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Even and odd functions

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Hey guys,
    I'm studying for my sigs and coms exam and I am having trouble with determining whether a function is even or odd.
    I understand that an even function satisfies f(-t) = f(t) and an odd function satisfies f(-t) = -f(t) Can I determine whether this function is even or odd by looking at it?

    2. Relevant equations

    { xsin(ax) dx

    3. The attempt at a solution
    I know I substitute x with -x but what do I do from there?
  2. jcsd
  3. Dec 10, 2008 #2
    Maybe i'm not reading your question right, but you said that you understand that a function is even when f(-t) = f(t), and odd when f(-t) = -f(t).

    If you have a graph of it, you should be able to tell whether the function is odd or even just by looking at the graph and taking t and -t. If all you're given is an equation, then choose a value that you can easily evaluate and see what happens on the other side of the axis.

    Or are you referring to the equation on #2?

    [tex]f(x) = \int xsin(ax)[/tex]

    I think that's what you meant on #2 anyway.
  4. Dec 10, 2008 #3


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    You try to either eliminate the "-" signs, or try to bring them outside the expression.

    Example #1:

    f(x) = x2
    f(-x) = (-x)2
    = x2 (eliminating the "-")​
    = f(x) ​
    so this is an even function

    Example #2:

    f(x) = x3
    f(-x) = (-x)3
    = -(x3) (bringing the "-" outside the expression)​
    = -f(x) ​
    so this is an odd function

    p.s. Welcome to PF :smile:
  5. Dec 10, 2008 #4
    Thanks for the replies guys and the warm welcome!! I think I figured it out. Yes we're given a diagram and I realized when the wave goes through the origin the function is odd and otherwise its even. I think!!
  6. Dec 10, 2008 #5
    That's not necessarily true. f(x^2) is an even function and f(x) is an odd function, but they both go through the origin. Since you're given the graph, take any value of x (besides the origin) in that graph and see where the function is, now look over to -x and see if you get the same value (even), or the opposite (odd).

    Edit: Or, you could imagine folding the the graph along the y-axis and see if the function lines up. If it does, then it's an even function. If not, then it's mostly an odd function. You could check if it's an odd function by "folding" the graph along the function y = x
  7. Dec 10, 2008 #6


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    That's not correct, as noted above. In fact the vast majority of functions you will deal with are neither odd nor even.
  8. Dec 11, 2008 #7
    As noted by Defennder, most functions are neither entirely even nor odd. Every function can be broken down into two parts, and even part and an odd part.

    If you take the (complex) Fourier transform of a function (when it exists), the real part of the transform will result from the even part of the function and the odd part of the function will give rise to the imaginary part of the transform.

    If you do a Fourier expansion of the function, the cosine terms (and the constant term) are the result of the even part of the function and the sine terms are the result of the odd part of the function.
  9. Dec 11, 2008 #8


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    You can also decompose a function into odd and even parts without using Fourier transforms:

    fodd(x) = ½[f(x) - f(-x)]

    feven(x) = ½[f(x) + f(-x)]
  10. Dec 11, 2008 #9
    Yes, Redbelly98, is most certainly correct about this simple decomposition into even and odd parts.
  11. Dec 11, 2008 #10
    Doesn't the even and oddity of a function also have to do with the Taylor Polynomial? For instance, cos(x) is even because the first term (f(0)) is 1, the second term f ' (0) = 0, f ''(0) = -1, so it's Taylor is 1-x^2/2!+x^4/4! ??
  12. Dec 11, 2008 #11
    This could become a chicken and egg thing, but the fundamental definitions of even and odd are those that were given in the first entry above by duckless. Everything else really follows from that, and it is probably best to stick with those simple expressions as the definitions, and then accept the rest as consequences.

    A Taylor series that contains only even powers will rather obviously produce an even function (such as the cosine), and one that contains only odd powers will produce an odd function. This is a consequence of the fact that, term by term, these functions are either even or odd, and therefore their sums are still even or odd.
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