History: the Basel problem
This problem was stated by Pietro Mengoli in 1644, and was solved by Euler almost a century later, in 1735. There is some question about the validity of his solution of the problem (or, at least, the validity of the solution we'll look at as presented by him!), and I'll address that later. The problem is to find some "analytical expression" or precise value for ∑_n=1^∞1/n². The Wikipedia article discusses the problem and supplies several solutions. The problem is notorious, and has attracted many solutions. Robin Chapman, an English mathematician, has assembled fourteen different proofs. You can also look at the very large number of references here.

Leonhard Euler
Euler was one of the greatest known mathematicians, and almost certainly the most prolific, writing incredible numbers of papers and books (including, as I mentioned, the first systematic calculus book). Here is a quote from another source:

Euler devises some clever approximations to get the [sum of the series] to six decimal places, 1.644934. Euler notes that to achieve such accuracy using direct calculation requires more than 1000 terms, so his estimate of the solution to the Basel problem was far more accurate than any available to his competitors. He later improved his estimate to 17 decimal places. Moreover, Euler was a genius at arithmetic, so he probably recognized this value as what will turn out to be the exact solution to the Basel problem, &Pi²/6. Euler didn't share this with the world, so he had a valuable advantage as he raced to solve the problem; he knew the answer.

The Vatter problems
By an amazing coincidence, Dr. Vincent Vatter, a former Rutgers math graduate student now at Dartmouth, visited me the morning before today's class. He asked me what I was teaching and when I told him today's topic he remarked that the topic was a collection of problems in his calculus book, still in manuscript form. He kindly allowed me to print the page containing the problems and then the class and I solved the problems. Here they are.

38. Derive the Maclaurin series for f(x)=sin(sqrt(x))/sqrt(x).

39. Find all the roots of f(x)=sin(sqrt(x))/sqrt(x), i.e., values of x for which f(x)=0.

40. If a polynomial of degree 3 has roots r₁, r₂, and r₃, then it is given by p(x)=c(x–r₁)(x–r₂)(x–r₃) for some constant c. By expanding this product, verify that
[1/r₁}+[1/r₂]+[1/r₃]=–{coefficient of x}/{constant term}.
(This fact is true for polynomials of any degree.)

41. Assume, as Euler did, that Exercise 40 holds for infinite series to show that
[1/r₁}+[1/r₂]+[1/r₃]+...=1/6
where r₁, r₂, r₃, ... are the roots from Exercise 39.
Conclude that ∑_n=1^∞1/n²=Π²/6.

42. The function f(x)=2–{1/(1–x)} has a single root, x=1/2. Derive its Maclaurin series and conclude that, contrary to Euler's assumption, Exercise 40 cannot be applied to infinite series.

I think the problems were mostly straightforward (people remembered in 42 to start with a geometric series).

I mentioned that Euler used similar techniques to find explicit values of ∑_n=1^∞1/n^EVEN, where EVEN is any even integer (the values have the form (rational number)·Π^EVEN, so that, for example, ∑_n=1^∞1/n¹²=(691/{638,512,875})Π¹²). There is no known nice form for the sum of ∑_n=1^∞1/n³, even though a lot of people have thought about it for a long time. I tried for while with no success.

So it was wrong, wasn't it?
So the proof we just "toured" was certainly incorrect. This is interesting. But the historical source I quoted earlier noted that Euler gave at least three other solutions of the Basel Problem, and the "proof" we just discussed is the one most remembered. Several of Euler's other proofs use techniques familiar to all 152 students, but the proofs are definitely much more intricate although quite ingenious, and they are not very appealing (although undoubtedly correct even by today's standards). It is interesting that an incorrect but insightful proof is the one which most people remember!

Further, Euler was sort of correct although power series are not "exactly" like long polynomials. There are ways of relating the roots of a power series to the coefficients of the power series. About 150 years later, another mathematician named Weierstrass fixed up Euler's proof using a factorization theorem for power series. One of the key assumptions made in Weierstrass's result rules out an example like the one given in problem 42 above (thank goodness!). Weierstrass, a great and famous mathematician, is less well known for being one of the lecturer's mathematical ancestors.

Rearranging a series
There are some other strange things that happen with series. Many of these were known to Euler, and he mostly disregarded them. For example, we know that the Alternating Harmonic Series,
1–1/2+1/3–1/4+1/5–1/6+1/7...
converges since it satisfies all of the hypotheses of the Alternating Series Test. Call its sum, S (actually its sum is ln(2) but we don't need this here). Now let's rearrange it.

Write a positive term followed by two negative terms, and keep doing this. So the rearranged series begins
1–1/2–1/4+1/3–1/6–1/8...
and notice that we'll never run out of either positive or negative terms since there are infinitely many of both of them. Now put parentheses around the results in this way:
(1–1/2)–1/4+(1/3–1/6)–1/8...
and realize that the numbers inside pairs of parentheses can be simplified:
1/2–1/4+1/6–1/8...
And then further realize that we could factor out 1/2:
(1/2)(1–1/2+1/3–1/4 ....)
which certainly shows that this series converges and its sum is (1/2)S.

Most people find this startling and at least moderately unpleasant. In fact, absolutely convergent series can be rearranged in any way, and rearrangement won't change the sum -- the resulting series will have the same sum as the original series. But any conditionally convergent series can be rearranged so that the sequence of partial sums does almost anything you want. For example, it can be rearranged so that the rearrangement converges to any selected number. (This is because that positive and negative parts of the series separately diverge, but a proof of all this is not really a part of this course.)

Convergence and infinite series can be rather subtle. But now we need to turn to preparing for the final exam.

Using series instead of L'H
I asked people to use Maclaurin series to compute

     (sin(3x)–3x)2
lim ---------------
x→0   (e2x–1–2x)3

Finding some terms of a Taylor series
I asked for the first four non-zero terms of the Maclaurin series for cos(3x)/(1+x⁴). This can be obtained by using series for cos(3x) (easy to get from the cosine series) and by realizing the 1/(1+x⁴) is the sum of a geometric series with first term 1 and ratio x⁴.

An integral Here's a problem from the textbook: write ∫0xln(1+t2)dt as a power series in x assuming |x|<1. How many terms of this series would be needed to get the value of the integral when x=1/3 to an accuracy of .001? (Of course, I can find an antiderivative of ln(1+t2) using integration by parts fairly easily.)

If you want to come to the review session ...
On Monday afternoon, I'll have a review session from 3 to 5 PM in Hill 525. You could prepare for it by filling out this neat quiz written by Dr. Julia Wolf. She gave it to her class and allowed half an hour. So give yourself half an hour, and then come to the review session with your results.

I don't want to kill the problems (from my pedagogical point of view!) by writing solutions to them in this diary. But I will give some amplification of the classroom discussions by displaying some appropriate graphs.

