text:
# Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$?

I am seeking a closed-form solution for this double sum:

$$\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?. \end{eqnarray*}$$

I will turn it into $$3$$ tough integrals in a moment. But first I will state some similar results:

$$\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{2}n+m)} &=& \frac{11}{8} \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{4}n+m)} &=& \frac{67}{32} \zeta(3) -\frac{G \pi}{2}. \\ \end{eqnarray*}$$

where $$G$$ is the Catalan constant. The last result took some effort ...

Now I know most of you folks prefer integrals to sums, so lets turn this into an integral. Using

$$\begin{eqnarray*} \frac{1}{n} &=& \int_0^1 x^{n-1} dx\\ \frac{1}{m} &=& \int_0^1 y^{m-1} dy\\ \frac{1}{3n+m} &=& \int_0^1 z^{3n+m-1} dz \\ \end{eqnarray*}$$ and summing the geometric series, we have the following triple integral $$\begin{eqnarray*} \int_0^1 \int_0^1 \int_0^1 \frac{z^3 dx dy dz}{(1-xz^3)(1-yz)}. \end{eqnarray*}$$

Now doing the $$x$$ and $$y$$ integrations we have $$\begin{eqnarray*} I=\int_0^1 \frac{\ln(1-z) \ln(1-z^3)}{z} dz. \end{eqnarray*}$$

Factorize the argument of the second logarithm ...

$$\begin{eqnarray*} I= \underbrace{\int_0^1 \frac{\ln(1-z) \ln(1-z)}{z} dz}_{=2\zeta(3)} + \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}$$

So if you prefer my question is ... find a closed form for:

$$\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}$$

Integrating by parts gives:

$$\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(z) \ln(1+z+z^2)}{1-z} dz + \int_0^1 \frac{(1+2z)\ln(z) \ln(1-z)}{1+z+z^2} dz. \end{eqnarray*}$$

and let us call these integrals $$I_2$$ and $$I_3$$ respectively.

All $$3$$ of these integrals are not easy for me to evaluate and any help with their resolution will be gratefully received.

• When I tried Wolfram ... it just told me $I_1 =1.10925$ , $I_2=0.713975$ and $I_3=0.395273$ ... at least it adds up ! Jul 15, 2019 at 21:19
• Mathematica gives the closed-form solution: $\frac{1}{54} \left(-6 \Gamma \left(\frac{4}{3}\right) \text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}\left(\left\{1,1,1,\frac{4}{3}\right\},\left\{2,2,\frac{4}{3}\right\},1\right)-6 \Gamma \left(\frac{5}{3}\right) \text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}\left(\left\{1,1,1,\frac{5}{3}\right\},\left\{2,2,\frac{5}{3}\right\},1\right)+12 \zeta (3)+2 \gamma \pi ^2+\pi ^2 \log (27)\right)$. Jul 15, 2019 at 21:19
• The sequence double sums are of the form $$\frac{1}{a}\sum _{m=1}^{\infty } \frac{H_{a m}}{n^2}$$ where $H_n$ is the Harmonic Number and $n=am$; $a$ being equal to $1$,$2$,$3$ or $4$. Jul 15, 2019 at 21:43
• Mathematica gives for the integral $$I=\frac{1}{216} \left(-216 \left(\text{Li}_3\left(\frac{\sqrt[6]{-1}}{\sqrt{3}}\right)+\text{Li}_3\left(-\frac{(-1)^{5/6}}{\sqrt{3}}\right)\right)+672 \zeta (3)+9 \log ^3(3)-15 \pi ^2 \log (3)+4 \pi \sqrt{3} \left(\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{1}{3}\right)\right)\right)$$ Jul 15, 2019 at 21:57
• @DonaldSplutterwit: This generalization may be of interest. Jul 23, 2019 at 4:30

A slightly different approach where I will make use of a particular Euler sum.

Let $$I = \int_0^1 \frac{\ln (1 - x) \ln (1 - x^3)}{x} \, dx.$$ Expanding the $$\ln (1 - x^3)$$ term gives \begin{align} I &= - \sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^{3n - 1} \ln (1 - x) \, dx\tag1 \end{align} Making use of the result (for a proof of this, see here) $$\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n}.$$ Re-indexing, namely $$n \mapsto 3n$$ gives $$\int_0^1 x^{3n - 1} \ln (1 - x) \, dx = -\frac{H_{3n}}{3n}.$$ Substitution of this result into (1) reduces our integral $$I$$ to the following Euler sum $$I = \frac{1}{3} \sum_{n = 1}^\infty \frac{H_{3n}}{n^2} = 3 \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2}.$$

For the Euler sum, since the series converges absolutely, terms in the sum may be rearranged. Doing so we have \begin{align} \sum_{n =1}^\infty \frac{H_{3n}}{(3n)^2} &= \frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots\\ &= \frac{2}{3} \left [\frac{3}{2} \frac{H_3}{3^2} + \frac{3}{2} \frac{H_6}{6^2} + \frac{3}{2} \frac{H_9}{9^2} + \cdots \right ]\\ &= \frac{2}{3} \left [\left (\frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots \right ) + \frac{1}{2} \left (\frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots \right ) \right ]\\ &= \frac{2}{3} \left [\left (-\frac{1}{2} \frac{H_1}{1^2} - \frac{1}{2} \frac{H_2}{2^2} + \frac{H_3}{3^2} - \frac{1}{2} \frac{H_4}{4^2} - \frac{1}{2} \frac{H_5}{5^2} + \frac{H_6}{6^2} - \cdots \right )\right.\\ & \qquad + \left. \frac{1}{2} \left (\frac{H_1}{1^2} + \frac{H_2}{2^2} + \frac{H_3}{3^2} + \frac{H_4}{4^2} + \cdots \right ) \right ]\\ &= \frac{2}{3} \sum_{n = 1}^\infty \frac{H_n}{n^2} \cos \left (\frac{2 \pi n}{3} \right ) + \frac{1}{3} \sum_{n = 1}^\infty \frac{H_n}{n^2} \tag2\\ &= \frac{2}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n^2} \left (e^{\frac{2 \pi i}{3}} \right )^n \end{align} Note in (2) the well-known result of $$\sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3)$$ has been used.

