- #1

WendysRules

- 37

- 3

- 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)##

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)##