Problem 1 b) Part b) brings up the function defined by the expression max(1,x). Here is a graph of this function on the interval [0,5].

Problem 1 a) And the answer to part a) makes me think of the two-variable function f(x,y)=max(x,y). Here is a graph (in three dimensions -- you'll appreciate it more next semester) of the function z=f(x,y) for x and y both between 0 and 2.

Problem 2 So here is a graph showning two curves on the interval [1,100]. One of them is ln(x) and the other is 210(x(1/210)–1). They overlay one another so closely that, to me, only one "curve" is consistently visible.
Problem 2 (another picture) Here are the same two functions on the interval [90,100]. The functions get separated more as the distance to 1 increases (similar to partial sums of most Taylor series). Now you can "see" the two curves and also maybe understand why the separation could not be seen on the scale of the graph above.
Problem 5 Here is a graph of the function e2cos(x)(cos(2sin(x)) on the interval [0,20]. You can compare this with the partial sum given in the problem statement. If this graph doesn't convince you that complex numbers exist (at least as much as the numbers sqrt(2) and Π "exist") then I don't know what argument would.
Problem 7 Here is a graph of J0, a Bessel function of the first kind on the interval [0,10].

Preparation for the final exam
Please see this file. I'll discuss whether you want a formal review session or office hours or whatever. The balance of class time this week will be devoted to further observations about power series.

A familiar series
We know 1/(1–x). It is the sum of a geometric series with first term 1 and ratio equal to x. So 1/(1–x)=∑_n=0^∞xⁿ. This equation is valid when |x|<1, or –1<x<1. Some remarkable things can be done with this series.

Logarithm
What the Maclaurin series of ln(x)? This is a trick question because y=ln(x) looks like this and the limit of ln(x) as x→0⁺ is –∞ so ln(x) can't have a Taylor series centered at 0. What's a good place to consider? Since ln(1)=0, we could center the series at 1. Most people would still like to compute near 0, though, so usually the function is moved instead! That is, consider ln(1+x) whose graph now behaves nicely at 0 so we can analyze it there.

If f(x)=ln(1+x), I want to "find" ∑_n=0^&infin[f⁽ⁿ⁾(0)/n!]xⁿ. Well, f(0)=ln(1+0)=ln(1)=0, so we know the first term. Now f´(x)=1/(1+x) so that ... wait, wait: remember to try to be LAZY.

Look at

 1
---
1+x

Computing with ln
What if x=–1/2 in the previous equation? Then ln(1–1/2)=ln(1/2)=–ln(2) and this is approximately –69314. A friend of mine has just computed &sum_n=1¹⁰[(–1)ⁿ⁺¹(–.5)ⁿ/n] and this turns out to be –.69306. We only get 3 decimal places of accuracy. It turns out that this series converges relatively slowly compared to the others we've already seen, which have the advantage of factorials in the denominator. So this series is usually not directly used for numerical computation, but other series related to it are used.

Book problem: 10.7, #9
Find the Maclaurin series of ln(1–x²).
Be LAZY. We know that
ln(1+x)=x–[x²/2]+[x³/3]–[x⁴/4]+[x⁵/5]+...=&sum_n=1^∞[(–1)ⁿ⁺¹xⁿ/n]. This
so we can substitute –x² for x and get
ln(1–x²)=–x²–[(–x²)²/2]+[(–x²)³/3]–[(–x²)⁴/4]+[(–x²)⁵/5]+...=&sum_n=1^∞[(–1)ⁿ⁺¹(–x²)ⁿ/n] and further
ln(1–x²)=–x²–[x⁴/2]–[x⁶/3]–[x⁸/4]–[x¹⁰/5]+...=–&sum_n=1^∞[(–1)ⁿ⁺¹x²ⁿ/n] (valid for |x|<1).

Computing a value of a derivative
I know that the degree 8 term in the Maclaurin series for ln(1–x²) is –[x⁸/4]. But it is also supposed to be (by abstract "theory") [f⁽⁸⁾(0)/8!]x⁸. This means "clearly" (!!!) that –1/4=[f⁽⁸⁾(0)/8!] and therefore f⁽⁸⁾(0)=–8!/4.

That's if you desperately wanted to know the value of the derivative. An alternate strategy would be to compute the 8^th derivative and evaluate it at x=0. Here is that derivative:

             6    8       4       
  10080 (28 x  + x  + 70 x  + 28 x  + 1)
- --------------------------------------
                   2 8
            (-1 + x )

arctan
Let me try to find a Taylor series centered at 0 (a Maclaurin series) for arctan. Well, the general Maclaurin series is ∑_n=0^∞[f⁽ⁿ⁾(0)/n!]xⁿ so we can just try to compute some derivatives and evaluate them at 0. Let's see:
n=0 f(x)=arctan(x) so f(0)=arctan(0)=0.
n=1 f´(x)=1/(1+x²) so f´(0)=Stop this right now! Why? Because this way is madness. Here is the 7^th derivative of arctan(x):

        6       4       2
720 (7 x  - 35 x  + 21 x  - 1)
------------------------------
              2 7
        (1 + x )

Instead look at

 1
----
1+x2

• arctan(x)=x–[x³/3]+[x⁵/5]–[x⁷/7]+[x⁹/9]...= ∑_n=0^∞[(–1)²ⁿ⁺¹x²ⁿ⁺¹/(2n+1)]

Computing π This series has been used to compute decimal approximations of π. For example, if x=1, arctan(1)=π/4, so this must be 1–1/3+1/5–1/7+... but the series converges very slowly (for example, the 1000th partial sum multiplied by 4 gives the approximation 3.1406 for π which is not so good for all that arithmetic!) . Here is a history of some of the classical efforts to compute decimal digits of π. You can search some of the known decimal digits of π here. There are more than a trillion (I think that is 1012) digits of π's decimal expansion known. Onward! The methods used for such computations are much more elaborate than what we have discussed.

In physics they say ...
Many of the force "laws" stated in physics are quadratic (second degree) and therefore it is not surprising that squares and square roots need to be computed frequently. What does sqrt(1+x) "look like" near 0? Well, in this case I will try a direct computation. If f(x)=sqrt(1+x) then ...

