Find a function which satisfies the relation $f(x+1)=(f(x)+1)^{1/2}$

Question

The original question was to find the value this function approaches as $x$ goes to infinity given that the limit exists. This is easy to figure and turns out to be $(1+5^{1/2})/2$.

I was though, thinking, if it were possible to come up with such a continuous function over non-negative numbers first and then find the limit. I did try a bit by assuming the function is continuous and taking the derivative and somehow managing a solvable differential equation but didn't reach anywhere. It'd be amazing to see if such a continuous function exists or not!!

$f : [0,\infty) \mapsto [0,\infty)$

$f(x+1)=(1+f(x))^{1/2}$

given $f(0)=0$

I guess that the function which you given is a sequence ${a_n}$ satisfying $a_{n+1}=\sqrt{a_n+1}$. Take the limit on both sides. We get $\lim_{n\to \infty}a_{n+1}=\sqrt{\lim_{n\to \infty}a_n+1}$. Let $S=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}a_n$. Then $S=\sqrt{S+1}$ and $S^2=S+1$ and $S=\frac{1\pm \sqrt{5}}{2}$. — bfhaha, Jul 01 '17 at 18:01
What is the domain of $f$ supposed to be? $\Bbb N$? $[0,\infty)$? $\Bbb R$? — Hagen von Eitzen, Jul 01 '17 at 18:09

score 1 · Answer 1 · answered Jul 01 '17 at 20:33

Assume $f$ is such a function. Then $f(x)\ge -1$ for all $x$ and $f(1)=1$.

Conversely, let $f_0\colon [0,1]\to[-1,\infty)$ be any continuous function with $f_0(0)=0$, $f_0(1)=1$. Then we can define recursively functions $f_n\colon [n,n+1]\to [-1,\infty)$ by letting $f_n(x)=\sqrt{f_{n-1}(x-1)+1}$. Then $f_n$ is continuous and we have $f_n(n+1)=\sqrt{f_n(n)+1}$. Therefore the function $f\colon [0,\infty)\to[-1,\infty)$ given by $f(x)=f_{\lfloor x\rfloor}(x)$ is continuous, has $f(0)=0$ and $f(x+1)=\sqrt{f(x)+1}$ for all $x$.

As $f_0$ is continuous, it is bounded, say $m_0\le f_0(x)\le M_0$ for all $x\in[0,1]$. And if $m_n\le f(x)\le M_n$ for $x\in[n,n+1]$, then $m_{n+1}:=\sqrt{m_n+1}\le f(x)\le \sqrt{M_n+1}=:M_{n+1}$ for $x\in[n+1,n+2]$. With $\phi:=\frac{\sqrt 5+1}2$, we may assume wlog that $m_0<\phi<M_0$. For $0\le x<y$, we have $$0<\sqrt{y+1}-\sqrt{x+1}=\frac{y-x}{\sqrt{y+1}+\sqrt{x+1}}<\frac{y-x}2.$$ Therefore (noting that $m_1\ge 0$), we have $m_n<\phi<M_n$ for all $n$ and $M_n-m_n\to 0$. We conclude that $\lim_{x\to\infty}f(x)=\phi$.

Find a function which satisfies the relation $f(x+1)=(f(x)+1)^{1/2}$

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