Power Series | College Math Teaching

May 25, 2021

Power series: review for inexperienced calculus teachers

This is “part 2” of a previous post about series. Once again, this is not designed for students seeing the material for the first time.

I’ll deal with general power series first, then later discuss Taylor Series (one method of obtaining a power series).

General Power Series Unless otherwise stated, we will be dealing with series such as $\sum a_n x^n$ though, of course, a series can expanded about any point in the real line (e. g. $\sum a_n (x-c)^n$ ) Note: unless it is needed, I’ll be suppressing the index in the summation sign.

So, IF we have a function $f(x) = \sum a_n x^n = a_0 + a_1 x + a_2 x^2 + ....a_k x^k + a_{k+1} x^{k+1}$ ..

what do we mean by this and what can we say about it?

The first thing to consider, of course, is convergence. One main fact is that the open interval of convergence of a power series centered at 0 is either:
1. Non existent …series converges at $x = 0$ only (e. g. $\sum_{k=0} (k!)x^k$ ; try the ratio test) or

2. Converges absolutely on $(-r, r)$ for some $r >0$ and diverges for all $|x| > r$ (like a geometric series) (anything possible for $|x| = r$ ) or

3. Converges absolutely on the whole real line.

Many texts state this result but do not prove it. I can understand as this result is really an artifact of calculus of a complex variable. But it won’t hurt to sketch out a proof at least for the “converge” case for $(-r, r)$ so I’ll do that.

So, let’s assume that $\sum a_n c^n$ converges (either absolute OR conditional ) So, by the divergence test, we know that the sequence of terms $a_n c^n \rightarrow 0$ . So we can find some index $M$ such that $n > M \rightarrow |a_n c^n| < 1$

So write: $|a_n x^n| = |a_n||c^n||| {x^n \over c^n} | < | {x^n \over c^n} |$ and $\sum | {x^n \over c^n} |$ converges absolutely for $|x| < |c|$ . The divergence follows from a simple “reversal of the inequalities (that is, if the series diverged at $x = c$ ).

Though the relation between real and complex variables might not be apparent, it CAN be useful. Here is an example: suppose one wants to find the open interval of convergence of, say, the series representation for ${1 \over 3+2x^2 }$ ? Of course, the function itself is continuous on the whole real line. But to find the open interval of convergence, look for the complex root of $3+2 x^2$ that is closest to zero. That would be $x = \sqrt{{3 \over 2}} i$ so the interval is $|x| < \sqrt{{3 \over 2}}$

So, what about a function defined as a power series?

For one, a power series expansion of a function at a specific point (say, $x =0$ ) is unique:

Say $a_0 + a_1 x + a_2 x^2 + a_3 x^3..... = b_0 + b_1 x + b_2 x^2 + b_3 x^3 ....$ on the interval of convergence, then substituting $x=0$ yields $a_0 = b_0$ . So subtract from both sides and get: $a_1 x + a_2 x^2 + a_3 x^3.. = b_1 x + b_2 x^2 + b_3 x^3 ...$ . Now assuming one can “factor out an x” from both sides (and you can) we get $a_1 = b_1$ , etc.

Yes, this should have property been phrased as a limit and I’ve yet to show that $f(x) = \sum a_n x^n$ is continuous on its open interval of absolute convergence.

Now if you say “doesn’t that follow from uniform convergence which follows from the Weierstrass M test” you’d be correct…and you are probably wasting your time reading this.

Now if you remember hearing those words but could use a refresher, here goes:

We want $|f(x) -f(t)| < \epsilon$ for $x, t$ sufficiently close AND both within the open interval of absolute convergence, say $(-r, r)$ Choose $s < r$ where $|s| > max(|x|, |t|)$ and $M$ such that ${\epsilon \over 4} > \sum_{k=M} |a_k s^k|$ and $\delta >0$ where $|x-t| < \delta \rightarrow {\epsilon \over 2 } >| \sum_{k=0} ^{M-1} a_k x^k - \sum_{k=0} ^{M-1} a_k t^k |$ (these are just polynomials).

The rest follows: $|f(x) -f(t)| = | \sum_{k=0} ^{M-1} a_k x^k - \sum_{k=0} ^{M-1} a_k t^k + \sum_{k=M} |a_k x^k| - \sum_{k=M} |a_k t^k| | \leq | \sum_{k=0} ^{M-1} a_k x^k - \sum_{k=0} ^{M-1} a_k t^k | + 2 | \sum_{k=M} |a_k s^k| | \leq \epsilon$

Calculus On the open interval of absolute convergence, a power series can be integrated term by term and differentiated term by term. Neither result, IMHO, is “obvious.” In fact, sans extra criteria, for series of functions, in general, it is false. Quickly: if you remember Fourier Series, think about the Fourier series for a rectangular pulse wave; note that the derivative is zero everywhere except for the jump points (where it is undefined), then differentiate the constituent functions and get a colossal mess.

