# How to show the approximation of $1-T_{1 / L}(1 / \delta)^{-2}$ when $L$ is large and $\delta$ is small?

I met this approximation problem in eq(2) of this paper, stating that the approximation is $$\left(\frac{\log (2 / \delta)}{L}\right)^{2}$$.

As stated in the question, $$T_L(x)$$ is the $$L_{th}$$ Chebyshev polynomial of the first kind and can be formulated as $$T_{n}(x)= \begin{cases}\cos (n \arccos x) & \text { if }|x| \leq 1 \\ \cosh (n \operatorname{arcosh} x) & \text { if } x \geq 1 \\ (-1)^{n} \cosh (n \operatorname{arcosh}(-x)) & \text { if } x \leq-1\end{cases}$$ When $$\delta$$ is small, I think we should choose the middle formula above for $$T_L$$, and then $$1-T_{1 / L}(1 / \delta)^{-2}=1-\frac{1}{\cosh \left[ \frac{1}{L}\mathrm{arcosh} \left[ \frac{1}{\delta} \right] \right] ^2}$$. But then I don't know how to approximate this formula into $$\left(\frac{\log (2 / \delta)}{L}\right)^{2}$$ especially confused with how can I get $$\log$$ in the approximation. Do you have any idea?

• The inverse cosh of a large positive number x is basically the same as $\log(2x)$, in actuality it will be a little bit smaller. That seems like a start.
– Ian
Commented Dec 4, 2021 at 9:41

My answer is based on the help of one comment of @lan. With some fixed small $$\delta$$, we have $$\mathrm{arcosh} \left( \frac{1}{\delta} \right) \approx \log \left( \frac{2}{\delta} \right)$$. And since we mainly consider how things scale with large $$L$$(computation complexity mainly consider so). So we have $$\frac{1}{L}\mathrm{log} \left( \frac{1}{\delta} \right)$$ is small number when $$L$$ is large, so we have taylor expansion of $$\cosh(x)=1+x^2/2+O(x^4)$$, and hence $$1-\frac{1}{\cosh \left( \frac{1}{L}\mathrm{arc}\cosh \left( \frac{1}{\delta} \right) \right) ^2}\approx 1-\frac{1}{\left( 1+\frac{\frac{1}{L^2}\log \left( \frac{2}{\delta} \right) ^2}{2}+O\left( \frac{1}{L^4}\log \left( \frac{2}{\delta} \right) ^4 \right) \right) ^2}\approx 1-\frac{1}{1+\frac{1}{L^2}\log \left( \frac{2}{\delta} \right) ^2}=\frac{\log ^2\left( \frac{2}{\delta} \right)}{L^2+\log ^2\left( \frac{2}{\delta} \right)}\approx \frac{\log ^2\left( \frac{2}{\delta} \right)}{L^2}$$