Taylor series and variable substitutions

In summary, the conversation discusses variable substitutions and how they can be seen as function compositions. The use of different functions and their Taylor series is also discussed, with the conclusion that the second Taylor series is actually for a composite function. The conversation also mentions the use of the ratio test to determine the radius of convergence for a power series.
  • #1
WendysRules
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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)##
 
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  • #2
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 .
 
  • #3
WendysRules said:
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:
WendysRules said:
##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##:
$$g(\theta) = \sec^2\theta \approx 1+\theta^2+\frac{2}{3}\theta^4+ \dots $$
 
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  • #4
PeroK said:
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.
 
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Related to Taylor series and variable substitutions

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point. It is used to approximate a function and can be used to find the value of a function at any point within its domain.

2. How is a Taylor series calculated?

A Taylor series is calculated using the derivatives of a function at a specific point. The general formula for a Taylor series is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ..., where a is the point at which the derivatives are evaluated.

3. What is a variable substitution in a Taylor series?

A variable substitution in a Taylor series is when a new variable is introduced in place of the original variable in the series. This is often done to simplify the series or to expand the series to a different point.

4. Why is a variable substitution useful in a Taylor series?

A variable substitution can make a Taylor series easier to work with, as it can simplify the series and make it easier to manipulate. It can also be used to expand the series to a different point, allowing for more accurate approximations of the function.

5. How can Taylor series and variable substitutions be applied in real-world situations?

Taylor series and variable substitutions are commonly used in physics, engineering, and other fields to approximate functions and make calculations more manageable. They can also be used in computer science to optimize algorithms and in economics to model complex systems.

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