Differentiation: most texts avoid the direct proof (with good reason; it is a mess) but it can follow from analysis results IF one first shows that the absolute (and uniform) convergence of $\sum a_n x^n$ implies the absolute (and uniform) convergence of $\sum n a_n x^{n-1}$

So, let’s start here: if $\sum a_n x^n$ is absolutely convergent on $(-r, r)$ then so is $\sum a_n nx^{n-1}$ .

Here is why: because we are on the open interval of absolute convergence, WLOG, assume $x > 0$ and find $s > 0$ where $\sum a_n (x+s)^n$ is also absolutely convergent. Now, note that $(x+s)^n = x^n + s nx^{n-1} + ...s^n > snx^{n-1}$ so the series $s \sum a_n nx^{n-1}$ converges by direct comparison on $(-r, r)$ which establishes what we wanted to show.

Of course, this doesn’t prove that the expected series IS the derivative; we have a bit more work to do.

Again, working on the open interval of absolute convergence, let’s look at:

$lim_{x \rightarrow t} {f(x) -f(t) \over x-t } = lim_{x \rightarrow t} {1 \over x-t} (\sum a_n x^n -\sum a_n t^n)$

Now, we use the fact that $\sum a_n n x^{n-1}$ is absolutely convergent on $(-r, r)$ and given any $\epsilon > 0$ we can find $M > 0$ so that $n > m \rightarrow \sum_{n} a_k k s^{k-1}$ for $s$ between $x$ and $t$

So let’s do that: pick $\epsilon >0$ and then note:

$lim_{x \rightarrow t} {1 \over x-t} (\sum a_n x^n -\sum a_n t^n) =$

$lim_{x \rightarrow t} {1 \over x-t} (\sum_{k=0}^{n-1}a_k( x^k-t^k) ) +\sum_{k=n} a_k (x^k-t^n) ) =$

$lim_{x \rightarrow t} \sum_{k=0}^{n-1} a_k {x^k-t^k \over x-t} + \sum_{k=n} a_k {x^k-t^n \over x-t}$

$= \sum_{k=1}^{n-1} a_k k x^{k-1} + lim_{x \rightarrow t} \sum_{k=n} a_k {x^k-t^n \over x-t}$

Now apply the Mean Value Theorem to each term in the second term:

$= \sum_{k=1}^{n-1} a_k k x^{k-1} + lim_{x \rightarrow t} \sum_{k=n} a_k k(s_k)^{k-1}$ where each $s_k$ is between $x$ and $t$ . By the choice of $n$ the second term is less than $\epsilon$ , which is arbitrary, and the first term is the first $n$ terms of the “expected” derivative series.

So, THAT is “term by term” differentiation…and notice that we’ve used our hypothesis …almost all of it.

Term by term integration

Theoretically, integration combines easier with infinite summation than differentiation does. But given we’ve done differentiation, we can then do anti-differentiation by showing that $\sum a_k {x^{k+1} \over k+1}$ converges and then differentiating.

But let’s do this independently; it is good for us. And we’ll focus on the definite integral.

$\int^b_a \sum a_k x^k dx =$ (of course, $[a,b] \subset (-r, r)$ )

Once again, choose $n$ so that $\sum_{k=n} a_k x^k < \epsilon$

Then $\int^b_a \sum a_k x^k dx = \int^b_a \sum_{k=0}^{n-1} a_kx^k dx + \int^b_a \sum_{k=n} a_k x^k dx$ and this is less than (or equal to)

$\sum_{k=0}^{n-1} \int^b_a a_kx^k dx + \epsilon (b-a)$

But $\epsilon$ is arbitrary and so the result follows, for definite integrals. It is an easy exercise in the Fundamental Theorem of Calculus to extract term by term anti-differentiation.

Taylor Series

Ok, now that we can say stuff about a function presented as a power series, what about finding a power series representation for a function, PROVIDED there is one? Note: we’ll need the function to have an infinite number of derivatives on an open interval about the point of expansion. We’ll also need another condition, which we will explain as we go along.

We will work with expanding about $x = 0$ . Let $f(x)$ be the function of interest and assume all of the relevant derivatives exist.

Start with $\int^x_0 f'(t) dt = f(x) -f(0) \rightarrow f(x) =f(0) + \int ^x_0 f'(t) dt$ which can be thought of as our “degree 0” expansion plus remainder term.

But now, let’s use integration by parts on the integral with $u = f'(t), dv = dt, du = f''(t) and v = (t-x)$ (it is a clever choice for $v$ ; it just has to work.

