Find a power series representation for the function and determine the interval of convergence. See first post for the problem, second post will be work i have so far.

Question

anonymous · Answer

$$f(x)=\frac{ 5 }{ 2-4x ^{2} }$$

anonymous · Answer

$$\large \frac{ 5 }{ 2 }\frac{ 1 }{ 1-2x ^{2} }=\sum_{n=0}^{\infty} \frac{ 5 }{2 } (2x ^{2})^{n}$$

This is where i'm not sure if I am doing this right. All the examples I could find did not have the a different constant to multiply back in.

terenzreignz · Answer

Don't worry, I don't really see any errors...

terenzreignz · Answer

I suppose you'll have no problem with finding the interval of convergence?

anonymous · Answer

well i had found it and realized i forgot the 5/2 do i just multiply the 5/2 to the 2x^2 and solve?

terenzreignz · Answer

No, the constant 5/2 has nothing to do with the interval of convergence.

It's just a constant factor after all, so it won't affect convergence one bit.

Now, as for the interval of convergence, have you found it?

anonymous · Answer

well if that does not interfere then I have:

$$\large \frac{ -1 }{ \sqrt{2} }\le x \le \frac{ 1 }{\sqrt{2} }$$

terenzreignz · Answer

nu-uh :)
Drop the equality, you know full well that if x is equal to any of those two endpoints, the resulting series diverges :P

anonymous · Answer

sorry the paper had it without the equal, was using those helping someone else:

$$\frac{ -1 }{ \sqrt{2} }<x <\frac{ 1 }{ \sqrt{2}  }$$ 

so that would be right?

terenzreignz · Answer

Yup. It converges whenever x is within that interval. Good job ^_^

anonymous · Answer

ok thank you