Taylor series for $\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}}$

Question

Let $$f(x) =\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}} = A_0 + A_1(x-1)+A_2(x-1)^2+\cdots$$ I got the following:

• $A_0 = (1+\sqrt{5})/2$
• $A_1 = 1/5$
• $A_2=-1/25$
• $A_3 = -1/168$

The values seem correct for $A_0, A_1, A_2$ but only a good approximation for $A_3$. Is it possible to find a simple iterative formula that gives all the coefficient $A_k$'s? What is the exact value for $A_3, A_4$ and $A_5$?

Note

$\lim_{x\rightarrow 0^+} f(x) = \sqrt{2}$. That limit is not equal to $1$, despite what it looks like at first glance. A Taylor approximation around $x=0$ seems much more challenging (if at all possible) than around $x=1$.

Answer 1

Let $f_n(z) = \sqrt{z^n + \sqrt{z^{n+1} + \sqrt{\ldots}}}$. Thus $f_n(z)^2 = z^n + f_{n+1}(z)$. Since we're expanding around $z=1$, it may help to write $z = 1+t$. Letting $f_n(z) = a_0(n) + a_1(n) t + a_2(n) t^2 + \ldots$ and working formally, we have

$$ (a_0(n) + a_1(n) t + a_2 (n) t^2 + \ldots)^2 = (1+t)^n + a_0(n+1)+a_1(n+1)t + a_2(n+1) t^2 + \ldots $$

Equating coefficients of each power of $t$:

$$ \eqalign{a_0(n)^2 &= 1 + a_0(n+1)\cr 2 a_0(n) a_1(n) &= n + a_1(n+1)\cr 2 a_0(n) a_2(n) + a_1(n)^2 &= \frac{n^2-n}{2} + a_2(n+1)\cr 2 a_0(n) a_3(n) + 2 a_1(n) a_2(n) &= \frac{n^3}{6} - \frac{n^2}{2}+\frac{n}{3} + a_3(n+1)\cr etc} $$ The first equation is consistent with $a_0(n)$ all being equal to a root of $z^2 = z+1$, presumably $(1+\sqrt{5})/2$.
Assuming that is the case, the second equation is consistent with $$a_1(n) = \dfrac{n}{\sqrt{5}} + \dfrac{1}{5}$$ Assuming that is the case, the third equation is consistent with $$ a_2(n) = \dfrac{3\sqrt{5}}{50} n^2 + \left(\frac{1}{25}-\frac{\sqrt{5}}{10}\right) n - \frac{1}{25} $$ Assuming that is the case, the fourth equation is consistent with $$ a_3(n) = {\frac {7\,\sqrt {5}{n}^{3}}{750}}+ \left( -{\frac{3}{250}}-{\frac {3 \,\sqrt {5}}{50}} \right) {n}^{2}+ \left( -{\frac{9}{250}}+{\frac {22 \,\sqrt {5}}{375}} \right) n+{\frac{1}{125}}-{\frac {4\,\sqrt {5}}{625 }} $$ So I think your $A_0$ to $A_2$ are correct, and $A_3$ should be ${\frac{1}{125}}-{\frac {4\,\sqrt {5}}{625 }} $. At the next steps, I get $A_4 = {\frac{34}{3125}}+{\frac {7\,\sqrt {5}}{625}}$ and $A_5 = -{\frac{353}{15625}}-{\frac {1138\,\sqrt {5}}{78125}}$.