Question

Suppose that f(x)=sum_{n=0}^{infty} for all of x. If f is an odd function, show that c sub n =0 for all even numbers n.

Answer 1

Suppose that \[f(x)=\sum_{n=0}^{\infty} \] for all of x. If f is an odd function, show that c sub n=0 for all even numbers n.

Answer 2

so f(x) = 0?

Answer 3

There is no f(x) after the summation notation. Just exactly how I typed it here. It's asking if we have the summation from n=0 to infinity of any odd function, then show that c of n = 0 for all even numbers of n.

Answer 4

What?

Answer 5

Is this function a constant with respect to \(x\)?

Answer 6

yeah that makes no sense.

Answer 7

Well at least I don't feel as bad not knowing how to even begin. :(

Answer 8

Here's the copy of my worksheet....

Answer 9

no the n=0 on the bottom and the infinity on top

Answer 10

\[f(x)=\sum_{n=0}^{\infty}c _{n} x^n\]

Answer 11

ok
Then how do I show that c sub n = 0 for all even numbers n

Answer 12

Proof by contradiction.

Answer 13

When you plug in \(-x\), then you'll see it is different from \(-f(x)\).

Answer 14

And even power of \(n\) will get rid of the negative.

Answer 15

Suppose that \[f(x) = \sum_{n = 0}^{\infty}c _{n} x ^n\] for all x.
Show that if f is an odd function, \[c _{0} = c _{2}=...=0\]
for even \[c _{1}=c _{3}=...=0\]

Answer 16

This is the question...yes, I need to prove that if it's an odd function all even numbers of n will make c sub n = 0. Can someone show me how to prove that?