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What type of functions can satisfy f(x)=f(a-x)?

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Given a function such that [itex]f(x)=f(a-x)[/itex] where a is some constant, what can we say about this function? This isn't actually homework; so please forgive the fact that I'm being a little vague

    2. Relevant equations

    None.

    3. The attempt at a solution

    I think that a periodic function (with period a) which is also even satisfies this but is there something else that can also satisfy this condition? Not really sure how to proceed in proving either that this is the only answer or to show an alternative.
     
  2. jcsd
  3. Oct 20, 2011 #2

    lurflurf

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    All you can say about the function is it is symmetric about x=a/2.
    What are the domain and range?
    If domain were real numbers for example we could define f to be an arbitrary function when x<=a/2 then define f when x>a using the given functional equation.
    or just take an arbitrary function g and define
    f=(1/2)g(x)+(1/2)g(a-x)=(1/2)g(a/2+(x-a/2))+(1/2)g(a/2-(x-a/2))
     
    Last edited: Oct 20, 2011
  4. Oct 20, 2011 #3
    The domain and range are the set of real numbers.

    Thank you for the reply lurflurf. But a function which is symmetric about a should be of the form f(a+x) = f(a-x). Not quite the same as mine, right?

    The second part looks fine but what is that function g(x)+g(a-x) for an arbitrary g?
     
    Last edited: Oct 20, 2011
  5. Oct 20, 2011 #4

    lurflurf

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    That should have been symmetric about x=a/2

    If
    f(x)=(1/2)g(x)+(1/2)g(a-x)
    then
    f(a-x)=(1/2)g(a-x)+(1/2)g(x)=f(x)
    in fact f=g for functions of the type desired

    or write any function f
    f(x)=(1/2)[f(x)+f(a-x)]+(1/2)[f(x)-f(a-x)]
    if
    (1/2)[f(x)-f(a-x)]=0
    then f is already a function of your type
    regardless
    (1/2)[f(x)+f(a-x)] is a function of your type and in some sense the function of your type most like f
     
  6. Oct 20, 2011 #5

    Ray Vickson

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    There are many functions that are NOT symmetric about x = a/2, but satisfy the OPs equation. All you can say is that the function is periodic, with period a.

    RGV
     
  7. Oct 20, 2011 #6
    Ray, could you post an example? Thank you both for replying
     
  8. Oct 20, 2011 #7

    Ray Vickson

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    Sorry: I mis-read the original as f(x-a) instead of your f(a-x). So, the statement regarding symmetry about x = a/2 is correct: the function _is_ symmetric. And: it is not (necessarily) periodic. Mea culpa.

    RGV
     
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