Maxim wrote: >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 ?)
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.
In article <vz6xjmdri49p@legacy>, Maxim <MOsadc...@nes.ru> wrote: >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 ?)
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 Rusin wrote: > In article <vz6xjmdri49p@legacy>, Maxim <MOsadc...@nes.ru> wrote:
>>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 ?)
> 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
We may show that if it exists m |f(x)= m^[x](x) xth iterate of m(x) equation y(f(x))=y(x)+x is equivalent to y(m(x)= y(x) + 1 (sci.math 09 june ). So with f(x)=exp(x)-1 you have to find m^[x](x)= exp(x)-1 ,an harder task than solving or computing m(x) from m(m(x))= exp(x)-1 ,
Very often we may build y(x)as a function point to point . Here y(f(x))=y(x)+x then from y(x0) , y(f(x0))=y(x0)+x0 .... in fact we've got:y(f^[n](x0)=y(f^[n-1](x0))+f^[n-1](x0) or y(f^[n](x0)=y(x0)+x0 +f(x0)+f^[2](x0)+.... f^[n](x0) stands for f(f(f(f(f(f (x0)..))) f n times. In our case we must choose x0 # 0 (to avoid y(0)=y(0)+0 a loop!). two consequences: value y(x0) is 'free', we obtain a'gruyère',that is to say a holy swiss cheese:function is known only in points x0, f(x0),....f^[n](x0).
