# Taylor series and variable substitutions

• I
• WendysRules

#### WendysRules

TL;DR Summary
Variable substitutions
I'm currently typing up some notes on topics since I have free time right now, and this question popped into my head.

Given a problem as follows:

Find the first five terms of the Taylor series about some ##x_0## and describe the largest interval containing ##x_0## in which they are analytic.

The function is ##f(x) = \frac{1}{1-x^2}## and ##x_0 = 0##. I was wondering, however, if we could make a substitution where ##x = \sin \theta## where we can now construct the new function ##f(\theta) = \sec^2 \theta## where ##0 = \sin \theta_0 \rightarrow \theta_0 = 0, \pi## where we restrict ##0 \leq \theta < 2 \pi##

But if we compute out the two Taylor series we would get (choosing ##\theta_0 = 0##...)
##f(x) = \frac{1}{1-x^2} \approx 1+x^2+x^4+...## and ##f(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+...##

So... what gives? It even seems like it isn't even an exact equation either! It seems as if ##\frac{1}{1-x^2} \approx \sec^2x## on ##(-1,1)##

If you have a formula for the general nth coefficient of the power series, say an = h(n), then you want to find the limit (if it exists) of the absolute value of the ratio of the (n+1)st term of the power series to the nth term.

That is, the limit of the ratio |an+1 zn+1 / an zn|, if it exists, should be less than 1 for the power series to exist. Writing that equation will tell you the radius of convergence within which the original power series converges to an analytic function.

E.g., if a_n = 2^n, then setting the ratio |an+1 zn+1 / an zn| < 1 gives
|2z| < 1 which leads to |z| < 1/2. So this is the interval on which this power series converges to an analytic function.

If that ratio doesn't approach a limit then you probably need the more involved "root test" as discussed in this article: https://en.wikipedia.org/wiki/Radius_of_convergence .

Summary:: Variable substitutions

I'm currently typing up some notes on topics since I have free time right now, and this question popped into my head.

Given a problem as follows:

Find the first five terms of the Taylor series about some ##x_0## and describe the largest interval containing ##x_0## in which they are analytic.

The function is ##f(x) = \frac{1}{1-x^2}## and ##x_0 = 0##. I was wondering, however, if we could make a substitution where ##x = \sin \theta## where we can now construct the new function ##f(\theta) = \sec^2 \theta## where ##0 = \sin \theta_0 \rightarrow \theta_0 = 0, \pi## where we restrict ##0 \leq \theta < 2 \pi##

But if we compute out the two Taylor series we would get (choosing ##\theta_0 = 0##...)
##f(x) = \frac{1}{1-x^2} \approx 1+x^2+x^4+...## and ##f(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+...##

So... what gives? It even seems like it isn't even an exact equation either! It seems as if ##\frac{1}{1-x^2} \approx \sec^2x## on ##(-1,1)##
One issue is that you are using ##f## to denote two different functions. If we say that ##f(x) = \frac{1}{1- x^2}##, and ##x = \sin \theta##, then we can define:
$$g(\theta) = f(\sin \theta) = \frac{1}{1- \sin^2 \theta} = \sec^2 \theta$$
And you can see that ##g## is not the same function as ##f##.

For example, where you have:
##f(x) = \frac{1}{1-x^2} \approx 1+x^2+x^4+...## and ##f(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+...##

This is wrong. By definition of the function ##f## we have:
$$f(\theta) = \frac{1}{1-\theta^2} \approx 1+\theta^2+\theta^4+ \dots$$
The second Taylor series is actually for the composite function ##g##:
$$g(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+ \dots$$

• hutchphd, WendysRules and Klystron
The second Taylor series is actually for the composite function ##g##:
$$g(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+ \dots$$
Ah, how foolish of me! Thanks for linking the two ideas for me. Variable substitutions are truly just function compositions. I've never thought about it that way.

• PeroK