So now we have: $f(x) = f(0) + |^x_0 f'(t)(t-x) - \int^x_0f''(t) (t-x) dt = f(0) + xf'(0) - \int^x_0f''(t) (t-x) dt$

It looks like we might run into sign trouble on the next iteration, but we won’t as we will see: do integration by parts again:

$u =f''(t), dv = (t-x), du = f'''(t), v = {1 \over 2} (t-x)^2$ and so we have:

$f(x) =f(0) + xf'(0) -( |^x_0 f''(t) {1 \over 2}(t-x)^2 -{1 \over 2} \int^x_0 f'''(t)(t-x)^2 dt$

This turns out to be $f(0) +xf'(0) +{1 \over 2} f''(0)x^2 + \int^x_0 f'''(t) (t-x)^2 dt$ .

An induction argument yields

$f(x) = \sum^n_{k=0} f^{(k)}{1 \over k!}x^k +{1 \over n} \int^x_0 f^{(n+1)}(t) (t-x)^n dt$

For the series to exist (and be valid over an open interval) all of the derivatives have to exist and;

$lim_{n \rightarrow \infty} {1 \over n!} \int^x_0 f^{(n+1)}(t) (t-x)^n dt = 0$ .

Note: to get the Lagrange remainder formula that you see in some texts, let $M = max \{|f^{(n+1)} (t) \}$ for $t \in [0,x]$ and then $| {1 \over n!} \int^x_0 f^{(n+1)}(t) (t-x)^n dt | \leq {1 \over n!} M \int^x_0 (t-x)^n dt| = {M \over (n+1)!}| |^x_0(t-x)^{(n+1)} | =M{x^{(n+1)} \over (n+1)!}$

It is a bit trickier to get the equality $f^{(n+1)}(\eta) {x^{n+1} \over (n+1)!}$ as the error formula; it is a Mean Value Theorem for integrals calculation.

About the remainder term going to zero: this is necessary. Consider the classic counterexample:

$f(x) = \begin{cases} e^{-1 \over x^2} & \text{ for} x \neq 0 \\ 0 & \text{ otherwise} \end{cases}$

It is an exercise in the limit definition and L’Hopital’s Rule to show that $f^{(k)} (0) = 0$ for all $k \in \{0, 1, 2, ... \}$ and so a Taylor expansion at zero is just the zero function, and this is valid at 0 only.

As you can see, the function appears to flatten out near x =0 $ but it really is NOT constant.

Note: of course, using the Taylor method isn’t always the best way. For example, if we were to try to get the Taylor expansion of ${1 \over 1+x^2}$ at $x = 0$ it is easier to use the geometric series for ${1 \over 1-u}$ and substitute $u= -x^2$ ; the uniqueness of the power series expansion allows for that.

May 21, 2021

Introduction to infinite series for inexperienced calculus teachers

Filed under: calculus, mathematics education, pedagogy, Power Series, sequences, series — oldgote @ 1:26 pm

Let me start by saying that this is NOT: this is not an introduction for calculus students (too steep) nor is this intended for experienced calculus teachers. Nor is this a “you should teach it THIS way” or “introduce the concepts in THIS order or emphasize THESE topics”; that is for the individual teacher to decide.

Rather, this is a quick overview to help the new teacher (or for the teacher who has not taught it in a long time) decide for themselves how to go about it.

And yes, I’ll be giving a lot of opinions; disagree if you like.

What series will be used for.

Of course, infinite series have applications in probability theory (discrete density functions, expectation and higher moment values of discrete random variables), financial mathematics (perpetuities), etc. and these are great reasons to learn about them. But in calculus, these tend to be background material for power series.

Power series: $\sum^{\infty}_{k=0} a_k (x-c)^k$ , the most important thing is to determine the open interval of absolute convergence; that is, the intervals on which $\sum^{\infty}_{k=0} |a_k (x-c)^k |$ converges.

We teach that these intervals are *always* symmetric about $x = c$ (that is, at $x = c$ only, on some open interval $(c-\delta, c+ \delta)$ or the whole real line. Side note: this is an interesting place to point out the influence that the calculus of complex variables has on real variable calculus! These open intervals are the most important aspect as one can prove that one can differentiate and integrate said series “term by term” on the open interval of absolute convergence; sometimes one can extend the results to the boundary of the interval.

Therefore, if time is limited, I tend to focus on material more relevant for series that are absolutely convergent though there are some interesting (and fun) things one can do for a series which is conditionally convergent (convergent, but not absolutely convergent; e. g. $\sum^{\infty}_{k=1} (-1)^{k+1} {1 \over k}$ .

Important principles: I think it is a good idea to first deal with geometric series and then series with positive terms…make that “non-negative” terms.

Geometric series: $\sum ^{\infty}_{k =0} x^k$ ; here we see that for $x \neq 1$ , $\sum ^{n}_{k =0} x^k= {1-x^{n+1} \over 1-x }$ and is equal to $n+1$ for $n = 1$ ; to show this do the old “shifted sum” addition: $S = 1 + x + x^2 + ...x^n$ , $xS = x+x^2 + ...+x^{n+1}$ then subtract: $S-xS = (1-x)S = 1-x^{n+1}$ as most of the terms cancel with the subtraction.

