1
vote
2answers
19 views

Calculate ch(0.2) to the nearest 0.01

Help me calculate ch(0.2) to the nearest 0.01. I tried to rewrite ch as a series but I still don't know how to evaluate it and what to do with factorial Help me please. it's very important
1
vote
3answers
51 views

Showing that $\sinh(\mathrm{e}^z)$ is entire

I am attempting to show that $\sinh(\mathrm{e}^z)$, where $z$ is a complex number, is entire. The instructions of the problem tell me to write the real component of this function as a function of $x$ ...
1
vote
0answers
54 views

$\lim\limits_{n\to\infty}\operatorname{arsinh}(1 + \operatorname{arsinh}(2 + \operatorname{arsinh}(3 + \dots\operatorname{arsinh}(n+\dots)\dots)))=?$

Does the limit $$\lim\limits_{n\to\infty}\operatorname{arsinh}(1 + \operatorname{arsinh}(2 + \operatorname{arsinh}(3 + \operatorname{arsinh}(4+\dots\operatorname{arsinh}(n+\dots)\dots))))$$ exist ...
1
vote
3answers
49 views

Is there a means of analytically proving the following identity?

Okay, so before I begin, my background is more in the world of applied, rather than pure, mathematics, so this question is motivated by a physics problem I'm looking at just now. Mathematically, it ...
2
votes
0answers
73 views

Solution of nonlinear waves( breathers)

The sine-Gordon equation is known as $$\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} + \sin u = 0,$$ Can you please derive the equation which is known as breather equation ...
9
votes
2answers
288 views

Series $\sum\limits_{n=1}^\infty \frac{1}{\cosh(\pi n)}= \frac{1}{2} \left(\frac{\sqrt{\pi}}{\Gamma^2 \left( \frac{3}{4}\right)}-1\right)$

I was playing around with Mathematica and found that $$\sum_{n=1}^\infty\frac1{\cosh(\pi n)} = \frac12\left(\frac{\sqrt{\pi}}{\Gamma \left(\tfrac34\right)^2}-1\right)$$ Does anybody know how to ...
2
votes
1answer
164 views

Definite integral involving hyperbolic cosine

I have had no experience so far with hyperbolic functions so any help will be appreciated. This is on the chapter of complex integration but I would especially appreciate it if you could turn this ...