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

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 $$x=1$$.

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 $$A_5 = -{\frac{353}{15625}}-{\frac {1138\,\sqrt {5}}{78125}}$$.