Monday, April 13  (Lecture #21) 

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 ∑a_{n}. Suppose we think
that a_{n} "resembles" cr^{n}. Well, if we take the
n^{th} root of cr^{n}, we get
(cr^{n})^{1/n}=c^{1/n}(r^{n})^{1/n}=c^{1/n}r^{n·(1/n)}=c^{1/n}r. Now if
n→∞, we have already seen that a
sequence like {c^{1/n}} has limit 1. So
(cr^{n})^{1/n}→r
as n→∞. So we can hope that the asymptotic behavior of
(a_{n})^{1/n} as n→∞ can help analyze the
convergence of ∑a_{n}.
Statement of the Root Test
The Root Test Consider the series ∑_{n=0}^{∞}a_{n}, and the limit lim_{n→∞}a_{n}^{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 n^{th} 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=1}^{∞}n^{n}x^{n} converge?
The powers of n signal to me that probably I should try the Root
Test. Here a_{n} is n^{n}x^{n}. We can't just
discard the absolute value here, but we can push it through to the x
because everything is multiplied. So:
n^{n}x^{n}^{1/n}=(n^{n}x^{n})^{1/n}=n^{n·(1/n)}x^{n·(1/n)}=nx.
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=1}^{∞}n^{n}x^{n}
converges exactly when x=0.
Even less silly example
Let's try this: for which x's does
∑_{n=1}^{∞}x^{n}/n^{n}
converge? I hope you will see the resemblance and contrast with the
previous computation:
x^{n}/n^{n}^{1/n}=(x^{n}/n^{n})^{1/n}=x^{n·(1/n)}n^{n·(1/n)}=x/n.
In this example, there aren't any special cases. For any x,
lim_{n→∞}x/n=0=L. Since L<1 always, the series
∑_{n=1}^{∞}x^{n}/n^{n}
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=0}^{∞}n!x^{n} and
∑_{n=0}^{∞}x^{n}/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=1}^{∞}x^{n}/n converge? We used
the Ratio Test, and a_{n+1}/a_{n} 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=1}^{∞}1^{n}/n=∑_{n=1}^{∞}1/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=1}^{∞}x^{n}/n^{2}
converge? Again, the Ratio Test, and a_{n+1}/a_{n}
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=1}^{∞}1^{n}/n^{2}=∑_{n=1}^{∞}1/n^{2}. This is a pseries with p=2>1, so it converges.
If x=–1, the series becomes
∑_{n=1}^{∞}(–1)^{n}/n^{2}. But this series converges absolutely (it gives us the pseries 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=1}^{∞}n x^{n} 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=1}^{∞}n=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 x_{0} (a fixed number) is an
infinite series of the form
∑_{n=0}^{∞}c_{n}(x–x_{0})^{n}
where the x is a variable and the c_{n} are some collection of
coefficients. It is sort of like an infinite degree
polynomial. Usually I (and most people) like to take x_{0} 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 x_{0} always has an interval of
convergence with the center of that interval equal to
x_{0}. 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 series  converges for x in  and has
radius of con vergence equal to  Pictures: convergence in red and divergence in green 

∑_{n=1}^{∞}n^{n}x^{n}  [0]  0  
∑_{n=1}^{∞}x^{n}/n^{n}  (–∞,+∞)  ∞  
∑_{n=1}^{∞}x^{n}/n  [–1,1)  1  
∑_{n=1}^{∞}x^{n}/n^{2}  [–1,1]  1  
∑_{n=1}^{∞}n x^{n}  (–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=0}^{∞}a_{n} (x–5)^{n}
centered at x_{0}=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
x_{0}=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:
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=0}^{∞}a_{n} (x–x_{0})^{n} has some positive radius of convergence, R, and suppose that f(x) is the sum of this series inside its radius of convergence.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.Differentiation The series ∑_{n=0}^{∞}n a_{n} (x–x_{0})^{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}^{∞}[a_{n}/(n+1)] (x–x_{0})^{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.
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=1}^{∞}n^{2}/5^{n}?
To compute
∑_{n=1}^{∞}n^{2}/5^{n},
look carefully at the individual terms. We're adding up
n^{2}/5^{n}=(n^{2})(1/5^{n}). In the
context of sequences and series, the (1/5^{n}) part might make
me think of ∑_{n=1}^{∞}x^{n}, 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=1}^{∞}x^{n}, 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 x^{n} to n·something may push you (I hope!) to
differentiate. So f´(x) is
∑_{n=1}^{∞}nx^{n–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=1}^{∞}nx^{n–1} when x<1.
Step 1 Differentiate.
But I need two n's (n^{2}) 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=1}^{∞}nx^{n–1}=∑_{n=1}^{∞}nx^{n}
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=1}^{∞}nx^{n}. This is just
like the first step: we differentiate. The derivative of the series is
∑_{n=1}^{∞}n^{2}x^{n–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=1}^{∞}n^{2}x^{n–1}=∑_{n=1}^{∞}n^{2}x^{n}
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=1}^{∞}n^{2}x^{n} 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 nx^{n} was, and how to use it to find
∑_{n=1}^{∞}n/5^{n}.
I actually would have liked to give the following as the QotD but
there wasn't time:
Discuss the convergence of
&sum_{n=1}^{∞}(–1)^{n}x^{n}/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) 

