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madmike159
Dec7-10, 03:07 PM
For functions when doing Fourier Transforms, when a function is odd bn = 0 and when a function is even an = 0.

Are all functions odd or even, or are there some cases where they're both / neither?

Hechima
Dec7-10, 03:23 PM
An even function has the property:
f(-x)=f(x)

An odd function has the property:
f(-x)=-f(x)

Suppose you have a function f(x) that is neither odd nor even.

Now, define a function g(x)=\frac{f(x)+f(-x)}{2}.

So that, g(-x)=\frac{f(-x)+f(-(-x))}{2}=\frac{f(x)+f(-x)}{2}.

This means that g(x) is an even function, regardless of the properties of f(x)!

Similarly, let h(x)=\frac{f(x)-f(-x)}{2}.

It's easy to show that h(x) must be odd.

So, functions may be broken up into even and odd components.

madmike159
Dec7-10, 03:30 PM
Ok thanks.

HallsofIvy
Dec8-10, 08:29 AM
For example, the function f(x)= x+ 1 has the property that f(2)= 3 while f(-2)= -1. Since neither f(2)= f(-2) nor f(2)= -f(-2) is true, this function is neither even nor odd.
In fact, most functions are neither even nor odd.

In order that a function be both even and odd, we would have to have both f(x)= f(-x) and f(x)= -f(-x) for all x. That is the same as saying f(x)= -f(-x)= -f(x) so that 2f(x)= 0 for all x. The is only true for the "zero" function, f(x)= 0 for all x.

Outlined
Dec12-10, 04:30 PM
the exp function is not odd nor even.