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Tag Archives: difference
Square difference formula for polylgoarithm proof
$$\mathrm{Li}_{\,n}(z) + \mathrm{Li}_{\,n}(z) = 2^{1n} \,\mathrm{Li}_{\,n}(z^2) $$ $$\textit{proof}$$ 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^{1n} … Continue reading
Posted in Dilogarithm, Polylogarithm
Tagged difference, dilogarithm, formula, polylgoarithm, proof, recurrence, square
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Dilogarithm difference formula proof
$$\mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z1} \right) = – \frac{1}{2} \log^2 (1z) \,\,\,\, \, z<1$$ $$ \textit{proof} $$ Start by the following $$\mathrm{Li}_2 \left(\frac{z}{z1} \right) = \int^{\frac{z}{z1}}_0 \frac{ \log(1t)}{t}\, dt$$ Differentiate both sides with respect to $z$ $$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z1} \right) = \frac{1}{(z1)^2}\left( \frac{ \log … Continue reading
Dilogarithm functional equation proof
$$\mathrm{Li}_2(z) + \mathrm{Li}_{2}(1z) = \frac{\pi^2}{6}\log(z) \log(1z) \,\,\,\, ,\,0<z<1$$ $$\textit{proof}$$ Start by the following $$\mathrm{Li}_2\left(z\right) = \int^{z}_0 \frac{\log(1t)}{t} \, dt $$ Now integrate by parts to obtain $$\mathrm{Li}_2\left(z\right)= \int^z_0 \frac{\log(t)}{1t} \, dt \log(z) \log(1z) $$ By the change of variable \(t=1x … Continue reading
Posted in Dilogarithm, Polylogarithm
Tagged difference, dilogarithm, formula, functional, proof
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Digamma difference formula proof
$$\psi(1x)\psi(x)=\pi \cot(\pi x) $$ $$\textit{proof}$$ We know by the reflection formula that $$\Gamma(x)\Gamma(1x)=\pi \csc(\pi x) $$ Now differentiate both sides $$\psi(x)\Gamma(x)\Gamma(1x)\psi(1x)\Gamma(x)\Gamma(1x)=\pi^2 \csc(\pi x)\,\cot(\pi x) $$ Which can be simplified $$\Gamma(x)\Gamma(1x)\left(\psi(1x)\psi(x)\right)=\pi^2 \csc(\pi x)\,\cot(\pi x) $$ Further simplifications using ERF results in $$ … Continue reading