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 100^{th} 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=0}^{100}(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:
50^{101}/(101!)+50^{102}/(102!)+50^{103}/(103!)+...<50^{101}/(101!)+50^{101}/(101!)[1/2]+50^{101}/(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 ∑a_{n}, then we may hope
that somehow a_{n} resembles cr^{n}, a term of a
geometric series. We further hope that a_{n+1} will
resemble cr^{n+1}. But then if the "resemblance" is good
enough, we might hope that the quotient,
a_{n+1}/a_{n}, will be like
(cr^{n+1})/(cr^{n}), 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=0}^{∞}a_{n}, and the limit lim_{n→∞}a_{n+1}/a_{n}. 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
a_{n}=(50)^{n}/n!, the next term, a_{n+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:
 50^{n+1} 50^{n+1}      (n+1)! (n+1)! (50^{n+1})n! 50(n!) 50(n!) 50  =  =  =  =  =   50^{n}  50^{n} (50^{n})(n+1)! (n+1)! (n+1)n! n+1      n!  n! Step A Step B Step C Step D Step E Step FLet 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 a_{n+1}/a_{n}. 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
50^{n+1}=50^{n}·50, we can cancel
50^{n} 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 lim_{n→∞}a_{n+1}/a_{n}=lim_{n→∞}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 e^{50}. 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=1}^{∞}n^{2}/5^{n}. 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
a_{n} is n^{2}/5^{n} and
a_{n+1}=(n+1)^{2}/5^{n+1}. So:
 (n+1)^{2} (n+1)^{2}      5^{n+1}  5^{n+1} (n+1)^{2}5^{n} (n+1)^{2}  =  =  =   n^{2}  n^{2} n^{2}5^{n+1} n^{2}5      5^{n}  5^{n}
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=1}^{∞}n^{2}/5^{n}, we can compute lim_{n→∞}a_{n+1}/a_{n}= lim_{n→∞}[(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}^{∞}n^{20}/(1.01)^{n}. The 20^{th} term in this series is about 2.6·10^{19}. That's BIG. Does this series converge? Well, the Ratio Test applies. If similar algebra is done, then a_{n+1}/a_{n} 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: {condi}vergence all depends on the infinite tail  you can't think about the "first few" terms. Here is a little more numerical information. If a_{n}=n^{20}/(1.01)^{n}, then a_{1,000}=4.7·10^{40} (approximately and this is even bigger) and a_{10,000}=6.1·10^{16} and a_{100,000}=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=0}^{∞}7^{2n}/(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
a_{n}=7^{2n}/(n!)^{2} then
a_{n+1}=7^{2(n+1)}/((n+1)!)^{2}. Parentheses
are your friends so use many of them in computations and you likely
will make fewer errors!
 7^{2(n+1)}  7^{2(n+1)}      ((n+1)!)^{2}  ((n+1)!)^{2} 7^{2(n+1)}(n!)^{2}  =  =   7^{2n}  7^{2n} 7^{2n}((n+1)!)^{2}      (n!)^{2}  (n!)^{2}But 7^{2(n+1)}=7^{2n+2}=7^{2n}7^{2} and so part of that cancels with 7^{2n}. Analysis of the factorials can be more confusing, but here it is:
Again, this is not a "random" series. The sum of ∑_{n=0}^{∞}7^{2n}/(n!)^{2} is close to a value of a Bessel function, J_{0}(14). The series for J_{0}(14) is actually ∑_{n=0}^{∞}(–1)^{n}7^{2n}/(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=1}^{∞}1/n and the series
∑_{n=1}^{∞}1/n^{2}. 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
pseries with p=1) diverges and the other series (a pseries 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 a_{n+1}/a_{n} 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=1}^{∞}x^{n}/(3^{n}+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.
Important facts about absolute value
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 a_{n}=x^{n}/(3^{n}+n), then a_{n}=x^{n}/(n+3^{n}) 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 a_{n+1} is similarly x^{n+1}/(n+1+3^{n+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+3^{n}) (n+3^{n})  = x· x^{n}(n+1+3^{n+1}) (n+1+3^{n+1})We need to analyze the behavior of the somewhat complicated quotient (n+3^{n})/(n+1+3^{n+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 3^{n}/3^{n+1}, and this is 1/3.
Formal analysis Look at (n+3^{n})/(n+1+3^{n+1}) and divide the top and bottom by 3^{n}. The result is ([n/3^{n}]+1)/([n/3^{n}]+[1/3^{n}]+[3^{n+1}/3^{n}]) which is ([n/3^{n}]+1)/([n/3^{n}]+[1/3^{n}]+3). What about [n/3^{n}] as n→∞? We will use L'Hôpital's Rule since this is again ∞/∞. Remember that A^{B}=e^{B ln(A)}, so that the quotient [n/3^{n}]=[n/e^{n ln(3)}]. The derivative of the top (with respect to n) is 1, and the derivative of the bottom with respect to n is e^{n 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, lim_{n→∞}[n/3^{n}]= lim_{n→∞}[1/e^{n ln(3)}ln(3)]= lim_{n→∞}[1/3^{n}ln(3)]=0. So (wow!) lim_{n→∞}(n+3^{n})/(n+1+3^{n+1})=lim_{n→∞}([n/3^{n}]+1)/([n/3^{n}]+[1/3^{n}]+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}^{∞}x^{n}/(3^{n}+n) when x=3. This is ∑_{n=1}^{∞}3^{n}/(3^{n}+n). The n^{th} term is 3^{n}/(3^{n}+n) and if we divide the top and bottom by 3^{n} we see that the n^{th} term is 1/(1+[n/3^{n}]). But we saw that as n→∞, n/3^{n}→0 so that a_{n}→1. Any series which converges must have its n^{th} 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}^{∞}[2^{2n}/{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) 

