# College Math Teaching

## September 9, 2014

### Chebyshev polynomials: a topological viewpoint

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 $T_4(x), T_5 (x)$

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

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. ðŸ™‚

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.

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?

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.