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 $1<f(0)<2$
  • $f(x)$ is strictly increasing
  • $f(x)$ is non-negative
  • 1
    $\begingroup$ Does $f(x)f(x)$ just mean $f(x)^2$? $\endgroup$ – mjqxxxx Dec 28 '12 at 4:11
  • $\begingroup$ Yes! For some reason I thought it would be less ambiguous that way. $\endgroup$ – Mr. G Dec 28 '12 at 4:18
  • $\begingroup$ When you write $\lim_{x\to\infty}f(x)$, do you mean $\lim_{x\to+\infty}f(x),\lim_{x\to-\infty}f(x)$, or both ? This is an important distinction here $\endgroup$ – Ewan Delanoy Dec 28 '12 at 4:41
  • $\begingroup$ I mean that $f(x)$ goes to positive infinity as $x$ goes to positive infinity. $\endgroup$ – Mr. G Dec 28 '12 at 4:52
  • $\begingroup$ Also see my question here, which generalizes this problem. $\endgroup$ – flonk Jul 25 '14 at 9:32

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.

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  • $\begingroup$ Wow, that's amazing! Actually, it's possible for me to impose that $1<f(0)<2$. There are a few other known facts that I didn't mention because I didn't think they were important, but I realize know that they probably are to people who know much more about this sort of thing than I do. I've put them all in the original post. $\endgroup$ – Mr. G Dec 28 '12 at 6:21
  • $\begingroup$ @Mr.G : I updated my answer to meet all your additional conditions. $\endgroup$ – Ewan Delanoy Dec 28 '12 at 8:12
  • $\begingroup$ Is there another condition I could find that would force a unique $f(0)$ (short of finding the $f(x)$ at another point?) I know that $f(0)$ has one single value but from what I'm understanding of your answer, the information contained in the functional equation alone (plus the auxiliary facts) isn't enough to extract it. $\endgroup$ – Mr. G Dec 28 '12 at 8:42
  • $\begingroup$ @Mr.G : well, it looks like there’s a piece of information you forgot to add. How exactly do you know that “f(0) has one single value” ? $\endgroup$ – Ewan Delanoy Dec 28 '12 at 11:01
  • $\begingroup$ $f(x) = \sqrt{x+1+\sqrt{x+2+\sqrt{x+3+...}}}$ My idea was to convert it to a functional equation to try to solve it and get it to look like something more recognizable. $\endgroup$ – Mr. G Dec 28 '12 at 21:16

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