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This difference equation $$ f(x)^2 = 1 + x f(x + 1) $$ can pop up when looking at a famous problem posted by Ramanujan.

This equation is equivalent to the following infinitely nested radical expression $$ f(x) = \sqrt{1 + x \sqrt{1 + (x+1) \sqrt{1 + (x+2)\sqrt{\cdots}}}} $$ assuming that we take the positive branch of all of the radicals.

Ok, what I would like to prove or disprove is whether the only analytic (or a more relaxed criterion) solution is $f(x) = 1+x$. You can easily enough show that if $f$ is differentiable, then $f(N) = 1+N$ for all integers $N\in\mathbb{Z}$.

You can also show that for all integers $$ f'(N) = a N 2^N + 1 $$ for some $a\in\mathbb{R}$ by solving the recurrence relation that appears after taking the derivative of the above difference equation. It is clear that $f(x) = 1+x$ is a solution, but I am trying to figure out if it is the only solution.

Bill Dubuque
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Todd Sierens
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    What work have you yourself undertaken. Please clarify more specific questions as to what you don't understand, or where you are stuck. This is not a "solve these questions for me" sort of site. Please add sufficient context. – amWhy Feb 09 '23 at 21:23
  • Check this posting and the answers therein. – Sangchul Lee Feb 09 '23 at 21:31
  • It doesn't seem so easy to show $f(N)=1+N$ for integers. – aschepler Feb 09 '23 at 22:51
  • @amWhy I feel like I have included a fair amount of context. Essentially while looking at various sources on stackexchange and otherwise a few times I came across the claim that says you can show that $f=1+x$ if $f$ is analytic, but I do not see how. I showed that I can prove that it is true for integers, and I also derived a necessary expression for the first derivative of the solution evaluated at integers, but that is where my leads run dry, hence why I created a post. – Todd Sierens Feb 10 '23 at 14:40
  • @aschepler

    By setting $x = 0$, and $x=-1$ you find that $f(0) = 1$ and $f(-1) = 0$. By taking the implicit derivative and looking at $x=0$ and $x=-1$ again, you find that $f'(0) = 1$ and thus $f(1)=2$ the rest follow by induction.

     – Todd Sierens Feb 10 '23 at 14:42
  • I think that I might have a path forward here. You can show that if $f$ is polynomial, then it must be $f(x)=1+x$, and I wonder if something like Weierstrass approximation theorem can take us the rest of the way through a proof by contradiction. – Todd Sierens Feb 10 '23 at 16:48
  • I don't find that $f(0)=1$. $f(0) \in {1,-1}$. – aschepler Feb 13 '23 at 18:25

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let $g : (0,1] \to \Bbb R$ be some arbitrary function. Define $$f(x) = \begin{cases}1&x = 0\\g(x)&x\in(0,1]\\\dfrac{f^2(x-1) - 1}{x-1} & x\in(1,\infty)\end{cases}$$ Where the last is applied inductively. If desired, the recursion can also be adapted to define $f$ on $(-\infty, 0)$. Because of the recursion, the relation $f^2(x) = 1+xf(x+1)$ holds everywhere for $x$.

Thus you see, there are uncountably many functions satisfying the equation. Even if we require $f$ to be infinitely differentiable, this will still be true. We just need $g$ to be infinitely differentiable and specify a relation between its behavior approaching $0$ and approaching $1$.

It is only analycity - specifically about $x = 1$ - that is a strong enough requirement to actually tame this beast.

The nested square roots expression either converges to a specific function satisfying the recursion formula, or else it diverges. It is not fully equivalent to the recursion formula.

Paul Sinclair
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  • This is interesting. Can you come up with an infinitely differentiable example (specifically near $x=1$ seems very tricky) – Todd Sierens Feb 10 '23 at 23:51
  • I'm thinking that $g(x) = x+1+\exp{\left(-\tfrac{1}{x^2(x-1)^2}\right)}$ might be able to do it. I'm not able to play with it now, but it's hard to see why it wouldn't work as a counter example. – Todd Sierens Feb 11 '23 at 00:39
  • You already know one infinitely differentiable example: $g(x) = 1 + x$ leads to $f(x) = 1 + x$. More generally, my claims about smooth functions are indeed based on the bump function. As long as $g$ behaves like $1+x$ at both ends, you can bump it to do whatever you want in the middle. – Paul Sinclair Feb 11 '23 at 00:46