5
$\begingroup$

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 $x=1$.

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.