Function	Value at x=0
f(x)=(1+x)1/2	1
f´(x)=(1/2)(1+x)–1/2	1/2
f´´(x)=(1/2)(–1/2)(1+x)–3/2	(1/2)(–1/2)
f(3)(x)=(1/2)(–1/2)(–3/2)(1+x)–5/2	(1/2)(–1/2)(–3/2)
I forget but look at the pattern.	(1/2)(–1/2)(–3/2)(–5/2)

So this Taylor series looks like
(1+x)^1/2=1+(1/2)x+[(1/2)(–1/2)/2]x²+[(1/2)(–1/2)(–3/2)/6]x³+[(1/2)(–1/2)(–3/2)(–5/2)/24]x⁴+...
where those other strange numbers come from the factorials, of course. Well, how might this be used in physics. Suppose you are trying to analyze sqrt(1+w). If |w| is very small, well then I bet that sqrt(1+w) is quite close to sqrt(1+0) which is 1. But the value will not equal 1 if w is not 0. What sort of "first order" estimate would I believe in? I bet that sqrt(1+w) is approximately 1+(1/2)w for small w. I also believe that the error will be (roughly) proportional to the size of w² (that's the Error Bound again). For many applications, knowing this is enough. But what if I wanted more accuracy, and I wanted an estimate which was correct to "second order terms". I bet this would take sqrt(1+w) and then the estimate would be 1+(1/2)w–(1/8)w², with an error which would be (roughly) proportional to w³. Depending on the application you were interested in, the estimate would be a bigger and bigger partial sum of the Taylor series.

Why do I insist on writing the coefficients of the series in the silly way done above? Why not just multiply out and write that? Well, if I do the result may be rather deceptive. It would be
1+(1/2)w–[1/8]w²+[1/16]w³–[5/128]w⁴+...
(I think we got the coefficient of the fourth degree term wrong in class when we "simplified" -- I'm sorry!) so if I accidentally saw only the first 4 terms I might think there is some obvious pattern to the series. Actually, the pattern is more complicated, as the coefficient of w⁴ showns. There is an abbreviation which is used (binomial coefficients) but the numbers are complicated.

Binomial series with m=1/3
One of Newton's most acclaimed accomplishments was the description of the Maclaurin series for (1+x)^m. Here is more information. Here I'll specialize by analyzing what happens when m=1/3. (In class we looked at 1/2.) We'll use a direct approach by taking lots of derivatives and trying to understand ∑_n=0^∞[f⁽ⁿ⁾(0)/n!]xⁿ.

• n=0 f(x)=(1+x)^1/3 so that f(0)=(1+0)^1/3=1.

• n=1 f´(x)=(1/3)(1+x)^–2/3 so that f´(0)=(1/3)(1+0)^–2/3=(1/3).

• n=2 f´´(x)=(1/3)(–2/3)(1+x)^–5/3 so that f´´(0)=(1/3)(–2/3)(1+0)^–5/3=(1/3)(–2/3).

• n=3 f⁽³⁾(x)=(1/3)(–2/3)(–5/3)(1+x)^–8/3 so that f⁽³⁾(0)=(1/3)(–2/3)(–5/3)(1+0)^–8/3=(1/3)(–2/3)(–5/3).

• n=4 f⁽⁴⁾(x)=(1/3)(–2/3)(–5/3)(–8/3)(1+x)^–11/3 so that f⁽⁴⁾(0)=(1/3)(–2/3)(–5/3)(–8/3)(1+0)^–11/3=(1/3)(–2/3)(–5/3)(–8/3).

1+(1/3)x+[(1/3)(–2/3)/2]x²+[(1/3)(–2/3)(–5/3)/2·3]x³+[(1/3)(–2/3)(–5/3)(–8/3)/2·3·4]x⁴+...

I'll come back to the general ideas later, but let's see how to use this in various ways.

Naive numerical use
So you want to compute (1.05)^1/3? This is a doubtful assumption, and anyway wouldn't you do a few button pushes on a calculator? But let's see:
Suppose we use the first two terms of the series and let x=.05:
     (1+.05)^1/3=1+(1/3)(.05)+Error
What's interesting to me is how big the Error is. Of course, we have the Error Bound, which states that |Error|≤[K/(n+1)!]|x–a|ⁿ⁺¹. Here, since the top term in the approximation is the linear term, we have n=1. And a, the center of the series, is 0, and x, where we are approximating, is .05. Fine, but the most complicated part still needs some work. K is an overestimate of the absolute value of the second derivative of f on the interval connecting 0 and .05. Well (look above!) we know that f⁽²⁾(x)=(1/3)(–2/3)(1+x)^–5/3. We strip away the signs (not the sign in the exponent, since that means something else!). We'd like some estimate of the size of (2/9)(1+x)^–5/3 on [0,.05]. Well, it is exactly because of the minus sign in the exponent that we decide the second derivative is decreasing on the interval [0,.05] and therefore the largest value will be at the left-hand endpoint, where x=0. So plug in 0 and get 1/9(1+0)^–5/3=2/9. This is our K. So the Error Bound gives us [(2/9)/2!](.05)², which is about .00027. We have three (and a half!) decimal digits of accuracy in the simple 1+(1/3)(.05) estimate.

What if we wanted to improve this estimate? Well, we can try another term. By this I mean use 1+(1/3)(.05)+[(1/3)(–2/3)/2](.05)² as an estimate of (1.05)^1/3. How good is this estimate? Again, we use the Error Bound: |Error|≤[K/(n+1)!]|x–a|ⁿ⁺¹. Now n=2 and a=0 and x=.05, and K comes from considering f⁽³⁾(x)=(1/3)(–2/3)(–5/3)(1+x)^–8/3. We need to look at (10/27)(1+x)^–8/3 on [0,.05]. The exponent is again negative (what an accident not -- these methods are actually used and things should be fairly simple!) and therefore the function is again decreasing and an overestimate is gotten by looking at the value when x=0, so (10/27)(1+x)^–8/3 becomes (10/27)(1+0)^–8/3=(10/27). Hey, [K/(n+1)!]|x–a|ⁿ⁺¹ in turn becomes [(10/27)/3!](.05)³, about .000008, even better.

Approximating a function on an interval
People usually use the partial sums considered above in a more sophisticated way. For example, consider replacing (1+x)^1/3 by T₂(x)=1+(1/3)x+[(1/3)(–2/3)/2!]x²=1+(x/3)–(x²/9) anywhere on the interval [0,.05]. I like polynomials, because they can be computed just by adding and multiplying. The function (1+x)^1/3 has this irritating and weird exponent, that I, at least, can't readily estimate. What about the error? The Error Bound declares that an overestimate of the error is [K/3!]|x–0|³. Now if 0<x<.05, then the largest x³ can be is (.05)³. What about K? Again, we look at the third derivative with the signs (not the exponent sign!) dropped. This is (10/27)(1+x)^–8/3 which we are supposed to estimate for any x in [0,.05]. But since the third derivative is still decreasing (–1/3<0) again the K is gotten by plugging in x=0. Hey: the estimate is the same as what we had before, about .000008. Below are some pictures illustrating what's going on.

