Why does f(-x) = f(x) force only even powers in a power series?

[linda300]

Hey guys,

I've been trying to work out this question,

http://img189.imageshack.us/img189/2954/asdagp.jpg

so the identity theorem is just that if the power series = 0 then the coefficient of the series must be zero.

Im having trouble seeing how that negative has any influence over the n in the x^n term, to make the x's in powers of either odd or even.

If you have f(-x) = f(x) then the series would be like

Ʃa (x-x_o)^n = Ʃa (-x-x_o)^n = Ʃ(-1)^n a (x+x_o)^n

So how does that make the powers only even? Is there somthing crusial that I am missing?

Thanks

 

[tiny-tim]

welcome to pf!

hey linda! welcome to pf!

forget x_o …

"even" and "odd" mean about x = 0 ​

does that make it easier?

 

[linda300]

Thanks!

Yea that helps, so then

Ʃa (x)^n = Ʃa (-x)^n = Ʃ(-1)^n a(x)^n

So is the trick that the only way this can be true is if all the odd powers of n arn't there since the left side will have + a x, + a x^3,.. and the right-a x,- a x^3,... (for a_{n odd} ) which can only be true if they are zero a_{n odd} = 0?

 

[tiny-tim]

yes

but it's a lot easier if you combine it into one series …

∑ { axⁿ - a(-x)ⁿ } = 0 ​

 

[linda300]

Cool,

Thanks a lot!