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.

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.

2019年12月16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.

2022年7月27日 · $\begingroup$ The reason why we take the positive square root for $\cosh$ is partially that $\cosh\ge0$ and it's probably inherent to the proof you're reading, but it should be noted that $\sinh^{-1}x$ has the explicit formula $\ln\left(x+\sqrt{x^2+1}\right)$, so you could just compute $\cosh\sinh^{-1}(x)$ directly in terms of elementary functions. $\endgroup$

2021年1月2日 · $\begingroup$ Well okay, but the 1968 edition (the only one I have immediate access to) actually has $\cosh^{-1}$ rather than $\operatorname{arcosh}$ ... so in light of what you say, perhaps Spiegel is insisting that the negative branch is …

2024年3月21日 · With the regular trig functions, if I ever end up with something like $\operatorname{trig}_1(\operatorname{arctrig}_2(f(x))$, where $\text{trig}_1$ and $\text{trig}_2$ are two arbitrary trigonometric

You can simplify cosh 1 to $$ cosh(1) = 1/2 (1/e+e^1) $$ And use the following approximation of e (good for 9 decimal digits): $$ \frac{271\,801}{99\,990} $$ Alternatively you can approximate e yourself using the series: $$ e=1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots $$ You will only have to develop the series for a few steps until the ...

2018年2月19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.

Evaluate $$\lim_{t\to\infty} (\cosh x)^{1/x}.$$ I tried to use L'Hopital's but I think I made a mess of the differentiation, and the differentiation doesn't seem like it'll help much. calculus limits

2015年12月23日 · The trick is to recognize the $1/\sqrt x$ singularity in the derivative of the function of interest and transform the series into a series in $\sqrt x$.