates numbers which (ideally!) get closer and closer to the root desired, even though there may be no exact algebraic formula for this root. Definite integrals (except in classrooms!) can only rarely be computed exactly. Instead, there are usually collections of approximations which are computed, and this collection of approximations is used to give a "value" for the definite integral. In this part of the course (really this is the last major part of the course!) we will study sequences and accompanying ideas.

Formal definition of a sequence
A sequence is a real-valued function whose domain is the positive integers. I will begin with what I hope are some rather simple examples but the later examples will be complicated enough for everyone.

Example and notation
Well, one example of a sequence is a sequence whose nth term or value or element (all these words are used!) is 1/n. So this is the function f(x)=1/x when x is a positive integer. The usual notation for this sequence is {1/n}. Yeah, a different notation is used -- usually the sequence "function" doesn't get shown. The formula is enclosed by the braces { and }. The letters that are used inside the braces are usually n and m and p and q. If needed, we might refer to the formula with a subscript. So, for example, we might write an=1/n. Therefore, a5=1/5 and a17=1/17. Also an+1=1/(n+1) and a5m=1/(5m). Notice especially the last two equations which might look strange. The stuff "down" in the subscript is the argument to the function which defines the sequence. Please try not to get confused. For example, if an=1/n, then a4=1/4, but a4+1=a5=1/5 and a4+1=(1/4)+1=5/4.
People also frequently "define" functions by listing their first few members. So this sequence might look like 1, 1/2, 1/3, 1/4, 1/5, ... and the "..." is supposed to indicate that the reader should now recognize the pattern. To me this is another use of the strange mathematical word "clearly". Clearly sometimes people will recognize the pattern, but also clearly many times there will be difficulty.
Maybe sometimes we might think of a picture of the sequence, but this really has limited use as you will see. To the right is a picture of the first 6 elements of this sequence, sitting on the number line.

Clearly (!) as n gets large, 1/n gets closer and closer to 0. This behavior is abbreviated, not too surprisingly, by limit notation. So here we would write limn→∞1/n=0. What could this mean in more precise language? Well, one implication that occurs is that as n increasing, 1/n gets close to 0. So maybe limn→∞an=L means that, as n gets large, an should get close to L. L is called the limit of the sequence.

Another example
This example will look initially quite silly, but thinking about it is useful. So the sequence is {(–1)n}. As a list, the elements of the sequence are (–1)1=–1, (–1)2=1, (–1)3=–1, (–1)4=1, etc. This sequence has only two distinct values, and these values depend on the parity (even/oddness) of its argument, n. When n is odd, the sequence value is –1, and when n is even, the value is 1. A rudimentary picture of the sequence is shown to the right. Notice that there is something missing from this sequence -- the dynamic aspect as the sequence wiggles and hops left and right. Every element of the sequence is in the picture and the picture isn't very helpful to me at all. Does this sequence converge? Historically an answer to this question wasn't obvious. People finally decided that the answer should be no. The only numbers that are authentic candidates for the limit of this sequence are 1 and –1. If we were going to use sequences as ways of getting better and better approximations to a root of an equation, then saying that the root is maybe 1 or maybe –1 is probably not the best answer. People usually want one specific answer. So the definition of limit, even if we want only to consider it informally, needs to be stated slightly differently.

Better definition of convergence
We'll say that limn→∞an=L is true if for n large enough, an gets close and stays close to L.
Certainly {–1)n for certain values of n is close to 1 (heck, it is equal to 1, and I don't know how much closer it could be). But also we can always find even larger values of n (large enough odd n) so that (–1)n is not at all close to 1. So the sequence {(–1)n} does not converge and does not have any limit. Another word is used: the sequence {(–1)n} diverges. The problem is that, although the sequence is sort of tame in that it never gets too large positive or too large negative, it wiggles and never sort of stabilizes.
There is a formal definition of limit in section 10.1 of the text. Later in your professional career, as you do more computations, you may likely need to work with that definition. The word "close" for example, is used to mean |an–L|, and usually people want to be able to control the size of |an–L| by selecting some number N so that when n>=N, then the size of |an–L| is "small" (and this is some number to be specified in practice). Now this is too detailed for Math 152 and a first visit to the definition. So back to the examples.

More examples

• {(1/2)n}. As a list, we can look at 1/2, 1/4, 1/8, 1/16, .... and I bet that this sequence converges, and its limit is 0: limn→∞(1/2)n=0.
• {10n}. This is 10, 100, 1000, 10000, etc. I think that this sequence diverges. It gets too big. It is unbounded.
• {(–1/2)n}. This sequence is more challenging. The first few elements are –1/2, +1/4, –1/8, +1/16, etc. Here the signs change, so the sequence wiggles, but the magnitude (the absolute value) of the terms goes to 0. So in spite of the wiggling, I think that this sequence converges and its limit is 0: limn→∞(–1/2)n=0.

Algebra and sequences
Some straightforward limit facts still are true, such as:
If limn→∞an=L1 and if limn→∞bn=L2, then limn→∞an+bn=L1+L2 and limn→∞an·bn=L1·L2.
Things like this are in the textbook, and I'd like to concentrate my attention on more subtle behavior.

{51/n}
Let's look at the sequence {51/n}. The first few terms, in decimal form, are these: 5., 2.236067977, 1.709975947, 1.495348781, 1.379729661, 1.307660486, 1.258498951. A picture of the first 6 of these is to the right. Maybe things here are not totally clear. In fact, one of the reasons I want to discuss this example is that things are not clear. We made some preliminary observations: certainly any root of 5 would have to be positive, in fact any root would have to be bigger than 1. And also any root would have to be less than 5. So what we know is that 1<51/n<5. The numbers in the sequence are "trapped" inside the interval from 1 to 5. I asked the class if just knowing this information was enough to guarantee that the sequence converged. After some thought, one student came up with the following example: {3+(–1)n}. Again depending on parity, the values of this sequence are either 2 (n odd) or 4 (n even). This sequence does not converge although its values are inside the interval from 1 to 5.

Comments on boundedness and convergence of sequences
If a sequence convergences, then the numbers in the sequence are bounded. (The reason is, essentially, that if there is convergence, all but a few of the terms are close to the limit of the sequence. Then the whole sequence is trapped near the limit and finitely many other numbers, and that can be put inside a bounded interval.)
The converse of the statement above is not true in general. I mean: If a sequence is bounded, then it may not converge. We've already seen several examples of this statement.

Here is a very brief discussion of some of the words used with logical implications.

If f(x) is a continuous function, and if {bn} is a convergent sequence with limn→∞bn=L, then the sequence {f(bn)} converges, and limn→∞f(bn)=f(L).
This is true because the bn's get close to L, and continuous functions take close inputs to close outputs, so an=f(bn) is close to f(L).
In the example {51/n} we know that {(1/n)ln(5)} converges and its limit is 0. Since the exponential function is continuous, {e(1/n)ln(5)} is a convergent sequence and its limit is e0=1. That is, limn→∞51/n=1.

A more intricate example
Consider the sequence {n1/n}. This is a bit like the preceding example. It has the form BASEEXPONENT. Here the base is n, and is growing. The exponent is 1/n, and is shrinking. Which one "wins"? That is, is this sequence not bounded, and not converging? Is there some other kind of behavior? Again, taking logs helps. So since ln(n1/n)=(1/n)ln(n)=ln(n)/n, we need to analyze limn→∞ln(n)/n. As n grows, the top and the bottom separately grow and go to ∞. This limit is eligible for L'Hôpital's Rule. (If you don't check the eligibility, you are bound to make a serious mistake sometime!) So: limn→∞ln(n)/n=limn→∞(1/n)/1=0 Therefore just as before, the limit of the original sequence is e0=1.
Here are the first 7 elements of the sequence: 1., 1.414213562, 1.442249570, 1.414213562, 1.379729661, 1.348006155, 1.320469248. These don't signal to me immediately that the limit is 1. Computation can be misleading! I mentioned in class that more complicated examples could easily be written where the contest of base vs. exponent is not easy to decide.

• {(50}n}. This sequence diverges (it is certainly unbounded).
• {(1/50}n}. This sequence converges (its terms approach 0 rapidly).
• Here is a rather sophisticated example: {(50)n/n!}. The top is bigger and bigger powers of 50, while the bottom is n!. Both the top and the bottom grow. Understanding the asymptotic behavior of this sequence as n→∞ is asked to decide which grows faster, the top or the bottom. I mentioned in class that this bizarre looking sequence is not random, and really does occur in applications. Here is some numerical evidence (the first 7 terms, rounded to 10 digit accuracy):
50, 1250, 20833.33333, 260416.6667, 2604166.667, 2.170138889·107, 1.550099206·108.
If you want more evidence, than the 20th is about 4·1015, and that's quite big to me. But, in fact, this sequence converges. It is important not to be deceived by what happens for small n's. We are interested in the asymptotics as n grows really, really big.

 Some students noticed something like the following argument. Take n to be an integer larger than 100. (You will soon see why I selected 100: it is because it makes some computations and comparisons easier). Then```/ (50)n \ (50)(50)(50) ··· (50) | ----- | = ---------------------- \ n! / 1·2·3·4·5···(n–1)n ``` Look at the product of the first 100 fractions above. This is some enormous number (the top is (50)100 and the bottom is 100! The valueof a100 is about 8·1011 actually. How many terms are left over? The terms from 101 up to n, and that's n–100 (n minus 100) terms. Each of them has 50 on top, and a number bigger than 100 on the bottom. So the leftover terms all are less than 1/2. That means: {(50)n/n!}< {(50)100/100!}(1/2)n–100. The powers of 1/2 drive down the sequence values when n is bigger than 100. I chose 100 because it would be nice to get some definite number like 1/2, so you can see things decreasing. In fact, when n=200, the value of the sequence element is about 7.89·10–36, quite small. I claim that the following result is correct, but is definitely not "clear": limn→∞{(50)n/n!}=0. Squeezing a limit out ... The formality behind the previous argument is a version of the Squeeze Theorem, which goes something like this. If I have three sequences, {an} and {bn} and {cn} and if I know for all n's that an≤bn≤cn and if I know that limn→∞an=L and limn→∞cn=L, then the middle sequence {bn} also converges, and its limit is L. The preceding example had an=0 and bn=(50)n/n! and cn=[(50)100/100!](1/2)n–100. I don't think the argument is obvious.

QotD
Well, I asked people to do problem 12 in section 10.1. This shouldn't be difficult.

Monday, March 23 (Lecture #15)
The previous QotD This was
Find the solution of y´=y2x3 which goes through (2,1). Be sure to write the solution as y=some function of x.
Volunteers in both lectures did this problem on the board. In the first lecture, these were Mr. Stebbins and Mr. Weisser. In the second lecture, they were Mr. Boemo and Mr. Fishbein.

