Question on Radius of Convergence for values of x, when f(x) is x^2

[gat0man]

Homework Statement

This is not so much an entire problem I need help with but just a part.

It is a power series where after you do the ratio test, you end up with |4x^(2)| < 1, so |x^(2)| < 1/4.

Since the radius of convergence is |x-a| < R, I end up with -1/4 < x^(2) < 1/4, but because you cannot take the square root of a negative number, I get 0 <= x < 1/2

So how would I describe the Radius of Convergence in this case? Thanks in advance.

Homework Equations

|x-a| < R (but in this case after the ratio test you end up with 4x^(2) < 1)

The Attempt at a Solution

See what I wrote in AEDIT: You can delete this post, I was just spacing on some primary algebra :(

|x^2| < 1/4 -----> |x| < 1/2, -1/2 < x < 1/2 so radius of convergence is 1/2

 

[Dick]

Are you really claiming that -1/4<x^2<1/4, means 0<=x<1/2 so x must be greater than or equal to zero?? That's not true. x=(-1/4) works in your original inequality just fine. Now you tell me, where did you go wrong?

 

[gat0man]

Are you really claiming that -1/4<x^2<1/4, means 0<=x<1/2 so x must be greater than or equal to zero?? That's not true. x=(-1/4) works in your original inequality just fine. Now you tell me, where did you go wrong?

See my edit :p I'm tired. Realized what I was doing wrong

 

[Dick]

See my edit :p I'm tired. Realized what I was doing wrong

Sure. A tired problem. Good job solving your own problem.