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A functional equation

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex] a.b,c,d [/itex] be real numbers such that [itex] a ≠ b [/itex] and [itex] c ≠ 0 [/itex], find f:R->R for which this statement holds:

    [tex] af(x+y) + bf(x-y) = cf(x) + dy [/tex] , for all x,y real numbers.



    2. Relevant equations

    Well this is a functional equation, that I know. I have less experience with those type of euqations.


    3. The attempt at a solution

    I tried to plug in some values:

    for x = y: [tex] af(2x)=cf(x)+dx-bf(0) [/tex]

    for x = y = 0 :[tex] af(0) + bf(0) = cf(0) [/tex]

    I don't know what to do next, and how could these facts can help me. Can anyone help me solve this problem, please :D ?
     
  2. jcsd
  3. Dec 29, 2013 #2

    pasmith

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    If you set [itex]y = -x[/itex] in the functional equation, you will get a second linear equation which [itex]f(x)[/itex] and [itex]f(2x)[/itex] must satisfy.

    Do these linear simultaneous equations have a unique solution? Are the resulting expressions for [itex]f(x)[/itex] and [itex]f(2x)[/itex] consistent, or do you have to impose further conditions on [itex]a[/itex], [itex]b[/itex], [itex]c[/itex] and [itex]d[/itex]?

    This tells you that if [itex]a + b = c[/itex] then [itex]f(0)[/itex] is arbitrary, and if [itex]a + b \neq c[/itex] then [itex]f(0) = 0[/itex].
     
  4. Dec 29, 2013 #3

    Ray Vickson

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    Unless f is identically 0 we must have a+b = c. Just put y = 0 in the functional equation to see why.
     
  5. Dec 31, 2013 #4

    pasmith

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    The interesting case is [itex]a + b \neq c[/itex] and [itex]d \neq 0[/itex].
     
  6. Dec 31, 2013 #5

    Ray Vickson

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    If ##a+b \neq c## then ##f(x) \equiv 0##. To see this, put y = 0 to get
    [tex] (a+b) f(x) = c f(x)[/tex]
    which is supposed to hold for all values of x. If ##f(x_0) \neq 0## then ##a+b = c##.
     
  7. Jan 1, 2014 #6
    Thank you all for your hints!!!! I managed to solved. You guys are the best!!!
     
  8. Jan 1, 2014 #7

    Ray Vickson

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    Just for interest: what is your solution?
     
  9. Jan 2, 2014 #8
    First, I noticed a subtle "symmetry" in the equation, I don't know how to call it otherwise. So I plugged -y instead of y and I got the system:

    [tex] af(x+y) + bf(x-y) = cf(x) +dy [/tex]
    [tex] af(x-y) + bf(x+y) = cf(x) - dy [/tex]

    Then, I played a little with the system. I won't post the whole the whole steps because it's only a little algebra. But, I will tell you what I tought: I wanted to get rid of f(x+y). So I mutliplyed the first equation by b and the second by a, the I substracted the second equation from the first, and it led me to:

    [tex] (b^2-a^2)f(x-y) = c(b-a)f(x) + d(b+a)y [/tex] .

    Next I plugged x , instead of y , and I got that:

    [tex] f(x) = \frac{a+b}{c}f(0)+\frac{d(a+b)}{c(a-b)}x [/tex]

    That function f(x) verifies the initial conditions. And that's it.. :D
     
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