College Math Teaching

April 28, 2023

Taylor Polynomials without series (advantages, drawbacks)

Filed under: calculus, series, Taylor polynomial., Taylor Series — oldgote @ 12:32 am

I was in a weird situation this semester in my “applied calculus” (aka “business calculus”) class. I had an awkward amount of time left (1 week) and I still wanted to do something with Taylor polynomials, but I had nowhere near enough time to cover infinite series and power series.

So, I just went ahead and introduced it “user’s manual” style, knowing that I could justify this, if I had to (and no, I didn’t), even without series. BUT there are some drawbacks too.

Let’s see how this goes. We’ll work with series centered at c =0 (expand about 0) and assume that f has as many continuous derivatives as desired on an interval connecting 0 to x .

Now we calculate: \int^x_0 f'(t) dt = f(x) -f(0) , of course. But we could do the integral another way: let’s use parts and say u = f'(t), dv = dt \rightarrow du = f''(t), v = (t-x) . Note the choice for v and that x is a constant in the integral. We then get f(x) -f(0)=\int^x_0 f'(t) dt = (f'(t)(t-x)|^x_0 -\int^x_0f''(t)(t-x) dx . Evaluation:

f(x) =f(0)+f'(0)x -\int^x_0f''(t)(t-x) dx and we’ve completed the first step.

Though we *could* do the inductive step now, it is useful to grind through a second iteration to see the pattern.

We take our expression and compute \int^x_0f''(t)(t-x) dx by parts again, with u = f''(t), dv =t-x \rightarrow du =f'''(t), v = {(t-x)^2 \over 2!} and insert into our previous expression:

f(x) =f(0)+f'(0)x - (f''(t){(t-x)^2 \over 2!}|^x_0 + \int^x_0 f'''(t){(t-x)^2 \over 2!} dt which works out to:

f(x) = f(0)+f'(0)x +f''(0){x^2 \over 2} + \int^x_0 f'''(t){(t-x)^2 \over 2!} dt and note the alternating sign of the integral.

Now to use induction: assume that:

f(x) = f(0)+f'(0)x +f''(0){x^2 \over 2} + ....f^{(k)}(0){x^k \over k!} + (-1)^k \int^x_0 f^{(k+1)}(t) {(t-x)^k \over k!} dt

Now let’s look at the integral: as usual, use parts as before and we obtain:

(-1)^k (f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!}|^x_0 - \int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt ). Taking some care with the signs we end up with

(-1)^k (-f^{(k+1)}(0){(-x)^{k+1} \over (k+1)! } )+ (-1)^{k+1}\int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt which works out to (-1)^{2k+2} (f^{(k+1)}(0) {x^{k+1} \over (k+1)!} )+ (-1)^{k+1}\int^x_0 f^{(k+2)}(t) {(t-x)^{k+1} \over (k+1)!} dt .

Substituting this evaluation into our inductive step equation gives the desired result.

And note: NOTHING was a assumed except for f having the required number of continuous derivatives!

BUT…yes, there is a catch. The integral is often regarded as a “correction term.” But the Taylor polynomial is really only useful so long as the integral can be made small. And that is the issue with this approach: there are times when the integral cannot be made small; it is possible that x can be far enough out that the associated power series does NOT converge on (-x, x) and the integral picks that up, but it may well be hidden, or at least non-obvious.

And that is why, in my opinion, it is better to do series first.

Let’s show an example.

Consider f(x) = {1 \over 1+x } . We know from work with the geometric series that its series expansion is 1 -x +x^2-x^3....+(-1)^k x^k + .... and that the interval of convergence is (-1,1) But note that f is smooth over [0, \infty) and so our Taylor polynomial, with integral correction, should work for x > 0 .

So, nothing that f^{(k)} = (-1)^k(k!)(1+x)^{-(k+1)} our k-th Taylor polynomial relation is:

f(x) =1-x+x^2-x^3 .....+(-1)^kx^k +(-1)^k \int^x_0 (-1)^{k+1}(k+1)!{1 \over (1+t)^{k+2} } {(t-x)^k \over k!} dt

Let’s focus on the integral; the “remainder”, if you will.

Rewrite it as: (-1)^{2k+1} (k+1) \int^x_0 ({(t -x) \over (t+1) })^k {1 \over (t+1)} dt .

