What is the Interval of Convergence for the Power Series of f(x) = 2/(1 - x^2)?

In summary, the problem is that the author does not understand how to solve an inequality involving the absolute value, and as such cannot determine the interval of convergence.
  • #1
Bashyboy
1,421
5

Homework Statement


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

Homework Equations





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?
 
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  • #2
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?
 
  • #3
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.
 
  • #4
There's a very easy simplification of |x2|. Alternatively, you can expand an inequality like |y| < a (where a > 0) as -a < y < a.
 
  • #5
So:

-1 < x^2 < 1

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

+/- i < x < +/- 1
?
 
  • #6
Based on the use of x rather than z, I assumed we were dealing with reals here. Was that a wrong assumption?
 
  • #7
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 [itex] \pm i < x < \pm 1 [/itex]. Was this the solution that you had alluded to as being obvious?
 
Last edited:
  • #8
Bashyboy said:
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?
 
  • #9
Would it be (-1, 1)?
 
  • #10
Bashyboy said:
Would it be (-1, 1)?

Looks good to me!

Think about it like this: if -1<x<1 then surely x^2<1 right?
 
  • #11
Well, certainly my textbook is wrong for saying the interval of convergence is (1, 1). Thank you both for your help.
 

FAQ: What is the Interval of Convergence for the Power Series of f(x) = 2/(1 - x^2)?

What is a power series representation?

A power series representation is an infinite series of the form ∑n=0^∞ an(x-c)n, where an are coefficients, x is the variable, and c is the center of the series. This representation is used to approximate functions with polynomials and can be used to find values of the function at any point within its radius of convergence.

What is the radius of convergence in a power series representation?

The radius of convergence is the distance from the center of the power series representation in which the series converges. It is typically denoted by R and can be found by using the ratio test or the root test.

How is a power series representation used to approximate functions?

A power series representation is used to approximate functions by finding the sum of an infinite number of terms of the series. By increasing the number of terms used, the accuracy of the approximation increases. This is especially useful for functions that are difficult to evaluate directly.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a power series representation centered at any point, while a Maclaurin series is a power series representation centered at 0. This means that the coefficients of a Maclaurin series have a specific formula that can be used to find them, while the coefficients of a Taylor series may need to be found using derivatives of the function.

How is a power series representation related to the derivatives of a function?

A power series representation is closely related to the derivatives of a function through its coefficients. The coefficients of a power series can be found by taking derivatives of the function at the center of the series. This means that the power series representation can be used to find the derivatives of a function at any point within its radius of convergence.

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