• Separate: y–2dy=x3dx
• Integrate: ∫y–2dy=∫x3dx
• The general solution is in implicit form as –y–1=[x4/4]+C
• Initial condition leads to constant of integration: –(1)–1=[24/4] so C=–5.
• The transition from implicit description to explicit description –y–1=[x4/4]–5 so y–1=–[x4/4]+5 and y=1/(–[x4/4]+5).

I mentioned the discouraging statistics about people who made algebraic errors when going from implicit to explicit description of the solution. I was unhappy. Anyone who believes that 1/(A+B) and (1/A)+(1/B) are likely to be the same is invited to give me two half-dollars ([1/2]+[1/2]) and I will give them in exchange a quarter (1/[2+2]). You've got to think about what you write -- everyone makes mistakes, but learn to check for them, and fix them. Please! At the end of this lecture is my candidate for the QotD. It is a problem similar to the one just discussed. You can try it and see if you get the correct answer.

Modeling this story to solve the problem
Declaring that the quantity FROG is directly proportional to the quantity TOAD is language meaning that FROG=KTOAD for some positive constant K. So if T represents the temperature of the hot object, then Newton's Law of Cooling implies that RATE OF CHANGE OF T=K(difference between T and 30). With the usual notation, realizing that "rate of change" is a derivative, and using T in oC and t for time measured in minutes since the object was brought into the room, Newton's Law of Cooling becomes dT/dt=K(T–30) with some unknown constant, K. What about the other information in the story? I've collected the information in a possible graph of the temperature of the object to the right. Our guess for the solution curve (the description of the temperature as time changes) is in magenta. It should tend to 30 as t→∞.

Solving the model equation
dT/dt=K(T–30) is separable, so we get (I'm beginning to skip steps) ∫[1/(T–30)]dT=∫K dt so that ln(T–30)=Kt+C. We have two constants, K and C, to be determined. But we know two "chunks" of information about the temperature, (10,80) and (20,40). Therefore (plugging in):
(10,80) gives us ln(50)=K(10)+C and (20,40) gives us ln(10)=K(20)+C.
We have two linear equations in two unknowns. We can solve them. Use your favorite method, or:
Double the first equation and subtract the second to get 2ln(50)–ln(10)=C. I will simplify using log properties, so C=2ln(50)–ln(10)=ln({50}2)–ln(10)=ln(2500)–ln(10)=ln(250).
Let's put this value of C in the first equation. We have ln(50)=K(10)+ln(250) so 10K=ln(50)–ln(250)=ln(50/250)=ln(1/5) and K=(1/10)ln(1/5).

The solution and the solution curve
The implicit form of the equation is therefore ln(T–30)=(1/10)ln(1/5)t+ln(250). Let's exponentiate, and use exponentiation properties and log properties:
T–30=e(1/10)ln(1/5)t+ln(250)=e(1/10)ln(1/5)teln(250)=eln[(1/5)(1/10)t](250)=250(1/5)(1/10)t, and, finally, T=250(1/5)(1/10)t+30. Whew! Please note that on an exam I would not require so much "simplification" but when you go out and work on such problems, people usually like the answers written in such "simple" ways. Since T(t)=250(1/5)(1/10)t+30, the initial temperature, T(0), was 250+30=280. The temperature after one hour is T(60) (60 minutes=1 hour) and that is 250(1/5)6+30 which is 30.016 (no, I did not compute this!). Also please note that as t→∞, since (1/5) is a positive number less than 1, 250(1/5)(1/10)t→0 so that T(t)→30 as we had guessed.
To the right is a graph of T(t)=250(1/5)(1/10)t+30 in magenta with dashed black lines at 80, 40, and 30. The 80 and 40 lines cross at 10 and 20, respectively, and the curve overlays the 30 line as t grows. The curve seems to cross the t=0 line at T=280.

Bacteria again ...
The differential equation dy/dx=Ky with K constant (so the solutions can be written as y=CeKx) is widely used for modeling drug levels in blood and radiocarbon dating (K negative) and for modeling growth of bacteria, yeast, fungus, etc. when K is positive. x is usually time. How valid is this? Well, if we had E. coli which would "divide every twenty minutes" we should be 50 feet under E. coli even if the darn things started only a few weeks ago. In fact, for such growth there are limits to nourishment and other factors which retard growth as the population increases. So people have considered many differential equations which model growth. Let me discuss a very simple case of a widely accepted model equation.

A better model: the logistic equation
Consider the differential equation dy/dx=y(2–y). If y is close to 0 but positive, then the right-hand side seems to be about "2-ish" multiplying y. So this is very much like dy/dx=2y, exponential growth. If y is close to 2 but less than 2, then dy/dx is (really small number) multiplying y, so y will grow very slowly. I will try to solve a specific initial value problem for this equation, which is an example of the Logistic Differential Equation. The numbers are selected to make the algebra as easy as possible.

The initial value problem will consist of the differential equation dy/dx=y(2–y) and the initial condition y(0)=1.

An explicit form of the solution is y=[2e2x]/[1+e2x]. This may be in a form that "real people" can understand.
A picture of this curve is shown to the right. The curve does pass through (0,1). For t negative, the curve does sort of look exponential. When t is large enough (t=0 in this case), the concavity of the curve flips (it is always increasing!) and as t→∞, y→2. 2 is sometimes called the carrying capacity of this "system". This function is sometimes called (a version of) the logistic function.

But what happens if ...
What if we wanted to look at a different, even just slightly more complicated model, say dy/dx=y(y–1)(y–2) etc. Then separating etc. is possible. But trying to convert the implicit description of the solution to an explicit description, one that can be analyzed more easily, is essentially impossible unless we are very lucky.

A different kind of reasoning
There is a different way to study such equations, one where geometric reasoning is used instead of lots of algebraic computations. When this sort of reasoning applies, getting asymptotic information is usually much easier than one would think possible. The tool I will discuss for studying differential equations is called the direction field in your textbook (in others it is called the slope field). I will use this tool to get an approximate idea of the shape and behavior of what are called solution curves or integral curves of the differential equation. These are the graphs of particular solutions of the differential equation.

Direction fields
Let's look at the differential equation y´=x2–x–4y2. I can't "solve" this but let me tell you just a small amount of information about the solution near the point (3,1). A solution curve passing through the point (3,1) will have its slope determined by the differential equation. I mean that y´ when x=3 and y=1 will be the value of x2–x–4y2 when x=3 and y=1. This is 32–3–4·12=9–3–4=2. So the tangent line to the curve at (3,1) will have slope 2. Well, let's think about it. I'll draw a thickened-up line segment of slope 2 at (3,1) and consider some curves going through that point.

 Curve #1 I've drawn a little piece of a curve, tangent to the chunk of line (it is formally called a line element, I think). I think this curve, which is increasing, might possibly be a solution to this differential equation near this point. A possibility Curve #2 Here is another curve going through (3,1). I believe that this could also be the solution curve. O.k.: it is concave down, and I surely don't know what the concavity of the solution curve is at (3,1) without more study. But in my casual way, I have two possible candidates for the solution curve. Certainly it is also increasing near (3,1) because the slope at (3,1), which is 2, is positive. A possibility Curve #3 Now this candidate is really a different sort of curve. This curve does pass through (3,1) but it is decreasing, and its slope is certainly not 2 at (3,1). This is a losing candidate (!) and certainly can't be a little piece of a solution curve. Not possible!

I don't think that this "reasoning" is profound. I am merely asserting that I can eliminate and (tentatively) accept certain candidate curves as good candidates for solution curves. Now I will extend the reasoning, by drawing lots of line segments at lots of points and see what my brain's visual processing power can tell me about the curves. In fact, (and this is somewhat amazing to me) with some practice, most people can "see" the curves quite clearly. So here are some examples. The direction fields were drawn by computer, but the curves were (badly) drawn by me. At least what's here is clearer than what happened in class!