Now this integral really isn’t that hard to do, if we use an algebraic trick:

Rewrite ({(t -x) \over (t+1) })^k = ({(t+1 -x-1) \over (t+1) })^k = (1-{(x+1) \over (t+1) })^k

Now the integral is a simple substitutions integral: let u = 1-{(x+1) \over (t+1) } \rightarrow du = (x+1)( {1 \over (t+1)})^2 dt so our integral is transformed into:

(-1) ({k+1 \over x+1}) \int^0_{-x} u^{k} du = (-1) {k \over (k+1)(x+1)} (-(-x)^{k+1}) = (-1)^{k+1} {k+1 \over (k+1)(x+1)} x^{k+1} =(-1)^{k+1}{1 \over (x+1)}x^{k+1}

This remainder cannot be made small if x \geq 1 no matter how big we make k

But, in all honesty, this remainder could have been computed with simple algebra.

{1 \over x+1} =1-x+x^2....+(-1)^k x^k + R and now solve for R algebraically .

The larger point is that the “error” is hidden in the integral remainder term, and this can be tough to see in the case where the associated Taylor series has a finite radius of convergence but is continuous on the whole real line, or a half line.

May 25, 2021

Power series: review for inexperienced calculus teachers

Filed under: infinite series, Power Series, series, Taylor Series — oldgote @ 5:59 pm

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 10, 2021

Series convergence tests: the “harder to use in calculus 1” tests may well be the most useful.

Filed under: algebra, calculus, infinite series, linear albegra, matrix algebra, sequences, series, Taylor Series — oldgote @ 3:17 pm

I talked about the root and ratio test here and how the root test is the stronger of the two tests. What I should point out that the proof of the root test depends on the basic comparison test.

And so..a professor on Twitter asked:

https://twitter.com/MathProfPeter/status/1385606921156284429

Of course, one proves the limit comparison test by the direct comparison test. But in a calculus course, the limit comparison test might appear to be more readily useful..example:

Show \sum {1 \over k^2-1} converges.

So..what about the direct comparison test?

As someone pointed out: the direct comparison can work very well when you don’t know much about the matrix.

One example can be found when one shows that the matrix exponential e^A where A is a n \times n matrix.

For those unfamiliar: e^A = \sum^{\infty}_{k=0} {A^k \over k!} where the powers make sense as A is square and we merely add the corresponding matrix entries.

What enables convergence is the factorial in the denominators of the individual terms; the i-j’th element of each A^k can get only so large.

But how does one prove convergence?

The usual way is to dive into matrix norms; one that works well is |A| = \sum_{(i,j)} |a_{i,j}| (just sum up the absolute value of the elements (the Taxi cab norm or l_1 norm )

Then one can show |AB| \leq |A||B| and |a_{i,j}| \leq |A| and together this implies the following:

For any index k where a^k_{i,j} is the i-j’th element of A^k we have:

| a^k_{i,j} | \leq |A^k| \leq |A|^k

It then follows that | [ e^A ]_{i,j} | \leq \sum^{\infty}_{k=0} {|A^k |\over k!} \leq \sum^{\infty}_{k=0} {|A|^k \over k!} =e^{|A|} . Therefore every series that determines an entry of the matrix e^A is an absolutely convergent series by direct comparison. and is therefore a convergent series.

September 9, 2014

Chebyshev polynomials: a topological viewpoint

Filed under: applied mathematics, calculus, numerical methods, optimization, Taylor Series — Tags: , , , , — collegemathteaching @ 3:47 pm

Chebyshev (or Tchebycheff) polynomials are a class of mutually orthogonal polynomials (with respect to the inner product: f \cdot g = \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} f(x)g(x) dx ) defined on the interval [-1, 1] . Yes, I realize that this is an improper integral, but it does converge in our setting.

These are used in approximation theory; here are a couple of uses:

1. The roots of the Chebyshev polynomial can be used to find the values of x_0, x_1, x_2, ...x_k \in [-1,1] that minimize the maximum of |(x-x_0)(x-x_1)(x-x_2)...(x-x_k)| over the interval [-1,1] . This is important in minimizing the error of the Lagrange interpolation polynomial.

2. The Chebyshev polynomial can be used to adjust an approximating Taylor polynomial P_n to increase its accuracy (away from the center of expansion) without increasing its degree.

The purpose of this note isn’t to discuss the utility but rather to discuss an interesting property that these polynomials have. The Wiki article on these polynomials is reasonably good for that purpose.

Let’s discuss the polynomials themselves. They are defined for all positive integers n as follows:

T_n = cos(n acos(x)) . Now, it is an interesting exercise in trig identities to discover that these ARE polynomials to begin with; one shows this to be true for, say, n \in \{0, 1, 2\} by using angle addition formulas and the standard calculus resolution of things like sin(acos(x)) . Then one discovers a relation: T_{n+1} =2xT_n - T_{n-1} to calculate the rest.

