I have searched the internet for methods on solving functional equations, unfortunately, most of them consist mainly of substituting values for the variables or on guessing solutions. I think those methods sound like either an impossible task due to the infinite number of possible substitutions or a leap of faith due also to the infinite number of guesses there can be.
How could one for example solve: $$f(1+x)=1+f(x)^2$$ I have tried many different ideas on this, from trying guesses to applying integrals, to eliminating the $1$ by calculating $f(2+x)$ and subtracting equations, then trying to simplify the answer through sums or products, but nothing seems to work.
I think this is a rather interesting subject because $f(x)=f(x-1)^2$ has a solution that is very easy to find but somehow adding a $1$ makes it a much harder problem. Are there ways to solve this equation or even a polynomial functional equation?

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    $\begingroup$ For this example you can take f to be anything on [0,1), then the rule determines what f is on [1,2), etc... $\endgroup$ – mathematician Jul 11 '16 at 15:32
  • $\begingroup$ But can it be solved? Can $f$ be expressed explicitely? $\endgroup$ – Guacho Perez Jul 11 '16 at 16:48
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    $\begingroup$ There are so many solutions that you have to be more specific about what you're looking for. For example, the equation has no polynomial solutions (Let $f(x)= a_nx^n+...+a_0$, you get a $2n$ degree polynomial equals an $n$ degree one, contradiction). You could plug in f(x)= a power series, and derive some relationship between the coefficients. $\endgroup$ – mathematician Jul 11 '16 at 17:10
  • $\begingroup$ Perhaps the difference between your easily-solvable and unsolvable examples can be traced back to the differences between a homogeneous (easy case) and inhomogeneous (hard case) difference equation? I think your two examples are not exactly difference equations, but maybe you can find some insight from that point of view? More generally, I recently stumbled onto this website. $\endgroup$ – jjstankowicz Jul 18 '16 at 5:22
  • $\begingroup$ With @mathematician comments, the problem as stated is solved. If you add some smoothness condition, such as ask the function to be continuos or differentiable, I guess maybe by trying to set the "initial condition" on [0,1) of the same smootheness, and some "boundary condition" on the 0 and the 1 to match continuity/differentiability in the integers, that should be steps toward solution of that problem ,but the stated one in question, pointed how to solve it. $\endgroup$ – Santropedro Sep 21 '17 at 23:58

As for iterating polynomials, there are two general forms of polynomials that we currently know how to iterate. They are polynomials of the form $$P(x)=a(x-b)^d+b$$ and $$T(x)=\cos(k\arccos(x))$$ and yes, I know the second one doesn't look like a polynomial, but for integer values of $k$, it is a polynomial where the inverse cosine is defined, and can be extended to other values (accurately, for our purposes) using the hyperbolic cosine function. For more info, see Chebyshev Polynomials.

The iteration formulas for each of these is given by $$P^n(x)=a^{\frac{b^n-1}{b-1}}(x-b)^{d^n}+b$$ and $$T^n(x)=\cos(k^n\arccos(x))$$

In case you were wondering how this applies to your problem of finding $f$ given that $$f(1+n)=(Q\circ f)(n)$$ where $Q$ is a polynomial, here's how: if you assign a value for $f(0)$, say $y_0$, then you can say that $$f(n)=(Q^n\circ f)(0)=Q^n(y_0)$$ which allows you to find $f:\mathbb Z\to\mathbb Z$.


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