I would like to show the absolute convergence of the coefficients of any series expansion of the function $f(x)=\sqrt{-\log(x)}$ in the form $f(x_0+x)= h(g(x_0+x) = \sum_l c_l x^l$
More formally, let $x_0 \in [-1, 1]$, $r \in (0,2]$, $\delta \in (0,r]$. Defined $f : [x_0 -r - \delta, x_0 +r+\delta] \mapsto \mathbb{R}$ such that for all $x \in [-r-\delta, r+\delta]$, we have that $$f(x_0 + x) = \sum_{k=0}^\infty \alpha_k x^k $$
In our case we can pick
- $x_0 =1$, r=$1-\beta$, and $\delta=\beta/2$, for a small $\beta >0$, or
- $x_0 = 1/2$, $r=1/2-\beta$, and$\delta = \beta/2$
Find $B$ such that $\sum_{k=0}^\infty (r+\delta)^k|\alpha_k| \leq B$
As it is easier for me, I focus on the Taylor expansion. Apparently, it's hard even for wolfram alpha to find a series expansion of $f(x_0 + x)$, so we have to find more ingenious ways. After introducing some preliminary results, I propose four possible directions and their drawbacks.
Let's state the following useful expansions: $$\sqrt{x} = h_{x_0=1/2}(x) = \sum_n^\infty 2^{-1/2+n}\left(-\frac{1}{2} + x\right)^n{1/2 \choose n} $$
$$log(1/x) = g_{x_0=1/2}(x) = \sum_{n> 1} \frac{(-2)^n (-1/2 +x)^n}{n} + \log(2)$$ $$log(1/x) = g_{x_0=1/2}(x_0+x) = \sum_{n> 1} \frac{(-2)^n (x)^n}{n} + \log(2)$$
If we center the polynomial expansion of $\sqrt{x}$ in $-log(1/2)=g(x_0)$, than the expansion becomes: $$ \sqrt{x} = h_{x_0=\log(2)))}(x) = \sum_n {1/2 \choose n}(x-\log(2))^n \log^{1/2-n}(2) $$
Also recall that: $$ {1/2 \choose l} = \frac{(-1)^{l-1}}{l2^{2l-1}}\frac{(2l-2)!}{((l-1)!)^2} $$ and that $\log(1/x) = -\log(x)$
Attempt 1:
We start by defining, as usual, $$ \sqrt{x} = h(x) = \sum_k a_k (x_0-x)^k$$ and analogously, but shifting, $$log(x_0 +x) = g(x_0 +x) = \sum_k a_k x^k$$
So the composition of the two should be:
$$h(g(x_0+x)) = \sum_l a_k \left(\sum_k b_k (x)^k \right)^l \tag{1}\label{dioporc}$$
that is:
$$f(x_0+x)=h(g(x_0+x))=\sum_n {1/2 \choose n}(g(x_0+x)-\log(2))^n \log^{1/2-n}(2)= \sum_n {1/2 \choose n}(g(x_0+x)-\log(2))^n \log^{1/2-n}(2)=$$ $$ \sum_n {1/2 \choose n}\left(\sum_{n> 1} \frac{(-2)^n (x)^n}{n} + \log(2) -\log(2)\right)^n \log^{1/2-n}(2) $$
$$ =\sum_n {1/2 \choose n}\left(\sum_{n> 1} \frac{(-2)^n (x)^n}{n} \right)^n \log^{1/2-n}(2) $$ Attempt 2:
We start by defining: $$ h(x)= \sum_k a_k (x - g(x_0))^k $$ $$ g(x) = \sum_l b_l (x-x_0)^l $$
$$h(g(x)) = \sum_k a_k \left[\sum_l b_l(x-x_0)^l - g(x_0)\right]^k \tag{2}\label{diocane}$$
This brings us to the following expansion: $$\sqrt{-\log(x)} = \sum_n \frac{(-1)^n (2n)!}{(1-2n)(n!)^{2}4^n}\left[ \frac{1}{ln (2)} \sum_l \frac{(-2)^l(x-1/2)^l}{l} \right]^n \label{sbajata}$$
The expansion for $\sqrt{-\log(x_0+x)}$ follows by substituting $( x-(1/2) ) \mapsto x$ in the previous equation.
Attempt 3:
Obtaining the Taylor expansion of $f(x)$ without using the composition. For this, we need to find some derivatives of $f(x0)$ and evaluate them at the $x_0$ of choice.
- $\frac{d}{dx} = -\frac{1}{2x(-log(x))^{1/2}} $
- $\frac{d^2}{dx^2}= -\frac{2log(x)+1}{4x^2(-log(x))^{3/2}} $
- $\frac{d^3}{dx^3}= \frac{-8\log^2(x) - 6\log(x)-3}{8x^3(-log(x))^{5/2}} $
- $\frac{d^4}{dx^4}= -\frac{48\log^3(x)+44\log^2(x)+36\log(x)+15}{16x^4(-\log(x))^7/2}$
and so on. From these, it's hard for me do devise a closed formula of the form $\sum_l c_l(x_0-x)^l$. But maybe this is the only feasible direction.
Attempt 4
Use the polynomial approximation of the square root of an infinite series. We can use this tool because the Taylor expansion of $\sqrt{x}$ in $1/2$ is just an infinite series. More about this here (still working on this).
Question:
I don't think that the approach 1 and 2 are correct, as the Taylor expansion of $\sqrt{x}$ is only valid approximation of $\sqrt{x}$ in the interval $(0,1)$, which does not contain che codomain of $\log(1/x)$.
What is the way to have a good approximation of the function? I have the feeling that the expansions in Eq. \ref{diocane} or \ref{dioporc} are not correct.