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 n^{th} 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 a_{n}=(–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.
S_{1}=1; S_{2}=1–1=0;
S_{3}=1–1+1=1; S_{4}=0, etc.
It isn't too difficult to see that S_{even integer}=0 and
S_{odd 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 onedimensional 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,
a_{n}=(–1)^{n}(1+{1/n}). And now we looked at the
{condi}vergence of ∑_{n=1}^{∞}(–1)^{n+1}(1+{1/n}).
The partial sums are more complicated and more interesting.
S_{1}=2; S_{2}=2–(1{1/2})–{1/2}=.5; S_{3}=2–(1{1/2})+{1{1/3})=11/6=1.8333; S_{4}=2–(1{1/2})+(1{1/3})–(1{1/4})={7/12}=.58333This is where I stopped in class, but, golly, I have a friend who could compute S_{17} 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
From S_{1} to S_{2}, we move left since the
second term in the series is negative. From S_{2} to
S_{3} we move right, because the third term in the series is
positive. But notice that we don't get to S_{1}. 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.
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
n^{th} 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 a_{n} 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.
s_{1}=1; s_{2}=1–(1/2)–1/2=.5; s_{3}=1–(1/2)+{1/3)=5/6=.8333; s_{4}=1–(1/2)+(1/3)–(1/4)=7/12=.58333Here'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 thatThis 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.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.
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(36n^{7}–2n^{2})+2sin(55n+88) ∑_{n=1}^{∞}  2^{n}Here the sign distribution of the top of the fraction defining a_{n} is quite complicated. These are the first 20 signs:
Does this series converge?
Please notice that with a few modifications, the corresponding question can be answered very easily. Look at:
7cos(36n^{7}–2n^{2})+2sin(55n+88) ∑_{n=1}^{∞}  2^{n}Absolute values signs have been put around the cosine and sine functions. Now the series has all nonnegative 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 2^{n}. Therefore each term of this series is at most 9/2^{n}. 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 oldfashioned folding
carpenter's rulers. If we have an infinite series ∑_{n=1}^{∞}a_{n}, we could
consider the associated series ∑_{n=1}^{∞}a_{n},
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}^{∞}a_{n}
converges, then ∑_{n=1}^{∞}a_{n} must converge
also (and, actually, ∑_{n=1}^{∞}a_{n}≤∑_{n=1}^{∞}a_{n}).
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=1}^{∞}a_{n} which has ∑_{n=1}^{∞}a_{n}
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=1}^{∞}a_{n} converges and ∑_{n=1}^{∞}a_{n} diverges is called conditionally convergent. The alternating harmonic series is conditionally convergent.
Another example
Consider ∑_{n=1}^{∞}{sin(5n+8)}^{37}/n^{5}. 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}/n^{5} has terms which are all smaller than ∑_{n=1}^{∞}1/n^{5} (a pseries
with p=5>1, so it must converge). The comparison test asserts that
∑_{n=1}^{∞}(sin(5n+8)^{37}/n^{5} converges, and therefore ∑_{n=1}^{∞}{sin(5n+8)}^{37}/n^{5}
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}/n^{5}, say to an accuracy of +/.00001? Well, I could split this series up into S_{N}+T_{N}, a partial sum plus an infinite tail, and try to select N so that T_{N}<.00001. Once I do that, well then I just (have a computer) compute the corresponding S_{N}. So how can I get N with T_{N}<.00001? Here is a way. I know that T_{N}=∑_{n=N+1}^{∞}{sin(5n+8)}^{37}/n^{5}. This is an infinite series. I bet (using the result of a preceding paragraph that T_{N}=∑_{n=N+1}^{∞}{sin(5n+8)}^{37}/n^{5}≤∑_{n=N+1}^{∞}1/n^{5} again since the biggest that sine can be absolute value is 1, and 1^{37}=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/x^{5}]dx and we can evaluate the improper integral. It is (see the link for computations!) exactly 1/[4N^{4}]. If we want this to be less than .00001=1/10^{5} 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 S_{13}=∑_{n=1}^{13}{sin(5n+8)}^{37}/n^{5} 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 (nonnegative, really) terms converges.
QotD
The question was something like this:
The series ∑_{n=1}^{∞}((6n sin(5n–7)+8)/n^{17}
converges. Explain why!
Wednesday, April 1  (Lecture #18) 