The cos(n acos(x)) definition allows for some properties to be calculated with ease: the zeros occur when acos(x) = \frac{\pi}{2n} + \frac{k \pi}{n} and the first derivative has zeros where arcos(x) = \frac{k \pi}{n} ; these ALL correspond to either an endpoint max/min at x=1, x = -1 or local max and mins whose y values are also \pm 1 . Here are the graphs of T_4(x), T_5 (x)

cheby4

cheby5

Now here is a key observation: the graph of a T_n forms n spanning arcs in the square [-1, 1] \times [-1,1] and separates the square into n+1 regions. So, if there is some other function f whose graph is a connected, piecewise smooth arc that is transverse to the graph of T_n that both spans the square from x = -1 to x = 1 and that stays within the square, that graph must have n points of intersection with the graph of T_n .

Now suppose that f is the graph of a polynomial of degree n whose leading coefficient is 2^{n-1} and whose graph stays completely in the square [-1, 1] \times [-1,1] . Then the polynomial Q(x) = T_n(x) - f(x) has degree n-1 (because the leading terms cancel via the subtraction) but has n roots (the places where the graphs cross). That is clearly impossible; hence the only such polynomial is f(x) = T_n(x) .

This result is usually stated in the following way: T_n(x) is normalized to be monic (have leading coefficient 1) by dividing the polynomial by 2^{n-1} and then it is pointed out that the normalized T_n(x) is the unique monic polynomial over [-1,1] that stays within [-\frac{1}{2^{n-1}}, \frac{1}{2^{n-1}}] for all x \in [-1,1] . All other monic polynomials have a graph that leaves that box at some point over [-1,1] .

Of course, one can easily cook up analytic functions which don’t leave the box but these are not monic polynomials of degree n .

September 20, 2013

Ok, have fun and justify this…

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

1233403_10151855199510419_2048104904_n

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”).

October 10, 2011

The Picard Iterates: how they can yield an interval of existence.

Filed under: advanced mathematics, differential equations, numerical methods, numerical solution of differential equations, Taylor Series, uniqueness of solution — collegemathteaching @ 9:33 pm

One of the many good things about my teaching career is that as I teach across the curriculum, I fill in the gaps of my own education.
I got my Ph. D. in topology (low dimensional manifolds; in particular, knot theory) and hadn’t seen much of differential equations beyond my “engineering oriented” undergraduate course.

Therefore, I learned more about existence and uniqueness theorems when I taught differential equations; though I never taught the existence and uniqueness theorems in a class, I learned the proofs just for my own background. In doing so I learned about the Picard iterated integral technique for the first time; how this is used to establish “uniqueness of solution” can be found here.

However I recently discovered (for myself) what thousands of mathematicians already know: the Picard process can be used to yield an interval of existence for a solution for a differential equation, even if we cannot obtain the solution in closed form.

The situation
I assigned my numerical methods class to solve y'= t + y^2 with y(0)=1 and to produce the graph of y(t) from t = 0 to t = 3 .

There is a unique solution to this and the solution is valid so long as the t and y value of the solution curve stays finite; note that \frac{\partial }{\partial y}f(t,y)=2y.

So, is it possible that the y values for this solution become unbounded?

Answer: yes.
What follows are the notes I gave to my class.

Numeric output seems to indicate this, but numeric output is NOT proof.

To find a proof of this, let’s turn to the Picard iteration technique. We
know that the Picard iterates will converge to the unique solution.

y_{0}=1

y_{1}=1+\int_{0}^{t}x+1dx=\frac{1}{2}t^{2}+t+1

y_{2}=1+\int_{0}^{t}x+(\frac{1}{2}x^{2}+x+1)^{2}dx=

y_{2}=\frac{1}{20}t^{5}+\frac{1}{4}t^{4}+\frac{2}{3}t^{3}+\frac{3}{2}t^{2}+t+1

The integrals get pretty ugly around here; I used MATLAB to calculate the
higher order iterates. I’ll show you y_{3}

y_{3}=\frac{49}{60}t^{5}+\frac{13}{12}t^{4}+\frac{4}{3}t^{3}+\frac{3}{2}t^{2}+t+1+O(t^{11})

where O(t^{11}) means assorted polynomial terms from order 6 to 11.

Here is one more:

y_{4}=\frac{17}{12}t^{5}+\frac{17}{12}t^{4}+\frac{4}{3}t^{3}+\frac{3}{2}t^{2}+t+1+O(t^{23})

We notice some patterns developing here. First of all, the coefficient of
the t^{n} term is staying the same for all y_{m} where m\geq n.

