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A Functional equation f(x)^n f(a - x) = 1

  1. Mar 24, 2016 #1
    Hi,

    In a problem I have been working on for a while now I have found that I want to find the function satisfying the functional relation

    f(x)n f(a - x) = 1

    for n = 1 I believe I have proven that f(x) = x/(a - x). On this page is an answer I do not quite understand. One of the prerequisits for f(x) in my problem was that f(0) = 0, f(a/2) = 1 and f(a) = ∞. I have difficulties seeing that the answer provided will satisfy f(0) = 0.

    Apart from this, I have made no progress what so ever in adressing the case where n ≠ 1. In particular I have failed to find solutions for when n = 2.

    The problem is this: I want to make a scale mapping where f(x) represents the value and x represents the position of the scale such that the graph of 1/xn is a straight line from positions (0, a) through ((a/2, a/2) to (a, 0), representing the values (0, ∞) through (1, 1) to (∞, 0).

    How can I go about finding f(x) for n ≠ 1?

    Oh, and I don't have any problem with not getting to solve it myself - I you know the answer, please just tell me!
     
  2. jcsd
  3. Mar 24, 2016 #2

    Ssnow

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    For ##n=1## the general solution is of the form ## e^{\Phi(x,a-x)}## where ##\Phi## is antisymmetric so ##f(x)f(a-x)=e^{\Phi(x,a-x)+\Phi(a-x,x)}=e^{\Phi(x,a-x)-\Phi(x,a-x)}=1##, so your equation is satisfied. After it specify the solution in the case where ##\Phi(x,a-x)=C(2x-a)##.


    For ##n=2## I don't see other solutions than ##f(x)=1## ...

    You can assume that ##f## admits a Taylor expansion of the form ##f(x)=a_ {0}+a_{1}x+ a_{2}x^2+...## you can put this into your equation and use a math program in order to find relations with coefficients ... You will obtain an infinite system of equations to solve ...
     
  4. Mar 25, 2016 #3
    f(x) = 1 does not satisty f(0) = 0. Nor does any exponential solution.... Your suggestion with taylor expansions seem interesting though, I will attempt it when I have some more time.
     
  5. Mar 25, 2016 #4

    Ssnow

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    ok, before I spoke about the problem in general I didn't consider the boundary condition. So in this case the exponential is not your solution ... If you want this condition you can search directly a Taylor (or better Laurent (with negative exponents)) expansion of this kind ##f(x)=a_{1}x+a_{2}x^{2}+\cdots ##
     
  6. Mar 25, 2016 #5

    mfb

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    I would try a Laurent expansion around x=a. We know f(0)=0 so there Taylor and Laurent would be identical. This could limit the function too much.

    The approach with an antisymmetric Φ should still work, if you let Φ go to -infinity if its first argument goes to zero.
     
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