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1d
comment Identity of $\coth $ using Fourier series
Similar work with $a:=iz$.
1d
comment Asymptotic expansion of integral with hyperbolic functions
@Braten: well $\;(-1)^k\binom{-1/2}{k}=\binom{2k}{k}/4^k\;$ (Mma writes this as $\,\binom{-1/2}{k}\,$ from the $1/\sqrt{1+x^2}$ expression) so that $$\binom{-1/2}{k}\binom{-1/2}{n-k}=\binom{2k}{k}\binom{2n-2k}{n-k}/(-4)^n$$ but the $\binom{2n-2k}{n-k}$ term should be a problem for you... Did you try a numerical check?
1d
revised Asymptotic expansion of integral with hyperbolic functions
n+1 -> 2n
1d
comment Asymptotic expansion of integral with hyperbolic functions
@Braten: I added the explicit formula in s and will let you (generously?) do the explicit formula in r (this should be a double or triple sum for every n). Fine continuation
1d
revised Asymptotic expansion of integral with hyperbolic functions
explicit formula for s and first terms in r
1d
comment Asymptotic expansion of integral with hyperbolic functions
Thanks @Braten! The Maclaurin Series for $\sinh(r)$ should do the job but the expression should not be so clear (the polynomials in $a$ : $3a^2+1,\;15a^4+10a^2+3,\cdots$ may be written as a sum of terms $a^{2k}(a^2+1)^{2(n-k)}$ divided by a central binomial up to some powers of two I think... Writting the result with $\sinh(r)$ expansed should give polynomials too but a 'closed form' may be harder to obtain).
2d
revised Asymptotic expansion of integral with hyperbolic functions
added 4 characters in body
2d
answered Asymptotic expansion of integral with hyperbolic functions
Jul
19
comment Continued fraction estimation of error in Leibniz series for $\pi$.
Concerning the error in Gregory's series I found this $1989$ article by the Borweins and Dilcher rather neat "Pi, Euler Numbers, and Asymptotic Expansions" (if you didn't know it...). Cheers,
Jul
9
comment Procedure converting decimals to rationals.
This thread may help.
Jul
7
comment Regularization of a (divergent) cosine series
Reference to get $\pi\delta(x)-\frac 12$ (a slight modification of Sangchul Lee's comment concerning distribution theory).
Jul
6
comment Is it possible to integrate this Riemann zeta function ratio so that I can produce this graph?
So that the limitations of the $f(t)$ formula may be the same as what you propose : we still need a precise evaluation of $\zeta$ to localize it's zeros! Not sure that this answers your "obstruction to reverse" question...
Jul
6
comment Is it possible to integrate this Riemann zeta function ratio so that I can produce this graph?
Let's start with $f(t)$ from MO. Riemann found that $N(t)$ (the number of zeros with imaginary part $<\,t$) was very near $\dfrac {\theta(t)}{\pi}+1$. As exposed here the counting function you represent is $\ \displaystyle f(t)=\frac 1{\pi}\rm{Arg}\;\zeta(1/2+it)+\frac {\theta(t)}{\pi}+1$. The jump of $1$ at (most of) the zeros of $\zeta(1/2+it)$ comes from the $\pi$ variation of the $(\pi/2,\pi/2]$ normalized argument of $\zeta$!
Jul
6
comment Is it possible to integrate this Riemann zeta function ratio so that I can produce this graph?
... bringing as back to the initial thread about all this so probably not so helpful sorry...
Jul
6
comment Is it possible to integrate this Riemann zeta function ratio so that I can produce this graph?
You may find a zeta zeros counting function in this thread. More details in this França & LeClair paper.
Jul
3
comment $L$-function, easiest way to see the following sum?
For an algebraic method see here, for an analytic solution see here.
Jun
28
comment How do I calculate the values of $\zeta(0.5+ie^x)$ for large $x$ ?
The paper by Gourdon and Sebah "Numerical evaluation of the Riemann Zeta-function" and this thread may help.
Jun
22
comment Numerical precision of arctan function
@YvesDaoust: I was just adding the same suggestion (for the case $x^2+y^2\ll z^2$ at least)... :-)
Jun
22
comment Numerical precision of arctan function
(answer converted in a comment) Concerning $\theta$ and $\phi$ the expressions should be : $$\theta = \arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\\ \phi = \arctan\left(\frac yx\right)$$ (the last one by dividing $y$ and $x$, the first one using $\cos(\theta)=\dfrac zr$). Supposing your code correct the problem was probably explained by Victor Liu : $\;\arccos\,$ may have a large relative error while evaluating $\arccos(1-\epsilon)$. Computing acos(1-0.5e-8)*1e4, acos(1-0.5e-10)*1e5, acos(1-0.5e-12)*1e6 using gcc on a mac I obtained 0.9999999974, 1.0000000414, 1.0000444493 0.9996002812.
Jun
13
comment How can i represent 3D space using 4x4 matrix?
See for example here for the direct generalization.