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
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    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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    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
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    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.
     
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