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Question Number 127020 by bramlexs22 last updated on 26/Dec/20
super nice ! show that ζ(6) = (π^6 /(945))
$$\:\:{super}\:{nice}\:! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\zeta\left(\mathrm{6}\right)\:=\:\frac{\pi^{\mathrm{6}} }{\mathrm{945}} \\ $$
Commented by liberty last updated on 26/Dec/20
hahaha very nice
$${hahaha}\:{very}\:{nice}\: \\ $$
Answered by Olaf last updated on 26/Dec/20
ζ(2k) = (((−1)^(k−1) B_(2k) (2π)^(2k) )/(2(2k)!)) For k = 3 : ζ(6) = (((−1)^2 B_6 (2π)^6 )/(2×6!)) = (2/(45))B_6 π^6 B_6 = (1/(42)) ζ(6) = (2/(45))×(1/(42))π^6 = (π^6 /(945))
$$\zeta\left(\mathrm{2}{k}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {B}_{\mathrm{2}{k}} \left(\mathrm{2}\pi\right)^{\mathrm{2}{k}} }{\mathrm{2}\left(\mathrm{2}{k}\right)!} \\ $$$$\mathrm{For}\:{k}\:=\:\mathrm{3}\:: \\ $$$$\zeta\left(\mathrm{6}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} {B}_{\mathrm{6}} \left(\mathrm{2}\pi\right)^{\mathrm{6}} }{\mathrm{2}×\mathrm{6}!}\:=\:\frac{\mathrm{2}}{\mathrm{45}}{B}_{\mathrm{6}} \pi^{\mathrm{6}} \\ $$$${B}_{\mathrm{6}} \:=\:\frac{\mathrm{1}}{\mathrm{42}} \\ $$$$\zeta\left(\mathrm{6}\right)\:=\:\frac{\mathrm{2}}{\mathrm{45}}×\frac{\mathrm{1}}{\mathrm{42}}\pi^{\mathrm{6}} \:=\:\frac{\pi^{\mathrm{6}} }{\mathrm{945}} \\ $$
Answered by Dwaipayan Shikari last updated on 26/Dec/20
ζ(2n)=(((−1)^(n+1) (2π)^(2n) )/(2(2n)!))B_(2n) ζ(6)=(((2π)^6 )/(2.6!)).(1/(42))=(π^6 /(15.63))=(π^6 /(945)) ζ(8)=(π^8 /(9450)) ζ(10)=(π^(10) /(93555)) ....
$$\zeta\left(\mathrm{2}{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\mathrm{2}\pi\right)^{\mathrm{2}{n}} }{\mathrm{2}\left(\mathrm{2}{n}\right)!}{B}_{\mathrm{2}{n}} \\ $$$$\zeta\left(\mathrm{6}\right)=\frac{\left(\mathrm{2}\pi\right)^{\mathrm{6}} }{\mathrm{2}.\mathrm{6}!}.\frac{\mathrm{1}}{\mathrm{42}}=\frac{\pi^{\mathrm{6}} }{\mathrm{15}.\mathrm{63}}=\frac{\pi^{\mathrm{6}} }{\mathrm{945}} \\ $$$$\zeta\left(\mathrm{8}\right)=\frac{\pi^{\mathrm{8}} }{\mathrm{9450}} \\ $$$$\zeta\left(\mathrm{10}\right)=\frac{\pi^{\mathrm{10}} }{\mathrm{93555}} \\ $$$$…. \\ $$
Commented by Dwaipayan Shikari last updated on 26/Dec/20
((sinπx)/(πx))=Π^∞ (1−(x^2 /n^2 )) log(sinπx)−log(πx)=Σ_(n=1) ^∞ log(1−(x^2 /n^2 )) π((cosπx)/(sinπx))−(π/(πx))=Σ_(n=1) ^∞ ((−((2x)/n^2 ))/(1−(x^2 /n^2 )))⇒πcot(πx)−(1/x)=Σ_(n=1) ^∞ ((2x)/(x^2 −n^2 )) ⇒πxcot(πx)=1−2Σ_(n=1) ^∞ ((x^2 /n^2 )/(1−(x^2 /n^2 )))⇒πxcot(πx)=1−2Σ_(n=1) ^∞ Σ_(k=1) ^∞ ((x/n))^(2k) ⇒πxcot(πx)=1−2Σ_(k=1) ^∞ ζ(2k)x^(2k) ⇒πxi(((e^(πxi) +e^(−πxi) )/(e^(πxi) −e^(−πxi) )))=1−2Σ^∞ ζ(2k)x^(2k) ⇒πxi+((2πxi)/(e^(2πxi) −1))=1−2Σ_(k≥1) ^∞ ζ(2k)x^(2k) ⇒1−Σ_(k≥0) ^∞ (β_(2k) /(2(2k!)))(2πix)^(2k) ζ(2k)x^(2k) =(β_(2k) /(2(2k!)))(2πix)^(2k) ⇒ζ(2k)=(−1)^(k+1) ((β_(2k) (2π)^(2k) )/(2(2k!)))