The sum can now be found by making use of the following generating function (for a simple proof of this result, see here) $$\sum_{n=1}^\infty\frac{H_n}{n^2}x^n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3).$$ Setting $$x = e^{\frac{2 \pi i}{3}}$$ gives \begin{align} \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2} &= \frac{2}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \left [\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) + \ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right.\\ & \qquad \left. + \frac{1}{2} \ln (e^{\frac{2 \pi i}{3}}) \ln^2 (1 - e^{\frac{2 \pi i}{3}}) + \zeta (3) \right ]\\ &= \frac{5}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \left [\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) + \ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right.\\ & \qquad \left. + \frac{\pi i}{3} \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ]\tag3 \end{align} Now, since (this part is tedious, but readily doable) \begin{align} \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= -\frac{4}{9} \zeta (3)\\ \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{\pi^2}{18} \ln 3 + \frac{13}{18} \zeta (3)\\ \operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi}{6} \ln 3\\ \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) \end{align} The Euler sum in (3) thus becomes $$\sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2} = \frac{5}{9} \zeta (3) + \frac{2 \pi^3}{81 \sqrt{3}} - \frac{\pi}{27 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ),$$ so that we finally arrive at the following value for the integral (and hence your double sum) of

$$\int_0^1 \frac{\ln (1 - x) \ln (1 - x^3)}{x} \, dx = \frac{5}{3} \zeta (3) + \frac{2 \pi^3}{27 \sqrt{3}} - \frac{\pi}{9 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )$$

• (+1) Good. In general, as stated very often in the book, (Almost) Impossible Integrals, Sums, and Series, use that $\sum_{k=1}^{\infty} \frac{1}{k(k+n)}=\frac{H_n}{n}$. Thus, you may reduce much the size of the solution. You get directly $3 \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2}$. Jul 16, 2019 at 13:27
• @omegadot would it be possible to tell me how to find (by hand) those real parts of the polylogarithms? Jul 20, 2019 at 20:36
• You may be interested in this generalization. Jul 23, 2019 at 4:29
• @Zacky - I calculate, by hand, the real parts of all those quantities I used in my answer in an addendum here. Jul 23, 2019 at 8:29

$$\boxed{I=\int_0^1 \frac{\ln(1-x) \ln(1-x^3)}{x}dx=\frac53\zeta(3) +\frac{2\pi^3}{27\sqrt 3} -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)}$$ As mentioned in the question we have: $$I=\int_0^1 \frac{\ln^2(1-x)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=2\zeta(3)+J$$

We can make use of the following series: $$-\frac12 \ln(1-2x\cos t+x^2)=\sum_{n=1}^\infty \frac{\cos(nt)}{n} x^n,\quad |x|<1, t\in \mathbb R$$ $$\Rightarrow J=\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=-2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n}\int_0^1 \ln(1-x) x^{n-1}dx$$ $$=2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n^2}H_n=2\Re \left(\sum_{n=1}^\infty \frac{z^n}{n^2}H_n\right),\quad z=e^{\frac{2\pi i}{3}}$$ Using the following generating function: $$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$ And by plugging in the values found in this post yields the announced result, as we obtain: $$J=\int_0^1 \frac{\ln(1-x)\ln(1-x+x^2)}{x}dx=\frac{2\pi^3}{27\sqrt 3}-\frac13\zeta(3) -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)$$

• One word: yikes (+1) Jul 17, 2019 at 14:51

Not a full answer, but another interesting expression for the series.

Let's introduce a function:

$$S(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m (3n +m)}$$

Assuming $$|x|<1$$ and $$|y|<1$$ we avoid any convergence problems, and can use partial fractions:

$$S(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m^2}-\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ m^2(n+ \frac{1}{3} m)}$$

$$S(x,y)=-\log(1-x) \text{Li}_2(y) -x \sum_{m=1}^\infty \frac{y^m}{ m^2} \Phi \left(x,1,\frac13 m+1 \right)$$

Let's use the integral representation of the Lerch transcendent:

$$\Phi \left(x,1,\frac13 m+1 \right)= \int_0^\infty \frac{e^{-(1+\frac13 m)t} ~dt}{1-x e^{-t}}$$

Summation under the integral gives us:

$$S(x,y)=-\log(1-x) \text{Li}_2(y) -x \int_0^\infty \text{Li}_2 \left(y e^{-t/3} \right) \frac{dt}{e^t-x}$$

So we can assume:

$$S=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x) -x \int_0^\infty \text{Li}_2 \left(x e^{-t/3} \right) \frac{dt}{e^t-x} \right]$$

Which does seem to work numerically, though of course is quite hard to evaluate symbolically.

Numerical check:

In[22]:= x=9999999/10000000;
y=9999999/10000000;
N[-Log[1-x]PolyLog[2,y],10]-x NIntegrate[PolyLog[2,y Exp[-t/3]]/(Exp[t]-x),{t,0,Infinity},WorkingPrecision->10]
Out[24]= 1.29484017

Compare to exact expression:

In[25]:= N[(1/216)*(-15*Pi^2*Log[3] + 9*Log[3]^3 + 4*Sqrt[3]*Pi*(-PolyGamma[1, 1/3] +
PolyGamma[1, 2/3]) - 216*(PolyLog[3, (-1)^(1/6)/Sqrt[3]] +
PolyLog[3, -((-1)^(5/6)/Sqrt[3])]) + 672*Zeta[3]), 10]
Out[25]= 1.2948652620+0.*10^-11 I

It may be that $$x=y$$ is not the best choice for the limit. For example, we can assume $$x=y^a$$ where $$a$$ is some real number. A good choice may lead to a better numerical convergence or even a closed form.

