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I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?

So far I've got only trivial values, which are $$f(1)=\sqrt{1+\sqrt{1+\cdots\sqrt{1}}}=\frac{\sqrt{5}+1}{2}$$ $$f(4) = 3$$

Second one follows from $$2^n+1 = \sqrt{4^n+(2^{n+1}+1)} = \sqrt{4^n+\sqrt{4^{n+1}+(2^{n+2}+1)}} = \sqrt{4^n+\sqrt{4^{n+1}+\cdots}}$$

I have managed to compute several derivatives in $x_0=1$ by using chain rule recursively on $f_n(x) = \sqrt{x^n + f_{n+1}(x)}$, namely:

\begin{align*} f^{(1)}(1) &= \frac{\sqrt{5}+1}{5}\\ f^{(2)}(1) &= -\frac{2\sqrt{5}}{25}\\ f^{(3)}(1) &= \frac{6\sqrt{5}-150}{625}\\ f^{(4)}(1) &= \frac{1464\sqrt{5}+5376}{3125}\\ \end{align*}

These gave me Taylor expansion around $x_0=1$

\begin{align*} T_4(x) &= \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}+1}{5} (x-1) - \frac{\sqrt{5}}{25} (x-1)^2 + \frac{6\sqrt{5}-150}{3750} (x-1)^3 \\ &\ \ \ \ \ + \frac{61\sqrt{5}+224}{3125} (x-1)^4 \end{align*}

However this approach seems to be useful only very closely to the $x=1$. I am looking for something more general in terms of any $x$, but with my limited arsenal I could not get much further than this. Any ideas?

This was inspiring but kind of stopped where I did http://integralsandseries.prophpbb.com/topic168.html

Edit: Thanks for the answers, i will need to go through them, looks like the main idea is to divide by $\sqrt{2x}$, so then I am getting $$\frac{\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}}{\sqrt{2x}} = \sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{16x}+\sqrt{\frac{1}{256x^4}+.‌​..}}}}$$ Then to make expansion from this. This is where I am not yet following how to get from this to final expansion.

Sil
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3 Answers3

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You can obtain an expansion in negative powers of $x$, valid for large $x$, by trying to find $$\frac{f(x)}{\sqrt{2x}}$$. If you pull the $\frac{1}{\sqrt{2x}}$ into the square roots (note that it gets raised to progressively higher powers as it is pulled into more inner square roots) you get an expansion of the form $$f(x) = \sqrt{2x} + \frac{\sqrt{2}}{8} + \frac{\kappa}{\sqrt{x}} + \cdots$$ where calculating $\kappa$ is straightforward but a bit messy (the value I get is $-\frac{3\sqrt{2}}{32}$ but numerically it looks to be about $-\frac{1}{18}$). Even the first order expression (without the $\frac{\kappa}{\sqrt{x}}$ term) is already accurate to an error of smaller than $0.0052$ for all $x>4$.

Mark Fischler
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In the same spirit as Mark Fischler's answer, setting $x=\frac{y^2}2$, for large values of $x$, Taylor expansion is $$y+\frac{1}{4 \sqrt{2}}-\frac{5}{64 }\frac 1 {y}+\frac{85}{256 \sqrt{2} }\frac 1 {y^2}-\frac{1709}{8192 }\frac 1 {y^3}+\frac{6399}{32768 \sqrt{2} }\frac 1 {y^4}+O\left(\frac{1}{y^5}\right)$$ which would converge quite fast.

Claude Leibovici
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For small $x$ and $n>4$,

$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots+\sqrt{x^{n}}}}}= x^{n/2^{n}}+\frac{1}{2}x^{1-n/2^{n}}+\frac{1}{8}x^{2-n/2^{n}}+\ldots$$

Thus $$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}} \approx 1+\frac{x}{2}+\frac{x^{2}}{8}$$

For large $x$,

$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}}= \sqrt{2x}+\frac{1}{4\sqrt{2}}-\frac{5}{64\sqrt{2x}}+\frac{85}{512x\sqrt{2}}-\ldots$$

Ng Chung Tak
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  • Chug Tak sir, How to derive these expressions? – Venkat Amith Mar 10 '21 at 05:29
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    For small $x$, and observe the pattern of the Taylor series for different values of $n$. For large $x$, just find the Taylor series about infinity. These can be easily obtained by using Mathematica or other CAS. – Ng Chung Tak Mar 11 '21 at 05:30