dy/dx=(1/20)x2
This is the direction field This is the direction field with
some solution curves.
Discussion
I had Maple draw a bunch of direction fields. I admit that the instructions took a bit of practice, but once I understood them it wasn't too difficult to produce the pictures, and I honestly tell you that there was a great deal less computation than if I had requested any approximations to solution curves numerically. There are 12·12=144 line elements in each direction field picture. The rather strange numbers which appear in the differential equations (for example, the (1/20) in this equation) were chosen so that the tilts of the direction field elements would be easier to see in these pictures.
This differential equation can be solved easily and its solutions are (1/20)(1/3)x3+C. But I want to look at the curves and not the formulas. Here to the left is the direction field without any decoration. To the right is my attempt to draw by "hand" curves which, whenever they touch a line element, have the line elements as segments of tangent lines. I think you can see that the solution curves are increasing, and that the concavity is down on the left and up on the right. That's all I want from this example.
dy/dx=(1/10)xy
This is the direction field This is the direction field with
some solution curves.
Discussion
O.k., algebraically this differential equation is still something we can handle fairly easily. This is a separable equation, and we can solve it fairly easily. Let me take the 144 pieces of the direction field and try to draw some curves which, when they touch any part of the direction field, will have that tiny line segment looking like it is tangent to the curve. To me these words more make things harder to understand. Draw some curves!
The curves I "drew" are concave up when they are above the x-axis, and they are concave down when they are below the x-axis. I feel that I understand how initial conditions would "evolve" (?) forward and backward. People frequent thing that x is a variable representing time. So an initial condition located in the upper halfplane, above the x-axis, evolves forward (the future) and gets very very large. Also, in the past, it came from something very very large. That's all I want here: just approximate qualitative information about solutions.
dy/dx=(1/30)(y+2)(y–2)
This is the direction field This is the direction field with
some solution curves.
Discussion
I can still solve this algebraically, I think (I haven't tried). This is a variant on the logistic equation which has a nicely worked out solution above. Here the numbers are a bit more random, though. The direction field tells me more in a short amount of time than working algebraically would tell me in a long time.
Look at the picture of the direction field, and then look at the solution curved indicated. There is a great deal going on in the picture of the curves, and let me try to pick out some features.
• First, there are some horizontal lines which are solutions. How can I detect them? If y=Constant is a solution to some complicated differential equation dy/dx=F(x,y), then because the horizontal line always has dy/dx=0, it had better be true the the complicated function F(x,y) is 0 when we plug "Constant" in for y. This may seem complicated to you, but here look at the right-hand side: (1/30)(y+2)(y–2). When is this equal to 0? Without being silly, I think I can write that clearly this occurs for y=–2 and y=+2. The horizontal lines defined by these questions are then solutions to the differential equation dy/dx=(1/30)(y+2)(y–2) because both sides are 0. These solutions don't change with "time" if we think of x as time. Here is some vocabulary which is used when discussing these solutions: they are called equilibrium solutions or steady-state solutions.
• Now I want to discuss other (non-equilibrium) solutions for this differential equation. Consider first solutions near the equilibrium solution y=–2. If we take initial conditions (x0,y0) with y0 not equal to –2 but near –2, soemthing sort of interesting happens. The curves above y=–2 get moved down as we go futureward (x increases). They get moved down towards –2, and as close to –2 as you want (yes, a limit indeed). And if y0 is below or less than –2, the curves below y=–2 move up towards –2. More precisely, if we take an initial condition (x0,y0) with y0 close to –2, the resulting solution y=f(x) has the property that limx→∞f(x)=2. Nearby solution curves converge to the equilibrium solution y=–2. This situation has its own name: y=–2 is called a stable equilibrium. There are many physical situations (mechanical, chemical, etc.: many!) where learning about stable equilibriums (equilibria?) is very important. In real life, determining an exact measurement (y=–2) may be very difficult. It is far better to know that, hey, starting something off near –2 will have the same long-term effect as starting it at –2. This is useful and neat.
• Let's look more closely at the other equilibrium solution, y=2, and its neighboring solution curves. Suppose we start near but above y=2. If you look at the picture, you'll see that the direction field pushes (?) the curve away from y=–2 (these curves go to ∞). Similarly, if we start near but below y=2, the curves get pushed away from y=2. The lower curves go towards y=–2, which right now is not interesting: I want to concentrate are things near y=2. So if this differential equation described a physical situation, I would now know that if my initial condition (x0,y0) had y0 exactly at 2, then on the solution curve, I would have y0 always equation to 2 as x0 increased (the future). But if I didn't get y0 exactly equal to 2, I have no ability to predict the future. Maybe y will tend to ∞ or maybe to –2. Here's the vocabulary, which shouldn't be surprising: y=2 is called an unstable equilibrium.
• Physical situation
 Stable equilibrium Think of a ball at the bottom of a bowl. It will just sit there, position not changing. If you jiggle the bowl a bit, then the ball will move, but eventually (if you stop jiggling) it will go towards the bottom again. This "bottom" situation is a stable equilibrium. Unstable equilibrium Turn the bowl over and put the ball on top of the bowl. You may conceive of a situation where you haved place the ball so carefully on the top that is it totally balanced and still, motionless. Ideally the situation would go on forever -- equilibrium. But now jiggle the ball in any way, with a very small jiggle, in any direction. The ball will roll off. This is unstable equilibrium.
dy/dx=(1/30)y2(y–2)
This is the direction field This is the direction field with
some solution curves.
Discussion
O.k.: I definitely would not want to try to "solve" this equation algebraically and get explicit solutions. But the direction field approach combined with simple reasoning allows me to find equilbrium solutions and even to detect which are stable and which are unstable, and this can be very useful in practice.
I diagnose equilibrium solutions by considering (1/30)y2(y–2)=0. The roots are 0 and 2. So y=0 and y=2 are the only equilibirum solutions for this differential equation. Yes, I admit it: this example is very carefully chosen so that the method works, but the ideas are sufficiently nice so that the method will work even for examples which aren't so "pretty".
What about stable and unstable? Again, look at the direction field, and then examine the solution curves which are drawn. Both of the equilibrium solutions are unstable. Nearby solution curves are not "sucked into" the equilibrium solution. Yes, it is true that curves above y=0 get pulled towards y=0, but to be an equilibrium solution, both sides (up and down) must have solution curves which are attracted to the equilibrium, and this does not occur here. Again, both of these equilibrium solutions are unstable.
I really fouled up time allocation in this lecture. Somehow I miscalculated. I didn't get a chance to show this last picture and discuss it in either class.
I also didn't get a chance to give a QotD. So the last picture I wanted to show you follows along with some discussion, and so does the QotD, which you can do if you wish. A solution is also provided via a link so you can compare your solution with mine.
dy/dx=(1/20)(x+y2)
This is the direction field This is the direction field with
some solution curves.
Discussion
O.k., the last example for this method is really nearly random. The right-hand side, (1/20)(x+y2), is a low-degree polynomial, but I don't know how to solve the equation (see the remark below if you are curious, though!). There are no equilibrium solutions because there is no constant C which makes (1/20)(x+C2) equal to 0 for all values of x. So here's a complicated situation. But still, consider the direction field, which is easy to have drawn, needing very little computational effort. And consider the curves I drew. I bet that the following occurs: if y=f(x) is a solution to this differential equation, then there will be exactly one critical point on the solution curve, and this critical point will be an absolute minimum. To the right of the solution curve, the function will increase. In fact, it will explode in finite time, that is, there will be some Q so that limx→Qf(x)=∞, but I don't think the picture necessarily tells that. What about to the left of the solution curve? I bet that somewhere the concavity of the solution changes, and it becomes concave down, and that the curve sort of tapers off to a sort of flat situation.
Everything predicted here is actually correct. It would be rather difficult (but possible, I think, but I wouldn't want to do it!) to confirm all this algebraically. I think the direction field idea is worth using as a tool when analyzing differential equations, especially when some quick asymptotic information is wanted.

 The real thing Much computation is needed to produce an "explicit" solution to dy/dx=(1/20)(x+y2). To the right is a picture of the solution curve which goes through (0,0). The function defining the curve can't be written in terms of the standard functions you know. The rather simple-looking picture is the result of a bunch of Airy functions combined in very strange ways. I am not inventing all this! The situation is very complicated. Here is the solution in detail, written by a silicon friend of mine:``` 2/3 2/3 50 x (2/3) 50 x (1/3) (1/6) AiryAi(1, - -------) 150 + 5 AiryBi(1, - -------) 20 3 100 100 1/5 --------------------------------------------------------------------- 2/3 2/3 50 x (2/3) 50 x (1/6) AiryAi(- -------) 3 + AiryBi(- -------) 3 100 100``` I don't understand this complicated formula at all. But I do understand the direction field and the solution curves.

QotD
Find the solution of y´=(ex/3)/y which satisfies y(0)=2. Be sure to write the equation as y=some function of x. Here is a solution which you should look at after you try the problem yourself.

Wednesday, March 11 (Lecture #14)
What's a differential equation?
A differential equation is an equation relating an unknown function and one or more derivatives of the unknown function with some other functions.

Examples
y´=x2 Done in 151; to be reviewed here.
y´=5y Done in 151; to be reviewed here.
y´=y2 We'll discuss this here.
y´=xy We'll discuss this here.
y´=x+y To be discussed in your differential equations course.
y´´=–y To be discussed in your differential equations course. This equation governs simple harmonic motion (the movement of an ideal vibrating spring).
y´=x2+y2 This can't be "solved" in terms of standard functions.

Order of a differential equation
This is the highest numbered derivative which occurs in the differential equation. In these two lectures we will look only at first order equations. All of the examples above are first order, except one (the simple harmonic motion equation) which is second order. Please note that the differential equations course which most of you will take will indeed study equations of higher order, and that these do occur. Some standard vibrating beam equations studied in mechanical engineering are fourth order, and many of the equations in physics and chemistry are second order.

Story #1
Probably we have all been told that bacteria (usually) reproduce by, say, binary fission. This is more or less correct, and more or less the fact means that the rate of increase of bacteria at any time is directly proportional to the number of bacteria at that time. So twice as many bacteria "now" means that twice as many bacteria are being born now. This is certainly dreadfully simplified, but this approximation works in many circumstances. I wondered, when I first heard this fact, why, if, say, E. coli doubles rather rapidly, shouldn't the world be covered very soon by a layer of E. coli which is 40 feet thick? In fact,

A single cell of the bacterium E. coli would, under ideal circumstances, divide every twenty minutes.
(From Michael Crichton (1969) The Andromeda Strain, Dell, N.Y. p.247)
But of course anything growing so rapidly in the real world (mold in a petri dish) enters a situation where the growth challenges the ability of the environment to support the thing. Most environments have a carrying capacity -- some sort of upper limit to the amount of the thing which can live in the environment. Differential equations can model this sort of situation fairly well, combined with the "exponential growth". But exponential growth was studied last semester, and the equation y´=5y sort of models unrestricted exponential growth.

Story #2
We start with an 800 gallon tank of pure water. It is being filled with a fluid at 50 gallons per minute, and these 50 gallons contain 5 lbs of salt. At the same time, 50 gallons per minute of the solution in the tank is being drained. How much salt is in the tank at any time? How much salt would you expect to be in the tank after a long time?

Let's construct a differential equation which models the salt in the tank. We'll call S(t) the number of pounds of salt in the tank at time t. How much salt is being added? Well, 5 pounds per minute. How much salt is being taken away? This is more subtle, and we had some discussion of our assumptions during class. The simplest analysis, which we will do here, is to assume that the tank contents are mixed well: it is homogeneous. The situation with a large real tank might not match this, of course. But, actually, real containers and tanks sometimes have mixing devices installed to try to match this assumption. Well, if there are S(t) pounds of salt in the tank at time t, and if the tank holds 800 gallons, and if 50 gallons are taken out, then the proportion of 50/800 of the salt is taken out: [1/(16)]S(t). Now we put things together.

dS/dt, the rate of change of the salt, is +5[1/(16)]S(t). The differential equation is dS/dt=5–[1/(16)]S. We also shouldn't forget that we start with no salt at all in the tank: S(0)=0.
Prediction? What should happen over the long term to the amount of salt in the tank? It starts out at 0, and then increases ... to what? Well, a guess is that the amount of salt in the tank should increase to 80 pounds, which is the same as the salt concentration (1 pound per 10 gallons) incoming. We will see how to solve the differential equation and check this prediction. The solution is discussed here.

Solution of a differential equation
A solution of a differential equation is a function which, when it and all of its relevant derivatives are inserted into the differential equation, makes the equation true for all values of the domain variable. I know this may seem long-winded, but I hope the discussion and examples which follow will shown why such elaboration is necessary.

An example: y´=x2
Well, we know how to solve y´=x2: just integrate. We did this repeatedly last semester (and even this semester). The solutions are y=(1/3)x3+C, where C is any constant. There are infinitely many solutions. A few of them are shown to the right.
The blue curve has C=1: y=(1/3)x3+1. It is the solution curve which goes through (0,1).
The red curve has C=3: y=(1/3)x3+3. It is the solution curve which goes through (0,3).
The green curve has C=–2: y=(1/3)x3–2. It is the solution curve which goes through (0,–2).
The solution curves are just vertical translates, up and down, of each other. They are all the same shape, have the same domain, etc. This situation is rather straightforward, as you will see.

General solution; particular solution
There is some special vocabulary used. The differential equation y´=x2 has the general solution f(x)=(1/3)x3+C. When C has a specific value, then the function is called a particular solution. So f(x)=(1/3)x3–2 is a particular solution, and it is the only particular solution which passes through (0,–2). The specification (0,–2) is called an initial condition. That comes from the physical situation where x represents time, and we think that the y-value corresponding to the given x-value represents a certain starting place. Sometimes people write y(0)=–2 as the initial condition. That can confuse me. The combination of an initial condition and a differential equation is called an initial value problem.

