I Taylor series and variable substitutions

[WendysRules]

TL;DR

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 $(-1,1)$

 

[zinq]

If you have a formula for the general nth coefficient of the power series, say a_n = 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 |a_n+1 zⁿ⁺¹ / a_n 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 |a_n+1 zⁿ⁺¹ / a_n 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 .

 

[PeroK]

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 $g$:
$$g(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+ \dots $$

 

[WendysRules]

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.