Using the dilogarithm properties, we have:

$$F(x,y)=-x \int_0^\infty \text{Li}_2 \left(y e^{-t/3} \right) \frac{dt}{e^t-x}=x \int_0^\infty \int_0^1 \frac{\log(1-e^{-t/3} y u) du dt}{u (e^t-x)}$$

Let's change the variable:

$$e^{-t}=v \\ t=- \log v$$

$$F(x,y)=x \int_0^1 \int_0^1 \frac{\log(1- y u v^{1/3}) du dv}{u (1-x v)}$$

Let's take:

$$y=x^{1/3}$$

We have:

$$F(x,y)=x \int_0^1 \int_0^1 \frac{\log(1- u (xv)^{1/3}) du dv}{u (1-x v)}$$

$$v=w/x$$

$$F(x,y)=\int_0^x \int_0^1 \frac{\log(1- u w^{1/3}) du dw}{u (1-w)}$$

$$F(x,y)=-\int_0^x \frac{\text{Li}_2 (w^{1/3}) dw}{1-w}$$

So, there's a neater expression for the limit:

$$\color{blue}{S=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x^{1/3}) -\int_0^x \frac{\text{Li}_2 (w^{1/3}) dw}{1-w} \right]}$$

This allows a simple generalization:

$$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m (an +m)}=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x^{1/a}) -\int_0^x \frac{\text{Li}_2 (w^{1/a}) dw}{1-w} \right]$$

Which checks out numerically with the examples from the OP.

I wonder if we can somehow use L'Hospital here to deal with the integral and get a closed form for the limit.

Integration by parts could also work.

• In fact, integration by parts will immediately give us the same integral as in the OP Jul 15, 2019 at 23:46

This is an addendum to the answer I gave here. As requested by @Zacky, I will show how the real parts of those quantities considered in my solution, particularly those containing polylogarithms, can be extracted by hand. The quantities in question are: \begin{align} \operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi}{6} \ln 3\tag1\\ \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= -\frac{4}{9} \zeta (3)\tag2\\ \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{\pi^2}{18} \ln 3 + \frac{13}{18} \zeta (3)\tag3\\ \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )\tag4 \end{align}

As we are dealing with complex valued functions, in all cases the principal value is taken.

Preliminaries

$$\ln \left (e^{\frac{2 \pi i}{3}} \right ) = \frac{2 \pi i}{3}.$$

The first quantity

It is routine to show that: $$\ln \left (1 - e^{\frac{2 \pi i}{3}} \right ) = \frac{1}{2} \ln 3 - i \frac{\pi}{6}.$$ Thus \begin{align} i \ln^2 \left (1 - e^{\frac{2 \pi i}{3}} \right ) &= i \left (\frac{1}{2} \ln 3 - i \frac{\pi}{6} \right )^2\\ &= i \left (\frac{1}{4} \ln^2 3 - \frac{\pi^2}{36} - i\frac{\pi}{6} \ln 3 \right )\\ &=\frac{\pi}{6} \ln 3 + \frac{i}{4} \ln^2 3 - \frac{i \pi^2}{36}, \end{align} and we immediately see that $$\boxed{\operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] = \frac{\pi}{6} \ln 3}$$

The second quantity

\begin{align} \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{1}{n^3} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^3} \cos \left (\frac{2 \pi n}{3} \right )\\ &= -\frac{1}{2} \cdot \frac{1}{1^3} - \frac{1}{2} \cdot \frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{2} \cdot \frac{1}{4^3} - \frac{1}{2} \cdot \frac{1}{5^3} + \frac{1}{6^3} - \cdots\\ &= \frac{3}{2} \left (\frac{1}{3^3} + \frac{1}{6^3} + \frac{1}{9^3} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right )\\ &= \frac{3}{2 \cdot 3^3} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right )\\ &= \frac{1}{18} \sum_{n = 1}^\infty \frac{1}{n^3} - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^3}\\ &= -\frac{4}{9} \sum_{n = 1}^\infty \frac{1}{n^3} \end{align} giving $$\boxed{\operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) = -\frac{4}{9} \zeta (3)}$$

Later we will have a need for its imaginary part so let's calculate it now.

\begin{align} \operatorname{Im} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Im} \sum_{n = 1} \frac{1}{n^3} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^3} \sin \left (\frac{2 \pi n}{3} \right )\\ &= \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 1,4,7,\ldots}}^\infty \frac{1}{n^3}}_{n \, \mapsto \, 3n + 1} - \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 2,5,8,\ldots}}^\infty \frac{1}{n^3}}_{n \, \mapsto \, 3n + 2}\\ &= \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 1)^3} - \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 2)^3}\\ &= \frac{\sqrt{3}}{54} \sum_{n = 1}^\infty \frac{1}{(n + \frac{1}{3})^3} - \frac{\sqrt{3}}{54} \sum_{n = 1}^\infty \frac{1}{(n + \frac{2}{3})^3}\\ &= -\frac{\sqrt{3}}{54} \cdot \frac{1}{2!} \psi^{(2)} \left (\frac{1}{3} \right ) + \frac{\sqrt{3}}{54} \cdot \frac{1}{2!} \psi^{(2)} \left (\frac{2}{3} \right )\\ &= \frac{\sqrt{3}}{108} \left [\psi^{(2)} \left (1 - \frac{1}{3} \right ) - \psi^{(2)} \left (\frac{1}{3} \right ) \right ]\\ &= \frac{\sqrt{3}}{108} \cdot \pi \left. \frac{d^2}{dz^2} \cot (\pi z) \right |_{z = \frac{1}{3}}\\ &= \frac{\sqrt{3}}{108} \cdot \frac{8 \pi^2}{3 \sqrt{3}}\\ &= \frac{2 \pi^2}{81}. \end{align} Thus we see that $$\boxed{\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}} ) = -\frac{4}{9} \zeta (3) + \frac{2 \pi^2}{81}i}$$