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,
nonnegative (≥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 nonnegative)
terms? Well, the sequence of partial sums is
increasing, since we're just adding more and more nonnegative
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=1}^{∞}1: 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 nonnegative 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.
A first use of comparison
I considered the series
∑_{n=1}^{∞}1/(2^{n}+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/(2^{n}+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/(2^{n}+n)<1/n always. So the partial sums for ∑_{n=1}^{∞}1/(2^{n}+n) are all less than the partial sums for ∑_{n=1}^{∞}1/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/(2^{n}+n)<1/2^{n} always. Now we are comparing the series ∑_{n=1}^{∞}1/(2^{n}+n) with ∑_{n=1}^{∞}1/2^{n}. 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/(2^{n}+n) are also bounded by 1, and therefore must converge.
It converges!
The series ∑_{n=1}^{∞}1/(2^{n}+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=1}^{∞}1/(2^{n}+n)
converges. Certainly, since all the terms are positive, I guess the
sum will be positive. Also, since we compared this series to
∑_{n=1}^{∞}1/2^{n}, 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=1}^{∞}1/(2^{n}+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=1}^{∞}1/(2^{n}+n)=S_{N}+T_{N}
where S_{N}=∑_{n=1}^{N}1/(2^{n}+n)
and T_{N}=∑_{n=N+1}^{∞}1/(2^{n}+n). If we can find
some nice specific value of N so that T_{N} is guaranteed to
be less than .001, then we will know that the corresponding finite
sum, S_{N}, will be within .001 of the true value of the sum
of the whole infinite series. So what can we do? I will
overestimate T_{N}:
T_{N}=∑_{n=N+1}^{∞}1/(2^{n}+n)<∑_{n=N+1}^{∞}1/2^{n}.
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+1}^{∞}1/2^{n} is a geometric series. Be a bit careful in deducing c and r here, please. The first term, c, is 1/2^{N+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)=[1/2^{N+1}]/(1–1/2)=[1/2^{N+1}]/(1/2)=[2/2^{N+1}]/(2/2)=1/2^{N}. 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, T_{N}, which we want to estimate, is positive and less than 1/2^{N}. If we want T_{N}<.001, then we can force it to be less by choosing N so that 1/2^{N}<.001. Let's see: when N=10, I think we saw that 1/2^{10}=1/(1,024)<.001=1/(1,000).
Therefore S_{10}=∑_{n=1}^{10}1/(2^{n}+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 nonnegative)
terms. We have the following situation:
Suppose we know that 0<a_{n}≤b_{n} for all n's.So convergence is "inherited" downward and divergence is "inherited" upward.
If ∑_{n=1}^{∞}b_{n} converges, then ∑_{n=1}^{∞}a_{n} converges.
If ∑_{n=1}^{∞}a_{n} diverges, then ∑_{n=1}^{∞}b_{n} diverges.
No information is obtained if we know only either that the smaller series converges or the larger series diverges.
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+1/2^{5}+1/3^{5}+1/4^{5}+...=∑_{n=1}^{∞}1/n^{5}?
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 a_{n}=1/n^{5}, think of this
quantity as an area of a rectangle whose width is 1 and whose height
is 1/n^{5}. Put this rectangle on the xyplane so that its
upper righthand corner is at the point (n,1/n^{5}). The
rectangles will all fit together as shown in the graph to the
right. The corners are all on the curve y=1/x^{5}. Everything
is arranged so things work. Now look: the improper integral
∫_{x=1}^{∞}[1/x^{5}]dx is larger than
a_{2}+a_{3}+a_{4}+... . I left out
a_{1} 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/x^{5}]dx=lim_{A→∞}∫_{x=1}^{x=A}[1/x^{5}]dx=lim_{A→}–1/4x^{4}_{x=1}^{x=A}=lim_{A→∞}(–1/4A^{4})–(–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, a_{1}=1. We know that
∑_{n=1}^{∞}1/n^{5}<1+∫_{x=1}^{∞}[1/x^{5}]dx=5/4.
It converges!
The partial sums are all bounded above by 5/4. Since
∑_{n=1}^{∞}1/n^{5} 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=1}^{∞}1/n^{5}=S_{N}+T_{N},
where S_{N}=∑_{n=1}^{N}1/n^{5} and
T_{N}=∑_{n=N+1}^{∞}1/n^{5}. I
will try to overestimate the infinite tail, T_{N}, by
something convenient, and then force it to be less than .001.
Here is how to overestimate this T_{N}. Look at the picture to
the right. The curve y=1/x^{5} is again there, and I am
interested in what happens for x's bigger than N. I put the boxes
representing T_{N} (remember, this starts with
1/(N+1)^{5}) under this portion of the curve. So (very very
tricky!) I can make this estimate:
T_{N}<∫_{x=N}^{∞}[1/x^{5}]dx=lim_{A→∞}∫_{x=N}^{x=A}[1/x^{5}]dx=ETC.=1/[4N^{4}].
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/[4N^{4}]<.001. We decided if N=4, then 1/[4(4^{4})]=1/(1,024)<1/1,000. So the fourth infinite tail will be less than one onethousandth. And if we want the sum of the series to 3 decimal places, we just need to compute S_{4}=∑_{n=1}^{4}1/n^{5} 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.
pseries in general
The other collection of examples you need to know are the pseries.
Suppose p is a positive number. Then the pseries is
∑_{n=1}^{∞}1/n^{p}=1+1/2^{p}+1/3^{p}+1/4^{p}+.... 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 pseries is not a geometric series!
We just discussed the pseries 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
pseries.
A divergent series
Take p=1/2. The pseries 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 S_{N} I have placed the boxes over the
curve. Then the upper lefthand corners of the boxes are on the
curve. And to get all of the area representing S_{N} 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=1}^{x=N+1}[1/sqrt(x)]dx<S_{N}. Now I
can "easily" compute this integral (well, more easily than I can
compute the partial sum!). Here:
∫_{x=1}^{x=N+1}[1/sqrt(x)]dx=2sqrt(x)_{x=1}^{x=N+1}=2sqrt(N+1)–2sqrt(1)=2sqrt(N+1)–2.
This is an underestimate of S_{N}. If I want to force S_{N} 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≥51^{2}=2601. So N should be at least 2600. Some computed partial sums are listed below.
N  10  100  1,000  2,000  2,500  2,600 

S_{N}  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=1}^{∞}1/(n2^{n}) converge? (This is
problem #19 in section 10.3 of the textbook.)
Solution Since 0≤1/(n2^{n})≤1/2^{n} for
all positive integer n, and since
∑_{n=1}^{∞}1/2^{n} converges
(geometric series with absolute value of ratio equal to 1/2, less than
1) the series ∑_{n=1}^{∞}1/(n2^{n})
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) 

