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Question Number 157323 by MathSh last updated on 22/Oct/21

$$\mathrm{Prove}\mathrm{that}:$$$${\mathrm{\Psi }}_1\left(\frac{1}{8}\right)+{\mathrm{\Psi }}_2\left(\frac{5}{8}\right)=32\mathrm{G}+4{\pi }^2$$$$\mathrm{\Psi }-\mathrm{trigamma}\mathrm{function}$$$$\mathrm{G}-\mathrm{catalan}\mathrm{constant}$$

Answered by mindispower last updated on 22/Oct/21

$$\mathrm{\Psi }\left(\frac{p}{q}\right)=\frac{{\pi }^2}{2{sin}^2\left(\frac{p\pi }{q}\right)}+2q\underset{k=1}{\sum ^{\left[\frac{q-1}{2}\right]}}sin\left(\frac{2\pi kp}{q}\right){Cl}_2\left(\frac{2\pi k}{q}\right)$$$${Cl}_2.Clausfunction$$$${\mathrm{\Psi }}_1\left(\frac{1}{8}\right)=\frac{{\pi }^2}{2{sin}^2\left(\frac{\pi }{8}\right)}+16\underset{k=1}{\sum ^3}sin\left(\frac{\pi k}{4}\right){Cl}_2\left(\frac{\pi k}{4}\right)$$$$=\frac{2{\pi }^2}{2-\sqrt{2}}+16(sin\left(\frac{\pi }{4}\right){Cl}_2\left(\frac{\pi }{4}\right)+{Cl}_2\left(\frac{\pi }{2}\right)+sin\left(\frac{\pi }{4}\right){cl}_2\left(\frac{3\pi }{4}\right))$$$${\mathrm{\Psi }}_1\left(\frac{5}{8}\right)=\frac{2{\pi }^2}{\sqrt{2}+2}+16(sin\left(\frac{10\pi }{8}\right){Cl}_2\left(\frac{\pi }{4}\right)+sin\left(\frac{20\pi }{8}\right){Cl}_2\left(\frac{\pi }{2}\right)+sin\left(\frac{30\pi }{8}\right){Cl}_2\left(\frac{3\pi }{4}\right))$$$$=\frac{2{\pi }^2}{2+\sqrt{2}}+16(-sin\left(\frac{\pi }{4}\right){Cl}_2\left(\frac{\pi }{4}\right)+{Cl}_2\left(\frac{\pi }{2}\right)-sin\left(\frac{\pi }{4}\right){Cl}_2\left(\frac{3\pi }{4}\right))$$$${\psi }_1\left(\frac{1}{8}\right)+{\psi }_1\left(\frac{5}{8}\right)$$$$=\frac{2{\pi }^2}{2+\sqrt{2}}+\frac{2{\pi }^2}{2-\sqrt{2}}+32{Cl}_2\left(\frac{\pi }{2}\right)$$$$=4{\pi }^2+32{Cl}_2\left(\frac{\pi }{2}\right)$$$${Cl}_2\left(x\right)=\sum _{n⩾1}\frac{sin\left(nx\right)}{n^2},{Cl}_2\left(\frac{\pi }{2}\right)=\sum _{n⩾1}\frac{sin\left(n\frac{\pi }{2}\right)}{n^2}$$$$sin\left(n\pi \right)=0,\mathrm{\forall }n\in \mathbb{Z}\Rightarrow$$$${Cl}_2\left(\frac{\pi }{2}\right)=\sum _{n⩾0}\frac{sin(n\pi +\frac{\pi }{2})}{{(2n+1)}^2}=\sum _{n⩾0}\frac{{(-1)}^n}{(2n+1)}=\beta \left(2\right)=G$$$$\iff \psi \left(\frac{1}{8}\right)+\psi \left(\frac{5}{8}\right)=4{\pi }^2+32G$$$$$$

Commented by MathSh last updated on 22/Oct/21

$$\mathrm{Perfect}\mathrm{dear}\mathbf{Ser},\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

Answered by qaz last updated on 22/Oct/21

$${\psi }^{\prime }\left(\frac{1}{8}\right)+{\psi }^{\prime }\left(\frac{5}{8}\right)$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+\frac{1}{8})}^2}+\frac{1}{{(\mathrm{n}+\frac{5}{8})}^2}$$$$=64\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(8\mathrm{n}+1)}^2}+\frac{1}{{(8\mathrm{n}+5)}^2}$$$$=-64\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1({\mathrm{x}}^{8\mathrm{n}}+{\mathrm{x}}^{8\mathrm{n}+4})\mathrm{lnxdx}$$$$=-64{\int }_0^1(1+{\mathrm{x}}^4)\mathrm{lnx}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{8\mathrm{n}}\mathrm{dx}$$$$=-64{\int }_0^1\frac{(1+{\mathrm{x}}^4)\mathrm{lnx}}{1-{\mathrm{x}}^8}\mathrm{dx}$$$$=-32{\int }_0^1(\frac{1}{1-{\mathrm{x}}^2}+\frac{1}{1+{\mathrm{x}}^2})\mathrm{lnxdx}$$$$=-32\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1({\mathrm{x}}^{2\mathrm{n}}+{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}})\mathrm{lnxdx}$$$$=32\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathrm{n}+1)}^2}+\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2}$$$$=32(1-2^{-2})\zeta \left(2\right)+32\mathrm{G}$$$$=4{\pi }^2+32\mathrm{G}$$

Commented by MathSh last updated on 22/Oct/21

$$\mathrm{Perfect}\mathrm{dear}\mathbf{Ser},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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