The third quantity

In order to evaluate this quantity we will make use of the following result (for a proof of this result, see here) $$\operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}. \qquad (*)$$ If we set $$z = e^{\frac{2 \pi i}{3}}$$, note that $$1 - \frac{1}{z} = \overline{1 - z}.$$ Thus $$\operatorname{Li}_3 \left (1 - \frac{1}{z} \right ) = \operatorname{Li}_3 (\overline{1 - z}) = \overline{\operatorname{Li}_3 (1 - z)},$$ allowing us to write $$\operatorname{Li}_3 \left (1 - z \right ) + \operatorname{Li}_3 \left (1 - \frac{1}{z} \right ) = \operatorname{Li}_3 (1 - z) + \overline{\operatorname{Li}_3 (1 - z)} = 2 \operatorname{Re} \operatorname{Li}_3 (1 - z).$$ So ($$*$$) can be rewritten as $$\operatorname{Re} \operatorname{Li}_3 (1 - z) = \frac{1}{2} \zeta (3) + \frac{1}{12} \ln^3 (z) + \frac{\pi^2}{12} \ln (z) - \frac{1}{4} \ln^2 (z) \ln (1 - z) - \frac{1}{2} \operatorname{Li}_3 (z).$$ Setting $$z = e^{\frac{2 \pi i}{3}}$$, one has \begin{align} \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{1}{2} \zeta (3) - \frac{1}{12} \cdot \frac{8 i\pi^3}{27} + \frac{\pi^2}{12} \cdot \frac{2 \pi i}{3} + \frac{1}{4} \cdot \frac{4 \pi^2}{9} \left (\frac{1}{2} \ln 3 - \frac{i \pi}{6} \right )\\ & \qquad - \frac{1}{2} \left (-\frac{4}{9} \zeta (3) + \frac{2 i \pi^3}{81} \right )\\ &= \frac{13}{18} \zeta (3) + \frac{\pi^2}{18} \ln 3 \end{align} Thus $$\boxed{\operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) = \frac{13}{18} \zeta (3) + \frac{\pi^2}{18} \ln 3}$$

The fourth quantity

To evaluate this quantity we will make use of Euler' reflexion formula, namely $$\operatorname{Li}_2 (z) + \operatorname{Li}_2 (1 - z) = \zeta (2) - \ln (z) \ln (1 - z).$$ Setting $$z = e^{\frac{2 \pi i}{3}}$$ we see that $$\operatorname{Li}_2 (1 - e^{\frac{2\pi i}{3}}) = \zeta (2) - \frac{2 \pi i}{3} \ln (1 - e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) \qquad (**)$$

Finding $$\operatorname{Li}_2 (e^{\frac{2 \pi i}{3}})$$. For the real part: \begin{align} \operatorname{Re} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{1}{n^2} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} \cos \left (\frac{2 \pi n}{3} \right )\\ &= -\frac{1}{2} \cdot \frac{1}{1^2} - \frac{1}{2} \cdot \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{2} \cdot \frac{1}{4^2} - \frac{1}{2} \cdot \frac{1}{5^2} + \frac{1}{6^2} - \cdots\\ &=\frac{3}{2} \left (\frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right )\\ &= \frac{3}{2 \cdot 3^2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right )\\ &= \frac{1}{6} \sum_{n = 1}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= -\frac{1}{3} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= -\frac{1}{3} \zeta (2), \end{align} giving $$\boxed{\operatorname{Re} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) = -\frac{\pi^2}{18}}$$

While for the imaginary part: \begin{align} \operatorname{Im} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Im} \sum_{n = 1} \frac{1}{n^2} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} \sin \left (\frac{2 \pi n}{3} \right )\\ &= \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 1,4,7,\ldots}}^\infty \frac{1}{n^2}}_{n \, \mapsto \, 3n + 1} - \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 2,5,8,\ldots}}^\infty \frac{1}{n^2}}_{n \, \mapsto \, 3n + 2}\\ &= \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 1)^2} - \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 2)^2}\\ &= \frac{\sqrt{3}}{18} \sum_{n = 1}^\infty \frac{1}{(n + \frac{1}{3})^2} - \frac{\sqrt{3}}{18} \sum_{n = 1}^\infty \frac{1}{(n + \frac{2}{3})^2}\\ &= \frac{\sqrt{3}}{18} \left [\psi^{(1)} \left (\frac{1}{3} \right ) -\psi^{(1)} \left (1 - \frac{1}{3} \right ) \right ]\\ &= \frac{\sqrt{3}}{18} \left [\psi^{(1)} \left (\frac{1}{3} \right ) - \left \{\frac{4 \pi^2}{3} - \psi^{(1)} \left (\frac{1}{3} \right ) \right \} \right ]\\ &= \frac{1}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) - \frac{2}{9 \sqrt{3}} \pi^2 \end{align} Thus we see that $$\boxed{\operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) = -\frac{\pi^2}{18} + \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) - \frac{2i \pi^2}{9 \sqrt{3}}}$$

So on making use of Euler's reflexion formula, we see that $$\operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) = \frac{\pi^2}{9} - \frac{\pi i}{3} \ln 3 + \frac{2i}{9\sqrt{3}} - \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ).$$ So at last \begin{align} \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \operatorname{Re} \left [\left \{\frac{1}{2} \ln 3 - \frac{\pi i}{6} \right \} \right. \times \\ & \qquad \left. \left \{\frac{\pi^2}{9} - \frac{\pi i}{3} \ln 3 + \frac{2i}{9\sqrt{3}} - \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) \right \} \right ]\\ &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ). \end{align} Thus $$\boxed{\operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] = \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )}$$

I can't comment,but I have a closed form about the sum

$$\sum_{n=1}^{\infty}\frac{H_{\alpha n}}{n^2}$$

It equals to $$\frac{1}{2}\left(\alpha^{2}+\frac{3}{\alpha}\right) \zeta(3)-\frac{\pi}{2 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(1)}\left(\frac{k}{\alpha}\right)$$ where $$\alpha\in\mathbb{N}^{+}$$