So a_{2}=sqrt(1+5)=sqrt(6). And a_{3}=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 nonnegative 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
a_{n}<100, then a_{n}+5<100+5=105, so that
sqrt(a_{n}+5)<sqrt(105). But the lefthand number is
a_{n+1} and that is less than sqrt(105), which is less than
100. So I have proved that all of the a_{n}'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 a_{n}<a_{n+1} then I can add 5 to both
sides and get a_{n}+5<a_{n+1}+5. Also square
rooting is increasing (remember the graph!) so that
sqrt(a_{n}+5)<sqrt(a_{n+1}+5). But this inequality
is exactly a_{n+1}<a_{n+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 {a_{n}} 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
a_{n+1}=sqrt(a_{n}+5) to find the limit quite nicely.
So if I know that lim_{n→∞}a_{n}=L then certainly (since {a_{n+1}}
is just about the same sequence, the numbers are all shoved along one
place) I know that lim_{n→∞}a_{n+1} is L also. So look:
If a_{n+1}=sqrt(a_{n}+5) take
lim_{n→∞}. The result is L=sqrt(L+5). Square both
sides, so L^{2}=L+5. Then L^{2}–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).
Now look at the picture below!
Don't trust numbers always!
I have a simple formula for another sequence. Here are the 10digit
decimal approximations of the first 7 terms: 
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:
Metaphor? An infinite series ∑_{n=1}^{∞}a_{n} can be thought of as ∑_{n=1}^{N}a_{n}+∑_{n=N+1}^{∞}a_{n}. So there is the N^{th} partial sum plus the "other" terms of the series, an infinite tail. These other terms I may sometimes call T_{N}. 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 a_{n}'s are 1, then the infinite series is
1+1+1+1+1+... and the partial sums are, well: S_{1}=1 and
S_{2}=1+1=2 and S_{3}=1+1+1=3 and ... well, I hope you
are convinced that S_{N}=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=1}^{∞}1/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, S_{1}=1 and S_{2}=1+1/2=3/2 and S_{3}=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.
N  S_{N} 

1  1.00000 
2  1.50000 
3  1.83333 
10  2.92897 
100  5.18738 
1,000  7.48547 
10,000  9.78761 
100,000  12.09015 
1,000,000  14.37273 
So what happens?
What can we conclude about the convergence of this series? The serious
answer is nothing. We have weighed the body, S_{some
numbers}, but we have no idea what the size of T_{these
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:
S_{N}<S_{N+1}. I will just look at some special
partial sums.
I wrote these lines on the board:
S_{1}=1=2/2
S_{2}=1+1/2=3/2
S_{4}=1+1/2+1/3+1/4>1+1/2+(1/4+1/4)=1+1/2+2(1/4)=4/2
S_{8}=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:
S_{16}=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: S_{2k}>=(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 S_{2200} is bigger than 100.
How long would it take to add up this sort of partial
sum? Well, 2^{200} is a large number. It is about
1.6·10^{60}. And, let's see, I would hope I could do
about 10^{11} additions and divisions in a second (that is an
overestimate). And there are about 1/3 of 10^{8} seconds in a
year. So there would be maybe 1/3 of 10^{19} additions and
divisions in a year. The age of the universe is sometimes estimated to
be about 20 billion years, or 2·10^{10} years. So
... in one "universe age" we could compute maybe
2/3·10^{29} terms (this is a cruddy computational
model, by the way). So all we would need is about ... uhhh
... 10^{31} 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+cr^{2}+cr^{3}+...=∑_{n=1}^{∞}cr^{n–1}.
Actually, I just noticed that the text writes this as ∑_{n=0}^{∞}cr^{n}, 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
S_{N}=∑_{n=1}^{N}cr^{n–1}. Unlike
almost any other series, it is possible to find an explicit formula
for this S_{N}. Here is the idea, which I think is usually
shown in some high school (and maybe even some middle school!) math or
physics courses:
S_{N}=c+cr+cr^{2}+cr^{3}+...+cr^{N}
Multiply by r:
rS_{N}=cr+cr^{2}+cr^{3}+cr^{4}+...+cr^{N+1}
Subtract the second equation from the first:
S_{N}–rS_{N}=c+(all the inside terms
cancel!)–cr^{N+1}
Factor the lefthand side:
(1–r)S_{N}=c–cr^{N+1}.
Now solve for S_{N}: if r is not equal to 1, S_{N}=[(c–cr^{N+1})/(1–r)].
Convergent geometric series
Now if r<1, powers of r, that is, the sequence {r^{n}},
must approach 0. So in the formula for S_{N}, when r<1, I
know that r^{N+1}→0 and S_{N}→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=1}^{∞}cr^{n–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 pattern may convince you that the total area is the sum of
Questions about some other geometric quantities involving these
squares can be asked. For example,
The pattern may convince you that the total perimeter is the sum of

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 eats  Igor eats 

1  1/2  1/4 
2  1/8  1/16 
3  1/32  1/64 
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.
Answer
The answer is NO. If the harmonic series were
c+cr+cr^{2}+... then (using just the initial segment
1+1/2+1/3) we see that r=cr/c=(1/2)/1=1/2 and
r=(cr^{2}/cr)=(1/3)/(1/2)=2/3. Since 1/2 is not 2/3, the
harmonic series is not a geometric series.
Comment on student answers
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=1}^{∞}10^{n}=10+100+1,000+10,000+... is a geometric series, and it certainly does not converge.
Wednesday, March 25  (Lecture #16) 