(1+x)1/3 & T2(x) on [0,.05]	(1+x)1/3 & T2(x) on [0,2]
Comments Yes, there really really are two curves here. One, (1+x)1/3, is green and one, T2(x)=1+(x/3)–(x2/9), is red. But the pixels in the image overlay each other a lot, because the error, .000008, makes the graphs coincide on the [0,.05] interval. There are only a finite number of positions which can be colored, after all!	Comments This graph perhaps shows more about what's going on. The domain interval has been changed to [0,2]. The K in the Error Estimate is not changed, but the x's now range up to 2. So [K/3!]x3 becomes as a worst case estimate [(10/27)/3!]23, which is about .49. You can see T2(x) revealed (!) as just a parabola opening downward (hey, 1+(x/3)–(x2/9) has –1/9 as x2 coefficient!). The two curves are close near x=0, and then begin to separate as x grows larger.

Improving the approximation
The whole idea is maybe by increasing the partial sum (going to T_n(x)'s with higher n's) we will get better approximations. This is only usable if the approximations are easy to compute (nice polynomials with simple coefficients) and if the error estimates are fairly easy to do. This actually occurs so that people use these ideas every day.

Binomial series in general
Suppose m is any number (yes: any number). Then

• (1+x)^m=1+mx+[m(m–1)/1·2]x²+[m(m–1)(m–2)/1·2·3]x³+[m(m–1)(m–2)(m–3)/1·2·3·4]x⁴+..., valid for |x|<1.

The list ...

• e^x=1+x+{x²/2}+{x³/6+...=∑_n=0^∞[xⁿ/n!] valid for all x.
• sin(x)=x–[x³/3!]+[x⁵/5!]–[x⁷/7!]+[x⁹/9!]...=∑_n=0^∞(–1)ⁿ⁺¹x²ⁿ⁺²/(2n+1)! valid for all x.
• cos(x)=1–[x²/2!]+[x⁴/4!]–[x⁶/6!]+[x⁸/8!]–[x¹⁰/10!]...=∑_n=0^∞(–1)ⁿ⁺¹x²ⁿ/(2n)! valid for all x.
• 1/(1–x)=1+x+x²+x³+...=∑_n=0^∞xⁿ valid for |x|<1.
• ln(1+x)=x–[x²/2]+[x³/3]–[x⁴/4]+[x⁵/5]+...=∑_n=1^∞[(–1)ⁿ⁺¹xⁿ/n] valid for |x|<1.
• arctan(x)=x–[x³/3]+[x⁵/5]–[x⁷/7]+[x⁹/9]...= ∑_n=0^∞[(–1)²ⁿ⁺¹x²ⁿ⁺¹/(2n+1)] valid for |x|<1 (the problem is that 1/(1+x²) behaves badly when x is +/–i and so the series can't have a radius of convergence bigger than 1).
• (1+x)^m=1+mx+[m(m–1)/1·2]x²+[m(m–1)(m–2)/1·2·3]x³+[m(m–1)(m–2)(m–3)/1·2·3·4]x⁴+..., valid for |x|<1.

Pascal's triangle
I very rapidly discussed the magical assembly of numbers called Pascal's triangle. There's a long wikipedia article about it, of course, and here's another link which has pictures of more old documents mentioning the triangle. I tried to explain some of the uses of the numbers which occur, and how the Taylor expansion above generalizes the result for (1+x)^{INTEGER POWER}.

1/(1+x)²
More a part of this course was my effort to compute that Taylor series for 1/(1+x)² using as many different techniques as we could imagine. I think we discussed the methods that follow.

The Binomial Theorem Take m=–2 and try to understand the binomial coefficients which result. Well:
The –2 over 0 binomial coefficient has 0! downstairs, and this is 1. On the top we have zero terms multiplied together, and (by special understanding, to make the notation work!) this is defined also to be 1 ("the empty product is 1" is the what people say). The result is 1.
The –2 over 1 binomial coefficient has 1!=1 downstairs, and on top has just –2. So the result is –2.
The –2 over 2 binomial coefficient has 2!=2 downstairs, and on top has just –2 multiplied by –2–1=–3. So the result is 6 divided by 2, or 3.
The –2 over 3 binomial coefficient has 3!=6 downstairs, and on top has –2 multiplied by –2–1=–3 multiplied by –2–2=–4. So the result is –24 divided by 6, or –4.
By now people mostly saw the pattern, and therefore 1/(1+x)²=1–2x+3x²–4x³+...

Multiply two series I know from the geometric series result that 1/(1+x) is the same as 1–x+x²–x³+... because the first term is 1 and the common ratio is –x. The sum is therefore 1/(1–{–x})=1/(1+x). So then 1/(1+x)² should be the same as (1–x+x²–x³+...)². Now we tried to discuss how to think about the multiplication. The only way to get a constant term is to multiply 1 times 1. There are two ways to get a degree 1 in x term: multiply 1 times x and also multiply x by 1. Both of these have a negative sign, so the result is –2x. How about x³? We can multiply 1 times x², and also x² times 1, and also –x times –x. The result is 3x². We also even computed the x³ term, and we got –4x³. Etc. (!!) Therefore 1/(1+x)² must be 1–2x+3x²–4x³+... again.

Differentiate That is, the original geometric series for 1/(1+x) is 1–x+x²–x³+... and we can differentiate. The derivative of 1/(1+x) is –1/(1+x)², and the derivative of the series is 0–1+2x–3x²+... and we get our result again if we multiply by 1. Wow.

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

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

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

And algebraically ...
It is also true that inside the radius of convergence, power series can be manipulated just like l-o-n-g polynomials so terms can be interchanged in any fashion, etc. So everything works inside the radius of convergence where power series converge absolutely.

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 and nicely as power series.

If a function has a power series then ...
Suppose I know that f(x) is equal to a sum like A+B(x–x₀)+C(x–x₀)²+D(x–x₀)³+E(x–x₀)⁴+... and I would like to understand how the coefficients A and B and C and D etc. relate to f(x). Here is what we can do.