For example,let $$\alpha=1,2,3,4,5,6$$,then we get \begin{aligned} &\sum_{n=1}^{\infty} \frac{H_{n}}{n^2}=2\zeta(3)\\ &\sum_{n = 1}^{\infty} \frac{H_{2n}}{n^2} = \frac{11}{4}\zeta(3)\\ &\sum_{n=1}^{\infty} \frac{H_{3n}}{n^2}=5\zeta(3)- \frac{\pi}{6\sqrt{3} } \left [ \psi^{(1)} \left ( \frac{1}{3} \right ) -\psi^{(1)} \left ( \frac{2}{3} \right ) \right ]\\ &\sum_{n=1}^{\infty} \frac{H_{4n}}{n^2}=\frac{67}{8} \zeta(3)-2\pi G\\ &\sum_{n=1}^{\infty} \frac{H_{5n}}{n^2}=\frac{64}{5} \zeta(3)-\frac{\pi}{10\sqrt{5}} \left\{ \sqrt{5+2\sqrt{5} } \left [ \psi^{(1)}\left ( \frac{1}{5} \right ) - \psi^{(1)}\left ( \frac{4}{5} \right ) \right ] + \sqrt{5-2\sqrt{5} } \left [ \psi^{(1)}\left ( \frac{2}{5} \right ) - \psi^{(1)}\left ( \frac{3}{5} \right ) \right ] \right\}\\ &\sum_{n=1}^{\infty} \frac{H_{6n}}{n^2}=\frac{73}{4}\zeta(3)-\frac{\pi}{12} \left \{ \sqrt{3}\left [ \psi^{(1)}\left ( \frac{1}{6} \right )- \psi^{(1)}\left ( \frac{5}{6} \right ) \right ] +\frac{1}{\sqrt{3} } \left [ \psi^{(1)}\left ( \frac{1}{3} \right )- \psi^{(1)}\left ( \frac{2}{3} \right ) \right ] \right \} \end{aligned}

Proof!!!!!
Consider function $$f(z)=\frac{\pi\cot(\pi z)(\gamma+\psi^{(0)}(-\alpha z))}{z^2} ,a\in\mathbb{N}^+$$
By the residue theorem!!!!!
\begin{aligned} &\mathrm{Res}(f,-n)=\frac{H_{\alpha n}}{n^2} -\frac{1}{\alpha n^3}\\ &\mathrm{Res}(f,0)=-\alpha^2\zeta(3)\\ &\mathrm{Res}(f,n)=-\frac{2}{\alpha n^3} +\frac{H_{\alpha n}}{n^2} \\ &\mathrm{Res}(f,n/\alpha)=\cfrac{\pi\alpha\cot\left(\frac{\pi n}{\alpha} \right)}{n^2} \end{aligned}
Then
$$\sum_{n=1}^{\infty}\frac{H_{\alpha n}}{n^2} =\frac{1}{2} \left ( \alpha^2+\frac{3}{\alpha} \right ) \zeta(3) -\frac{\pi\alpha}{2} \sum_{n=1,n\not{{\equiv}}0(\mathrm{mod}\alpha)}^{\infty} \frac{\cot\left ( \frac{\pi n}{\alpha} \right ) }{n^2}$$
And \begin{aligned} &\sum_{n=1,n\not{{\equiv}}0(\mathrm{mod}\alpha)}^{\infty} \frac{\cot\left ( \frac{\pi n}{\alpha} \right ) }{n^2}\\&= \frac{\cot\left ( \frac{\pi}{\alpha} \right ) }{1^2} +\frac{\cot\left ( \frac{2\pi}{\alpha} \right ) }{2^2} +\frac{\cot\left ( \frac{3\pi}{\alpha} \right ) }{3^2}+... +\frac{\cot\left ( \frac{(\alpha-1)\pi}{\alpha} \right ) }{(\alpha-1)^2}\\ &+\frac{\cot\left ( \frac{\pi}{\alpha} \right ) }{(\alpha+1)^2} +\frac{\cot\left ( \frac{2\pi}{\alpha} \right ) }{(\alpha+2)^2} +\frac{\cot\left ( \frac{3\pi}{\alpha} \right ) }{(\alpha+3)^2}+... +\frac{\cot\left ( \frac{(\alpha-1)\pi}{\alpha} \right ) }{(2\alpha-1)^2}+...\\ &=\cot\left ( \frac{\pi}{\alpha} \right ) \left [ \frac{1}{1^2}+\frac{1}{(\alpha+1)^2}+\frac{1}{(\alpha+2)^2}+\frac{1}{(\alpha+3)^2}+... \right ]\\ &+\cot\left ( \frac{2\pi}{\alpha} \right ) \left[\frac{1}{2^2}+ \frac{1}{(2\alpha+1)^2}+\frac{1}{(2\alpha+2)^2}+\frac{1}{(2\alpha+3)^2}+... \right ]+...\\ &=\frac{1}{\alpha^2} \sum_{k=1}^{\alpha-1} \cot\left ( \frac{\pi k}{\alpha} \right ) \psi^{(1)}\left (\frac{k}{\alpha} \right ) \end{aligned}
So, $$\boxed{\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{2}}=\frac{1}{2}\left(\alpha^{2}+\frac{3}{\alpha}\right) \zeta(3)-\frac{\pi}{2 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(1)}\left(\frac{k}{\alpha}\right)}$$

We also have,$$\boxed{\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{4}}=\frac{1}{2}\left(\alpha^{4}+\frac{5}{\alpha}\right) \zeta(5)-\frac{\pi^2\alpha^2}{6}\zeta(3) -\frac{\pi}{12 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(3)}\left(\frac{k}{\alpha}\right)}$$
They can make some interesting integrals,like \begin{aligned} &\int_{0}^{1} \frac{\ln^2(1-x)}{x} \text{d}x=2\zeta(3)\\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x)}{x} \text{d}x=-\frac{5}{8}\zeta(3)\\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x^2)}{x} \text{d}x=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G \\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x+x^2)}{x}\text{d}x=-\frac{1}{3} \zeta(3)- \frac{\pi}{18\sqrt{3} } \left [ \psi^{(1)}\left ( \frac{1}{3} \right )-\psi^{(1)}\left ( \frac{2}{3} \right )\right ]\\ \\ & \int_{0}^{1} \frac{\text{Li}_2^2(x)}{x}\text{d}x=\frac{\pi^2}{3}\zeta(3)-3\zeta(5)\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^2)}{x}\text{d}x=\frac{\pi^2}{4}\zeta(3)-\frac{37}{16}\zeta(5)\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^3)}{x}\text{d}x=\frac{2\pi^2}{9}\zeta(3)-\frac{124}{27}\zeta(5) +\frac{\pi}{324\sqrt{3} } \left [ \psi^{(3)}\left ( \frac{1}{3} \right )-\psi^{(3)}\left ( \frac{2}{3} \right )\right ]\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^4)}{x}\text{d}x=\frac{5\pi^2}{24}\zeta(3)-\frac{1029}{128}\zeta(5)+2\pi\beta(4) \end{aligned}

