# Power Series Representation

## Homework Statement

I am asked to find a power series for the function f(x) = 2/(1 - x^2), centered at 0.

## The Attempt at a Solution

The only part I can't determine is the interval of convergence. I get stuck on the step |x2| < 1. What am I to do next?

haruspex
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Are you saying you know the convergence domain is |x2| < 1, and you want to express that as a domain for x? Umm... isn't it obvious?

Not really, either wise I wouldn't have posed the question in the first place. I thought of taking the square root of both sides of the inequality, but wasn't entirely sure what the result would be.

haruspex
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There's a very easy simplification of |x2|. Alternatively, you can expand an inequality like |y| < a (where a > 0) as -a < y < a.

So:

-1 < x^2 < 1

+/- sqrt(-1) < x < +/- sqrt(1)

+/- i < x < +/- 1
?

haruspex
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Based on the use of x rather than z, I assumed we were dealing with reals here. Was that a wrong assumption?

No, this problem does not entertain numbers other than reals. You said that I could expand the inequality involving the absolute, of which I did; henceforth, I solved for x, of which clearly resulted in a rather odd looking inequality, namely $\pm i < x < \pm 1$. Was this the solution that you had alluded to as being obvious?

Last edited:
haruspex
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So:

-1 < x^2 < 1

+/- sqrt(-1) < x < +/- sqrt(1)
There are two problems with that step.
When operating with inequalities, you must be particularly careful about changing signs unwittingly. You have assumed sqrt(x^2) = x. The sqrt function is defined to return the principal value, i.e. the non-negative square root. Thus sqrt(x^2) = |x|, and the square roots of x^2 are ±|x|.
Secondly, it makes no sense to take sqrt(-1) if we're dealing with reals. Isn't the inequality -1 < x^2 somewhat redundant? Can you think of a tighter bound?

Would it be (-1, 1)?

Would it be (-1, 1)?
Looks good to me!

Think about it like this: if -1<x<1 then surely x^2<1 right?

Well, certainly my textbook is wrong for saying the interval of convergence is (1, 1). Thank you both for your help.