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There's a problem I've been working on for awhile that involves some hefty functional equations. For example, I may have something along the lines of

$ f(x)f(x) =x+1+f(x+1) $

I've tried several different methods of attack (the farthest I ever got was probably with a power series which didn't yield a recurrence relation) but it never amounts to much. As if that wasn't bad enough, I don't actually know any values of $f(x)$, other than that $\lim_{x\to\infty}f(x)=\infty$. The thing is, I don't actually care about $f(x)$, I only want to know $f(0)$ (analytically) but I can never seem to get two equations for a given point.

I was wondering if anyone had any ideas on how to solve this, or even just some insight into whether or not it can be solved. Thanks!

Edit: Additional facts

  • It can be required that $1
  • $f(x)$ is strictly increasing
  • $f(x)$ is non-negative
  • 0
    Also see my question [here](http://math.stackexchange.com/questions/864262/nonlinear-inhomogeneous-recurrence-fx2-fx1sx-to-find-nested-radical), which generalizes this problem.2014-07-25

1 Answers 1

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Your constraints are not strong enough to impose a unique value to $f(0)$. In fact, I show below that there for every value $a$ between $1.8$ and $2$ there is a solution with $f(0)=a$ (in fact for a given $a$ there are uncoutably many such solutions, as our construction will show).

Given a function $\phi : [0,1[ \to {\mathbb R}$, there is a unique solution $f: [0,+\infty[ \to {\mathbb R}$ to the functional equation that coincides with $\phi$ on $[0,1[$. Indeed, we will have

$ \begin{array}{lcl} f(x)&=&\phi(x-1)^2-x=\phi_1(x) \ \ (\text{for }\ x\in[1,2[) \\ f(x)&=&\phi_1(x-1)^2-x=\phi_2(x) \ \ (\text{for }\ x\in[2,3[) \\ \end{array}\tag{1} $ and so on.

Now, let $a \in [1.8,2]$ and let $\phi : [0,1] \to {\mathbb R}$ be a continuous, strictly increasing map satisfying the boundary conditions $\phi(0)=a, \phi(1)=a^2-1$ (so for example you could take $\phi$ affine, or trigonometric, etc).

Then the $f$ uniquely defined by (1) will also be continuous and strictly increasing (in particular, $f$ will be positive). So all we need to show is that $f \to +\infty$ at $+\infty$ ; since $f$ is strictly increasing, it suffices to show that $f(n) \to +\infty$ when $n$ is an integer.

We have $ \begin{array}{lcl} f(1)&=&f(0)^2-1 \geq 1.8^2-1=2.24 \\ f(2)&=&f(1)^2-2 \geq 2.24^2-2=3.0176 \geq 3 \\ f(3)&=&f(2)^2-3 \geq 3^2-2=6 \\ \end{array}\tag{1} $

We can then use mathematical induction : let us show that $f(k) \geq k$, for every $k\geq 3$.

The $k=3$ case has just been checked. Suppose that the hypothesis is true for some $k\geq 3$. We have

$f(k+1)=f(k)^2-(k+1) \geq k^2-(k+1)=k-2+(k-1)^2 \geq k-2+4=k+2 \geq k+1 $

So the hypothesis holds on the next level. This shows that $\lim_{+\infty}(f)=+\infty$ as wished.

  • 0
    I find it very puzzling, that $f(0)$ can be chosen freely, because any solution $f(x)$ should equal $f(x)=\sqrt{x+1+\sqrt{x+2+\sqrt{x+3+...}}}$ (this follows by iterative application of the equation) and hence $f(0)=\sqrt{1+\sqrt{2+\sqrt{3+...}}}$ should follow also straight from the equation. Where is my mistake?2014-07-25