Tricks for estimating $ \lim_{x\rightarrow 0} \frac{d}{dx} \bigl(-\frac{1}{x} \ln\bigl(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \bigr) \bigr)$

Question

I'm trying to find a Taylor approximation of $ f(x) =-\frac{1}{x} \ln\left(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \right) $ at $x = 0$.

For the derivation part Wolfram returns a quite a tedious expression to the problem, taking the limit of which does result in the answer I'm looking for.

As I found this problem in a textbook which is not focused on honing derivation/approximation skills, I'm curious, whether there are some known tricks that would help easying up the calculation, or is this just a technical exercise?

Apologies if this is a dumb question.

Answer 1

$\begin{array}\\ f(x) &=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\ &=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+O(x^2)) (-bx+O(x^2))}{-x+O(x^2)} \right)\\ &=-\frac{1}{x} \ln\left(1 + \dfrac{abx^2+O(x^3)}{-x+O(x^2)} \right)\\ &=-\frac{1}{x} \ln\left(1 - abx+O(x^2) \right)\\ &=\frac{1}{x}(abx+O(x^2))\\ &=ab+O(x)\\ \end{array} $

Take more terms (e.g., $e^{-x} = 1-x+x^2/2+O(x^3) $) to get more precision.

Here is a try to get one more term.

$\begin{array}\\ f(x) &=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\ &=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+a^2x^2+O(x^3)) (-bx+b^2x^2/2+O(x^3))}{-x+x^2/2+O(x^3)} \right)\\ &=-\frac{1}{x} \ln\left(1 + x^2\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-x+x^2/2+O(x^3)} \right)\\ &=-\frac{1}{x} \ln\left(1 + x\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-1+x/2+O(x^2)} \right)\\ &=-\frac{1}{x} \ln\left(1 - x\dfrac{ab-(a^2b+ab^2)x+O(x^2)}{1-x/2+O(x^2)} \right)\\ &=-\frac{1}{x} \ln\left(1 - x(ab-(a^2b+ab^2)x+O(x^2))(1+x/2+O(x^2)) \right)\\ &=-\frac{1}{x} \ln\left(1 - abx(1-(a+b)x+O(x^2))(1+x/2+O(x^2)) \right)\\ &=-\frac{1}{x} \ln\left(1 - abx(1-(a+b+1/2)x+O(x^2)) \right)\\ &=-\frac{1}{x} \left( - abx(1-(a+b+1/2)x+O(x^2)) +((a+b+1/2)x+O(x^2))^2/2\right)\\ &=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)+O(x))^2/2\right)\\ &=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)^2+O(x))/2\right)\\ &=-\frac{1}{x} \left( -x\left(ab+ab(a+b+1/2)x+O(x^2) +x((a+b+1/2)^2+O(x))/2\right)\right)\\ &=-\frac{1}{x} \left( -x\left( ab +x(ab(a+b+1/2)+(a+b+1/2)^2\right)/2+O(x^2)\right)\\ &= ab +x(ab(a+b+1/2)+(a+b+1/2)^2)/2+O(x^2)\\ &= ab +x(ab(a+b+1/2))(1+(a+b+1/2))/2+O(x^2)\\ &= ab +xab(a+b+1/2)(a+b+3/2)/2+O(x^2)\\ \end{array} $

You better check my algebra, because Prob(error) > 1/e.

Answer 2

Well, considering the Taylor expansion of $$ln(1+x)=x-\frac{x^2}{2}+o(x^2)\,\,\,\,\,\text{ for }x\to0$$ since $$\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\to 0\,\,\,\,\,\text{for }x\to 0$$ (it can be seen by taylor expand the exponential up to the first order) then you get: $$f(x)=-\frac{1}{x}\left[\left(\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\right)-\frac{1}{2}\left(\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\right)^2+...\right]$$ Then you taylor expand the exponential: $$e^x=1+x+o(x)\,\,\,\text{ for }x\to 0$$ and get: $$f(x)=-\frac{1}{x}\left[\left(\frac{(-ux)(-vx)}{-x}\right)-\frac{1}{2}\left(\frac{(-ux)(-vx}{-x}\right)^2+...\right]=-\frac{1}{x}\left[-uvx-\frac{1}{2}u^2v^2x^2+...\right]$$ You can clearly expand the exponential up to second order to get more terms.