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Function, f, with domain (-infinity, + infinity)

  1. Jan 30, 2010 #1
    I was working on the following problem from a textbook. The textbook has no answer. I have included my solution - I am not sure whether it is correct Any ideas and or solutions? (guidance)

    Suppose that f is any function with domain (-infinity, +infinity)

    a) Does the function g defined by g(x) = f(x) + f(-x) have any special symmetry?

    My solution

    A function f which satisfies:
    f(-x) = -f(x) is symmetrical about the horizontal axis. This is also:
    f(x) + f(-x) = 0
    or we could substitute values of x giving non zero solutions for:
    f(x) + f(-x) = g(x) (some other function)

    Therefore g(x) = f(x) + f(-x) has symmetry about horizontal axis?

    Any guidance appreciated.

  2. jcsd
  3. Jan 30, 2010 #2
    No. f(x)+f(-x) is an even function: it's symmetric about the vertical axis.
  4. Jan 30, 2010 #3
    Try and work backwards: take an arbitrary function [tex]g[/tex] defined on [tex]\mathbb{R}[/tex] and try to create a function [tex]f[/tex] s.t. [tex] f(x) + f(-x) = g(x)[/tex] for all x. What assumptions must be made about [tex]g[/tex] in order to do this?
  5. Jan 30, 2010 #4


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    Just use the definition of odd and even functions for this type of problem. Without making any assumptions about f(x) (apart from the given domain) simply write g(-x) and compare it with g(x). That is,

    g(x) = f(x) + f(-x) : given

    So g(-x) = f(-x) + f(-(-x))
    = f(-x) + f(x)
    = g(x).

    And g(-x) = g(x) is the definition of an even function.
    Last edited: Jan 30, 2010
  6. Jan 31, 2010 #5
    I figured it was an even function after I had posted this.

    Thanks to both some_dude and uart for posting two different approaches to the problem. I understand this problem a lot better and have solved some similar now.

    Thanks again,
  7. Feb 1, 2010 #6


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    To test yourself, a new question: do the same for g(x)=f(x)-f(-x).
  8. Feb 2, 2010 #7


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    By the way, f(x)+ f(-x) and f(x)- f(-x) are almost the "even and odd parts" of the function f(x). e(x)= (f(x)+ f(-x))/2 and o(x)= (f(x)- f(-x))/2 are even and odd functions that add to give f(x). Of course, if f(x) is an even function to begin with, o(x) will be 0 and e(x)= f(x). If f(x) is an odd function, e(x)= 0 and o(x)= f(x).

    The exponential function, ex, is neither even nor odd. Its even and odd parts are [itex]\frac{e^x+ e^{-x}}{2}= cosh(x)[/itex] and [itex]\frac{e^x- e^{-x}}{2}= sinh(x)[/itex].
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