Step 0 Since f(x)=A+B(x–x₀)+C(x–x₀)²+D(x–x₀)³+E(x–x₀)⁴+... if we change x to x₀ we get f(x₀)=A. All the other terms, which have powers of x–x₀, are 0.
Step 1a Differentiate (or, as I said in class, d/dx) the previous equation which has x's, not x₀'s. Then we have f´(x)=0+B+2C(x–x₀)¹+3D(x–x₀)²+4E(x–x₀)³+...
Step 1b Plug in x₀ for x and get f´(x₀)=B. All the other terms, which have powers of x–x₀, are 0.
Step 2a Differentiate (or, as I said in class, d/dx) the previous equation which has x's, not x₀'s. Then we have f´´(x)=0+0+2C+3·2D(x–x₀)¹+4·3E(x–x₀)²+...
Step 2b Plug in x₀ for x and get f´´(x₀)=2C, so C=[1/2!]f⁽²⁾(x₀). All the other terms, which have powers of x–x₀, are 0.
Step 3a Differentiate (or, as I said in class, d/dx) the previous equation which has x's, not x₀'s. Then we have f⁽³⁾(x)=0+0+0+3·2·1D+4·3·2E(x–x₀)¹+...
Step 3b Plug in x₀ for x and get f⁽³⁾)(x₀)=3·2·1D=3!C so D=[1/3!]f⁽³⁾(x₀). All the other terms, which have powers of x–x₀, are 0.
ETC. Continue as long as you like. What we get is the following fact: if f(x)=∑_n=0^∞a_n(x–x₀)ⁿ then a_n=[f⁽ⁿ⁾(x₀)/n!]. This is valid for all non-negative integers, n. Actually, this formula is one of the reasons that 0! is 1 and the zeroth derivative of f is f itself. With these understandings, the formula works for n=0.
What this means is

If a function is equal to a power series, that power series must be the Taylor series of the function.

I hope you notice, please please please, that the partial sums of the Taylor series are just the Taylor polynomials, which we studied earlier.

Usually I'll take x₀=0, as I mentioned so that (x-x₀)ⁿ becomes just xⁿ. Then the textbook and some other sources call the series the Maclaurin series but I am too lazy to remember another name. A useful consequence of this result (it will seem sort of silly!) is that if a function has a power series expansion, then it has exactly one power series expansion (because any two such series expansions are both equal to the Taylor series, so they must be equal). This means if we can get a series expansion using any sort of tricks, then that series expansion is the "correct one" -- there is only one series expansion. I'll show you some tricks, but in this class I think I will just try some standard examples which will work relatively easily.

e^x
I'll take x₀=0. Then all of the derivatives of e^x are e^x, and the values of these at 0 are all 1. So the coefficients of the Taylor series, a_n, are [f⁽ⁿ⁾(x₀)/n!]=1/n!. The Taylor series for e^x is therefore ∑_n=0^∞[1/n!]xⁿ.

e^–.3
Let's consider the Taylor series for e^x when x=–.3. This is ∑_n=0^∞[1/n!](–.3)ⁿ. What can I tell you about this? Well, for example, my "pal" could compute a partial sum, say ∑_n=0¹⁰[1/n!](–.3)ⁿ. The result is 0.7408182206. That's nice. But what else do we know? Well, this partial sum is T₁₀(–.3), the tenth Taylor polynomial for e^x centered at x₀=0, and evaluated at –.3. The Error Bound gives an estimation of |T₁₀(–.3)–e^–.3|. This Error Bound asserts that this difference is at most [K–|.3–0|¹¹/11!], where K is some overestimate of the absolute value of the 11^th derivative of e^x on the interval between –.3 and 0. Well, that 11^th derivative is also e^x. And we know that e^x is increasing (exponential growth after all!) so that for x's in the interval [–.3,0], e^x is at most e⁰=1, and we can take that for K. So the Error Bound is 1|–.3–0|¹¹/11!. Now let's look at some numbers:
|–.3|¹¹=0.00000177147 and 11!=39,916,800, and their quotient is about 4·10^–14.
This means that e^–.3 and T₁₀(–.3) agree at least to 13 decimal places. Indeed, to 10 decimal places, e^–.3 is reported as 0.7408182206, the same number we had before. Wow? Wow!

Let's change 10 to n and 11 to n+1. Then |T_n(–.3)–e^–.3| is bounded by K|–.3–0|ⁿ⁺¹/(n+1)!. Here K=1 again because all of the derivatives are the same, e^x. Since 1|–.3–0|ⁿ⁺¹/(n+1)!→0 as n→∞ what do we know?

I think that the sequence {T_n(–.3)} converges, and its limit is e^–.3. Since this sequence of Taylor polynomial values is also the sequence of partial sums of the series ∑_n=0^∞[1/n!](–.3)ⁿ, I think that the series converges, and its sum is e^–.3. Therefore
e^–.3=∑_n=0^∞[1/n!](–.3)ⁿ.

e.7 We could try the same sequence of ideas with x=.7. First examine T10(.7). This is ∑n=010[1/n!](.7)n. To 10 decimal places, this is 2.0137527069. We have information from the Error Bound. It declares that |T10(.7)–e.7| is no larger than K|.7–0|11/11!. Here K is an overestimate of the 11th derivative, which is ex, on the interval [0,.7]. The exponential function is (still!) increasing, so the largest value is at x=.7. But I don't know e.7. I do know it is less than e1 which I hardly know also. I will guess that e<3. So a nice simple K to take is 3. Let me try that. The Error Bound is less than 3|.7–0|11/11!. Let's do the numbers here. 11!=39,916,800 (again) but .711=0.0197732674 (small, but not as small as |–.3|11). The Error Bound 3|.7–0|11/11! is about 2·10–9, not quite as small. Now e.7, to 10 decimal places, is 2.0137527074 and this is close enough to the sum value quoted before. Again, go to n and n+1: |Tn(.7)–e.7| is less than 3|.7–0|n+1/(n+1)!, and again, as n→∞ this goes to 0. Our conclusion is: The sequence {Tn(.7)} converges, and its limit is e.7. Since this sequence of Taylor polynomial values is also the sequence of partial sums of the series ∑n=0∞[1/n!](.7)n, I think that the series converges, and its sum is e.7. Therefore e.7=∑n=0∞[1/n!](.7)n.

e⁵⁰
Just one more example partly because we'll see some strange numbers. Let's look at T₁₀(50) which is ∑_n=0¹⁰[1/n!]50ⁿ. This turns out to be (approximately!) 33,442,143,496.7672, a big number. The Error Bound says that |T₁₀(50)–e⁵⁰| is less than K|50–0|¹¹/11! where K is the largest e^x can be on [0,50]. That largest number is e⁵⁰ because e^x is increasing. I guess e⁵⁰ is less than, say, 3⁵⁰, which is about 7·10²³. I'll take that for K. Now how big is that Error?
K|50–0|¹¹/11! still has 11! underneath but now the top is growing also. This is approximately 9·10³⁴, a sort of big number.

The situation for x=50 may look hopeless, but it isn't really. To analyze |T_n(50)–e⁵⁰| we need to look at K[(50)ⁿ⁺¹/(n+1)!]. Here the fraction has power growth on the top and factorial growth on the bottom. Well, we considered this before. I called it a "rather sophisticated example". Factorial growth is faster eventually than power growth. So this sequence will →0 as n→∞. A similar conclusion occurs:
e⁵⁰=∑_n=0^∞[1/n!](50)ⁿ.

