Taylor series of $\sqrt{\ln\left(\frac{1}{x}\right)}$

I am trying to compute the Taylor series of:

$\sqrt{\ln\left(\frac{1}{x}\right)}$

I have computed the derivatives and evaluated them in $x=1/e$ but I cannot find the formula for the sequence of the coefficients.

Any other ideas?

Thanks!

PS: The approximation should be valid over the interval $x \in [0,1]$

Solution 1:

This is a note on how you might compute the derivatives using implicit differentiation. It isn't really an answer, but was too long for a comment, and someone may see how to use this in an answer.

$$y=\left(-\ln(x)\right)^{\frac 12}$$

$$y'=\frac 12\left(-\ln(x)\right)^{-\frac 12}\cdot-\frac 1x=-\frac 1{2xy}$$

Whence $$2xyy'=-1$$ and $$2yy'+2xy'^2+2xyy''=0$$ From which you can cancel a factor two and keep going to compute higher derivatives.

Using $2xyy'=-1$ you can multiply through by $y'$ and obtain:

$$2yy'^2+2xy'^3-y''=0$$ or $$y''=2xy'^3+2yy'^2$$ Which you can differentiate as often as you like.

Solution 2:

Let's try to find the expansion about $x=1/2$. Also, note that $\ln(1/x) = -\ln(x)$. $$ f(x) = \sqrt{-\ln(x)};\quad f(1/2) = \sqrt{\ln(2)}\\ f'(x)= \frac{1}{2}(-\ln(x))^{-1/2}\cdot(-1/x); \quad f'(1/2) = -1/(\sqrt{\ln(2)}) $$ Whatever this Taylor Series is, it's only going to get worse from here. To get something workable, I think you'll have to settle for something that converges only over a part of $(0,1)$.