That is tedious to prove. But what is easier to show (and sufficient) is
that the coefficients for the t^{n} terms for y_{n} all appear to be
bigger than 1. This is important!

Why? If we can show that this is the case, then our ”limit” solution \sum_{k=0}^{\infty }a_{k}t^{k} will have an interval of convergence less than 1. Why? Substitute t=1 and see that the sum
diverges because the a_{k} not only fail to converge to zero, but they
stay greater than 1.

So, can we prove this general pattern?

YES!

Here is the idea: y_{m}=q(t)+p(t) where p(t) is a polynomial of order m
and q(t) is a polynomial whose terms all have order m+1 or greater.

Now put into the Picard process:

y_{m+1}=1+\int_{0}^{t}((q(x)+p(x))^{2}+xdx=

1+\int_{0}^{t}((q(x)^{2}+2p(x)q(x))dx+\int_{0}^{t}(p(x))^{2}+xdx

Note: all of the terms of y_{m+1} of degree m+1 or higher must come from
the second integral.

Now by induction we can assume that all of the coefficients of the
polynomial p(x) are greater than or equal to one.

When we ”square out” the polynomial, the coefficients of the new
polynomial will consist of the sum of positive numbers, each of which is
greater than 1. For the coefficients of the polynomial (p(x))^{2} of
degree m or higher: if one is interested in the k^{\prime }th
coefficient, one has to add at least k+1 numbers together, each of which
is bigger than one.

Now when one does the integration on these particular terms, one, of course,
divides by k+1 (power rule for integration). But that means that the
coefficient (after integration) is then greater than 1.

Here is a specific example:

Say p(x)=a+bx+cx^{2}+dx^{3}

Now p(x)^{2}=a^{2}+(ab+ab)x+(ac+ca+b^{2})x^{2}+(ad+da+bc+cb)x^{3}+\{O(x^{6})\}

Remember that a,b,c,d are all greater than or equal to one.

Now p(x)^{2}+x=a^{2}+(ab+ab+1)x+(ac+ca+b^{2})x^{2}+(ad+da+bc+cb)x^{3}+\{O(x^{6})\}

Now when we integrate term by term, we get:

\int_{0}^{t}(p(x))^{2}+xdx=a^{2}x+\frac{1}{2}(ab+ab+1)x^{2}+\frac{1}{3}(ac+ca+b^{2})x^{3}+\frac{1}{4}(ad+da+bc+cb)x^{4}+\{O(x^{7})\}

But note that ab+ab+1>2,ac+ca+b^{2}\geq 3, and ad+da+bc+cb\geq 4

Since all of the factors are greater than or equal to 1.

Hence in our new polynomial approximation, the order 4 terms or less all
have coefficients which are greater than or equal to one.

We can make this into a Proposition:

Proposition
Suppose p(x)=\sum_{j=0}^{k}a_{j}x^{j} where each a_{j}\geq 1.

If q(t)=\sum_{j=0}^{2k+1}b_{j}x^{j}=1+\int_{0}^{x}((p(t))^{2}+t)dt

Then for all j\leq k+1,b_{j}\geq 1.

Proof. Of course, b_{0}=1,b_{1}=a_{0}^{2}, and b_{2}=\frac{2a_{0}a_{1}+1}{2}

Let n\leq k+1.

Then we can calculate: (since all of the a_{n-1},a_{n-2},....a_{1} are
defined):

If n is odd, then b_{n}=\frac{1}{n}(2a_{0}a_{n-1}+2a_{1}a_{n-2}+...2a_{\frac{n-3}{2}}a_{\frac{n+1}{2}}+(a_{\frac{n-1}{2}})^{2})\geq \frac{1}{n}(2\ast \frac{n-1}{2}+1)=1

If n is even then b_{n}=\frac{1}{n}(2a_{0}a_{n-1}+2a_{1}a_{n-2}+....2a_{\frac{n-1}{2}}a_{\frac{n+1}{2}})\geq \frac{1}{n}(2\ast \frac{n}{2})=1

The Proposition is proved.

Of course, this possibly fails for b_{n} where n>k+1 as we would fail to
have a sufficient number of terms in our sum.

Now if one wants a challenge, one can modify the above arguments to show that the coefficients of the approximating polynomial never get ”too big”; that is, the coefficient of the k^{\prime }th order term is less than, say, k.

It isn’t hard to show that b_{n}\leq \max a_{i}^{2} where i\in\{0,1,2,...n-1\}

Then one can compare to the derivative of the geometric series to show that
one gets convergence on an interval up to but not including 1.

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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