• Only $2$ years late to the party ... But what a beautiful answer ... Thank you $\ddot \smile$ Jun 6, 2021 at 13:26


text:
# Had Heron Known Analytic Geometry…

In my previous two posts, “An Algebraic Proof of Heron’s Formula” and “An Alternative Derivation of Heron’s Formula,” I proved Heron’s formula for the area of a triangle with three given sides.  Based on Heron’s formula, we can now prove a theorem concerning the area of any triangle in a rectangle coordinate system, namely,

The area of a triangle with vertices at $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ in a rectangle coordinate system can be expressed as

$|\frac{1}{2}D|\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$

where $D$ is the determinant of matrix:

$\left(\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right)$

I offer the following proof:

Fig. 1

By Heron’s formula, the area of triangle in Fig. 1

$A=\sqrt{s (s-a) (s-b) (s-c)}\quad\quad\quad\quad\quad(2)$

where $a, b, c$ are three sides of the triangle and, $s=(a+b+c)/2$.

Therefore,

$A^2=s(s-a)(s-b)(s-c).$

where

$a^2=(x_2-x_3)^2+(y_2-y_3)^2,$

$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$

$c^2=(x_1-x_2)^2+(y_1-y_2)^2.$

Let $B=|\frac{1}{2} D|$,  we have

$B^2 ={|\frac{1}{2} D|}^2=(\frac{1}{2}D)^2.$

Compute $A^2-B^2$ using Omega CAS Explorer (see Fig. 2) , the result shows

$A^2-B^2=0.$

Fig. 2

Since $A> 0, B\geq 0$ implies

$A+B >0,$

$A^2-B^2=(A-B)(A+B)=0$ implies

$A-B=0$,

i.e.,

$A = B.$

Hence, (1) and (2) are equivalent.

I would like to learn any other alternative proof.

format: plain


text:
# How do you solve 3e^x+1=5?

Oct 20, 2016

x≈0.2877" to 4 dec. places"

#### Explanation:

Begin by expressing the equation in terms of ${e}^{x} .$

$3 {e}^{x} = 4 \Rightarrow {e}^{x} = \frac{4}{3}$

Take ln (natural logarithm) of both sides.

$\Rightarrow \ln {e}^{x} = \ln \left(\frac{4}{3}\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\log {a}^{x} \Leftrightarrow x \log a} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow x {\cancel{\ln e}}^{1} = \ln \left(\frac{4}{3}\right)$

rArrx=ln(4/3)≈0.2877" to 4 dec. places"


text:
Function GMP::Column::GetStatus(GMP, column)

# GMP::Column::GetStatus

The function GMP::Column::GetStatus returns the status of a column in a generated mathematical program.

GMP::Column::GetStatus(
GMP,            ! (input) a generated mathematical program
column          ! (input) a scalar reference or column number
)

## Arguments

GMP

An element in AllGeneratedMathematicalPrograms.

column

A scalar reference to an existing column in the matrix or the number of that column in the range $$\{ 0 .. n-1 \}$$ where $$n$$ is the number of columns in the matrix.

## Return Value

An element in the predefined set AllRowColumnStatuses. The set AllRowColumnStatuses contains the following elements:

• Active,

• Deactivated,

• Deleted,

• NotGenerated,

• PresolveDeleted.

Note

• This function will return ‘PresolveDeleted‘ only if the generated mathematical program has been created with GMP::Instance::CreatePresolved. Status ‘PresolveDeleted‘ means that the column was generated for the original generated mathematical program but deleted when the presolved mathematical program was created.

• Status ‘Deactivated‘ is not possible for columns.


text:
Program Listing for File InitFlags.h

Return to documentation for file (modules/holoviz/src/holoviz/InitFlags.h)

Copy
Copied!

/*
*
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
*
* Unless required by applicable law or agreed to in writing, software
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
*/

#pragma once

#include <cstdint>

namespace clara::holoviz
{

typedef enum
{
NONE       = 0x00000000,
FULLSCREEN = 0x00000001
} InitFlags;

} // namespace clara::holoviz


text:
## 5/22/2016

### cryptography15

Cryptography 15

Elliptic Curve

Elliptic curve is an equation with two variables and coefficients and it is a cubic equation like the following form:
$y^{2}+axy+by=x^{3}+cx^{2}+dx+e$
$y^{2}=x^{3}$

Let's plot $y^{2}=x^{3}+ax+b$ with set of points E(a,b) = E(1,1)

First we have element called O (zero point or point of infinity) and we have a rule in geometric tearm "If 3 points on an elliptic curve lie on a straight line, their sum is O."
We can drive some rules from this rule :
1- For any point P on elliptic curve P+O=P.
2- The negative of point P with the same X coordinate, so if P=(x,y) then -P=(x,-y).
3- To add two points P and Q,draw a straight line between them and the third intersection point with curve is -R, then these three points P+Q=-R and the positive value of P+Q is the mirror image of the third point.
If the line between P and Q is tangent to the curve then R=P or R=Q.
4- To double point P,draw the tangent line and find the intersection point S then 2P=-S.

Algebraic addition over Elliptic curve :

1- To add two points $P({X}_{p},{Y}_{p}) \ Q({X}_{q},{Y}_{q})$, then the slope of the line that join them $\Delta = \frac{{Y}_{q}-{Y}_{p}} {{X}_{q}-{Y}_{p}}$
Then we can express R (which R= P+Q)

${X}_{r} = \Delta ^{2} - {X}_{p}-{X}_{q}$
${Y}_{r} = -{Y}_{p}+\Delta({X}_{p}-{X}_{r})$

2- To double point P P+P=2P = R
${X}_{r} =(\frac{3{X}_{p}^{2}+a}{2{Y}_{p}})^{2}-2{X}_{p}$
${Y}_{p} =(\frac{3{x}_{p}^{2}+a} {2{Y}_{p}})({X}_{p}-{X}{r})-{Y}_{p}$

Elliptic Curve over finite field :

Elliptic Curve over finite field is using elliptic curve in which variables and coefficients are all restricted to elements of finite field.

