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Let $x$ be a real number and $f(x)$ a real analytic function such that $f(x+1) = P(f(x),x)$ where $P$ is a given real polynomial. Express $f(x)$ as an integral from $0$ to $\infty$.

As an example we have the $\Gamma$ function. http://mathworld.wolfram.com/GammaFunction.html

I'm looking for the general solution.

I was thinking about the recursions used to compute integrals of type $\int$ $f(x)^k \mathrm{d}x$.

Also hypergeometric functions crossed my mind.

But I am stuck.

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    How can you "let $f(x+1)=P(f(x),x)$", given that $f$ is already given? Do you mean "assume $f(x+1)=P(f(x),x)$?"2012-11-09
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    Also it is totally unclear what you mean by "express $f$ as an integral transform". I doubt $f(x)=\int_0^{\infty}f(x)e^{-t}\,dt$ is what you are looking for.2012-11-09
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    @ThomasAndrews maybe i should connect those two sentenses with 'such that'. @ fedja do you know the integral transform that defines the gamma function ? I did not suggest putting f(x) in the integral itself ... maybe I should remove transform ?2012-11-09
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    I edited. Hope it is better now.2012-11-09
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    Do you have a reason to expect that this will be possible in this general form?2012-11-09
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    @LukasGeyer well yes and no. There are many recursions known for hypergeo functions and for integrals such as those of a power(as I wrote in the OP). And many of those calculus/hypergeo identities have a history of getting more and more general identities. So that makes me optimistic. I am rather pessimistic in the complexity of the answer then about its existance. I think this is outside of most books and educations and I regret that , but I could be wrong ...2012-11-09
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    What do you mean by analytic? Do you mean entire? Do you mean analytic for all po sitive real $x$? The gamma function is not entire. You also cannot expect uniqueness. If P(f,x)=f, then any entire periodic function will work--try sines and cosines.2012-11-09
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    @nayrb I mean Coo. Although I must say I forgot about the poles of gamma ... Nontrivial cases are unique ... such as gamma ... Btw I did not ask for uniqueness ...2012-11-09
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    Real analytic means R->R and Coo at R.2012-11-09
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    I think you ask $f(x+1)=P(x)f(x)$ rather than $f(x+1)=P(f(x),x)$ . $f(x+1)=P(f(x),x)$ is a nonlinear equation an even its form of general solution is still unknown.2012-11-15
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    Then the answer is unknown. Do you have evidence that it is unknown ? Thanks.2012-11-17
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    The solution to $f(x+1)=P(f(x))$ is known. This can be expressed by a socalled superfunction of $P$. This is classical complex dynamics. In spite of that $f(x+1)=P(f(x),x)$ seems unsolved ... according to @doraemonpaul but I want the evidence for that claim.2013-01-21
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    If we replace/approximate $f(x+1)$ with a truncated Taylor series of $f(x+1)$ in terms of $f(x)$, we get a differential equation. Not sure if that works well though.2013-06-27

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