# Fourier Series - Odd and Even Functions

Discussion in 'Homework Help' started by Chelver6571, Oct 27, 2008.

1. ### Chelver6571 Thread Starter New Member

Oct 27, 2008
1
0
Trying to work through some coursework and not 100% of where to start.
I need to demonstrate that the following functions are odd, even or neither odd or even by comparing values.
The first function is f(x)=4x where 0<x<10, which by plotting the graph of f(x)=4x I think the function is odd as it is symmetrical around the origin. By comparing values then when x=2 then 4x = 8 and when x is -2 then 4x=-8 etc.
Second function f(x)=10x^6-x^2, again by plotting the waveform I think the function is odd using comparison values I'm not sure.
Final one I am at a loss, f(x)= {cos x where 0<x< ∏ 0 where ∏<x<2∏ and f(x)=f(x+2x) not sure where to start with this one.
Any help appreciated.
ps apologies about the input form of the last function, not my strongest point typing.

2. ### steveb Senior Member

Jul 3, 2008
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469
There are probably a few ways to do this, but the one that comes to mind is to use the basic definitions of even and odd functions.

Odd: f(-x) = - f(x)
Even: f(-x) = f(x)

For example if you have a sine function sin(x), ask what sin(-x) equals. It is a basic trig identity that sin(-x) = -sin(x). So, sin(x) is odd.

For cos(x), we know cos(-x) = cos(x), so it is even.

For a simple case f(x) = x^n where n is an integer:

f(-x) = -f(x) if n is odd
e.g f(x) = x^3 leads to f(-x) =(-x)^3=(-1)^3*x^3=-x^3

f(-x) = f(x) if n is even
e.g f(x) = x^2 leads to f(-x) =(-x)^2=(-1)^2*x^2=x^2

3. ### steveb Senior Member

Jul 3, 2008
2,431
469
I wanted to add a comment about comparing values. You can compare values to show that a function is not even, or is not odd, or is neither even nor odd. However, you can't use comparison of values to show that a function is even or is odd because you would need to do so for an infinite number of values.

This is just a basic logic issue, which you are probably already aware of, but I thought I would just mention it.