Now to show the geometric series converges, (convergence being the standard kind: $\sum^n_{k = 0} c_k = S_n$ the “n’th partial sum, then the series $\sum^{\infty}_{k = 0} c_k$ converges if an only if the sequence of partial sums $S_n$ converges; yes there are other types of convergence)

Now that we’ve established that for the geometric series, $S_n = {1-x^{n+1} \over 1-x }$ and we get convergence if $|x^{n+1}|$ goes to zero, which happens only if $|x| < 1$ .

Why geometric series: two of the most common series tests (root and ratio tests) involve a comparison to a geometric series. Also, the geometric series concept is used both in the theory of improper integrals and in measure theory (e. g., showing that the rational numbers have measure zero).

Series of non-negative terms. For now, we’ll assume that $\sum a_k$ has all $a_k \geq 0$ (suppressing the indices).

Main principle: though most texts talk about the various tests, I believe that most of the tests involved really involve three key principles, two of which the geometric series and the following result from sequences of positive numbers:

Key sequence result: every monotone bounded sequence of positive numbers converges to its least upper bound.

True: many calculus texts don’t do that much with the least upper bound concept but I feel it is intuitive enough to at least mention. If the least upper bound is, say, $b$ , then if $a_n$ is the sequence in question, there has to be some $N > 0$ such that $a_n > b-\delta$ for any small, positive $\delta$ . Then because $a_n$ is monotone, $b> a_{m} > b-\delta$ for all $m > n$

The third key principle is “common sense” : if $\sum c_k$ converges (standard convergence) then $c_k \rightarrow 0$ as a sequence. This is pretty clear if the $c_k$ are non-negative; the idea is that the sequence of partial sums $S_n$ cannot converge to a limit unless $|S_n -S_{n+1}|$ becomes arbitrarily small. Of course, this is true even if the terms are not all positive.

Secondary results I think that the next results are “second order” results: the main results depend on these, and these depend on the key 3 that we just discussed.

The first of these secondary results is the direct comparison test for series of non-negative terms:

Direct comparison test

If $0< c_n \leq b_n$ and $\sum b_n$ converges, then so does $\sum c_n$ . If $\sum c_n$ diverges, then so does $\sum b_n$ .

The proof is basically the “bounded monotone sequence” principle applied to the partial sums. I like to call it “if you are taller than an NBA center then you are tall” principle.

Evidently, some see this result as a “just get to something else” result, but it is extremely useful; one can apply this to show that the exponential of a square matrix is defined; it is the principle behind the Weierstrass M-test, etc. Do not underestimate this test!

Absolute convergence: this is the most important kind of convergence for power series as this is the type of convergence we will have on an open interval. A series is absolutely convergent if $\sum |c_k|$ converges. Now, of course, absolute convergence implies convergence:

Note $0 < |c_k| -c_k \leq 2|c_k|$ and if $\sum |c_k|$ converges, then $\sum |c_k|-c_k$ converges by direct comparison. Now note $c_k = |c_k|-(|c_k| -c_k) \rightarrow \sum c_k$ is the difference of two convergent series: $\sum |c_k| -\sum (|c_k|-c_k )$ and therefore converges.

Integral test This is an important test for convergence at a point. This test assumes that $f$ is a non-negative, non-decreasing function on some $[1, \infty)$ (that is, $a >b \rightarrow f(a) \geq f(b)$ ) Then $\sum f(n)$ converges if and only if $\int_1^{\infty} f(x)dx$ converges as an improper integral.

Proof: $\sum_{n=2} f(n)$ is just a right endpoint Riemann sum for $\int_1^{\infty} f(x)dx$ and therefore the sequence of partial sums is an increasing, bounded sequence. Now if the sum converges, note that $\sum_{n=1} f(n)$ is the right endpoint estimate for $\int_1^{\infty} f(x)dx$ so the integral can be defined as a limit of a bounded, increasing sequence so the integral converges.

Yes, these are crude whiteboards but they get the job done.

Note: we need the hypothesis that $f$ is decreasing (or non-decreasing). Example: the function $f(x) = \begin{cases} x , & \text{ if } x \notin \{1, 2, 3,...\} \\ 0, & \text{ otherwise} \end{cases}$ certainly has $\sum f(n)$ converging but $\int^{\infty}_{1} f(x) dx$ diverging.

Going the other way, defining $f(x) = \begin{cases} 2^n , & \text{ if } x \in [n, n+2^{-2n}] \\0, & \text{ otherwise} \end{cases}$ gives an unbounded function with unbounded sum $\sum_{n=1} 2^n$ but the integral converges to the sum $\sum_{n=1} 2^{-n} =1$ . The “boxes” get taller and skinnier.