In fact, e⁵⁰ is 5.18470552858707·10²¹ while the partial sum with n=100, ∑_n=0¹⁰⁰[1/n!](50)ⁿ has value 5.18470552777323·10²¹: the agreement is not too bad, relatively.

And generally for exp ...
It turns out that ∑_n=0^∞[1/n!]xⁿ converges for all x and its sum is always e^x. The way to verify this is what we just discussed. Most actual computation of values of the exponential function relies on partial sums of this series. There are lots of computational tricks to speed things up, but the heart of the matter is the Taylor series for the exponential function.

Sine
We analyzed sine's Taylor polynomials, taking advantage of the cyclic (repetitive) nature of the derivatives of cosine: sine→cosine→–sine→–cosine then back to sine. At x₀=0, this gets us a cycle of numbers: 0→1→0→–1→0 etc. The Taylor series for sine centered at 0 leads off like this:
x–[x³/3!]+[x⁵/5!]–[x⁷/7!]+[x⁹/9!]–...

It alternates in sign, it has only terms of odd degree, and each term has the reciprocal of an "appropriate" factorial (same as the degree) as the size of its coefficient. Using summation notation, which is convenient and compact, this series is ∑_n=0^∞[(–1)ⁿ/(2n+1)!]x²ⁿ⁺¹.

What happens to the error bound?
This is similar to what we did with exp. There are two claims: the series ∑_n=0^∞[(–1)ⁿ/(2n+1)!]x²ⁿ⁺¹ converges and the sum of the series is sin(x). Well, to see that this is true we investigate the difference between sin(x) and S_N, the N^th partial sum of the series. But S_N is the same as T_N(x), the N^th Taylor polynomial. And the error bound tells us that |sin(x)–T_N(x)|≤[K/(N+1)!]|x–0|ⁿ⁺¹. Just as before, [|x|^N+1/(N+1)!]→0 as N→∞. What about the K's? If they misbehave (get very big) that could make the whole estimate lousy. But in fact in this specific case, |K| is an overestimate on the size of some derivative of sine. But all of the derivatives of sine are +/–sine and +/–cosine, and these all are ≤1 in their absolute values. So, in fact, we're done. We have verified that the series converges and that sin(x) is its sum.

Cosine
We could duplicate this work for cosine, or, as I mentioned in class, be a bit cleverer. Since we know that sin(x)=∑_n=0^∞[(–1)ⁿ/(2n+1)!]x²ⁿ⁺¹ is true for all x, we could differentiate this equation. The result is cos(x)=∑_n=0^∞[(–1)ⁿ/(2n+1)!](2n+1)x²ⁿ. In fact, most people realize that (2n+1)/(2n+1)! is 1/(2n)! so that we have verified the equation cos(x)=∑_n=0^∞[(–1)ⁿ/(2n)!]x²ⁿ for all x. This is two facts: the series converges, and the sum of the series (the limit of the sequence of partial sums) is cos(x) for all x's.

A numerical example: cos(1/3) How close is 1–[(1/3)2/2!]+[(1/3)4/4!]–[(1/3)6/6!]+[(1/3)8/8!]–[(1/3)10/10!] to cos(1/3)? Here we sort of have two candidates because T10(1/3) is the same as T11(1/3) since the 11th degree term is 0. Error bound, n=10 So we have K|(1/3)–0|11/11!, where K is a bound on the size of the 11th derivative of cosine. Hey: I don't care much in this example, because I know that this derivative is +/–cosine or +/–sine, so that I can take K to be 1. Now it turns out that (1/3)11/11! is about 1.4·10–14. This is tiny, but ... Error bound, n=11 This is even better. So we have K|(1/3)–0|12/12!, where K can again be taken as 1 (this is easier than exp!) So (1/3)12/12! is about 4·10–15, even tinier. Hey, cos(1/3)=0.944956946314738 and T10(1/3)=0.944956946314734.

Cosine and sine estimates
For both cosine and sine, the estimates are easy because K for both can be taken to be 1.

Is success guaranteed?
The material in this lecture showed that exp and sine and cosine have Taylor series and their Taylor series converge everywhere and the sum of each Taylor series is always equal to the original function. This is a wonderful situation, and, for example, is essentially used to compute values of exp and sine and cosine using partial sums of the series (as I wrote above, there are some other "computational tricks to speed things up, but the heart of the matter is the Taylor series").

But these are the first and nicest and simplest examples. The situation is not always so easy. We will see a few functions where things don't work out. I can even think about one of them with you now. Consider tangent. Certainly if we take x₀ to be 0, we can differentiate tangent lots of times and "get" a Taylor series for tangent centered at 0. The reason I wrote "get" with quotes is that the coefficients of the Taylor series for tangent are well-known in the math world to be rather, well, rather irritating and difficult to predict and understand. So already there's a problem. How about convergence? Things also don't go well there because, if you remember tangent's graph, vertical asymptotes occur at odd multiples of Π/2. You can't expect that the series will converge, for example, at +/–Π/2 and, in fact, the radius of convergence turns out to be only Π/2 (this is not so obvious, actually). Most calculators I know compute values of tangent by computing values of sine and cosine and then dividing. This is easier than direct computation of tangent.

I asked a pal to compute tan^(j)(0) for j running from 0 to 20. My pal produced these numbers:

0, 1, 0, 2, 0, 16, 0, 272, 0, 7936, 0, 353792, 0, 22368256, 0, 1903757312, 0, 209865342976, 0, 29088885112832, 0

This is a very peculiar sequence. Here is a very abbreviated description of what's known about it. Incidentally, anyone with any mathematical curiosity (or even anyone who can count!) should go to that website and spend some wonderful time, like a kid in a toy store. Fun, fun, fun ...

A series for cos(x3) We can use the series we know to "bootstrap" and get other series. That is, we build on known results. For example, since we know that cos(x)=1–[x2/2!]+[x4/4!]–[x6/6!]+[x8/8!]–[x10/10!]... for all x, I bet cos(x3)=1–[(x3)2/2!]+[(x3)4/4!]–[(x3)6/6!]+[(x3)8/8!]–[(x3)10/10!]... which is 1–[x6/2!]+[x12/4!]–[x18/6!]+[x24/8!]–[x30/10!]... and this is much easier than trying to compute derivatives of cos(x3) which we would have to do to plug in the values of the derivatives in the Taylor series. The Chain Rule makes things horrible. For example, the fifth derivative of cos(x3) is –243sin(x3)x10+1620cos(x3)x7+2160sin(x3)x4–360cos(x3)x and that's fairly horrible. How about x2cos(x3)? Multiply the previous series by x2. The result is x2cos(x3)=x2–[x8/2!]+[x14/4!]–[x20/6!]+[x26/8!]–[x32/10!]... Wow? A question and its answer What are the first four non–zero terms of the Taylor series for x3e–2x centered at 0? Since ex=1+x+x2/2+x3/6+... (3! is 6) we know e–2x=1–2x+(–2x)2/2+(–2x)3/6+...=1–2x+4x2/2–8x3/6+...= =1–2x+2x2–4x3/3+... and then we multiply by x3. The answer is x3–2x4+2x5–4x6/3.

