How can I prove this equality and general derivative?

Question

$$ f_{n}(x) = \underbrace{\sqrt{x+\sqrt{x+ ...+ \sqrt{x}}}}_{n} \quad \quad \lim_{n\to\infty} \sum^{n}_{r=1}\frac{1}{2^{r}}\prod^{r-1}_{i=0}\frac{1}{f_{n-i}(x)} = \frac{1}{2f_{\infty}(x)-1} $$

I produced the left hand side by examining the pattern of the derivatives of $\, f_{n}(x)$ and the right hand side by using the property of $\, f_{\infty}(x)$ that: $\, f_{\infty}(x)=\sqrt{x + f_{\infty}(x)}$ and differentiating. I have no idea how to prove such a thing, nor do I know how to prove the general formula of the derivate (EDIT: I have now proved the general formula of the derivative but am still stumped by the limit problem!):

$$_{m}f_{n}(x) = \underbrace{\sqrt[m]{x+\sqrt[m]{x+ ... + \sqrt[m]{x}}}}_{n} \quad \quad _{m}f_{n}'(x) = \sum^{n}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (f_{n-i}(x))^{1-m}$$

Obviously given my second formula, I could generalise the first equality but, I thought the case $m=2$ came out quite nicely.

One last thing, is there any nicer notation I can use for the function rather than writing it out with the underbrace and whatnot?

Edit: Just realised I didn't mention it at all in the original post, the right-hand side is the infinite sum of a geometric progression:

$$\sum^{\infty}_{r=1}\bigg(\frac{1}{2f_{\infty}(x)}\bigg)^{r} = \frac{1}{2f_{\infty}(x)-1} $$

So the question can be rephrased as proving:

$$\lim_{n\to\infty} \sum^{n}_{r=1}\frac{1}{2^{r}}\prod^{r-1}_{i=0}\frac{1}{f_{n-i}(x)} = \sum^{\infty}_{r=1}\bigg(\frac{1}{2f_{\infty}(x)}\bigg)^{r}$$

Which seems intuitively true but, I do not know how to prove it rigorously.

Answer 1

It took me a while to see how you managed to find that formula, but indeed when you define

$$ _mf_n(x) = (x + {_mf_{n-1}}(x))^{1/m} $$

with $f_{n} = 0$ for $n \leq 0$, then it follows that

$$ _mf'_n(x) = \frac1m(x + {_mf'_{n-1}}(x))^{1/m - 1} (1 + {_mf'_{n-1}}(x)) = \frac1m(_mf_n(x))^{1 - m} (1 + {_mf'_{n-1}}(x)) $$

and through induction you can then show that

$$ _{m}f_{n}'(x) = \sum^{n}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}. $$

actually if we let $(_mf_{n}(x))^{1-m} = 0$ for $n \leq 0$ then we can make our lives slightly easier by writing this as

$$ _{m}f_{n}'(x) = \sum^{\infty}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}. $$

If we now let $_mf_{\infty}(x) = \lim_{n \to \infty} {_mf_n}(x)$ then we can show that for any $r$

$$ \lim_{n\to\infty} \frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m} = \left(\frac{(_mf_{\infty}(x))^{1-m}}{m} \right)^{r} $$

as long as $_mf_{\infty}(x)$ exists.

We can also show that $_mf_{n}(x)$ is monotonically increasing in $n$. Note that for all $x > 0$ we have ${_mf_{1}}(x) = x^{1/m} > 0 = {_mf_{0}}(x)$ and if $$ {_mf_{n}}(x) \geq {_mf_{n-1}}(x) $$ then $$ {_mf_{n+1}}(x) = (x + {_mf_{n}}(x))^{1/m} \geq (x + {_mf_{n-1}}(x))^{1/m} = {_mf_{n}}(x) $$ so ${_mf_{n}}(x)$ is always monotonically increasing for $x>0$. This allows us to use the monotone convergence theorem to conclude that for all $x>0$

$$ \begin{align} \lim_{n\to\infty} \sum^{\infty}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m} &= \sum^{\infty}_{r=1} \lim_{n\to\infty} \frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}\\ &= \sum^{\infty}_{r=1} \left(\frac{(_mf_{\infty}(x))^{1-m}}{m} \right)^{r}\\ &= \frac{1}{1 - \frac{(_mf_{\infty}(x))^{1-m}}{m}} - 1\\ &= \frac{(_mf_{\infty}(x))^{1-m}}{(_mf_{\infty}(x))^{1-m} - m} \end{align} $$

setting $m=2$ results in the special case you found.

You could also try to prove this by showing

$$ \lim_{n\to\infty} \lim_{h\to0} \frac{_{m}f_{n}(x+h) - {_{m}f_{n}(x)}}{h} = \lim_{h\to0} \lim_{n\to\infty} \frac{_{m}f_{n}(x+h) - {_{m}f_{n}(x)}}{h} $$

but showing that you're allowed to exchange the two limits is a bit involved, especially with no idea what the function looks like. So using the expression you found for $_{m}f_{n}'(x)$ might well be the easiest approach.