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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?

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    $\begingroup$ 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. $\endgroup$
    – Ian
    Commented Dec 4, 2021 at 9:41

1 Answer 1

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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 $$1-\frac{1}{\cosh \left( \frac{1}{L}\mathrm{arc}\cosh \left( \frac{1}{\delta} \right) \right) ^2}\approx 1-\frac{1}{\left( 1+\frac{\frac{1}{L^2}\log \left( \frac{2}{\delta} \right) ^2}{2}+O\left( \frac{1}{L^4}\log \left( \frac{2}{\delta} \right) ^4 \right) \right) ^2}\approx 1-\frac{1}{1+\frac{1}{L^2}\log \left( \frac{2}{\delta} \right) ^2}=\frac{\log ^2\left( \frac{2}{\delta} \right)}{L^2+\log ^2\left( \frac{2}{\delta} \right)}\approx \frac{\log ^2\left( \frac{2}{\delta} \right)}{L^2}$$

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  • $\begingroup$ I think you lost track of the -2 power in the original expression. $\endgroup$
    – Ian
    Commented Dec 4, 2021 at 15:19
  • $\begingroup$ @lan Thanks! I fixed the typo and make the progress more transparent. $\endgroup$
    – narip
    Commented Dec 5, 2021 at 0:28

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