Monthly Archives: January 2017

Signed Stirling numbers of first kind as coefficients

Signed Stirling numbers of the first kind We define the following $$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$ Prove the following $$(x)_n = x(x-1)(x-2)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k$$ $$\textit{proof}$$ We already proved that $$x^{(n)} = \sum_{k=0}^n \left[n\atop k\right] x^k$$ Which can be … Continue reading

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Generating number of the Stirling numbers of the first kind

Prove that $$x^{(n)} = x(x+1)(x+2)\cdots(x+n-1) = \sum_{k=0}^n \left[n\atop k\right]x^k$$ Or $$ \sum_{k=0}^n \left[n\atop k\right]x^k = \frac{\Gamma(x+n)}{\Gamma(x)}$$ $$\textit{proof}$$ By induction for \( n=0\) we have $$ \left[0\atop 0\right] = 1$$ Assume it is true for \( n \) then for \( … Continue reading

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Relation between harmonic numbers and Stirling numbers of the first kind

Prove that $$\left[n\atop 2\right] = H_{n-1}\Gamma(n)$$ By induction on \( n\) we have for \( n=2\) $$\left[2\atop 2\right] = H_{1}\times\Gamma(1) = 1$$ Assume that $$\left[k\atop 2\right] = H_{k-1}\Gamma(k)$$ Then by the recurrence relation $$\left[k+1\atop 2\right] = k\left[k\atop 2\right] + \left[k\atop … Continue reading

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Stirling numbers of the first kind special values proof

Prove that $$\left[n\atop 1\right] = \Gamma(n)$$ Use the recurrence relation $$\left[n+1\atop k\right] = n\left[n\atop k\right] + \left[n\atop k-1\right]$$ This implies that for \( k=1 \) $$\left[n+1\atop 1\right] = n\left[n\atop 1\right] + \left[n\atop0\right]$$ Now use that \( \left[n\atop 0\right] = 0 … Continue reading

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Special values of the dilgoarithm function

Prove that $$\mathrm{Li}_2\left( \frac{\sqrt{5}-1}{2} \right) = \frac{\pi^2}{10} – \log^2 \left( \frac{\sqrt{5}-1}{2}\right)$$ $$\textit{proof}$$ Use the following functional equation $$\mathrm{Li}_2 \left( \frac{z}{z-1} \right) + \frac{1}{2} \mathrm{Li}_2 (z^2) – \mathrm{Li}_2(-z) = -\frac{1}{2} \log^2 (1-z) $$ These are proved here and here  Now let … Continue reading

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Square difference formula for polylgoarithm proof

$$\mathrm{Li}_{\,n}(-z) + \mathrm{Li}_{\,n}(z) = 2^{1-n} \,\mathrm{Li}_{\,n}(z^2) $$ $$\textit{proof}$$ As usual we write the series representation of the LHS $$\sum_{k=1}^\infty \frac{z^k}{k^n}+\sum_{k=1}^\infty \frac{(-z)^k}{k^n}$$ Listing the first few terms $$z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right) $$ The odd terms will cancel $$2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots $$ Take \( 2^{1-n} … Continue reading

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Dilogarithm difference formula proof

$$\mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z-1} \right) = – \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1$$ $$ \textit{proof} $$ Start by the following $$\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = -\int^{\frac{z}{z-1}}_0 \frac{ \log(1-t)}{t}\, dt$$ Differentiate both sides with respect to $z$ $$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log … Continue reading

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Dilogarithm at 2

$$\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12}-\frac{1}{2}\log^2 \left(\frac{1}{2}\right) $$ $$ proof $$ Using the duplication formula proved here  $$\mathrm{Li}_2\left(z\right)+\mathrm{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\, $$ We can easily deduce that for \( z=\frac{1}{2}\) $$2\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$ It follows by dividing by 2.

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Dilogarithm functional equation proof

$$\mathrm{Li}_2(z) + \mathrm{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\, ,\,0<z<1$$ $$\textit{proof}$$ Start by the following $$\mathrm{Li}_2\left(z\right) = -\int^{z}_0 \frac{\log(1-t)}{t} \, dt $$ Now integrate by parts to obtain $$\mathrm{Li}_2\left(z\right)= -\int^z_0 \frac{\log(t)}{1-t} \, dt -\log(z) \log(1-z) $$ By the change of variable \(t=1-x … Continue reading

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Relation between Zeta and Dirichlet eta functions proof

$$\eta(s) = \left( 1-2^{1-s} \right) \zeta(s) $$ $$\textit{proof}$$ We will start by the RHS $$\left( 1-2^{1-s} \right) \zeta(s) = \zeta(s) – 2^{1-s} \zeta(s)$$ Which can be written as sums of series $$\sum_{n=1}^\infty \frac{1}{n^s} – \frac{1}{2^{s-1}}\sum_{n=1}^\infty \frac{1}{n^s}$$ $$\sum_{n=1}^\infty \frac{1}{n^s} – 2\sum_{n=1}^\infty … Continue reading

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