An example: y´=y2
Now let's change and consider y´=y2. I guess that the general solution is f(x)=1/(C–x). (I'll show you how to guess it also, very soon!) How could you check that my suggestion for a solution actually is a solution? Well, if f(x)=1/(C–x) then, since f(x)=(C–x)–1, we know that f´(x)=(–1)(C–x)–2(–1). The first –1 comes from the power, and the second –1 comes from the Chain Rule, so they cancel. But (C–x)–2, the derivative of the function, is actually the square of (C–x)–1, the original function. We have now verified that f(x)=1/(C–x). does solve y´=y2. Now let's look at some particular solutions.
The blue curve has C=1: y=1/(1–x). It is the solution curve which goes through (0,1). It has domain (–∞,1), and is increasing and concave up.
The red curve has C=1/3: y=1/({1/3}–x). It is the solution curve which goes through (0,3). It has domain (–∞,1/3), and is increasing and concave up.
The green curve has C=–1/2: y=1/(–{1/2}–x). It is the solution curve which goes through (0,–2). It has domain (–1/2,∞), and is increasing and concave down.
These solution curves are not just vertical translates of each other. Their domains are different (yeah, this matters in real life) and the solution curves have different shapes. The particular solutions "blow up" at different numbers. And this is still a fairly simple differential equation.

A big theorem and a joke
When I was young, so much younger than today, I was told the following BIG THEOREM about differential equations.

THEOREM
Suppose we have some function of two variables, F(x,y), and we are interested in the differential equation y´=F(x,y), and a solution going through the point (x0,y0). Then there always is a solution, and there is exactly one solution.
The theorem's name is the Existence and Uniqueness Theorem for solutions of differential equations. "Existence" because the theorem declares that there is a solution, and "Uniqueness" because the theorem declares there is exactly one solution. There are some mild "technical" conditions the function F(x,y) should satisfy, but almost everything you're likely to look at will be covered theoretically by this theorem. You will see this result later in your differential equations courses.

After I learned about this theorem, I thought that all this worry about differential equations was totally silly -- the theorem tells you everything. This is false. In practice, the theorem doesn't tell you how to compute or approximate solutions efficiently. It doesn't tell you what the domains of the solutions are (this is important in applications). It doesn't tell you the asymptotic behavior of the solutions (how much salt there is after a long time). These questions are very important, and they are the questions which need to be answered in practice. All this makes me think of my favorite math joke.

JOKE
Several people are in a hot-air balloon, trying to land over a fog-shrouded countryside at the end of a long day. The balloon dips down low and they see the ground faintly. Spotting a person, one of them calls down: "Where are we?" Some minutes later the wind is carrying them away and they hear faintly, "You're in a balloon!" One person in the balloon gondola says thoughtfully to the other, "It's so nice to get help from a mathematician." The other says, "How do you know that was a mathematician?" The first replies, "There are three reasons: it took a long time to get the answer, it was totally correct, and, finally, it was absolutely useless."
While I love mathematics, and I think math is beautiful and helpful, please remember the final sentence of this joke.

Separable equations
A separable first order differential equation is one which can be written in the following way: dy/dx=F(y)G(x). The right-hand side is a product of some function in y multiplied by some function in x. I'll describe a procedure which leads, in many cases, to a solution.

1. Separate: put all the y stuff on one side and put all the x-stuff on the other side. The equation becomes dy/F(y)=G(x)dx.
2. Integrate:∫dy/F(y)=∫G(x)dx. Of course, the practicality of this (at least "by hand") depends on the specific functions F and G.
3. Solve for y as a function of x. Again, this may or may not be practical. Examples will help you to understand.
This method works because it is implicit differentiation in reverse. I didn't discuss the reasoning because the examples right now are more important.

Some examples

• Consider y´=y2 which is dy/dx=y2. Here F(y)=y2 and G(x)=1. Well, the first step gives dy/y2=dx. Now integrate both sides (remember that 1/y2=y–2, please). We get –1/y=x+C. This can be solved for y fairly easily, and so y=1/(C–x). (Yes, if you are very alert you would have noticed that I changed –C to C, and everyone does that, because I can stuff the minus sign into the unknown constant C.) So at least I've shown you where my guess above comes from. I've forgotten what I did, but I think it was something like: let's find a solution of y´=x2y which goes through the point (1,2). So if dy/dx=x2y then dy/y=x2dx and then∫dy/y=∫x2dx so that ln(y)=(1/3)x3+C. If we want this to "go through" (1,2) then ln(2)=(1/3)13+C so that C=ln(2)–(1/3). And the particular solution is ln(y)=(1/3)x3+ln(2)–(1/3). Most people, especially in applications, prefer an explicit solution, so an attempt to solve for y is made. I can do this here (the situation is simple) but in general things may be too complicated. The equation ln(y)=(1/3)x3+ln(2)–(1/3) can be exponentiated to give y=e(1/3)x3+ln(2)–(1/3). Although for me on an exam this would be fine, I would not want to mislead you. Again, especially in applications, appropriate simplification is very useful. Here this is what would be done:
e(1/3)x3+ln(2)–(1/3)=e(1/3)x3eln(2)e–(1/3)=2e(1/3)[x3–1]. So the particular solution of y´=x2y satisfying the initial condition y(1)=2 is f(x)=2e(1/3)[x3–1]. Part of a graph of y=f(x) is shown to the right. The function increases very quickly when x is positive.
• Here is an important counterexample: y´=x2+y2 is a differential equation which is not separable. This is not totally obvious. Please let me try to convince you.

If x2+y2=F(y)G(x) is true, I will "explore" the possibilities using values of x and y. For example, if x=0 and y=0, we see that 0=F(0)G(0). Therefore either F(0)=0 or G(0)=0 (or both, of course). What if G(0)=0? Then try x=0 and y=1. The equation becomes 02+12=F(1)G(0), and this is 1=F(1)G(0). If G(0)=0, this is impossible! If the alternative holds, that is, F(0)=0, just insert x=1 and y=0 to get a contradiction.

To the right is a picture of the solution curve to y´=x2+y2 which goes through (1,2). The function defining the curve can't be written in terms of the standard functions you know. Quite a bit of computation is needed to produce the rather simple-looking picture (the darn picture needs a dozen Bessel functions combined in very strange ways).

Back to the salt tank ...
The differential equation dS/dt=5–[1/(16)]S is separable. There was some difficulty in convincing students of this. Look:
5–[1/(16)]S=(5–[1/(16)]S)(1), and 5–[1/(16)]S is a function of S alone, and 1 is a function of t alone.

Let's separate and solve. So dS/{5–[1/(16)]S}=dt, and the right-hand side integrates to t+C. The left-hand side is maybe a bit more intricate. You could substitute: w=5–[1/(16)]S so dw=–[1/(16)]dS and dS=–16dw. The result is –16ln(w)=–16ln(5–[1/(16)]S). I generally guess, get it wrong, and need to guess again. So after integrating we have –16ln(5–[1/(16)]S)=t+C. The initial condition here, a result of the tank originally being filled with pure water, is S(0)=0. So we can get C:
–16ln(5–[1/(16)]S)=t+C become –16ln(5–[1/(16)]0)=0+C and C is –16ln(5).

The solution is –16ln(5–[1/(16)]S)=t–16ln(5). Most people prefer a more explicit formulation, so we solve for S as a function of t.
Divide by –16: ln(5–[1/(16)]S)=–[1/(16)]t+ln(5).
Exponentiate: 5–[1/(16)]S=e–[1/(16)]t+ln(5).
Some algebra on the right: e–[1/(16)]t+ln(5)=e–[1/(16)]teln(5)=5e–[1/(16)]t.
Now get S: 5–[1/(16)]S=5e–[1/(16)]t becomes –[1/(16)]S=–5+5e–[1/(16)]t which turns into S=80–80e–[1/(16)]t.

What does the solution look like?
If we believe S(t)=80–80e–[1/(16)]t, let's check the initial condition:
S(0)=80–80e–[1/(16)]0=80–80·1=0. Good!
How about the long-range asymptotic behavior? That is, what happens when t gets very large (t→∞)?
If t→∞, then –[1/(16)]t→∞, so e–[1/(16)]t→0. The combination 80–80e–[1/(16)]t must therefore →80, which is what we expected. Let's see what the S(t) curve looks like. But here, unlike in class, I will try to explain the final picture.

 Here is 80e[1/(16)t for t between –50 and 0. This is part of an exponential growth curve, and it starts small and increase up to 80, the value at 0. It is concave up. Now I've flipped the curve across the vertical axis. This is 80e–[1/(16)t for t between 0 and 50. The curve is still concave up, but it is decreasing: since the constant is negative, this is exponential decay. I flipped the previous curve across the horizontal axis. This is a graph of –80e–[1/(16)t for t between 0 and 50. It is concave down and increasing, from 80 to near 0. We'll get the real picture of S(t) by translating this up 80.

To the right is a graph of S(t)=80–80e–[1/(16)t for t between 0 and 100. The dashed red horizontal line is at height 80, the asymptotic level of the salt in the tank. You can see that the salt starts at 0, the initial condition, and then increases and sort of curves underneath the line at height 80. The curve is concave down. The difference between 80 and S(t) becomes rather small as t grows.

QotD
Find the solution of y´=y2x3 which goes through (2,1). Be sure to write the solution as y=some function of x.

Monday, March 9 (Lecture #13)
MVT simplified: just Rolle's Theorem
I wrote this on the board before the class began. It is a result I'll use several times in the initial discussion. Here:
Suppose we have a differentiable function which is 0 at two distinct points. Then the derivative of the function will be 0 at least once between these two values.
Important and useful
Much of the remainder of the semester will be devoted to material which is extremely useful to everyone involved with engineering and science. Today is likely to be your first exposure to the ideas which are generally used to compute values of functions. It is almost certain that these ideas are at the center of how the calculators and computers you use compute (and graph) most functions. You should know something about these ideas.

When I taught this last year I jumped right into an example, and didn't show people any of the structure. The examples, even the easiest, can be intricate. I don't feel I was too successful here last year, so I want to try a different approach today. Some of what I say will not be in the textbook. I will begin with a totally unmotivated question. The question will look somewhat weird.

What's K?
Suppose f(x) is a differentiable function, a and b are numbers with a≠b, and we consider the equation:
f(b)=f(a)+f´(a)(b–a)+[f´´(a)/2](b–a)2+K(b–a)3.
Then there is some number K which makes this equation correct. Why? Because since a≠b, (b–a)3 isn't 0, and I could solve for K in this whole messy equation. So I want to investigate what K is, and get another way of writing K. To do this I will use the Mean Value Theorem repeatedly in the form quoted above.

Consider the function
G(x)=f(x)–(f(a)+f´(a)(x–a)+[f´´(a)/2](x–a)2+K(x–a)3).
What do I know about G(x)? Well, I know that G(a)=f(a)–(f(a)+f´(a)(a–a)+[f´´(a)/2](a–a)2+K(a–a)3) so that G(a)=0. I also know that G(b)=f(b)–(f(b)+f´(a)(b–a)+[f´´(a)/2](b–a)2+K(b–a)3). This is also 0 because K was chosen so that this is true. So G(a)=0 and G(b)=0. Therefore (MVT/RT above) there is some number in between a and b, I'll call it c1, so that G´(c1)=0.

