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

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

1. ### MaximGuest

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

2. ### Dave RusinGuest

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

3. ### Marc NardmannGuest

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

-- Marc Nardmann

Marc Nardmann, Jan 9, 2005
4. ### Daniel GeislerGuest

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