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 antiexams. 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 realvalued 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 n^{th}
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
a_{n}=1/n. Therefore, a_{5}=1/5 and
a_{17}=1/17. Also a_{n+1}=1/(n+1) and
a_{5m}=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 a_{n}=1/n, then a_{4}=1/4,
but a_{4+1}=a_{5}=1/5 and a_{4}+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.
Preliminary remarks about convergence
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 lim_{n→∞}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
lim_{n→∞}a_{n}=L means that, as n gets large,
a_{n} 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 lim_{n→∞}a_{n}=L is true if
for n large enough, a_{n} 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 a_{n}–L, and usually people want to
be able to control the size of a_{n}–L by selecting some
number N so that when n>=N, then the size of a_{n}–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
Algebra and sequences
Some straightforward limit facts still are true, such as:
If lim_{n→∞}a_{n}=L_{1} and
if lim_{n→∞}b_{n}=L_{2}, then
lim_{n→∞}a_{n}+b_{n}=L_{1}+L_{2}
and
lim_{n→∞}a_{n}·b_{n}=L_{1}·L_{2}.
Things like this are in the textbook, and I'd like to concentrate my
attention on more subtle behavior.
{5^{1/n}}
Let's look at the sequence {5^{1/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<5^{1/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 {b_{n}} is a convergent sequence with lim_{n→∞}b_{n}=L, then the sequence {f(b_{n})} converges, and lim_{n→∞}f(b_{n})=f(L).This is true because the b_{n}'s get close to L, and continuous functions take close inputs to close outputs, so a_{n}=f(b_{n}) is close to f(L).
A more intricate example
Consider the sequence {n^{1/n}}. This is a bit like the
preceding example. It has the form
BASE^{EXPONENT}. 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(n^{1/n})=(1/n)ln(n)=ln(n)/n, we need to analyze
lim_{n→∞}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:
lim_{n→∞}ln(n)/n=lim_{n→∞}(1/n)/1=0
Therefore just as before, the limit of the original sequence is
e^{0}=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.
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)nLook 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 a_{100} is about 8·10^{11} 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": lim_{n→∞}{(50)^{n}/n!}=0.
Squeezing a limit out ... If I have three sequences, {a_{n}} and {b_{n}} and {c_{n}} and if I know for all n's that a_{n}≤b_{n}≤c_{n} and if I know that lim_{n→∞}a_{n}=L and lim_{n→∞}c_{n}=L, then the middle sequence {b_{n}} also converges, and its limit is L.The preceding example had a_{n}=0 and b_{n}=(50)^{n}/n! and c_{n}=[(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) 

Find the solution of y´=y^{2}x^{3} 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.
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 halfdollars ([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.
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´=x^{2}–x–4y^{2}. 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
x^{2}–x–4y^{2} when x=3 and y=1. This is
3^{2}–3–4·1^{2}=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 thickenedup line segment of slope 2 at (3,1) and consider some
curves going through that point.
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!
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)x^{3}+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.  
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 xaxis, and they are concave down when they are below the xaxis. 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 xaxis, 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.  
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.  
 
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)y^{2}(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.  
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 righthand side, (1/20)(x+y^{2}), is a lowdegree 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+C^{2}) 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 lim_{x→Q–}f(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+y^{2}). 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 simplelooking 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 100I 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´=(e^{x/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) 

Examples
y´=x^{2} Done in 151; to be reviewed here.
y´=5y Done in 151; to be reviewed here.
y´=y^{2} 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´=x^{2}+y^{2} 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.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.
(From Michael Crichton (1969) The Andromeda Strain, Dell, N.Y. p.247)
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 longwinded, but I hope the discussion
and examples which follow will shown why such elaboration is
necessary.
An example: y´=x^{2}
Well, we know how to solve y´=x^{2}: just integrate. We
did this repeatedly last semester (and even this semester). The
solutions are y=(1/3)x^{3}+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)x^{3}+1. It
is the solution curve which goes through (0,1).
The red curve has C=3: y=(1/3)x^{3}+3. It is
the solution curve which goes through (0,3).
The green curve has C=–2:
y=(1/3)x^{3}–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´=x^{2} has the general solution
f(x)=(1/3)x^{3}+C. When C has a specific value, then the
function is called a particular solution. So
f(x)=(1/3)x^{3}–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
yvalue corresponding to the given xvalue 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´=y^{2}
Now let's change and consider y´=y^{2}. 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´=y^{2}. 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.
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.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 (x_{0},y_{0}). Then there always is a solution, and there is exactly one solution.
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.
While I love mathematics, and I think math is beautiful and helpful, please remember the final sentence of this joke.JOKE Several people are in a hotair balloon, trying to land over a fogshrouded 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."
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 righthand 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.
Some examples
If x^{2}+y^{2}=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 0^{2}+1^{2}=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´=x^{2}+y^{2} 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 simplelooking 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 righthand
side integrates to t+C. The lefthand 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)]t}e^{ln(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 longrange 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´=y^{2}x^{3} which goes
through (2,1). Be sure to write the solution as y=some function of x.
Monday, March 9  (Lecture #13) 

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
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 c_{1}, so that
G´(c_{1})=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´(c_{1})=0. Now
MVT/RT applied to the function G´(x) tells us that its derivative
is 0 somewhere between a and c_{1}. That is, there is
c_{2} between a and c_{1} so that
G´(c_{2})=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´(c_{2})=0. So
MVT/RT again applies to tell us that there is c_{3} between a
and c_{2} with G´´´(c_{3})=0. Wow.
The derivative of G´´(x):
G´´´(x)=f´´´(x)–(0+0+0+(3·2·1)(K)1).
Now we know when x=c_{3} this is 0. So we know that
f´´´(c_{3})–(3·2·1)K is
actually equal to 0. We can
solve this for K, and get
K=[f´´´(c_{3})/(3·3·1)].
c_{3} is between a and c_{2} and
c_{2} is between a and c_{1} and
c_{1} is between a and b so that c_{3} 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
T_{n}(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  1  2  3  4  5  6  7  8  9  10 

n!  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.
T_{n}(x)=∑_{j=0}^{n}[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 T_{8}(x) for sin(x) and a=0. So we need derivatives, and here they are:
Derivative #  Function  Value at a=0 

