# 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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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 Posted in Striling numbers of first kind | | Leave a comment ## 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 $$\left[k+1\atop 2\right] = k\left[k\atop 2\right] + \left[k\atop … Continue reading | Tagged , , , , , | Leave a comment ## Stirling numbers of the first kind special values proof Prove that$$\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 Posted in Striling numbers of first kind | Tagged , , , , , , | Leave a comment ## 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 Posted in Dilogarithm, Polylogarithm | Leave a comment ## 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 Posted in Dilogarithm, Polylogarithm | | 1 Comment ## 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 Posted in Dilogarithm, Polylogarithm | Tagged , , , | 1 Comment ## 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. Posted in Dilogarithm, Polylogarithm | Tagged , | Leave a comment ## 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 Posted in Dilogarithm, Polylogarithm | | 1 Comment ## 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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