Now let's compute the derivative of G(x). There are many letters around. I am differentiating with respect to x. With this in mind, I see:
G´(x)=f´(x)–(0+f´(a)1+[f´´(a)/2]2(x–a)+3K(x–a)2).
Now what? Well, G´(a)=f´(a)–(0+f´(a)1+[f´´(a)/2]2(a–a)+3K(a–a)2) is 0 because of the a–a's and because the f´(a)'s cancel. We also know that G´(c1)=0. Now MVT/RT applied to the function G´(x) tells us that its derivative is 0 somewhere between a and c1. That is, there is c2 between a and c1 so that G´(c2)=0.

Now let's compute the derivative of G´(x), again being careful.
G´´(x)=f´´(x)–(0+0+[f´´(a)/2]21+3·2K(x–a)).
Of course we consider G´´(a) which is f´´(a)–(0+0+[f´´(a)/2]21+3·2K(a–a)) and this is 0. Also G´(c2)=0. So MVT/RT again applies to tell us that there is c3 between a and c2 with G´´´(c3)=0. Wow.

The derivative of G´´(x):
G´´´(x)=f´´´(x)–(0+0+0+(3·2·1)(K)1).
Now we know when x=c3 this is 0. So we know that f´´´(c3)–(3·2·1)K is actually equal to 0. We can solve this for K, and get K=[f´´´(c3)/(3·3·1)]. c3 is between a and c2 and c2 is between a and c1 and c1 is between a and b so that c3 is itself between a and b.

What's going on?
Here is what we know: If
f(b)=f(a)+f´(a)(b–a)+[f´´(a)/2](b–a)2+K(b–a)3.
then there is a number c between a and b so that K=[f´´´(c)/(3·2·1)] (three of the ´ in this). This turns out to be the beginning of a marvelous and successful computational strategy.
So what's your problem with all this? You might claim to not understand what the heck is going on, and, more particularly, why any moderately sane person might want to go through these algebraic contortions. Yes I totally agree with you. But it turns out (sit here, look at the fireworks (?) that follow) this is extremely useful. There have been centuries (!!) of thought involved in preparing and using all this stuff -- it is really clever.

Taylor polynomials
This is copied from page 502 of your text. The Taylor polynomial of degree n for the function f(x) centered at a (wow, what a collection of words!) is
Tn(x)=f(a)+f´(a)(x–a)+[f´´(a)/2](x–a)2+[f(3)(a)/3!](x–a)3+...+[f(n)(a)/n!](x–a)n
There are a whole bunch of things to discuss. Let's see. First, if you've never seen it before, the appearance of the excitement mark, !. This is called a factorial. The value of the factorial of a positive integer is the product of the integer together with all of the integers less than it down to 1: n!=n(n–1)(n–2)(n–3)···(3)(2)(1). Here is a very brief table of factorials:

 n n! 1 2 3 4 5 6 7 8 9 10 1 2 6 24 120 720 5,040 40,320 362,880 3,628,800

The major thing you should notice right now is that the factorials grow very big very quickly. That's computationally important. I also should mention that most people define 0! to be 1. That's so certain formulas are easier to write (really). (It turns out to be possible to define factorials of other numbers. For example, in 251, you can define and compute (1/2)! -- wait for that.)

More notation is in such things as f(4)(a). This means the fourth derivative of f evaluated at a. So f´´(a) can also be written as f(2)(a), and even just f´(a) is f(1)(a). Again, in order to make writing certain formulas easier, most people think that f(0)(a), the zeroth derivative of f evaluated at a (so no derivatives are done!), should just be f(a).

If all of this notation is clear, then here's another, very compact way to write the Taylor polynomial.
Tn(x)=∑j=0n[f(j)(a)/j!](x–a)j.
I hope you can see where the zero factorial and zeroth derivative make this much easier to write.

Example 1
Let's get T8(x) for sin(x) and a=0. So we need derivatives, and here they are:

Derivative #FunctionValue at a=0
0sin(x)0
1cos(x)1
2–sin(x)0
3–cos(x)–1
4sin(x)0
5cos(x)1
6–sin(x)0
7–cos(x)–1
8sin(x)0

The numbers in the last column, together with the factorials, are the coefficients which build the Taylor polynomial. I emphasized with all of these examples that noticing patterns is the way to go. In the case of sine, the derivatives repeated every four. I am not cheating by choosing something excessively simple. Almost every function I know that arises in modeling physical and geometrical situations has patterns in its derivatives, and part of the fun (?) is finding these patterns. Well, now I can write T8(x):
T8(x)=[0/1](x–0)0+[1/1](x–0)1+[0/2](x–0)2+[–1/6](x–0)3+[0/24](x–0)4+[1/120](x–0)5+
+[0/720](x–0)6+[–1/5,040](x–0)7+[0/40,320](x–0)8
Only a nitwit or a very pedantic math instructor would write it that way. Most people would drop the 0 terms, change (x–0) to x, make a few other notational simplifications,and get
T8(x)=x–[1/6]x3+[1/120]x5+[1/5,040]x7
Then there are some questions.
What is T4(x)? It must be x–[1/6]x3.
What is T7(x)? It must be =x–[1/6]x3+[1/120]x5+[1/5,040]x7, in this case just the same as T8(x).
What is T10(x)? I bet it is the same as T9(x), and this would be T10(x)=x–[1/6]x3+[1/120]x5+[1/5,040]x7–[1/362,880]x9.

• Graph of sin(x) on the interval [–8,8].
How can we compute values of the sine function? Here's a graph, but even producing this graph requires a computer to get many values of sine and then use them to pick out and color certain pixels. We could construct a bunch of triangles, measure angles and lengths carefully, and then get some values of sine. This is, more or less, how things were done in Egypt and Greece. Certainly it is an intricate process and very difficult to implement. But hundreds of years later, a more systematic and accurate method was developed, first in India, and then in Western Europe. The history in India was long ignored. Here's a web reference, and a more detailed and readable account is in an article entitled Was Calculus Invented In India? by David Bressoud, in the College Math Journal: Volume 33, 2002. If you are at a Rutgers terminal, this link should get you directly to the article (you might as well take advantage of the fact that you're at a big-time university purchasing big-time web access to lots of stuff!).

• Graphs of sin(x) and T1(x)=x on the interval [–8,8].
The start of this game is to look again at the tangent line approximation which was discussed in calc 1. Let's try to approximate sin(x) locally near 0. The tangent line goes through (0,0) since sin(0)=0 (hey: one value of sine which is easy to know!) and it has slope 1 since the derivative of sine is cosine and cos(0)=1. For x's near 0, look at the picture, and realize that sin(x) will be approximately equal to x.
For example, sin(.05) is 0.04997916927, so that's close. And sin(.0005) is 0.0004999999792: hey, lots of agreement.
But what if we wanted sine's values for x's not so close to 0, and what if we also wanted some measurement of accuracy, some error analysis?

• Graphs of sin(x) and T3(x)=x–(1/6)x3 on the interval [–8,8].
Now I displayed a mysterious (?) cubic (third degree) polynomial. It turns out that this polynomial comes from a rather simple process, one that can be described easily and efficiently to a computer or calculator. In fact, it is a process which is simple enough so that computations can be done by hand, as they were for several centuries. The picture shows that the cubic polynomial stays closer to sin(x) in a larger interval than the linear approximation, x, does. I can't show you also a sort of microscopic picture, but I do assert that cubic polynomial is a better approximation to sin(x) than just x alone. I'll give some numbers.
As I remarked, sin(.0005) is 0.0004999999792 to 10 digit accuracy. Of course, the value of the polynomial "x" at .0005 is .0005. The value of the polynomial x–(1/6)x3 at .0005 is, actually, 0.0004999999792 to 10 digit accuracy. In fact, more extensive computation shows that they agree to 15 digits! I sincerely doubt that anyone in this class will need 15 digits of accuracy in sine computations.

• Graphs of sin(x) and T5(x)=x–(1/6)x3+(1/120)x5 on the interval [–8,8].
Now the polynomial degree is 5, and the graph shows that this polynomial again improves the accuracy of the approximation on a larger interval.
Just a little more numerical evidence (all 10 digit accuracy):
sin(1)=0.8414709848; x at x=1 is 1., of course; x–(1/6)x3 at x=1 is 0.8333333333, a bit closer; and x–(1/6)x3+(1/120)x5 at x=1 is 0.8416666666, so the error is down to less than .001.

• Graphs of sin(x) and T9(x)=x–(1/6)x3+(1/120)x5–(1/5040)x7+(1/362880)x9 on the interval [–8,8].
So I'm skipping a few polynomials and jumping up to the 9th degree approximation. The numbers may look a bit strange (you will get used to them!).
By now doing numerical computations is almost silly, but here is the value of the 9th degree polynomial at x=1: it is 0.8414710096. This agrees with the true value of sin(1) to 6 places (rounded). At x=.5, the approximation and true value agree to 10 decimal places.

Let me explain why T9(.5) and sin(.5) agree to so many decimal places. Well, we need a generalization of the first computation we did. Here is the result, a version of what's called Taylor's Theorem:

If Tn(x)=∑j=0n[f(j)(a)/j!](x–a)j, then the difference between this and f(x) is
[f(n+1)(c)/(n+1)!](x–a)n+1 for some c between a and x.
This is the error or remainder. In the case of sine and n=9, we need to estimate [f(n+1)(c)/(n+1)!](x–a)n+1 when a=0 and x=.5. Well, the 10th derivative of sine is sine or –sine or cosine or –cosine, and in any case the absolute value can't get bigger than 1. So an overestimate of the absolute value of the error:
[f(n+1)(c)/(n+1)!](x–a)n+1 when a=0, x=1/2, and n=9 becomes [1/(10!)](1/2)10 and this is [1/3,628,800](1/1,024) which is less than 10–10. That's where the 10 place accuracy comes from.

Example 2
We get T8(x) for f(x)=ex and a=0.

Derivative #FunctionValue at a=0
....blah.......blah.......blah.......blah.......blah..........blah...
any n you want!ex1
....blah.......blah.......blah.......blah.......blah..........blah...

T7(x)=1+x+[x2/2]+[x3/6]+[x4/24]+[x5/120]+[x6/720]+[x7/5,040]
It's supposed to be easy: this is a technique people actually use.

How good is T7(–.4) as an approximation to e–.4? Here the remainder is     [f(n+1)(c)/(n+1)!](x–a)n+1 when a=0, x=–.4, and n=7 becomes ec[1/(8!)](–.4)8. Now c is between –.4 and 0 and since ex is increasing, the biggest value occurs at c=0, where e0=1. (It is supposed to be easy, otherwise people would not use it!) So the error is at most [1/(8!)](–.4)8. This is less than .00000002. The "true value" of e–.4 is 0.670320046035639 the value of T7(–.4) is 0.670320030476191. I had a machine do these computations, of course. The reason I put quotes around "true value" is that the machine used Taylor polynomials to do the computation, of course.

