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Question Number 217314 by Tawa11 last updated on 09/Mar/25
How is ψ^2 (1) = − 2ζ(3) and ψ^2 ((1/2)) = − 14ζ(3) ???
$$\mathrm{How}\:\mathrm{is}\:\:\:\psi^{\mathrm{2}} \left(\mathrm{1}\right)\:\:=\:\:−\:\:\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\mathrm{and}\:\:\:\:\psi^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:\:=\:\:\:−\:\:\mathrm{14}\zeta\left(\mathrm{3}\right)\:\:\:\:\:\:\:??? \\ $$
Commented by mr W last updated on 10/Mar/25
with ψ^2 (x) do you mean (ψ(x))^2 or ψ^((2)) (x) ?
$${with}\:\psi^{\mathrm{2}} \left({x}\right)\:{do}\:{you}\:{mean}\:\left(\psi\left({x}\right)\right)^{\mathrm{2}} \:{or} \\ $$$$\psi^{\left(\mathrm{2}\right)} \left({x}\right)\:? \\ $$
Commented by Tawa11 last updated on 10/Mar/25
Polygamma function sir.
$$\mathrm{Polygamma}\:\mathrm{function}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 10/Mar/25
then the relationships are clear.
$${then}\:{the}\:{relationships}\:{are}\:{clear}. \\ $$
Commented by mr W last updated on 10/Mar/25
Commented by mr W last updated on 10/Mar/25
just take n=2
$${just}\:{take}\:{n}=\mathrm{2} \\ $$
Commented by Tawa11 last updated on 10/Mar/25
I have seen what I needed in the image you sent sir. I really appreciate. It really helped.
$$\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{what}\:\mathrm{I}\:\mathrm{needed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{image}\:\mathrm{you}\:\mathrm{sent}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{It}\:\mathrm{really}\:\mathrm{helped}. \\ $$
Commented by Tawa11 last updated on 10/Mar/25
Sir, is this a book? Can I get more functions. dealing with sum and integration.
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{this}\:\mathrm{a}\:\mathrm{book}? \\ $$$$\mathrm{Can}\:\mathrm{I}\:\mathrm{get}\:\mathrm{more}\:\mathrm{functions}. \\ $$$$\mathrm{dealing}\:\mathrm{with}\:\mathrm{sum}\:\mathrm{and}\:\mathrm{integration}. \\ $$
Commented by Tawa11 last updated on 10/Mar/25
Noted. Thanks sir.
$$\mathrm{Noted}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 10/Mar/25
i found it in internet. i have only the same internet which you also have.
$${i}\:{found}\:{it}\:{in}\:{internet}.\:{i}\:{have}\:{only} \\ $$$${the}\:{same}\:{internet}\:{which}\:{you}\:{also}\: \\ $$$${have}. \\ $$
Answered by mathmax last updated on 11/Mar/25
Ψ^′ (x)=Σ_(n=0) ^∞ (1/((n+x)^2 )) and Ψ^((2)) (x)=−2Σ_(n=0) ^∞ (1/((n+x)^3 )) ⇒ Ψ^((2)) (1)=−2Σ_(n=0) ^∞ (1/((n+1)^3 ))=−2Σ_(n=1) ^∞ (1/n^3 ) =−2ξ(3) Ψ^((2)) ((1/2))=−2Σ_(n=0) ^∞ (1/((n+(1/2))^3 ))=−16Σ_(n=0) ^∞ (1/((2n+1)^3 )) ξ(3)=Σ_(n=1) ^∞ (1/n^3 )=(1/8)Σ_(n=1) ^∞ (1/n^3 )+Σ_(n=0) ^∞ (1/((2n+1)^3 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^3 ))=(1−(1/8))ξ(3)=(7/8)ξ(3) ⇒Ψ^((2)) ((1/2))=−16.(7/8) ξ(3)=−14 ×ξ(3)
$$\Psi^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{x}\right)^{\mathrm{2}} }\:{and}\: \\ $$$$\Psi^{\left(\mathrm{2}\right)} \left({x}\right)=−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\Psi^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)=−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$$=−\mathrm{2}\xi\left(\mathrm{3}\right) \\ $$$$\Psi^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }=−\mathrm{16}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\xi\left(\mathrm{3}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{8}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)\xi\left(\mathrm{3}\right)=\frac{\mathrm{7}}{\mathrm{8}}\xi\left(\mathrm{3}\right) \\ $$$$\:\Rightarrow\Psi^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{16}.\frac{\mathrm{7}}{\mathrm{8}}\:\xi\left(\mathrm{3}\right)=−\mathrm{14}\:×\xi\left(\mathrm{3}\right) \\ $$
Commented by Tawa11 last updated on 11/Mar/25
Thanks sir, I really appreciate.
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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