Note: the above shows the integral and sum starting at 0; same principle though.

Now wait a minute: we haven’t really gone over how students will do most of their homework and exam problems. We’ve covered none of these: p-test, limit comparison test, ratio test, root test. Ok, logically, we have but not practically.

Let’s remedy that. First, start with the “point convergence” tests.

p-test. This says that $\sum {1 \over k^p}$ converges if $p> 1$ and diverges otherwise. Proof: Integral test.

Limit comparison test Given two series of positive terms: $\sum b_k$ and $\sum c_k$

Suppose $lim_{k \rightarrow \infty} {b_k \over c_k} = L$

If $\sum c_k$ converges and $0 \leq L < \infty$ then so does $\sum b_k$ .

If $\sum c_k$ diverges and $0 < L \leq \infty$ then so does $\sum b_k$

I’ll show the “converge” part of the proof: choose $\epsilon = L$ then $N$ such that $n > N \rightarrow {b_n \over c_n } < 2L$ This means $\sum_{k=n} b_k \leq \sum_{k=n} c_k$ and we get convergence by direct comparison. See how useful that test is?

But note what is going on: it really isn’t necessary for $lim_{k \rightarrow \infty} {b_k \over c_k}$ to exist; for the convergence case it is only necessary that there be some $M$ for which $M > {b_k \over c_k}$ ; if one is familiar with the limit superior (“limsup”) that is enough to make the test work.

We will see this again.

Why limit comparison is used: Something like $\sum {1 \over 4k^5-2k^2-14}$ clearly converges, but nailing down the proof with direct comparison can be hard. But a limit comparison with $\sum {1 \over k^5}$ is pretty easy.

Ratio test this test is most commonly used when the series has powers and/or factorials in it. Basically: given $\sum c_n$ consider $lim_{k \rightarrow \infty} {c_{k+1} \over c_{k}} = L$ (if the limit exists..if it doesn’t..stay tuned).

If $L < 1$ the series converges. If $L > 1$ the series diverges. If $L = 1$ the test is inconclusive.

Note: if it turns out that there is exists some $N >0$ such that for all $n > N$ we have ${c_{n+1} \over c_n } < \gamma < 1$ then the series converges (we can use the limsup concept here as well)

Why this works: suppose there exists some $N >0$ such that for all $n > N$ we have ${c_{n+1} \over c_n } < \gamma < 1$ Then write $\sum_{k=n} c_k = c_n + c_{n+1} + c_{n+2} + ....$

now factor out a $c_n$ to obtain $c_n (1 + {c_{n+1} \over c_n} + {c_{n+2} \over c_n} + {c_{n+3} \over c_{n}} +....)$

Now multiply the terms by 1 in a clever way:

$c_n (1 + {c_{n+1} \over c_n} + {c_{n+2} \over c_{n+1}}{c_{n+1} \over c_n} + {c_{n+3} \over c_{n+2}} {c_{n+2} \over c_{n+1}} {c_{n+1} \over c_{n}} +....)$ See where this is going: each ratio is less than $\gamma$ so we have:

$\sum_{k=n} c_k \leq c_n \sum_{j=0} (\gamma)^j$ which is a convergent geometric series.

See: there is geometric series and the direct comparison test, again.

Root Test No, this is NOT the same as the ratio test. In fact, it is a bit “stronger” than the ratio test in that the root test will work for anything the ratio test works for, but there are some series that the root test works for that the ratio test comes up empty.

I’ll state the “lim sup” version of the ratio test: if there exists some $N$ such that, for all $n>N$ we have $(c_n)^{1 \over n} < \gamma < 1$ then the series converges (exercise: find the “divergence version”).

As before: if the condition is met, $\sum_{k=n} c_n \leq \sum_{k=n} \gamma^k$ so the original series converges by direction comparison.

Now as far as my previous remark about the ratio test: Consider the series:

$1 + ({1 \over 3}) + ({2 \over 3})^2 + ({1 \over 3})^3 + ({2 \over 3})^4 +...({1 \over 3})^{2k-1} +({2 \over 3})^{2k} ...$

Yes, this series is bounded by the convergent geometric series with $r = {2 \over 3}$ and therefore converges by direct comparison. And the limsup version of the root test works as well.

But the ratio test is a disaster as ${({2 \over 3})^{2k} \over ({1 \over 3})^{2k-1} } ={2^{2k} \over 3 }$ which is unbounded..but ${({1 \over 3})^{2k+1} \over ({2 \over 3})^{2k} } ={1 \over (2^{2k} 3) }$ .

What about non-absolute convergence (aka “conditional convergence”)

Series like $\sum_{k=1} (-1)^{k+1} {1 \over k}$ converges but does NOT converge absolutely (p-test). On one hand, such series are a LOT of fun..but the convergence is very slow and unstable and so might say that these series are not as important as the series that converges absolutely. But there is a lot of interesting mathematics to be had here.