QotDbr> Suppose f(x)=x1/3 and a=8. Then f(8)=2. I asked people to find T3(x)=f(8)+f´(8)(x–8) +[f´´(8)/2](x–8)2 +[f´´´(8)/3!](x–8)3.

A silicon friend told me that the answer may be 2+1/12*(x–8)–(1/288)*(x–8)^2+(5/20736)*(x–8)^3.

Remainder or error estimate in the textbook
There are many ways of estimating the error between the function f(x) and the Taylor polynomial Tn(x). I think I've actually only ever used one estimate, which is given in your book on page 506 and is called the ERROR BOUND. The text also calls the error the remainder. Here it is:

 Suppose we want to consider f(x), maybe hard to compute, and the nth degree Taylor polynomial centered at a, Tn(x), which we hope is easier to compute. Then |Tn(x)–f(x)|≤K[|x–a|n+1/(n+1)!] where K is some overestimate of |f(n+1)(x)| between x and a.
The most important thing to notice here is the factorial "downstairs". The growth of the factorial as n increases is what frequently will make the error small. This actually happens in many, many examples.

Let me show you a consequence of this estimate in connection with the QotD. Here f(x)=x1/3 and T3(x)=2+(1/12)(x–8)–(1/288)(x–8)2+(5/20736)(x–8)3. To the right is a picture containing graphs of both y=x1/3 and y=T3(x) as x goes from 0 to 16.

Look at the curves. Near x=8, the curves look very, very close. A bit farther away, the curves separate. Indeed, the best way I can see which curve is which is that x1/3 goes through (0,0) so that must be the green curve.

I hope you can see that knowing the polynomial and using it instead of x1/3 might be advantageous, maybe, maybe (you'll see later but the idea is that polynomials are very easy to compute and other functions might not be).

To the right is another picture. Look at it very carefully please. It is also a graph of both y=x1/3 and y=T3(x) as x goes from 6 to 10. Both curves are displayed. But I can only see one curve. Maybe if I look really really really closely, maybe I can see some dots of two colors but I am not sure. How close are these curves?

Well, let's try to use the error estimate in the textbook. That error estimate is quoted above. So an overestimate of the absolute value of the difference between T3(x) and f(x) for x in [6,10] is
|Tn(x)–f(x)|≤K[|x–a|n+1/(n+1)!] where K is some overestimate of |f(n+1)(x)| between x and a.
How can we use this? Well, here f(x)=x1/3 and n=3 and a=8 and the x's involved are between 6 and 10. So we need to understand K[|x–a|n+1/(n+1)!]. This becomes K[|x–8|4/4!]. Of course 4! is 24. What about |x–8|? This is the distance from x to 8. If x is in [6,10], the largest distance to 8 is 2, so |x–8| is always ≤2. Good: now we know that the error is at most K[16/24] (because 24 is 16). What about K? To compute K we need f(4)(x). Well, here are f(x) and a few of its derivatives:
f(x)=x1/3;     f´(x)=(1/3)x–2/3;     f´´(x)=f(2)(x)=–(2/9)x–5/3;     f(3)(x)=(10/27)x–8/3;     f(4)(x)=–(80/81)x–11/3.
I am interested in the absolute value (the size) of the error, so I will forget the minus sign. Therefore I need to know how large (80/81)x–11/3 can get on [6,10]. Because of the negative sign in the exponent, this is a decreasing function of x, and its largest value is at the left-hand endpoint. So the biggest this can be is (80/81)6–11/3. O.k.: I give up -- I will finally use a calculator. This number is about (rounding up!) .0014, and I'll take that as my value for K.

Therefore the error is at most (.0014)(16/24) and this is about .00092. Now look carefully at the picture. Do you think you can see a difference in height of less than .001? The difference between two of the "hashmarks" on the vertical scale is .05, and the graphs of the functions are separated by one-fiftieth (that's 1/50) of that difference. I don't think that the pixels in the picture are small enough to show this!

The picture to the right shown here is on a different interval, from 6 to 6.25 for x. The vertical scale is also different. Each vertical hashmark here represents a height difference of .001, and now, in the part of this interval which is farthest away from 8, near 6, I hope you can indeed see two curves, separated by just a little bit. The error estimate actually gives useful quantitative information.

Wednesday, March 4 (Lecture #12)
I wrote a version of L'Hôpital's Rule on the board, since I would need it twice in the next few minutes. I copied it from the book (page 272).
L'H Suppose f(x) and g(x) are differentiable functions inside an interval containing a and that f(a)=g(a)=0. Also assume g (x) is not 0 for x near but not equal to a. Then
```     f(x)          f´(x)
lim ------ =  lim ------
x→a  g(x)     x→a  g´(x)
```
if the limit on the right-hand side exists. This result is also valid if both of the limits of f(x) and g(x) as x→a are +∞ or –∞.

It is very important that there is a quotient in the algebraic form. Also, the symbolic quotients 0/0 or ∞/∞ are sometimes called indeterminate forms.

x·ln(x)
I asked people for the graph of y=x·ln(x) on the interval [0,1]. Some people were willing to use their graphing calculators. A graph much like what is shown to the left was produced.

The graph seems to be a slightly assymetrical bump below the x-axis, hanging from (0,0) and (1,0). The "bottom" is at, about, –1/3 (near x=1/3, actually). If you "ask" your computer or calculator for the value of x·ln(x) at x=0, however, there will be some sort of complaint rather than a number. The machines have been advised that 0 is not in the domain of ln. But the graph certainly seems to indicate that (0,0) is there (wherever there is!). What's going on?

L'H
Really what the graph indicates is a suggested value for limx→0+x·ln(x). The guess for this is 0. To verify this guess (in a math course!) we will use L'Hôpital's Rule. As x→0+, we have a 0 multiplied by (–) ∞. L'Hôpital's Rule works on quotients, so we need to rearrange things algebraically. Here we go:
limx→0+x·ln(x)=rearranging=limx→0+[ln(x)/{1/x}]=using L'H=limx→0+[{1/x}/{–1/x2}]=rearranging back=limx→0+–x=0 so the limit is verified.

The integral
How much "area" is included in the bump above? More precisely, what is the definite integral of x·ln(x) from 0 to 1. The bump is below the x-axis so this integral should be negative. It sort of resembles a triangle with base [0,1] and altitude about 1/3. So a guess is that the area should be about –1/6. An actual graph of the function and of this approximating triangle is shown to the right. The function is concave down so it bulges beneath the triangle. And, in fact, the "point" of the triangle is above the graph. So there is more (absolute value!) area in the bump than in the triangle.

Let's compute. The area is ∫01x·ln(x)dx. This is officially an improper integral (it really is difficult to evaluate the integrand at x=0!). So first I'll compute ∫Q1x·ln(x)dx for Q small positive, and then let Q→0+. To use FTC we need an antiderivative of x·ln(x). Integration by Parts again works. If u=ln(x), then dv=x dx, and du=(1/x)dx and v=(1/2)x2. So:
∫x·ln(x)dx=(1/2)x2ln(x)–∫(1/2)x2(1/x)dx=(1/2)x2ln(x)–(1/4)x2+C
The definite integral is (1/2)x2ln(x)–(1/4)x2|Q1=((1/2)Q2ln(Q)–(1/4)Q2)((1/2)12ln(1)–(1/4)12).

L'H
I know that limQ→0+–(1/4)Q2=0 because Q2 is continuous and I can just "plug in". What about limQ→0+(1/2)Q2ln(Q)? Here we (officially!) need L'Hôpital's Rule again. So:
limQ→0+(1/2)Q2ln(Q)=rearranging=limQ→0+(1/2)[ln(Q)/{1/Q2}=using L'H=limQ→0+(1/2)[1/Q}/{–2/Q3}=rearranging back=limQ→0+(–1/4)Q2=0.

If you now put everything together (and don't lose track of the minus signs!) you can see:
01x·ln(x)dx converges and its value is –1/4. The –1/4 is certainly consistent with the estimate we made earlier, backed up by the red and green graph.

A real application ...
This strange function with its somewhat strange graph is actually related to a function used in applications. The function used in applications is actually a bit more complicated (sigh). Here is how to think about it.
 A graph of y=x·ln(x) on [0,1]. The function is concave up and below the x-axis. The area is –1/4, as just computed. Flip the curve over the x-axis. So this is a graph of y=–x·ln(x) on [0,1]. The function is concave down and above the x-axis. The area is +1/4, because now the region we're considering is above the x-axis. This is a more complicated flip. Replace x by 1–x. This is a flip which exchanges left and right because of the +/– change in the x multiplier. In fact, the left and right are exchanged, and the y-axis and x=1 are interchanged. So this is a graph of y=–(1–x)·ln(1–x) on [0,1]. The function is concave up and above the x-axis. The area is 1/4. Now add up the two previous functions. Here is a graph of y=–x·ln(x)–(1–x)·ln(1–x) on [0,1]. Yes, it is weird. This function officially has two strange behaviors, at both 0 and 1. The function is called the binary symmetric entropy function and it is used to study the amount of information flowing through a "channel" (you could think of a channel as a wire, and the information as 0's and 1's -- bits). The entropy function helps to analyze what happens in complicated situations where there may be interference (noise). This function has one bump, and its total area is 1/2.

Comment
You can't actually predict that the result of adding the two bumps would be just one symmetric bump. In fact, that result may not occur.
 Here is a graph of x[ln(x)]2 on [0,1]. Because of the square, the function is positive, and the bump is more pronounced. And this is a graph of x[ln(x)]2+ (1–x)[ln(1–x)]2. So two distinct bumps can show up.

Two silly (?) formulas
The object of this lecture is to tell you about two formulas, one for arc length and one for surface area (both discussed in section 8.1). I called the formulas silly because of their limited usefulness, at least limited in the sense that "hand computation" using FTC is not very practical. Both arc length and surface area will be revisited in calc 3, where much better perspectives can be given for both.

The philosophy behind the definite integral and its use
Maybe the formulas are not totally silly. Both of them are illustrations of how definite integrals can be used to compute various quantities. The procedure (which we have already used in various area and volume situations, and also with work) represents an attempt to compute "something" complicated:

1. Break up the complicated quantity into little pieces.
2. Approximate the little pieces by something simple.
3. Add up the little pieces, and take a limit.
With some luck and skill and ... whatever, the errors which occur will be small, and as the number of subdivisions or pieces or whatever grow, the total error will get small.

