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Question Number 217314 by Tawa11 last updated on 09/Mar/25

$$\mathrm{How}\mathrm{is}{\psi }^2\left(1\right)=-2\zeta \left(3\right)$$$$\mathrm{and}{\psi }^2\left(\frac{1}{2}\right)=-14\zeta \left(3\right)???$$

Commented by mr W last updated on 10/Mar/25

$$with{\psi }^2\left(x\right)doyoumean{\left(\psi \left(x\right)\right)}^{2}or$$$${\psi }^{\left(2\right)}\left(x\right)?$$

Commented by Tawa11 last updated on 10/Mar/25

$$\mathrm{Polygamma}\mathrm{function}\mathrm{sir}.$$

Commented by mr W last updated on 10/Mar/25

$$thentherelationshipsareclear.$$

Commented by mr W last updated on 10/Mar/25

Commented by mr W last updated on 10/Mar/25

$$justtaken=2$$

Commented by Tawa11 last updated on 10/Mar/25

$$\mathrm{I}\mathrm{have}\mathrm{seen}\mathrm{what}\mathrm{I}\mathrm{needed}\mathrm{in}\mathrm{the}\mathrm{image}\mathrm{you}\mathrm{sent}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.\mathrm{It}\mathrm{really}\mathrm{helped}.$$

Commented by Tawa11 last updated on 10/Mar/25

$$\mathrm{Sir},\mathrm{is}\mathrm{this}\mathrm{a}\mathrm{book}?$$$$\mathrm{Can}\mathrm{I}\mathrm{get}\mathrm{more}\mathrm{functions}.$$$$\mathrm{dealing}\mathrm{with}\mathrm{sum}\mathrm{and}\mathrm{integration}.$$

Commented by Tawa11 last updated on 10/Mar/25

$$\mathrm{Noted}.\mathrm{Thanks}\mathrm{sir}.$$

Commented by mr W last updated on 10/Mar/25

$$ifounditininternet.ihaveonly$$$$thesameinternetwhichyoualso$$$$have.$$

Answered by mathmax last updated on 11/Mar/25

$${\mathrm{\Psi }}^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+x)}^2}and$$$${\mathrm{\Psi }}^{\left(2\right)}\left(x\right)=-2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+x)}^3}\Rightarrow$$$${\mathrm{\Psi }}^{\left(2\right)}\left(1\right)=-2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^3}=-2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}$$$$=-2\xi \left(3\right)$$$${\mathrm{\Psi }}^{\left(2\right)}\left(\frac{1}{2}\right)=-2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+\frac{1}{2})}^3}=-16\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^3}$$$$\xi \left(3\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}=\frac{1}{8}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^3}$$$$\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^3}=(1-\frac{1}{8})\xi \left(3\right)=\frac{7}{8}\xi \left(3\right)$$$$\Rightarrow {\mathrm{\Psi }}^{\left(2\right)}\left(\frac{1}{2}\right)=-16.\frac{7}{8}\xi \left(3\right)=-14\times \xi \left(3\right)$$

Commented by Tawa11 last updated on 11/Mar/25

$$\mathrm{Thanks}\mathrm{sir},\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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