0  sin(x)  0 
1  cos(x)  1 
2  –sin(x)  0 
3  –cos(x)  –1 
4  sin(x)  0 
5  cos(x)  1 
6  –sin(x)  0 
7  –cos(x)  –1 
8  sin(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 T_{8}(x):
T_{8}(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
T_{8}(x)=x–[1/6]x^{3}+[1/120]x^{5}+[1/5,040]x^{7}
Then there are some questions.
What is T_{4}(x)? It must be
x–[1/6]x^{3}.
What is T_{7}(x)? It
must be
=x–[1/6]x^{3}+[1/120]x^{5}+[1/5,040]x^{7},
in this case just the same as T_{8}(x).
What is
T_{10}(x)? I bet it is the same as T_{9}(x), and this
would be
T_{10}(x)=x–[1/6]x^{3}+[1/120]x^{5}+[1/5,040]x^{7}–[1/362,880]x^{9}.
Let me explain why T_{9}(.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 T_{n}(x)=∑_{j=0}^{n}[f^{(j)}(a)/j!](x–a)^{j}, then the difference between this and f(x) isThis 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 10^{th} 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} for some c between a and x.
Example 2
We get T_{8}(x) for f(x)=e^{x} and a=0.
Derivative #  Function  Value at a=0 

....blah.......blah.......blah.......blah.......blah..........blah...  
any n you want!  e^{x}  1 
....blah.......blah.......blah.......blah.......blah..........blah... 
T_{7}(x)=1+x+[x^{2}/2]+[x^{3}/6]+[x^{4}/24]+[x^{5}/120]+[x^{6}/720]+[x^{7}/5,040]
It's supposed to be easy: this is a technique people actually use.
How good is T_{7}(–.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 e^{c}[1/(8!)](–.4)^{8}. Now c is between –.4 and 0 and since e^{x} is increasing, the biggest value occurs at c=0, where e^{0}=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 T_{7}(–.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)=x^{1/3} and a=8. Then f(8)=2. I asked people to find T_{3}(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 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)=x^{1/3} and T_{3}(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=x^{1/3} and y=T_{3}(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 x^{1/3} 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 x^{1/3} 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=x^{1/3} and y=T_{3}(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 T_{3}(x) and f(x) for x in [6,10] is
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 onefiftieth (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) 

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. Thenf(x) f´(x) lim  = lim  x→a g(x) x→a g´(x)if the limit on the righthand 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 xaxis, 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
lim_{x→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:
lim_{x→0+}x·ln(x)=rearranging=lim_{x→0+}[ln(x)/{1/x}]=using L'H=lim_{x→0+}[{1/x}/{–1/x^{2}}]=rearranging back=lim_{x→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 xaxis 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 ∫_{0}^{1}x·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 ∫_{Q}^{1}x·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)x^{2}. So:
∫x·ln(x)dx=(1/2)x^{2}ln(x)–∫(1/2)x^{2}(1/x)dx=(1/2)x^{2}ln(x)–(1/4)x^{2}+C
The definite integral is
(1/2)x^{2}ln(x)–(1/4)x^{2}_{Q}^{1}=((1/2)Q^{2}ln(Q)–(1/4)Q^{2})–((1/2)1^{2}ln(1)–(1/4)1^{2}).
L'H
I know that lim_{Q→0+}–(1/4)Q^{2}=0 because
Q^{2} is continuous and I can just "plug in". What about
lim_{Q→0+}(1/2)Q^{2}ln(Q)? Here we (officially!) need
L'Hôpital's Rule again. So:
lim_{Q→0+}(1/2)Q^{2}ln(Q)=rearranging=lim_{Q→0+}(1/2)[ln(Q)/{1/Q^{2}}=using L'H=lim_{Q→0+}(1/2)[1/Q}/{–2/Q^{3}}=rearranging back=lim_{Q→0+}(–1/4)Q^{2}=0.
If you now put everything together (and don't lose track of the minus signs!) you can see:
∫_{0}^{1}x·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 xaxis. The area is –1/4, as just computed. 

Flip the curve over the xaxis. So this is a graph of y=–x·ln(x) on [0,1]. The function is concave down and above the xaxis. The area is +1/4, because now the region we're considering is above the xaxis.  
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 yaxis 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 xaxis. 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.