What do we know?
If a function f(x) has a power series expansion centered at a, then f(x)=∑_n=0^∞[f⁽ⁿ⁾(a)/n!](x–a)ⁿ. The partial sums of this series are Taylor polynomials, and this is called a Taylor series. Almost all the time we and other people in the world take a to be 0. Then the series is also called a Maclaurin series. Examples so far include:

• e^x=1+x+{x²/2}+{x³/6+...=∑_n=0^∞[xⁿ/n!] valid for all x.
• sin(x)=x–[x³/3!]+[x⁵/5!]–[x⁷/7!]+[x⁹/9!]...=∑_n=0^∞(–1)ⁿ⁺¹x²ⁿ⁺²/(2n+1)! valid for all x.
• cos(x)=1–[x²/2!]+[x⁴/4!]–[x⁶/6!]+[x⁸/8!]–[x¹⁰/10!]...=∑_n=0^∞(–1)ⁿ⁺¹x²ⁿ/(2n)! valid for all x.

As I mentioned in class, I am not allowed to tell you why these series resemble each other because your heads might explode. e^ix=cos(x)+isin(x) The not-accidential resemblance will be discussed in your differential equation course. This can be verified, at least on some sort of symbolic level, by plugging ix into the Taylor series for exp, and then realizing that powers of i are periodic with period 4 (how curious -- where did we ever see than before?): i, i²=–1, i³=–i, i⁴=1, and then reorganizing the result to "find" cos(x)+isin(x).

I mentioned the logical difficulty of the request to "find the first four terms in the Maclaurin series" since maybe 0+x+x²+[x³/3] answers that question. Usually people want the first four non-zero terms and that was the question I answered.

Of course, in class I messed up this computation because I can't think standing up. (Well, I can't think sitting either but ...)

In honor of Mr. Gretzmacher I added a part b) to this question:
b) If f(x)=e^xsin(x), what is f⁽⁵⁾(0)?
There are several ways to answer this question. One would be to compute the fifth derivative. This is actually not too difficult, but let me be a little slick. I know that f(x) has one and only one power series centered at 0, and that series must be its Taylor series. Therefore the coefficient of x⁵ must be f⁽⁵⁾(0)/5!. But we know that coefficient. It is –1/30. So f⁽⁵⁾(0)/5!=–1/30 and f⁽⁵⁾(0)=–5!/30=–4. Isn't that cute?

Of course Mr. Levi nearly ruined this magical moment by exclaiming something about a workshop problem. Sigh. It turns out that this trick, used for nothing really too interesting here, has actually some more profound uses in such areas as probability (when studying what's called the moment generating function).

Book problem: 10.7, #14
Find the Maclaurin series of cos(sqrt(x)).
Since cos(x)=1–[x²/2!]+[x⁴/4!]–[x⁶/6!]+[x⁸/8!]–[x¹⁰/10!]...=∑_n=0^∞(–1)ⁿ⁺¹x²ⁿ/(2n)! I know that cos(sqrt(x))=1–[(sqrt(x))²/2!]+[(sqrt(x))⁴/4!]–[(sqrt(x))⁶/6!]+[(sqrt(x))⁸/8!]–[(sqrt(x))¹⁰/10!]...=∑_n=0^∞(–1)ⁿ⁺¹(sqrt(x))²ⁿ/(2n)! and so cos(sqrt(x))=1–[x/2!]+[x²/4!]–[x³/6!]+[x⁴/8!]–[x⁵/10!]...=∑_n=0^∞(–1)ⁿ⁺¹xⁿ/(2n)!, and please be LAZY.

Book problem: 10.7, #19
Find the Maclaurin series of (1–cos(x²))/x.
Since cos(x)=1–[x²/2!]+[x⁴/4!]–[x⁶/6!]+... I know that cos(x²)=1–[x⁴/2!]+[x⁸/4!]–[x¹²/6!]+... and 1–cos(x²)=[x⁴/2!]–[x⁸/4!]+[x¹²/6!]+... so that (1–cos(x²))/x=[x³/2!]–[x⁷/4!]+[x¹¹/6!]+...

If we had to, we could write this in summation form (I hope I get this correct!)
(1–cos(x²))/x=∑_n=0^∞(–1)ⁿ⁺¹x⁴ⁿ⁺³/(2(n+1))!

Note, please, that at least initially the "function" given, (1–cos(x²))/x, looks like it has bad behavior at 0 since it looks like division by 0. But the series approach just wipes that out. Many computations done using L'Hôpital's Rule also can be easily accomplished with Taylor series manipulation.

An integral
The function e^–x2 is extremely important in probability. Its integral is called the error function. Suppose we want to compute ∫_x=0^.5e^–x2dx. It can be proved that e^–x2 has no antiderivative which can be written in terms of familiar functions. How could we then compute this definite integral? Its value, a pal of mine tells me, is approximately 0.461281. Well, I could use the Trapezoid Rule or Simpson's Rule or ... Look at this:

    e^x=∑_n=0^∞xⁿ/n!
Substitute –x² for x.
    e^–x2=∑_n=0^∞(–x²)ⁿ/n!=∑_n=0^∞(–1)ⁿx²ⁿ/n!
Integrate.
    ∫_x=0^x=.5e^–x2dx=∫_x=0^x=.5∑_n=0^∞(–1)ⁿx²ⁿ/n! dx=∑_n=0^∞∫_x=0^x=.5(–1)ⁿx²ⁿ/n!=∑_n=0^∞(–1)ⁿx²ⁿ⁺¹/([2n+1]n!)|_x=0^x=.5
Evaluate.
The integral is∑_n=0^∞(–1)ⁿ(1/2)²ⁿ⁺¹/([2n+1]n!)

This series is alternating, and satisfies all the hypotheses of the Alternating Series Test. Any partial sum is within the accuracy of the first omitted term (look in the textbook or think about with the help of the Xmas tree picture to the right). Well, if I want 5 digit accuracy, I just need to find n so that (1/2)²ⁿ⁺¹/([2n+1]n!) is less than .00001, which is 1/(100,000).

If n=4, then (1/2)⁹/[9·24] is {1/(512)}·[1/(216)] which is 110,592. The sum from n=0 to 3, that is, S₃, is 0.461272 (accurate to +/–.00001 as desired).

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 elaborate and 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^∞nⁿxⁿ converge?

