I’m analyzing the security of a secret sharing scheme. One attempt I’m analyzing is “blind luck”. I return a random share and hope that noone notices. The probability $p$ of someone not noticing will be quite small, around $10^{-14}$ [it’s meant to be secure, after all]. I have a theoretical lower bound on $p$, but […]

Question Is it already known whether the $\zeta(4):=\sum_{n=1}^{\infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $\zeta(3)$ and $\zeta(2)$? $$\zeta(4)=\frac{36}{17}\sum_{n=1}^{\infty}\frac{1}{n^{4}\binom{2n}{n}}.\tag{1}$$ (a) In other words, does there exist a pair of functions […]

Any triangle can be divided into 4 congruent shapes: http://www.math.missouri.edu/~evanslc/Polymath/WebpageFigure2.png An equilateral triangle can be divided into 3 congruent shapes. Questions: 1) a triangle can be divided into 3 congruent shapes. Is it equilateral? 2) a shape in the plane can be divided into n congruent shapes for any positive integer n. What can it […]

Can anyone come up with a way to divide any given x by any given y without actually dividing? For example to add any given x to any given y without adding you would just do: $x-(-y)$ And to subtract any given x from any given y (that is, y-x) you could do: $y+xe^{iπ}$ *edit: […]

I have been working on an article at https://oeis.org/wiki/Table_of_convergents_constants where I posted a table of “convergents constants” (defined at https://oeis.org/wiki/Convergents_constant) for a few numbers. It would be nice to support the article with some quality analysis. Before June 9, 2011, was starting to extract and clearly define a pattern to these constants cf the article. […]

Consider the rather interesting and new evaluations for $_2F_1\left(\tfrac14,\tfrac34;\color{blue}{\tfrac{n}{n+1}};z\right)$, $$\begin{aligned} _2F_1\left(\tfrac14,\tfrac34;\color{blue}{\tfrac23};\tfrac{2^2\times3^3}{121}\right) &= \large\tfrac{\sqrt{33}}{3}\\[2mm] _2F_1\left(\tfrac14,\tfrac34;\color{blue}{\tfrac56};-\tfrac{135}{121}\right) &=\large\tfrac{\sqrt{33}}{10^{5/6}}\\[2mm] _2F_1\left(\tfrac14,\tfrac34;\color{blue}{\tfrac78};\tfrac{48}{49}\right) &= \tfrac{\sqrt7}3(1+\sqrt2)\\[2mm] _2F_1\left(\tfrac14,\tfrac34;\color{blue}{\tfrac9{10}};\tfrac{4}{5}\right) &=\large \tfrac1{5^{1/4}}\,\phi^{3/2}\end{aligned}$$ and golden ratio $\phi$, with the last a transformed version of Nemo’s answer. The transformation, $$_2F_1\left(\tfrac14,\tfrac34;c;\tfrac{4z(z-1)}{(1-2z)^2}\right)=\sqrt{1-2z}\,(1-z)^{1-c}\,_2F_1\left(-c+\tfrac32,\,c-\tfrac12;c;z\right)$$ allows it to be transformed to another form also with $a+b=-c+\tfrac32+c-\tfrac12 =1$. For example, the second one yields, […]

Why is $\zeta(2) = \frac{\pi^2}{6}$ almost equal to $\sqrt{e}$? Experimenting a bit I also found $\zeta(\frac{8}{3}) \approx e^\frac{1}{4}$, $\zeta(\frac{31}{9}) \approx e^\frac{1}{8}$ and $\zeta(\frac{141}{23}) \approx e^\frac{1}{64}$. I also figured out that $\zeta(x)$ approaches $e^{2^{-x}}$ but I’m not sure that helps explain why these almost-equalities exist. How to quantify how surprising these almost-equalities are, and what is […]

There are few known closed form for values of the dilogarithm at specific points. Sometimes only the real part or only the imaginary part of the value is known, or a relation between several different values is known: [1][2][3][4][5][6]. Discovering a new identity of this sort is always of a great interest. I numerically discovered […]

What are the best free user-friendly alternatives to Mathematica and Maple available online? I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds. Very basic computations can be carried out with Wolfram Alpha. What if one is […]

On a regular basis, one sees at MSE approximate numerology questions like Prove $\log_{{1}/{4}} \frac{8}{7}> \log_{{1}/{5}} \frac{5}{4}$, Prove $\left(\dfrac{2}{5}\right)^{{2}/{5}}<\ln{2}$, Comparing $2013!$ and $1007^{2013}$ or yet the classical $\pi^e$ vs $e^{\pi}$. In general I don’t like this kind of problems since a determined person with calculator can always find two numbers accidentally close to each other […]

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