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:
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 ∫_{a}^{b}sqrt(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 ∫_{a}^{b}sqrt(1+[f´(x)]^{2})dx which is ∫_{1}^{4}sqrt(1+[2/3]^{2})dx. The integrand is a constant, so the result is sqrt(1+[2/3]^{2})x_{1}^{4}=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(5^{2}–x^{2}). Now I need sqrt(1+[f´(x)]^{2}). So: f´(x)=(1/2)(5^{2}–x^{2})^{–1/2}2x using the Chain Rule. The 2's cancel, and we need to square the derivative, so: (f´(x))^{2}=(5^{2}–x^{2})^{–1}x^{2} but this is the same as x^{2}  5^{2}–x^{2}to which we must add 1: x^{2} 5^{2}–x^{2}+x^{2} 5^{2} 1 +  =  =  5^{2}–x^{2} 5^{2}–x^{2} 5^{2}–x^{2}Finally we supposed to take the square root of this result, so that the integral we need to compute is ∫_{–5}^{5} 5/sqrt(5^{2}–x^{2})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)_{–5}^{5}=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)x^{3}+x^{–1} (problem #3)
(x/4)^{4}+(1/{2x^{2}}) (problem #4)
x^{3/2} (problem #7)
(1/3)x^{3/2}–x^{1/2} (problem #8)
(1/4)x^{2}–(1/2)ln(x) (problem #9)
ln(cos(x)) (problem #10)
{e^{x}+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=x^{4} between x=2 and x=6. The problem actually asks only for the definite integral and adds but do not evaluate.
We consider ∫_{a}^{b}sqrt(1+[f´(x)]^{2})dx. Here a=2 and b=6, and since f(x)=x^{4}, f´(x)=4x^{3}. The answer to problem 1 is therefore ∫_{2}^{6}sqrt(1+16x^{6})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+16x^{6}) 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)=(e^{x}+e^{–x})/2
from x=–10 to x=10. Now
f´(x)=(e^{x}–e^{–x})/2. Now let's
square.
(e^{x})^{2} –2 +(e^{–x})^{2} (f´(x))^{2}=  4All 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 –2e^{x}e^{–x}. Now another subtle observation: 1=4/4. Therefore (look closely!)
(e^{x})^{2} –2 +(e^{–x})^{2} 4 + (e^{x})^{2} –2 +(e^{–x})^{2} 1+(f´(x))^{2}= 1+  =  4 4Now the top of that fraction is (e^{x})^{2} +2 +(e^{–x})^{2}. Realize that 2 is 2e^{x}e^{–}. Notice (not an accident!) that this top is actually a "perfect square". It is (e^{x}+e^{–x})^{2}. So the mysterious and almost always horrible sqrt(1+(f´(x))^{2}) becomes, in this case, exactly (e^{x}+e^{–x})/2. Wow. The arc length integral is ∫_{–10}^{10}[(e^{x}+e^{–x})/2]dx and this is (e^{x}–e^{–x})/2_{–10}^{10}, which is [(e^{10}–e^{–10})/2]–[(e^{–10}–e^{10})/2]=e^{10}–e^{–10}.
These functions and these graphs
The functions quoted from the problems in section 8.1 are mostly
rather silly. This one, (e^{x}+e^{–x})/2, is
not. Some pictures:
e^{x} Exponential growth 
e^{–x} Exponential decay 
(e^{x}+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 (e^{x}+e^{–x})/2 is called the hyperbolic cosine and is usually abbreviated cosh(x) (pronounced cosh ecks). Its derivative is (naturally!) the hyperbolic sine, (e^{x}–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 xaxis. 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 xaxis. 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 xaxis. (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 xaxis is ∫_{a}^{b}2Π f(x)sqrt(1+[f´(x)]^{2})dx. Notice that sqrt(1+[f´(x)]^{2})dx (this uses what we had for ds).
Sphere We need to compute ∫_{a}^{b}2Π 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(5^{2}–x^{2})dx. But f(x)=sqrt(5^{2}–x^{2}) so, wow! (yeah, wow) there is cancellation and the arclength becomes ∫_{–5}^{5}(2Π)5dx which does indeed work out to 100Π as it should.
More "truth" 
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) 

So far we have discussed computing standard definite integrals. Although certainly a definite integral like ∫_{a}^{b}f(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 xaxis, 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 ∫_{a}^{b}f(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/x^{2}. The integral ∫_{1}^{B}(1/x^{2})dx (I'm using B as an
abbreviation for BIG) can be computed directly:
∫_{1}^{B}(1/x^{2})dx=–1/x_{1}^{B}=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/x^{2})dx converges and its value is 1.
Toy example #2
Look at y=1/x. Consider the analogous integral ∫_{1}^{B}(1/x)dx (again think of B as
a BIG number). We compute it:
∫_{1}^{B}(1/x)dx=ln(x)_{1}^{B}=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(10^{2})=2ln(10)
is about 4.6 (that's 2·2.3), and ln(10^{3})=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 10^{5,000} is bigger than ... 10,000: so
there!). So the values of ∫_{1}^{B}(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 {condi}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 reallife phenomena are described using something called the
exponential probability distribution. For example, the probable lifetime 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 t_{1}<t_{2} are
later times, then then the probability of lightbulb failure between
the times t_{1} and t_{2} is proportional to ∫_{t1}^{t2}e^{–Ct}dt.
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 ∫_{0}^{∞}e^{–Ct}dt. Let me analyze this carefully.
Suppose A is a large positive number. Let's compute ∫_{0}^{A}e^{–Ct}dt=–(1/C)(e^{–Ct})]_{0}^{A}=–(1/C)e^{–CA}+(1/C)e^{0}=(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 ∫_{0}^{∞}e^{–Ct}dt 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 {expectationmeanaverage} 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 type  Observed lifespan  Proportion 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^{–Ct}dt. The appropriate weighted sum in this case
multiplied the proportion by t and adds it all up with an integral:
∫_{0}^{∞}tCe^{–Ct}dt. 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^{–Ct}dt using integration by parts. If u=t then
dv=Ce^{–Ct}dt and du=dt and v=–e^{–Ct}. Therefore
uv–∫v du is
–te^{–Ct}–∫–e^{–Ct}dt=–te^{–Ct}–(1/C)e^{–Ct}.
The definite integral from t=0 to t=A is
–te^{–Ct}–(1/C)e^{–Ct}]_{0}^{A}=–Ae^{–CA}–(1/C)e^{–CA}–{–(1/C)e^{0}}.
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:
lim_{t→∞}te^{–Ct}=lim_{t→∞}t/[e^{Ct}]=lim_{t→∞}1/[Cexp(Ct)]=0.
So the integral from 0 to A, as A→∞, approaches a limit,
which is –{–(1/C)e^{0}}=(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 ∫_{0}^{∞}t^{2}Ce^{–Ct}dt, 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 ∫_{0}^{1}(1/x^{3})dx. Here
f(x)=1/x^{3} 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
∫_{w}^{1}(1/x^{3})dx. We can do this with
FTC by finding an antiderivative. The integral is
–1/(2x^{2})]_{w}^{1}=
–1/(2·1^{2})–[–1/(2·1^{2})]=(1/2w^{2})–(1/2).
Now as w→0^{+}, certainly this→+∞. Therefore
the improper integral ∫_{0}^{1}(1/x^{3})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 ∫_{0}^{1}(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

Monday, February 23  (Lecture #10) 

Maintained by greenfie@math.rutgers.edu and last modified 2/26/2009.