Prime Curve over Zp :

Prime Curve is using a cubic equation in which the variables and coefficients take value from 0 to P-1.
Equation of elliptic curve over Zp :
$y^{2} mod(p)=(x^{3}+ax+b) mod(p)$

The rules for addition over ${E}_{p}(a,b)$:

1- If $P({X}_{p},{Y}_{p}) \ then \ -P({X}_{p},{-Y}_{p})$.
2- To add two points $P({X}_{p},{Y}_{p}) \ and \ Q ({X}_{q},{Y}_{q}) \ R= P+Q$
Then R =
${X}_{r}=(\lambda^{2}-{X}_{p}-{X}_{q})mod(p)$
${Y}_{r}=(\lambda({X}_{p}-{X}_{r})-{Y}_{p})mod(p)$

$\lambda = \begin{cases}(\frac{{}{Y}_{q}-{Y}_{p}}{{X}_{q}-{X}_{p}})mod(p)) & if\ p\neq q \\ \\ (\frac{3{X}_{p}^{2}+a}{2{Y}_{p}})mod(p) & \text if \ p=q (double\ point\ p) \end{cases}$

For example :

Let P=(3,10) Q=(9,7) in ${E}_{23}(1,1)$
$\lambda=\frac{7-10}{9-3}mod(23)\ =\frac{-1}{2}mod(23)=11$
${X}_{r}=(11^{2}-3-9)mod(23=13$
${Y}_{r}=(11(3-17)-10)mod(23)=20$
Then P+Q = (17,20)

To double point P 2P

For example :

Let P =(3,10)
$\lambda=(\frac{3(3^{2})+1}{2*10})mod(23)=6$
${X}_{r}=(6^{2}-3-3)mod(23)=7$
${Y}_{r}=(6(3-7)-10)mod(23)=12$
Then 2P = (7,12)

Elliptic curve over $GF(2^{m})$ :

$GF(2^{m})$ consists of $2^{m}$ elements with addition and multiplication can be defined over polynomials, which variables and coefficients are elements of $GF(2^{m})$.
The cubic equation for cryptographic applications for elliptic curve can take the following form :

$y^{2}+xy= x^{3}+ax^{2}+b$
For Example :

For g = 0010
$y^{2}+xy=x^{3}+g^{4}x^{2}+1 \ \ \ where \ a=g^{4} ,b=g^{0}$
$a= g^{4}, b = 1$  and points that satisfies this equation $(g^{5},g^{3})$
$=(9^{3})^{2}+(9^{5})(9^{3})=(9^{5})^{3}+(9^{4})(9^{5})^{2}+1$
$=9^{6}+9^{8}=9^{15}+9^{14}+1$
1100+0101 = 0001+1001+0001
1001 = 1001

The rules of addition over $GF(2^{m})$ :

1- If $P({X}_{p},{Y}_{p}) and Q({X}_{q},{Y}_{q}) R = P+Q$
Then
${X}_{r}=\lambda^{2}+\lambda+{X}_{p}+{X}_{q}+a$
${Y}_{r}=\lambda({X}_{p}+{X}_{r})+{X}_{r}+{Y}_{p}$
$\lambda=(\frac{{Y}_{q}+{Y}_{p}}{{X}_{q}+{X}_{p}})$

2- If $P({X}_{p},{Y}_{p}) R = 2P$
${X}_{r}=\lambda^{2}+\lambda+a$
${Y}_{r}={X}_{p}^{2}+(\lambda+{X}_{p})$
$\lambda={X}_{p}+\frac{{Y}_{p}}{{X}_{p}}$

Key exchange using elliptic curve:

First choose a large integer q OR a prime number P or integer of form $2^{m}$
Then choose base point $G =({x}_{1},{x}_{2})$ in Ep(a,b) whose order is a very large value n, The order n of point G is the smallest positive integer n such that nG=0.
The Key exchange can be done by the following:
1- User A select an integer ${n}_{A}$ less than n , This is A's private key, then compute public key ${P}_{a}={n}_{A} * G$
2- User B select ${n}_{B}$ and compute the public key ${P}_{b}$
3- User A compute the secret key $K={n}_{A} * {P}_{b}$
4- User B compute the secret key $K={n}_{B} * {P}_{a}$
Note : The secret key is a pair of numbers.

For Example:
P=211 in Ep(0,-4)
The equation $y^{2} = x^{3} - 4$
G = (2,2)
A's private key ${n}_{A}=121$
B's private key ${n}_{B}=203$
Then
A's public key ${P}_{a}={n}_{A} * G = 121(2,2) = (115,48)$
B's public key ${P}_{b}={n}_{B} * G = 203(2,2) = (130,203)$
A's secret key $K={n}_{A} * {P}_{b} = 121(130,203) = (161,69)$
B's secret key $K={n}_{B} * {P}_{a} = 203(115,48) = (161,69)$
Then the shared key k = (161,69)

Elliptic Curve encryption algorithm :

First we have to encode the message to be sent as x and y coordinates, and it is not easy because not all points are in ${E}_{p}(a,b)$.
As with the key exchange
- Choose a point G
- User A select a private key ${n}_{A}$
- User A calculate the public key ${P}_{a}={n}_{A} * G$
To send a message Pm to User B, User A choose a random integer K and then produce the ciphertext Cm by the following:
${C}_{m} = (KG.{P}_{m} + K{P}_{b})$
Note : User A masked the message Pm by adding (K.Pb) and nobody but user A knows k value.

For Example :
P=751 in ${E}_{p}(-1,188)$
The equation $y^{2} = x^{3} -x + 188$
G=(0,376) Pm = (562,201)
User A choose k=386
${P}_{b} = (201,5)$
Then K.G = 386(0,376) = (676,558)
${P}_{m} + K,{P}_{b}$
(562,201) + 386(201,5) = (385,328)
Then the ciphertext Cm
Cm = {(676,558),(385,328)}

## 5/16/2016

### cryptography14

Cryptography 14

Elgamal Cryptographic System

Elgamal cryptographic system is a public-key algorithm based on discrete logarithm.
Elgamal cryptographic system is used in digital signature standard (DSS).