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

Even less silly example
Let's try this: for which x's does ∑_n=1^∞xⁿ/nⁿ converge? I hope you will see the resemblance and contrast with the previous computation:
     |xⁿ/nⁿ|^1/n=(|x|ⁿ/nⁿ)^1/n=|x|^n·(1/n)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^∞xⁿ/nⁿ converges absolutely for all x's and therefore converges for all x's.

Comment: Root vs. Ratio
As I mentioned in class, I have an emotional preference for the Ratio Test that I can't explain. But I will admit that analyzing the two previous examples with the Ratio Test would be very difficult. However, the Ratio Test works exceptionally well when series have factorials (you'll see why there are lots of series with factorials in the next lecture). So series with similar results to the two previous examples which I'd examine with the Ratio Test would be ∑_n=0^∞n!xⁿ and ∑_n=0^∞xⁿ/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. Example 76 For which x's does ∑n=1∞xn/n converge? We used the Ratio Test, and |an+1/an| simplified fairly easily to |x|[(n+1)/n]. Now L'H or simple algebraic manipulation shows that ρ=|x|. So we get guaranteed absolute convergence and therefore convergence when |x|<1 and divergence when |x|>1. For |x|=1, we don't get any information. I'll write the answer using interval notation now: if x is in (–1,1), the series converges. If x is in (1,∞), the series diverges. If x is in (–∞,–1), the series diverges. There's no information for x=1 or x=–1. If you insist on knowing what happens at x=+/–1, let's "insert" these values of x into the series and investigate. If x=+1, the series becomes ∑n=1∞1n/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∞xn/n2 converge? Again, the Ratio Test, and |an+1/an| simplified fairly easily to |x|[n/(n+1)]. And again manipulation shows that ρ=|x|. So we have absolute convergence and therefore convergence when |x|<1 and divergence when |x|>1. For |x|=1, we don't get any information. As intervals: if x is in (–1,1), the series converges. If x is in (1,∞), the series diverges. If x is in (–∞,–1), the series diverges. There's no information for x=1 or x=–1. (Looks a lot the same, huh?) To see what happens at x=+/–1, put these values of x into the series and investigate the result directly (I don't know any other ways to do this). If x=+1, the series becomes ∑n=1∞1n/n2=∑n=1∞1/n2. This is a p-series with p=2>1, so it converges. If x=–1, the series becomes ∑n=1∞(–1)n/n2. But this series converges absolutely (it gives us the p-series just considered when the signs are stripped off) and therefore it must converve. And again, look at the "improved" picture to the right.

Because I have the brains of a toad, I forgot to do this example, which is logically necessary. This example shows that we can have divergence at both ends of the interval of convergence. Example 78 For which x's does ∑n=1∞n xn converge? The same limiting ratio is reported, and we get the same convergence/divergence/no information result, with the same initial picture. When x=1, the series is ∑n=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.

A challenge example
I asked students to give me an interval and to tell me whether to include or exclude endpoints. I was given (4,8]. I then declared that this interval was the collection of x's for which the series ∑_n=1^∞(-1)ⁿ(x-6)ⁿ/(2ⁿn) converged, and that it would diverge at all other x's. At the suggestion of Mr. Oakes we used the (sigh) Root Test where L=|x-6|/2. It worked but I think the Ratio Test would also have worked. We separately examined the endpoints x=4 and x=8 and got the correct answers.

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

What is a power series?
A power series centered at x₀ (a fixed number) is an infinite series of the form ∑_n=0^∞c_n(x–x₀)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 x₀ to be 0 because this just makes thinking easier.

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

Why is this true?
Here is an indication of what's going on. Let me take x₀ to be 0, since I am lazy and the logic is the same. So we will consider a power series centered at 0: ∑_n=0^∞c_nxⁿ. The series will always converge when x=0 (not much news!). But suppose this series converges for a number v which is not 0. Let me explore the consequences. (This will involve the innermost source of the "strength" of power series, and why they are used so much.)

We assume that ∑_n=0^∞c_nvⁿ converges. If an infinite series converges, the terms must go to 0 as n→∞. So lim_n→∞c_nvⁿ=0. In particular, I bet that the sequence of the individual terms, {c_nvⁿ}, is bounded. For n large enough, these all cluster "near" 0. More explicitly, There is some M>0, maybe a really large M, so that |c_nvⁿ|<M.

Now take any number w with |w|<|v|. I want to analyze the {con|di}vergence of ∑_n=0^∞c_nwⁿ. I will do this by comparing the individual terms with the same series with v instead of w (there isn't much else to do!). I'll take absolute values because I like absolute convergence. Signs make things more delicate and maybe a simple approach will work. Certainly since |w|<|v| we know |c_nwⁿ|<|c_nvⁿ|<M. Well, look at this:
     |c_nwⁿ|=|c_nwⁿ(v/w)ⁿ|=|c_nwⁿ||v/w|ⁿ=Mrⁿ.

Here r=|v/w| is a number which is less than 1. But this means if we consider the series ∑_n=0^∞c_nwⁿ and take absolute values, the terms will all be less than ∑_n=0^∞Mrⁿ, a convergent geometric series (since r<1). So convergence must "spread inward" for power series.

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∞nnxn	[0]	0
∑n=1∞xn/nn	(–∞,+∞)	∞
∑n=1∞xn/n	[–1,1)	1
∑n=1∞xn/n2	[–1,1]	1
∑n=1∞n xn	(–1,1)	1

These examples show that the interval of convergence of a power series may or may not include one or both of the endpoints of the interval. The reason for the number of examples was to show you, explicitly, that it is possible for the series to converge on neither or one or both of the boundary points. I wanted to show a "complete" collection of examples.
It turns out that behavior on the edge of the interval is probably only interesting (sigh) as an exam question in calc 2 (where it is frequently asked!) because of some results you'll be told about in a few lines.

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 x₀=5. You are told that the series converges when x=–1 and diverges when x=14. What can you say about the radius of convergence? For which x's must this series converge and for which x's must this series diverge? You are given no other information.

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

What we can conclude from this information is the following:

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

  Step A     Step B     Step C       Step D   Step E   Step F

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

|  72(n+1)   |     72(n+1) 
| --------  |    --------  
| ((n+1)!)2 |    ((n+1)!)2    72(n+1)(n!)2
|-----------| = ---------- = ------------
|   72n     |       72n       72n((n+1)!)2     
| -------   |    -------   
|  (n!)2    |      (n!)2    

 |x|n+1(n+3n)         (n+3n)
-------------- = |x|·---------
|x|n(n+1+3n+1)       (n+1+3n+1) 

      7cos(36n7–2n2)+2sin(55n+88)
∑n=1∞ ---------------------------
                  2n

      7|cos(36n7–2n2)|+2|sin(55n+88)|
∑n=1∞ --------------------------------
                   2n