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 Yearling
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Jul
27
accepted Asymptotic expansion of integral with hyperbolic functions
Jul
27
comment Asymptotic expansion of integral with hyperbolic functions
Raymond, I have got a slightly different closed form given by: $f(r)=\sum\limits_{n=0}^{\infty}(...)\sum\limits_{k=0}^{n} a^{2(n-k)}(a^2+1)^k\binom{-1/2}{k}\binom{-1/2}{n-k}$.
Jul
27
comment Asymptotic expansion of integral with hyperbolic functions
Nice calculation! Thank you very much. Is it possible to derive the asymptotic expansion in terms of the asymptotic scale ${r^n}$, i.e. $f(r)=\sum A_k r^k$ out of the asymptotic expansion $f(r)= \sum a_{2k+1}\sinh(r)^{2k+1}$ in a closed form?
Jul
26
asked Asymptotic expansion of integral with hyperbolic functions
Jul
14
comment Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
Thank you very much for your help! This works pretty well
Jul
14
accepted Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
Jul
14
revised Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
added 11 characters in body
Jul
14
comment Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
Thank you for your wonderful answer. The core problem is still not solved yet. How can I get the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ for $\ell \geq k$?
Jul
14
comment Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
Hello gammatester! That sounds like a nice argument. How did you compute the taylor series of $tanh(x)^{2k}$ out of the taylor series for $tanh(x)$?
Jul
14
asked Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$
Jun
19
revised Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
deleted 374 characters in body; edited title
Jun
19
awarded  Yearling
Jun
19
revised Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
added 394 characters in body
Jun
19
comment Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
@nospoon: actually the equality which I asked for is used in my book to prove the equality you wrote down. By the way, I'm interested in equations of the type $\sum\limits_{j=1}^{m-1}\frac{1}{sin^(2p)(\frac{j\pi}{m})}=(...)$ for any $p\in\mathbb{N}$ are there formulas known for that?
Jun
19
comment Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
@Gerry Myerson: I had this idea, but I could not prove it by this idea. But I will try it again, maybe I had a mistake in my calculations.
Jun
19
asked Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
Jun
18
comment Inequality with natural logarithm
Thank you very much! Very elegant. :)
Jun
18
accepted Inequality with natural logarithm
Jun
18
comment Inequality with natural logarithm
Thank you very much iadvd! :-)
Jun
18
comment Inequality with natural logarithm
@iadvd: Yes, this is of course equivalent. But I do not see, why this is easier?