y(f(x))=y(x)+x

Discussion in 'Math Research' started by Maxim, Jan 8, 2005.

  1. Maxim

    Maxim Guest

    Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
    function? (e.g., f(x)=exp(x)-1 ?)
     
    Maxim, Jan 8, 2005
    #1
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  2. Maxim

    Dave Rusin Guest

    What do you mean by "solve"?

    Exactly this example is treated as a prototype for a broad family of
    similar problems in a paper by Szekeres; he argues that there is one
    particularly optimal solution y and in an accompanying paper provides
    tables of numerical values. (He uses the analysis to describe a
    one-parameter family of functions f_s with f_s o f_t = f_{s+t} and
    in particular describes a functional "square root" f_{1/2} of f = f_1.)

    MR0141905 (25 #5302)
    Szekeres, G.
    Fractional iteration of exponentially growing functions.
    J. Austral. Math. Soc. 2 1961/1962 301--320. MSC section 39.99

    dave
     
    Dave Rusin, Jan 9, 2005
    #2
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  3. If f(0)=0 and f is differentiable at 0 with f'(0)=1, as in the case
    f(x)=exp(x)-1, then there is no solution y which is differentiable at 0.

    PROOF: Assume that y is differentiable at 0. Then so is y°f. For x not
    equal to 0, we have [(y°f)(x)-(y°f)(0)]/x - [y(x)-y(0)]/x = 1; so for x
    -> 0, we obtain 1+y'(0) = (y°f)'(0) = y'(f(0)) f'(0) = y'(0), a
    contradiction.


    -- Marc Nardmann
     
    Marc Nardmann, Jan 9, 2005
    #3
  4. See
    http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A052122
    for f(x) such that f(f(x)) = exp(x)-1.
    Daniel
     
    Daniel Geisler, Jan 11, 2005
    #4
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