So, let’s chat about these a bit.

We say $\sum c_k$ is conditionally convergent if the series converges but $\sum |c_k|$ diverges.

One elementary tool for dealing with these is the alternating series test:

for this, let $c_k >0$ and for all $k, c_{k+1} < c_k$ .

Then $\sum_{k=1} (-1)^{k+1} c_k$ converges if and only if $c_k \rightarrow 0$ as a sequence.

That the sequence of terms goes to zero is necessary. That it is sufficient in this alternating case: first note that the terms of the sequence of partial sums are bounded above by $c_1$ (as the magnitudes get steadily smaller) and below by $c_1 - c_2$ (same reason. Note also that $S_{2k+2} = S_{2k} -c_{2k+1} + c_{2k+2} < S_{2k}$ so the sequence of partial sums of even index are an increasing bounded sequence and therefore converges to some limit, say, $L$ . But $S_{2k+1} = S_{2k} + c_{2k+1}$ and so by a routine “epsilon-N” argument the odd partial sums converge to $L$ as well.

Of course, there are conditionally convergent series that are NOT alternating. And conditionally convergent series have some interesting properties.

One of the most interesting properties is that such series can be “rearranged” (“derangment” in Knopp’s book) to either converge to any number of choice or to diverge to infinity or to have no limit at all.

Here is an outline of the arguments:

To rearrange a series to converge to $L$ , start with the positive terms (which must diverge as the series is conditionally convergent) and add them up to exceed $L$ ; stop just after $L$ is exceeded. Call that partial sum $u_1$ . Note: this could be 0 terms. Now use the negative terms to go of the left of $L$ and stop the first one past. Call that $l_1$ Then move to the right, past $L$ again with the positive terms..note that the overshoot is smaller as the terms are smaller. This is $u_2$ . Then go back again to get $l_2$ to the left of $L$ . Repeat.

Note that at every stage, every partial sum after the first one past $L$ is between some $u_i, l_i$ and the $u_i, l_i$ bracket $L$ and the distance is shrinking to become arbitrarily small.

To rearrange a series to diverge to infinity: Add the positive terms to exceed 1. Add a negative term. Then add the terms to exceed 2. Add a negative term. Repeat this for each positive integer $n$ .

Have fun with this; you can have the partial sums end up all over the place.

That’s it for now; I might do power series later.

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January 18, 2014

Fun with divergent series (and uses: e. g. string theory)

Filed under: advanced mathematics, calculus, physics, Power Series, sequences of functions, series — Tags: Cesaro summation, divergent series — collegemathteaching @ 5:32 am

One “fun” math book is Knopp’s book Theory and Application of Infinite Series. I highly recommend it to anyone who frequently teaches calculus, or to talented, motivated calculus students.

One of the more interesting chapters in the book is on “divergent series”. If that sounds boring consider the following:

we all know that $\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}$ when $|x| < 1$ and diverges elsewhere, PROVIDED one uses the “sequence of partial sums” definition of covergence of sums. But, as Knopp points out, there are other definitions of convergence which leaves all the convergent (by the usual definition) series convergent (to the same value) but also allows one to declare a larger set of series to be convergent.

Consider $1 - 1 + 1 -1 + 1.......$

of course this is a divergent geometric series by the usual definition. But note that if one uses the geometric series formula:

$\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}$ and substitutes $x = -1$ which IS in the domain of the right hand side (but NOT in the interval of convergence in the left hand side) one obtains $1 -1 +1 -1 + 1.... = \frac{1}{2}$ .

Now this is nonsense unless we use a different definition of sum convergence, such as the Cesaro summation: if $s_k$ is the usual “partial sum of the first $k$ terms: $s_k = \sum^{n=k}_{n =0}a_n$ then one declares the Cesaro sum of the series to be $lim_{m \rightarrow \infty} \frac{1}{m}\sum^{m}_{k=1}s_k$ provided this limit exists (this is the arithmetic average of the partial sums).

(see here)

So for our $1 -1 + 1 -1 ....$ we easily see that $s_{2k+1} = 0, s_{2k} = 1$ so for $m$ even we see $\frac{1}{m}\sum^{m}_{k=1}s_k = \frac{\frac{m}{2}}{m} = \frac{1}{2}$ and for $m$ odd we get $\frac{\frac{m-1}{2}}{m}$ which tends to $\frac{1}{2}$ as $m$ tends to infinity.

Now, we have this weird type of assignment.

But that won’t help with $\sum^{\infty}_{k = 1} k = 1 + 2 + 3 + 4 + 5.....$ . But weirdly enough, string theorists find a way to assign this particular series a number! In fact, the number that they assign to this makes no sense at all: $-\frac{1}{12}$ .

What the heck? Well, one way this is done is explained here:

Consider $\sum^{\infty}_{k=0}x^k = \frac{1}{1-x}$ Now differentiate term by term to get $1 +2x + 3x^2+4x^3 .... = \frac{1}{(1-x)^2}$ and now multiply both sides by $x$ to obtain $x + 2x^2 + 3x^3 + .... = \frac{x}{(1-x)^2}$ This has a pole of order 2 at $x = 1$ . But now substitute $x = e^h$ and calculate the Laurent series about $h = 0$ ; the 0 order term turns out to be $\frac{1}{12}$ . Yes, this has applications in string theory!

Now of course, if one uses the usual definitions of convergence, I played fast and loose with the usual intervals of convergence and when I could differentiate term by term. This theory is NOT the usual calculus theory.

Now if you want to see some “fun nonsense” applied to this (spot how many “errors” are made….it is a nice exercise):

And read this to see exploding heads. 🙂

What is going on: when one sums a series, one is really “assigning a value” to an object; think of this as a type of morphism of the set of series to the set of numbers. The usual definition of “sum of a series” is an especially nice morphism as it allows, WITH PRECAUTIONS, some nice algebraic operations in the domain (the set of series) to be carried over into the range. I say “with precautions” because of things like the following:

1. If one is talking about series of numbers, then one must have an absolutely convergent series for derangements of a given series to be assigned the same number. Example: it is well known that a conditionally convergent alternating series can be arranged to converge to any value of choice.

2. If one is talking about a series of functions (say, power series where one sums things like $x^n$ ) one has to be in OPEN interval of absolute convergence to justify term by term differentiation and integration; then of course a series is assigned a function rather than a number.

So when one tries to go with a different notion of convergence, one must be extra cautious as to which operations in the domain space carry through under the “assignment morphism” and what the “equivalence classes” of a given series are (e. g. can a series be deranged and keep the same sum?)

This Phil Plait article started this post in motion for me and I got to it via 3-quarks daily.

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September 20, 2013

Ok, have fun and justify this…

Filed under: calculus, popular mathematics, Power Series, series, Taylor Series — Tags: math memes — collegemathteaching @ 7:59 pm

Ok, you say, “this works”; this is a series representation for $\pi$ . Ok, it is but why?

Now if you tell me: $\int^1_0 \frac{dx}{1+x^2} = arctan(1) = \frac{\pi}{4}$ and that $\frac{1}{1+x^2} = \sum^{\infty}_{k=0} (-1)^k x^{2k}$ and term by term integration yields:
$\sum^{\infty}_{k=0} (-1)^k \frac{1}{2k+1}x^{2k+1}$ I’d remind you of: “interval of absolute convergence” and remind you that the series for $\frac{1}{1+x^2}$ does NOT converge at $x = 1$ and that one has to be in the open interval of convergence to justify term by term integration.

True, the series DOES converge to $\frac{\pi}{4}$ but it is NOT that elementary to see. 🙂

Boooo!

(Yes, the series IS correct…but the justification is trickier than merely doing the “obvious”).

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May 2, 2012

Composition of an analystic function with a non-analytic one

Filed under: advanced mathematics, analysis, complex variables, derivatives, Power Series, series — collegemathteaching @ 7:39 pm

On a take home exam, I gave a function of the type: $f(z) = sin(k|z|)$ and asked the students to explain why such a function was continuous everywhere but not analytic anywhere.

This really isn’t hard but that got me to thinking: if $f$ is analytic at $z_0$ and NON CONSTANT, is $f(|z|)$ ever analytic? Before you laugh, remember that in calculus class, $ln|x|$ is differentiable wherever $x \neq 0$ .

Ok, go ahead and laugh; after playing around with the Cauchy-Riemann equations at bit, I found that there was a much easier way, if $f$ is analytic on some open neighborhood of a real number.

Since $f$ is analytic at $z_0$ , $z_0$ real, write $f = \sum ^ {\infty}_{k =0} a_k (z-z_0)^k$ and then compose $f$ with $|z|$ and substitute into the series. Now if this composition is analytic, pull out the Cauchy-Riemann equations for the composed function $f(x+iy) = u(x,y) + iv(x,y)$ and it is now very easy to see that $v_x = v_y =0$ on some open disk which then implies by the Cauchy-Riemann equations that $u_x = u_y = 0$ as well which means that the function is constant.

So, what if $z_0$ is NOT on the real axis?