Arc length
We're given a function, f(x), defined on the interval [a,b]. The quantity to be computed is the length of the graph, the curve y=f(x). This is called arc length. Here is the idea.
Break up [a,b] into many little subintervals, whose length we will call dx (or Δx). "Above" each little subinterval is a little piece of the curve. The usual name for a little piece of curve is ds. If you magnify the little piece, as shown, well, the result is almost a right triangle. The curve length is still somewhat curvy, but, well, maybe I can approximate it by a straight line segment. The resulting picture is just about a right triangle. dy is the change in y (the function) when the input variable, x, Pythagoras then declares that (ds)2 should be the same (really, approximately the same!) as (dx)2+(dy)2. Therefore ds=sqrt{(dx)2+(dy)2). Let's rewrite what's inside the square root:
(dx)2+(dy)2=(dx)2(1+{dy/dx}2).
So sqrt(=(dx)2(1+{dy/dx}2))=dx·sqrt(1+{f´(x)}2).

Now we should add up these pieces and take limits. In this context, this is all done by writing a definite integral. So the arc length formula is ∫absqrt(1+[f´(x)]2)dx. This is the official formula. Let's see how well it works with some examples.

Line segment
Maybe the simplest curve is a straight line segment. Let me "find" the length of the line segment joining (1,1) and (4,3). This should be the same as the distance from (1,1) to (4,3), which is (square root of the sum of the squares!) sqrt(13). Let's find this number using the calculus formula above.

We need a formula for the line segment. The slope will be (3–1)/(4–1) which 2/3. So f(x)=(2/3)x+something. What will the "something" be? Since the line should pass through (1,1), when we put x=1, the result should be 1. Therefore (2/3)(1)+something=1, so something is 1/3. The formula is f(x)=(2/3)x+(1/3). The derivative is f´(x)=(2/3). Now the arc length is ∫absqrt(1+[f´(x)]2)dx which is ∫14sqrt(1+[2/3]2)dx. The integrand is a constant, so the result is sqrt(1+[2/3]2)x|14=sqrt(1+[2/3]2)4–sqrt(1+[2/3]2)1=sqrt(1+[2/3]2)3. This is the same as sqrt(13).

 Circle Maybe the next curve to look at is a circle, but we need the graph of a function so let's try to find the arc length of a semicircle. Let's look at the upper semicircle, radius 5, center at (0,0). For this curve, f(x)=sqrt(52–x2). Now I need sqrt(1+[f´(x)]2). So: f´(x)=(1/2)(52–x2)–1/22x using the Chain Rule. The 2's cancel, and we need to square the derivative, so: (f´(x))2=(52–x2)–1x2 but this is the same as``` x2 ----- 52–x2```to which we must add 1:``` x2 52–x2+x2 52 1 + ----- = --------- = ------ 52–x2 52–x2 52–x2``` Finally we supposed to take the square root of this result, so that the integral we need to compute is ∫–55 5/sqrt(52–x2)dx. This should look slightly familiar. The trig substitution x=5sin(θ) makes this integral into ∫5θ dθ=5arcsin(x/5)+C. I am skipping the details because I've done many of these integrals already. Now evaluate the definite integral: 5arcsin(x/5)|–55=5arcsin(1)–5arcsin(–1), and (since I know arcsin(1)=Π/2 and arcsin(–1)=–Π/2) this works out to 5Π, which is indeed half the circumference of a circle of radius 5.

Problems in the book
These two curves work out fairly well. But let's look at section 8.1, and some of the problems there. The problems mostly have the form, "Find the length of the graph of the function defined by the following formula" and I think the instructions should be modified to read "the following absurd formula." Here are some of the formulas from there:
(1/12)x3+x–1 (problem #3)
(x/4)4+(1/{2x2}) (problem #4)
x3/2 (problem #7)
(1/3)x3/2–x1/2 (problem #8)
(1/4)x2–(1/2)ln(x) (problem #9)
ln(cos(x)) (problem #10)
{ex+e–x}/2 (problem #18)

Why didn't the book ask something simpler, instead of functions defined by such bizarre formulas? Let's see why. I will answer problem 1 of section 8.1, which asks for the arclength of y=x4 between x=2 and x=6. The problem actually asks only for the definite integral and adds but do not evaluate.

We consider ∫absqrt(1+[f´(x)]2)dx. Here a=2 and b=6, and since f(x)=x4, f´(x)=4x3. The answer to problem 1 is therefore ∫26sqrt(1+16x6)dx. What about evaluation? In the sense most often used in calculus courses, this integral can't be evaluated. That is, there is no antiderivative of sqrt(1+16x6) which can be written in terms of standard functions. This isn't because we're ignorant, but because it is impossible to do this. If you wanted to compute this arclength, you would need to use one of the numerical techniques.

The secret to the problems in the textbook which were quoted above is that all of the bizarre functions were selected so that sqrt(1+[f´(x)]2) becomes something which it is possible to integrate (in the sense of "find an antiderivative and use FTC"). I did problem #18. Here it is.

Section 8.1, problem #18
Let's find the arc length of f(x)=(ex+e–x)/2 from x=–10 to x=10. Now f´(x)=(exe–x)/2. Now let's square.

```          (ex)2  –2  +(e–x)2
(f´(x))2= ------------------
4```
All sorts of subtle things are going on here. Notice that (–e‐x)2 is written (e‐x)2 because the two minuses cancel. Also notice that –2 is really –2exe–x. Now another subtle observation: 1=4/4. Therefore (look closely!)
```               (ex)2  –2	   +(e–x)2   4 + (ex)2  –2  +(e–x)2
1+(f´(x))2= 1+ ------------------ =  ---------------------
4                     4```
Now the top of that fraction is (ex)2 +2 +(e–x)2. Realize that 2 is 2exe. Notice (not an accident!) that this top is actually a "perfect square". It is (ex+e–x)2. So the mysterious and almost always horrible sqrt(1+(f´(x))2) becomes, in this case, exactly (ex+e–x)/2. Wow. The arc length integral is ∫–1010[(ex+e–x)/2]dx and this is (ex–e–x)/2|–1010, which is [(e10–e–10)/2]–[(e–10–e10)/2]=e10–e–10.

These functions and these graphs
The functions quoted from the problems in section 8.1 are mostly rather silly. This one, (ex+e–x)/2, is not. Some pictures:

ex
Exponential growth
e–x
Exponential decay
(ex+e–x)/2
The average of the two

The curve shown in the third box is called a catenary, and it is the curve a uniform chain describes. The function (ex+e–x)/2 is called the hyperbolic cosine and is usually abbreviated cosh(x) (pronounced cosh ecks). Its derivative is (naturally!) the hyperbolic sine, (ex–e–x)/2, and is abbreviated sinh(x) (pronounced cinch ecks). Really.

"Truth"
The truth for arc length is that, more or less, the computability of the arc length integral using FTC is impossible almost all of the time! Therefore, from the elementary, student point of view, maybe this is all a waste of time. But, really, it isn't. As soon as you give me a definite integral and want to approximate the values, there are all sorts of strategies. So what's important is that arc length can be computed by a definite integral, and what's important for you to try to understand is the philosophy of going from the vague idea of arc length to the integral formula for the arc length. And that philosophy will now be displayed again as we get an integral formula for a certain type of surface area.

 Surface area Suppose we are again given a function y=f(x) defined on an interval [a,b]. I would like to "compute" (the quotes are because we will get a definite integral formula which will share the benefits and defects of the previous result) the surface area which results when the graph of y=f(x) is revolved around the x-axis. We will get our formula using the same philosophical approach. We can chop up [a,b] into many little pieces, each having length, say, dx. Then (the picture!) the little piece of arc length laying over dx, which we called ds, will be revolved around the x-axis. This gets us a sort of ribbon. What is the area of that ribbon? We won't be able to compute it exactly, but maybe we can approximate the area of the ribbon nicely. Well, we can take the magic scissors (hey: I was able to draw the darn scissors almost correctly this time!) and cut the ribbon and then, sort of, almost, lay it out flat. The result will sort of, almost, be a rectangle. What are the dimensions of this rectangle? One side is the length of the piece of arc, ds. The other side is the circumference of a circle whose radius is f(x), the height of that part of the curve away from the x-axis. (The reason for the repeated "sort of, almost" is that this is actually a distortion of the true value - the ribbon really would not lie flat, and the ribbon really would not be more than an approximate rectangle. I will try later to address these sorts of slight (?) distortions.) So a piece of the surface area is 2Π f(x) ds. We use a definite integral to get the total surface area and add everything up. The result for the area when the curve is revolved around the x-axis is ∫ab2Π f(x)sqrt(1+[f´(x)]2)dx. Notice that sqrt(1+[f´(x)]2)dx (this uses what we had for ds). Sphere Here is a result from a long time ago (thousands of years!): the surface area of a sphere of radius R is 4Π R2. (This is the area of four "great circles" of the sphere, circles made by intersecting a plane with the center of the sphere.) I would like to verify this result using the surface area formula. I'll use the same semicircle as before: f(x)=sqrt(52–x2), with a=–5 and b=5. Please note that revolving this semicircle around the x-axis gets the area of the whole sphere of radius 5, so that the answer should be 4Π(52). We need to compute ∫ab2Π f(x)sqrt(1+[f´(x)]2)dx. Notice that sqrt(1+[f´(x)]2)dx is what we called ds before, and we did compute ds in a previous example. We saw that ds was equal to 5/sqrt(52–x2)dx. But f(x)=sqrt(52–x2) so, wow! (yeah, wow) there is cancellation and the arclength becomes ∫–55(2Π)5dx which does indeed work out to 100Π as it should. More "truth" There are very few simple powers of x (x2 and x3 and maybe x1) which give me integrands in the surface area formula that I can find antiderivatives of. (That's a horrible sentence!) If I want to compute surface areas for almost any "random" function defined by a formula, I'll need to use numerical approximations.

Returning the exam
The exam was returned. Also available are an answer sheet and a discussion of the grades and grading.

Wednesday, February 25 (Lecture #11)
Improper integrals: what are they?
 So far we have discussed computing standard definite integrals. Although certainly a definite integral like ∫abf(x)dx can represent many different ideas, the most familiar instantiation is as an area (asssuming that f(x)>0 for x in [a,b]) bounded by the x-axis, x=a, x=b, and the graph of y=f(x). In most of the computations we've done, the function f has been rather "nice" -- differentiable, mostly, and only a few times has it had some discontinuities. In fact, there are many applications where this simplicity is made more complicated because the applications themselves demand a "stronger" kind of integral.

 For example, we might consider a situation where, say, b gets larger and larger and larger. Where, say, b→∞. I will give a real physical example of this at the next lecture. This integral will have a domain which is an infinite interval. It could be [a,∞) or (–∞,b] or even (–∞,∞). Such integrals also occur very frequently in statistics (and therefore they "infiltrate" almost all experimental sciences!). A completely naive interpretation of the area would declare that since the length of the base is infinite, the total area must somehow be forced to be infinite. More subtly, the height in certain cases will decrease fast enough so that the total integral can be thought of as finite.

