Is where a 'best' way to make a nested radical expansion for an analytic function? This way seems convenient:
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots=\sqrt{a_0^2+2a_0a_1x+(a_1^2+2a_0a_2)x^2+\cdots}=$$
$$=\sqrt{a_0^2+2a_0a_1x+(a_1^2+2a_0a_2)x^2 \sqrt{1+\cdots}}$$
I conjecture that this infinitely nested radical expansion has the same interval of convergence as the original Taylor series. Is this correct?
Also, the number of 'roots' we take into account gives us twice the number plus one of the correct terms in the Taylor series.
For example:
$$e^x=\sqrt{1+2x+2x^2\sqrt{1+\frac{4}{3}x+\frac{10}{9}x^2\sqrt{1+\frac{32}{25}x+\frac{681}{625}x^2\sqrt{1+\dots}}}}=$$
$$=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\dots$$
This expression converges for any $x$.
$$\frac{1}{1-x}=\sqrt{1+2x+3x^2\sqrt{1+\frac{8}{3}x+\frac{46}{9}x^2\sqrt{1+\frac{76}{23}x+\frac{4089}{529}x^2\sqrt{1+\dots}}}}=$$
$$=1+x+x^2+x^3+x^4+x^5+x^6+\dots$$
This expression converges for $|x|<1$.
This idea may seem pointless, since the coefficients in the radical expansion are very hard to calculate even for functions with 'simple' Taylor series.
But is it possible, that some function with 'complicated' Taylor series will have simple nested radical expansion of this kind?
For example, consider the easiest nested radical of this kind:
$$f(x)=\sqrt{1+x+x^2\sqrt{1+x+x^2\sqrt{1+x+x^2\sqrt{1+\dots}}}}=$$
$$=1+\frac{x}{2}+\frac{3x^2}{8}+\frac{x^3}{16}+\frac{11x^4}{128}-\frac{9x^5}{256}+\frac{27x^6}{1024}+\dots+$$
Though this kind of functions we can always find in closed form, if we assume the infinite nested radical converges.
$$f(x)=\sqrt{1+a_1x+a_2x^2\sqrt{1+a_1x+a_2x^2\sqrt{1+a_1x+a_2x^2\sqrt{1+\dots}}}}=$$
$$=\frac{a_2}{2}x^2+\sqrt{1+a_1x+\frac{a_2^2}{4}x^4}$$
I've never seen this topic discussed anywhere, so a reference would be nice. The only thing I've seen is Ramanujan nested radical, and it's usually presented as a funny trick, nothing more.