Elgamal Cryptographic algorithm:

There are prime number q and primitive root of q = a.
User A want to generate private and public keys as the following:

1- Generate a random integer ${X}_{a}$ such that 1 < ${X}_{a}$ < q-1.
2- Compute ${Y}_{a}=a^{{X}_{a}} mod(q)$.
${X}_{a}$ is A's private key and (q,a,${Y}_{a}$) is A's public key.

If user B Has A's public key then he can send him encrypted message as the following:
1- Represent the message as an integer M such that 0 <= M <= q-1, and longer messages are send as a sequence of blocks, each block has integer value less than q.
2- Choose a random number k such that 1 <= k <= q-1.
3- Calculate one-time key $K = ({Y}_{a})^{k} mod(q)$.
4- Encrypt the message M as a pair of integer(C1,C2)
${C}_{1} = a^{k} mod(q) \\\ {C}_{2} = KM mod(q)$.

User A can decrypt the message :

1- Recover the key by calculating $K ={C}_{1}^{{X}_{a}} mod(q)$.
2- Calculate $M = {C}_{2}*k^{-1} mod(q)$.

How user A can recover the K value to decrypt the message M?

K value is defined during the encryption process
$K = {Y}_{a}^{k} mod(q)$  ----> we can substitute ${Y}_{a}=a^{{X}_{a}} mod(q)$
Then $K = (a^{{X}_{a}} mod(q))^{k} mod(q)$
Then $K = a^{(k{X}_{a})} mod(q)$ ---> we substitute $a^{k} mod(q) = {C}_{1} ( {C}_{1} = a^{k} mod(q))$
Then $K = {C}_{1}^{{X}_{a}} mod(q)$.

Example :

In Elgamal public key scheme suppose prime number q = 19
and primitive root of 19 we choose a=10
User A generate private and public keys as the following :
1- User A choose ${X}_{a} = 5$.
2- ${Y}_{a} = a^{{X}_{a}} mod(q) = 10^{5} mod(19) = 13$

User A's public key = (19,10,3) and A's private key=5

User B want to send encrypted message M to user A with M value = 17

1- User B choose k=6.
2- $K = {Y}_{a}^{k} mod(q) = 3^{6} mod(19) = 7$.
Then ${C}_{1} = a^{k} mod(q) = 10^{6} mod(19) = 11$
${C}_{2} = KM\ mod(q) = 7*17 mod(19) = 5$

Then user B send${C}_{1} = 11 \ and \ {C}_{2} = 5$ to user A.
Then user A decrypt the message
1- User A calculate $K = {C}_{1}^{{X}_{a}} mod(q) = 11^{5} mod(19) = 7$
2- The $K^{-1}\ in \ GF(19) = K^{-1} mod(19) = 7^{-1} mod(19) = 11$
3- $M = ({C}_{2}K^{-1})mod(q) = (5*11) mod(19) = 17$ #.


text:
People’s PC

With the dot-com bubble a distant memory, recession-ravaged Silicon Valley insiders are wondering aloud about the next big thing. They haplessly sift the sands of social networks and they chase the chimera of wireless networks, but they ignore technology’s biggest opportunity that is staring them in the face.

It is what I call a Massputer a computer that costs $300 for the computing hungry masses in emerging economies like India, China and Brazil. Users of this massputer should be able to do basic tasks like writing documents, Internet surfing, email and perhaps some business-related tasks like data entry. There are nearly four billion people who live in these emerging markets and assuming that only 10 per cent of them can afford$300 it is still a market of 400 million. Currently, the PC troika of Intel, Microsoft and Dell typically charge $750 for every computer, which most people in these emerging markets find unaffordable. Even the so-called network computers cost an unaffordable$500-plus.

So this is a perfect opportunity for brave entrepreneurs. It is not an easy task, but it isn’t impossible. With free open source software like Linux and powerful low-cost processors made by companies like Texas Instruments, the ingredients are all right there for someone to build this Massputer. Sure, it isn’t as exotic as building a super fast optical system, but it can definitely be lucrative.

At $300 a machine, the market is going to split wide open. In India alone nearly 55 million such machines could be sold to schools, colleges, government, and small businesses, estimates Netcore Solutions, a Mumbai, India-based company. Multiply this many times, and you see the numbers add up. The total market opportunity for the lower priced machines is upwards of$16 billion. (This is a concept championed by Rajesh Jain.)

The $300 sticker is vital- it keeps the devices affordable, and at the same time allows the corporation selling this massputer makes a decent profit. Gizmos such as color televisions, washing machines, refrigerators and air conditioners were snapped up in large numbers once they were priced right around$300. The proliferation of mobile phones in the emerging world is proof that at the right and affordable price people everywhere will adopt the right technology. There was a time when a mobile phone cost $400 and a mere 10 million people had the service. Now more than 400 million phones will be sold this year and 1.4 billion people, many in not very rich countries, will make mobile calls. That is because the price of the phone at$100 is now affordable in these emerging countries. More users means the price-per-minute has come down as well. In short, everyone benefits.

In an article for MIT Sloan Management Review, Harvard Business School Professor Clayton Christensen described such breakthroughs as “disruptive innovations,” or a breakthrough that brings products to those “left out entirely or poorly served by existing products and who are therefore quite happy to have a simpler, more modest version of what is available in the high-end markets.”

For the established giants, there is little risk of cannibalizing their markets. After all US consumers are used to the ever-powerful machines, and have time and again shown little or no interest in such stripped down machines. Even General Motors, bowing to market forces makes tiny cars in Asia. Why can’t a PC maker do the same?

Many see the tip of the technology pyramid as being exotic. That was true in the era of the mainframe where volumes were low and the technology was complex. Today there is an inversion of this phenomenon where the most complex technology is produced in high volume at commodity prices. Current PC industry giants such as Microsoft, Intel and Dell are shackled to their gilded past and are ignoring this market reality. The days of $150 processors and$300 operating systems are gone. Like many before them, those who hang on to the past are destined to become part of it.

The social implications of Massputer cannot be underscored. Popularity of cell phones and text messaging promoted social revolutions, and peaceful protests in hitherto turbulent societies in Asia. Philippines comes to mind. I believe the availability of a Massputer connected to the Internet will help develop more educated, more informed and more open societies. If rest of the world has to embrace the principles of free markets, they need the tools. Massputer is a perfect example.

Massputer is the market of the future. Any takers?