There may also be a sort of defect (?) in the range of f. For simplicity, consider the situation where, although f is "nice" for x>a, as x→a+, f(x) gets larger and larger and larger: in fact, we might need to try to "integrate" or compute the value of ∫abf(x)dx even if f tends to ∞ as x→a.
Again, a first look could convince you that such "areas" need to be infinite, also, because the height is infinite. But, actually, the way f grows as x→a+ is what matters. It is possible to imagine that the growth of f is so controlled that the total approximating areas don't tend to ∞. And maybe, in such situations, we should be able to compute the integral.

I will first consider the "defect in the domain" case.

Toy example #1
Look at y=1/x2. The integral ∫1B(1/x2)dx (I'm using B as an abbreviation for BIG) can be computed directly:
1B(1/x2)dx=–1/x|1B=1–(1/B) (be careful of the signs!)
Now if B→∞, certainly 1–(1/B)→1. Then we will declare that the improper integral ∫1(1/x2)dx converges and its value is 1.

Toy example #2
Look at y=1/x. Consider the analogous integral ∫1B(1/x)dx (again think of B as a BIG number). We compute it:
1B(1/x)dx=ln(x)|1B=ln(B)–ln(1)=ln(B)–0
I wanted also to consider here the behavior as B→∞, but some students seemed to be confused (this is confusing!). What does happen to ln(B) when B gets large? If you only have a loose idea of the graph in your head, well the log curve might not look like it is increasing too fast. Well, it actually is not increasing very fast, but it is increasing. Look: ln(10) is about 2.3, so ln(102)=2ln(10) is about 4.6 (that's 2·2.3), and ln(103)=3ln(10) is about 6.9 (that's 3·2.3), etc. Here etc. means I can get values of ln as large as I want by taking ln's of large enough powers of 10 (hey, ln of 105,000 is bigger than ... 10,000: so there!). So the values of ∫1B(1/x)dx do not approach a specific number as B→∞. Therefore we say: the improper integral ∫1(1/x)dx diverges.

The distinction between converges (approaches a specific finite limit) and diverges (does not approach a specific finite limit) is the one that is important in applications and that motivates the distinct use of the two words.

Geometric constrast
I love pictures. I like computation, but I can barely tolerate (!) "algebra". But I introduced the actual definition of {con|di}vergent improper integrals with some algebra, and didn't draw any pictures. Why? Well, because pictures, while useful, don't help too much. Here are qualitative pictures of the two graphs.

Well, they are actually different graphs, sorta. But I wanted to emphasize, through sketching them not too carefully (and on different axes, without scales!) these curves both "start at" (1,1), and as they go right, are always positive, decreasing, concave up, with 0 as limits. My eyes, at least, can't tell that one of them (on the left?) has finite total area, and one of them (to the right!) has infinite total area. The difference is quite surprising.

The exponential probability distribution
Many real-life phenomena are described using something called the exponential probability distribution. For example, the probable life-time of a lightbulb could be described with it. See here for more information.

I will just discuss lightbulbs now. Here is a first attempt to be precise. If the probability that a light bulb will fail in t minutes is proportional to e–Ct then the lifespan of the bulb is said to have an exponential probability distribution. More specifically, if t=0 is NOW, and t1<t2 are later times, then then the probability of lightbulb failure between the times t1 and t2 is proportional to ∫t1t2e–Ctdt.
That is, that portion of the area to the right which is shaded blue represents the chance of a lightbulb burning out during that particular time interval.

What is the total probability from NOW to FOREVER?
Every (real!) lightbulb is going to fail some time. Probabilities of an event range from 0 to 1, where an event which is certain will have probability 1. Since any lightbulb will fail between t=0 (NOW) and t=∞ (FOREVER), we should consider the (improper) integral ∫0e–Ctdt. Let me analyze this carefully.
Suppose A is a large positive number. Let's compute ∫0Ae–Ctdt=–(1/C)(e–Ct)]0A=–(1/C)e–CA+(1/C)e0=(1/C)–(1/C)e–CA.
The antiderivative has a –(1/C) factor because when e–Ct is differentiated, the Chain Rule produces a multiplicative –C, and the –(1/C) cancels this. Now what happens to (1/C)–(1/C)e–CA as A gets large, A→∞? e–CA (with a negative sign, with A and C positive) must go to 0 (this is exponential decay). Therefore the improper integral ∫0e–Ctdt converges, and its value is 1/C. But, wait: since every lightbulb fails, shouldn't this be 1? Yes, surely. Let's fix this up. The key is "proportional to". We should multiply the function e–Ct by a constant so that the improper integral will turn out to be 1. The computation we just did shows that the constant should be C. So the probability distribution is actually Ce–Ct.

The {expectation|mean|average} of an exponential probability distribution
Let me try to discuss something a little bit harder. First, some background with (maybe) some easier ideas. We could imagine a population of, say, bugs. Maybe there are three types of bugs in my bug collection and I know the following information:

My bug collection
Bug typeObserved lifespanProportion in
my collection
Orange spotted bug 20 days 30%
Blue striped bug 50 days 50%
Red plaid bug 80 days 20%

What is the average lifespan of this bunch of bugs? Well, it isn't the average of the 3 lifespans (the sum of 20, 50, and 80 divided by 3 -- a rather naive computation) since that doesn't take into account the varying proportions of the bug types. If you think about it, the average in this case is 20·(.3)+50·(.5)+80·(.25): a sort of weighted sum. It is a sum of lifetimes multiplied by the proportion of the population. You should convince yourself that this is the correct number.
What about lightbulbs? What proportion of a lightbulb population will "die" at time t? Well, that proportion is about Ce–Ctdt. The appropriate weighted sum in this case multiplied the proportion by t and adds it all up with an integral: ∫0tCe–Ctdt. If we compute this integral, we maybe can get some idea of when an average lightbulb dies. This quantity is called the mean or the expectation.

We can compute ∫tCe–Ctdt using integration by parts. If u=t then dv=Ce–Ctdt and du=dt and v=–e–Ct. Therefore uv–∫v du is –te–Ct–∫–e–Ctdt=–te–Ct–(1/C)e–Ct. The definite integral from t=0 to t=A is –te–Ct–(1/C)e–Ct]0A=–Ae–CA–(1/C)e–CA–{–(1/C)e0}.
What happens at A→∞? Well, e–CA certainly goes to 0 (radioactive decay!). But the term Ae–CA has a sort of conflict. Although the exponential decays, certainly A→∞? Which factor "wins"? Exponential decay is faster than any degree of polynomial growth, actually, so the limit is 0. You certainly can see this with L'Hopital's rule:
limt→∞te–Ct=limt→∞t/[eCt]=limt→∞1/[Cexp(Ct)]=0.
So the integral from 0 to A, as A→∞, approaches a limit, which is –{–(1/C)e0}=(1/C). This is the average lifespan of a lightbulb. Incidentally, if we want to check that this is a valid model, we can look at sample lifespans, and this can be used to identify the value of the parameter C. The average lifespan written on a package I just examined is 750 hours. Other technical words which refer to average lifespan are mean and expectation.

Other things ...
People who study statistics are interested in more details about life and lightbulbs than are described here. For example, they may want to know how dispersed the lifespans are around the mean. That is, do all of the lightbulbs tend to die out right around 1/C, or is there a considerable amount of variability? Various numbers measure this, including variance and standard deviation. They all need computation of ∫0t2Ce–Ctdt, a different improper integral.

Another kind of improper integral
The integrals we've looked at are called improper because their domains are infinite. But there is another collection of integrals which are also labeled improper because something goes wrong in their ranges: the function to be integrated becomes infinite. Here are some simple examples.

Toy example #1
Let's look at ∫01(1/x3)dx. Here f(x)=1/x3 and surely as x→0+, f(x)→+∞. Here is the official way to analyze this improper integral. Suppose that w is a small positive number. Compute ∫w1(1/x3)dx. We can do this with FTC by finding an antiderivative. The integral is –1/(2x2)]w1= –1/(2·12)–[–1/(2·12)]=(1/2w2)–(1/2). Now as w→0+, certainly this→+∞. Therefore the improper integral ∫01(1/x3)dx diverges.

The picture accompanying this is, again, useless. It helps me to organize the computation, but does not show enough quantitative information to help decide convergence or divergence.

Toy example #2
Let's look at ∫01(1/sqrt(x))dx. This became the QotD. In fact, this integral converges and its value is 2.

Here is an extended example of how to use improper integrals in a physical setting. I've used this example in our 152 instantiations, but don't have time this semester.

Escape velocity
When I was very young, I read a science fiction novel by Robert Heinlein which stated "... the escape velocity from the Earth is 7 miles per second ..." and now I would like to sort of verify this using only well-known (?) facts and some big ideas of physics.
 The radius of the earth Well, a sketch of the (continental) United States is shown to the right. There are 4 time zones. The U.S. is about 3000+ (maybe 3200?) miles wide. Therefore one time zone is about 1000 miles wide (I think the Pacific time zone actually slops a bit into the ocean), and since there are 24 time zones around the world, the circumference of the world is ... uhh ... about 25,000 miles. Or so. And therefore the radius of the earth is that divided by 2Pi, and therefore the radius of the earth is about 4,000 miles.

• Newton and gravitation Two masses attract each other with a force whose magnitude is proportional to the product of the masses and inversely proportional to the square of the distance between them. Therefore, if I have a mass, m, and if the Earth has mass M, the magnitude of the force of gravity is GmM/r2. G is a constant.
• Work lifting up Suppose we want to lift a mass m from the surface of the earth to a distance R, where R is very large. Then the work done is force multiplied by distance. The force needed to act against gravity certainly changes with distance. So I will compute the work with calculus. If x is some distance between 4,000 and R, then the force is GmM/x2. If the distance is a little bit, say, dx, then the work dW needed is GmM/x2 dx. The total work, W, is ∫x=4,000RGmM/x2 dx which I can compute readily as –GmM/x]4,000R=GmM({1/4,000}–1/R). I think I did the minus signs correctly. Notice that as R→∞, this work →GmM/4,000: this is the most work you can do, to get anywhere in the universe (assuming the universe is empty except for the mass m and the earth, of course).
• Kinetic energy How much kinetic energy would we need to supply to the mass m so that it would equal the potential energy the mass would have if it were lifted out to anywhere in the universe? Well, kinetic energy is (1/2)mv2 and that potential energy we already computed is GmM/4,000. So (1/2)mv2=GmM/4,000, and thus v2=2GM/4,000. But what is GM?
• But F=ma On the earth, a, the acceleration of gravity, is 32 ft/(sec)2. Yes, this is an archaic system of measurement, but that's part of the fun. But also F=GmM/(4,000)2. So GmM/(4,000)2=32m. Therefore GM=(4,000)2·32.
• And the answer is ... v2=2GM/4,000=2[(4,000)2·32]/4,000=8,000·32= (256,000)/(5,280)=(approximately)50. The 5,280 came from converting feet to miles. Therefore v, the escape velocity, is about 7mps. I think this computation is so silly that it is cool.

• Monday, February 23 (Lecture #10)
This was our first exam.