Expanding $\ln(1+f(x))$ around $f(x)=0$ when we do not know whether there is an $x$ such that $f(x)=0$.

Question

I want to expand $\ln(1+f_T(x,\theta))$ about $1+f_T(x,\theta)=1$. What I have in mind is something like

$$ \ln(1+f_T(x,\theta))=\ln(1)+f_T(x,\theta)-\frac{1}{2} \frac{1}{1+\tilde{f}} f_T(x,\theta)^2 $$

where $\tilde{f}$ is the mean value between $0$ and $f_T(x,\theta)$. If it is going to be of any help, $f_T(x,\theta)=o_p(1)$ as $T\to\infty$. My question is: to do that expansion, do I need to ensure that there is indeed some $x^*$ and/or $\theta^*$ such that $f(x^*,\theta^*)=0$?

Many thanks in advance!!

Answer 1

You are just plugging in $f_T(x, \theta)$ for $x$ in the Taylor series for $\ln(1 + x)$:

$$ \ln(1 + x) = \int \frac{dx}{x + 1} = \int \sum_0^\infty (-1)^ix^i = \sum_0^\infty (-1)^i \frac{x^{i + 1}}{i + 1} = \sum_1^\infty (-1)^{i-1}\frac{x^i}{i} $$

So now you can just plug in $f_T(x, \theta)$:

$$ \ln(1 + f(x, \theta)) = \sum_1^\infty \frac{(-1)^{i - 1}f_T^i(x, \theta)}{i} $$

Answer 2

Apparently, what you are looking for is $$ \ln\big(1+f_T(x,\vartheta)\big)=f_T(x,\vartheta)-\frac{f_T^2(x,\vartheta)}{2}+\frac{f_T^3(x,\vartheta)}{3}+\cdots. $$