Again, we write $f(x + iy) = u(x,y) + iv(x,y)$ and we use $U_{X}, U_{Y}$ to denote the partials of these functions with respect to the first and second variables respectively. Now $f(|z|) = f(\sqrt{x^2 + y^2} + 0i) = u(\sqrt{x^2 + y^2},0) + iv(\sqrt{x^2 + y^2},0)$ . Now turn to the Cauchy-Riemann equations and calculate:
$\frac{\partial}{\partial x} u = u_{X}\frac{x}{\sqrt{x^2+y^2}}, \frac{\partial}{\partial y} u = u_{X}\frac{y}{\sqrt{x^2+y^2}}$
$\frac{\partial}{\partial x} v = v_{X}\frac{x}{\sqrt{x^2+y^2}}, \frac{\partial}{\partial y} v = v_{X}\frac{y}{\sqrt{x^2+y^2}}$
Insert into the Cauchy-Riemann equations:
$\frac{\partial}{\partial x} u = u_{X}\frac{x}{\sqrt{x^2+y^2}}= \frac{\partial}{\partial y} v = v_{X}\frac{y}{\sqrt{x^2+y^2}}$
$-\frac{\partial}{\partial x} v = -v_{X}\frac{x}{\sqrt{x^2+y^2}}= \frac{\partial}{\partial y} u = u_{X}\frac{y}{\sqrt{x^2+y^2}}$

From this and from the assumption that $y \neq 0$ we obtain after a little bit of algebra:
$u_{X}\frac{x}{y}= v_{X}, u_{X} = -v_{X}\frac{x}{y}$
This leads to $u_{X}\frac{x^2}{y^2} = v_{X}\frac{x}{y}=-v_{X}$ which implies either that $u_{X}$ is zero which leads to the rest of the partials being zero (by C-R), or this means that $\frac{x^2}{y^2} = -1$ which is absurd.

So $f$ must have been constant.

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January 29, 2011

Taylor Series: student misunderstanding

Filed under: advanced mathematics, calculus, Power Series, student learning, Taylor Series — collegemathteaching @ 10:05 pm

I am going to take a break from the Lebesgue stuff and maybe write more on that tomorrow.
My numerical analysis class just turned in some homework and some really have some misunderstanding about Taylor Series and Power Series. I’ll provide some helpful hints to perplexed students.

For the experts who might be reading this: my assumption is that we are dealing with functions $f$ which are real analytic over some interval. To students: this means that $f$ can be differentiated as often as we’d like, that the series converges absolutely on some open interval and that the remainder term goes to zero as the number of terms approaches infinity.

This post will be about computing such a series.
First, I’ll give a helpful reminder that is crucial in calculating these series: a Taylor series is really just a power series representation of a function. And if one finds a power series which represents a function over a given interval and is expanded about a given point, THAT SERIES IS UNIQUE, no matter how you come up with it. I’ll explain with an example:

Say you want to represent $f(x) = 1/(1-x)$ over the interval $(-1,1)$ . You could compute it this way: you probably learned about the geometric series and that $f(x) = 1/(1-x) = 1 + x + x^2 + x^3....+ x^k+.... = \sum_{i=0}^{\infty} x^i$ for $x \in (-1,1)$ .

Well, you could compute it by Taylor’s theorem which says that such a series can be obtained by:
$f(x) = f(0) + f^{'}(0)x + f^{''}())x^2/2! + f^{iii}(0)x^3/3! +.... + \sum_{k=0}^{\infty} f^{k}(0)x^k/k!$ If you do such a calculation for $f(x) = 1/1-x$ one obtains $f^{'} = (1-x)^2$ , $f^{''} = 2(1-x)^3$ , $f^{iii} = 3!(1-x)^4 ....$ and plugging into Taylor’s formula leads to the usual geometric series. That is, the series can be calculated by any valid method; one does NOT need to retreat to the Taylor definition for calculation purposes.

Example: in the homework problem, students were asked to calculate Taylor polynomials (of various orders and about $x=0$ ) for a function that looked like this:

$f(x) = 3x(sin(3x)) - (x-3)^2$ . Some students tried to calculate the various derivatives and plug into Taylor’s formula with grim results. It is much easier than that if one remembers that power series are unique! Sure, one CAN use Taylor’s formula but that doesn’t mean that one should. Instead it is much easier if one remembers that $sin(x) = x -x^3/3! + x^5/5! - x^7/7!......$ Now to get $sin(3x)$ one just substitutes $3x$ for $x$ and obtains: $sin(3x) = 3x -(x^3)3^3/3! + (x^5)3^5/5! -( x^7)3^7/7!......$ . Then $3x(sin(3x)) =9x^2 -(x^4)3^4/3! + (x^6)3^6/5! -( x^8)3^8/7!....$ and one subtracts off $(x-3)^2 = x^2 -6x +9$ to obtain the full power series: $-9 + 6x+ 8x^2 -(x^4)3^4/3! + (x^6)3^6/5! -( x^8)3^8/7!....= -9 + 6x+ 8x^2 +\sum_{k=2}^{\infty} (-1)^{k+1}x^{2k}3^{2k}/(2k-1)!$

Now calculating the bound for the remainder after $k$ terms is, in general, a pain. Sure, one can estimate with a graph, but that sort of defeats the point of approximating to begin with; one can use thumb rules which overstate the magnitude of the remainder term.

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