$\begingroup$ @Lucian I always took for granted that the "approximately" qualifier acted idempotently, so while we can distinguish between 'exact equality' and 'approximate equality', 'exact approximate equality' is the same thing as 'approximate approximate equality'. A world in which this is not true makes me want to stress vomit.

2015年12月17日 · After trying to derive the explicit formula, I gave up and tried to approximate it with the integral, which I was not able to calculate. Is there a way to get a formula or at least find the order of growth of this function when k-> $\infty $?

Also, a good way to find antilogs will be nice as well. I just realized that I can't compute decimal powers. $$\Large 10^{0.3010} = 10^{0.3}*10^{0.001} = \sqrt[10]{1000} * \sqrt[1000]{10} = \text{Calculator Required}$$ I checked " How to calculate a decimal power of a number" but alas, the thing which came closest to what I needed required a calculator for an intermediate step.

2018年12月12日 · The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $\,\zeta(s)\,$ using the finite sum $\;\displaystyle\sum_{n=1}^X\frac{1}{n^s}\;$ like Euler-Maclaurin seem to impose $\,X\,$ to be larger than $\dfrac{|t|}{2\pi}$ to be precise (as illustrated in this answer).

2015年1月18日 · There is a simple way of approximating $\coth$ by noticing that it is a logarithmic derivative. Since: $$ \frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1} $$ by the Weierstrass product for the (hyperbolic) sine function, we have: $$ \log\sinh z-\log z = \sum_{n=1}^{+\infty}\log\left(1+\frac{z^2}{\pi^2 n^2}\right),\tag{2}$$ so, by differentiating both sides ...

2015年8月12日 · To indicate approximate equality, one can use ≃, ≅, ~, ♎, or ≒. I need to indicate an approximate inequality. Specifically, I know A is greater than a quantity of approximately B. Is there a way to

2017年7月7日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2024年8月22日 · But the way these things are discussed makes me doubt my conclusion. For example the Wiki page says at some point that "It can be shown that the approximate point spectrum of a bounded multiplication operator equals its spectrum", and the qualifier "multiplication" here would

Approximate Riemann zeta function. Ask Question Asked 13 years, 1 month ago. Modified 9 years, 6 months ago.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.