1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 24, 2016 #1

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    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


    User Avatar
    2017 Award

    Staff: Mentor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted