Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Even plus odd functions

  1. Jan 9, 2015 #1

    {~}

    User Avatar

    I have been looking at my old calculus textbook because to my dismay I seem to have forgotten most of the calculus I learned. I am given 3 cases of ##(f+g)(x) ##.

    Case 1 both f and g are even:
    I know ##f(x) = f(-x) ## and ##g(x)=g(-x) ## for the domain of the function. I can reason by substitution that
    ##f(x)+g(x)=f(-x)+g(-x) ##
    ##(f+g)(x)=(f+g)(-x) ##
    ##(f+g)(x) ## is even. So far so good.

    Case 2 both f and g are odd:
    I found that if ##f(x)=-g(x)+c ## then
    ##(f+g)(x)=c ## which is even.
    Otherwise I think that ##(f+g)(x) ## would be odd though I don't know how to assert that.

    Case 3 f is even and g is odd:
    I think that other than is special case where one or both of our functions are zero for all x in the domain ##(f+g)(x) ## would neither be even or odd. I don't know how to prove this.

    I know that the notions of even an odd is defined in terms of sets rather than algebraically like I did here. I think if I understood sets better I might have more of a handle on this, I don't know. Hints?
     
  2. jcsd
  3. Jan 9, 2015 #2
    If ##f## and ##g## are odd:

    ##(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f+g)(x)##.
     
  4. Jan 9, 2015 #3

    {~}

    User Avatar

    I had something similar, but why does this look like it's always true when it isn't? Or is my special case wrong?
     
  5. Jan 9, 2015 #4
    What special case? ##-g(x)+c## is not odd in general.
     
  6. Jan 9, 2015 #5

    {~}

    User Avatar

    Thanks for pointing out my error. That made things less confusing.
     
  7. Jan 9, 2015 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Suppose that ##f## is even and ##g## is odd.

    If ##f+g## is even then ##f(x) + g(x) = f(-x) + g(-x)## for all ##x##. Since ##f(x) = f(-x)##, it follows that ##g(x) = g(-x)##. But ##g## is odd, so also ##g(x) = -g(-x)##. Therefore, ##g(-x) = -g(-x)##, or equivalently ##2g(-x) = 0##. This is true for all ##x##, so ##g## is the zero function.

    If ##f+g## is odd then ##f(x) + g(x) = -f(-x) - g(-x)## for all ##x##. Since ##g(x) = -g(-x)##, this is equivalent to ##f(x) = -f(-x)##. But ##f## is even, so also ##f(x) = f(-x)##. Therefore, ##f(-x) = -f(-x)##, or ##2f(-x) = 0##. This is true for all ##x##, so ##f## is the zero function.
     
    Last edited: Jan 9, 2015
  8. Jan 11, 2015 #7

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    In Case 3 you can not say anything about the sum of an even + odd function. Any function, h(x), can be represented as the sum of an even function and an odd function. So you can not state any conclusions about h.
    ( Define even(x) = (h(x) + h(-x))/2 and odd(x) = (h(x) - h(-x))/2. Prove that h(x) = even(x) + odd(x), that even(x) is even, and that odd(x) is odd.)
     
    Last edited: Jan 11, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Even plus odd functions
  1. Even or Odd Functions? (Replies: 6)

Loading...