$$\frac{{sin}\pi{x}}{\pi{x}}=\overset{\infty} {\prod}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${log}\left({sin}\pi{x}\right)−{log}\left(\pi{x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{log}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$\pi\frac{{cos}\pi{x}}{{sin}\pi{x}}−\frac{\pi}{\pi{x}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\frac{\mathrm{2}{x}}{{n}^{\mathrm{2}} }}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }}\Rightarrow\pi{cot}\left(\pi{x}\right)−\frac{\mathrm{1}}{{x}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\pi{xcot}\left(\pi{x}\right)=\mathrm{1}−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }}\Rightarrow\pi{xcot}\left(\pi{x}\right)=\mathrm{1}−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{x}}{{n}}\right)^{\mathrm{2}{k}} \\ $$$$\Rightarrow\pi{xcot}\left(\pi{x}\right)=\mathrm{1}−\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} \Rightarrow\pi{xi}\left(\frac{{e}^{\pi{xi}} +{e}^{−\pi{xi}} }{{e}^{\pi{xi}} −{e}^{−\pi{xi}} }\right)=\mathrm{1}−\mathrm{2}\overset{\infty} {\sum}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} \\ $$$$\Rightarrow\pi{xi}+\frac{\mathrm{2}\pi{xi}}{{e}^{\mathrm{2}\pi{xi}} −\mathrm{1}}=\mathrm{1}−\mathrm{2}\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} \Rightarrow\mathrm{1}−\underset{{k}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\beta_{\mathrm{2}{k}} }{\mathrm{2}\left(\mathrm{2}{k}!\right)}\left(\mathrm{2}\pi{ix}\right)^{\mathrm{2}{k}} \\ $$$$\zeta\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}} =\frac{\beta_{\mathrm{2}{k}} }{\mathrm{2}\left(\mathrm{2}{k}!\right)}\left(\mathrm{2}\pi{ix}\right)^{\mathrm{2}{k}} \Rightarrow\zeta\left(\mathrm{2}{k}\right)=\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\beta_{\mathrm{2}{k}} \left(\mathrm{2}\pi\right)^{\mathrm{2}{k}} }{\mathrm{2}\left(\mathrm{2}{k}!\right)}\:\: \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 26/Dec/20
very nice mr payan solution with explanation..
$${very}\:{nice}\:\:{mr}\:\:{payan} \\ $$$${solution}\:\:{with}\:{explanation}.. \\ $$
Answered by liberty last updated on 26/Dec/20
via Fourier sine series b_n =(2/π)∫_0 ^( π) (πx−x^2 )sin (nx) dx = (4/π).((1−(−1)^(n+1) )/n^3 ) b_n = { ((0 ; if n is odd)),(((8/(πn^3 )) ; if n even )) :} writting n=2k−1 for some positive integer k we get πx−x^2 ∼ Σ_(k=1) ^∞ ((8sin ((2k−1)x))/(π(2k−1)^3 )) (2/π)∫_0 ^( π) (πx−x^2 )^2 dx = Σ_(k=1) ^∞ ((8/(π(2k−1)^3 )))^2 simplifying this yields (π^4 /(15)) = ((64)/π^2 ) Σ_(k=1) ^∞ ((1/(2k−1)))^6 and Σ_(k=1) ^∞ (1/((2k−1)^6 )) = (π^6 /(960)) however Σ_(k=1) ^∞ (1/((2k−1)^6 )) = Σ_(n=1) ^∞ (1/n^6 ) − Σ_(n=1) ^∞ (1/(2n)) = (1−(1/2^6 ))Σ_(n=1) ^∞ (1/n^6 ) = ((63)/(64)) ζ(6) ζ(6) = ((64)/(63)).(π^6 /(960)) = (π^6 /(945)).
$${via}\:{Fourier}\:{sine}\:{series}\: \\ $$$$\:{b}_{{n}} =\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \left(\pi{x}−{x}^{\mathrm{2}} \right)\mathrm{sin}\:\left({nx}\right)\:{dx}\:=\:\frac{\mathrm{4}}{\pi}.\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} } \\ $$$$\:{b}_{{n}} \:=\:\begin{cases}{\mathrm{0}\:;\:{if}\:{n}\:{is}\:{odd}}\\{\frac{\mathrm{8}}{\pi{n}^{\mathrm{3}} }\:;\:{if}\:{n}\:{even}\:}\end{cases} \\ $$$$\:{writting}\:{n}=\mathrm{2}{k}−\mathrm{1}\:{for}\:{some}\:{positive}\:{integer}\:{k} \\ $$$${we}\:{get}\:\pi{x}−{x}^{\mathrm{2}} \:\sim\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{8sin}\:\left(\left(\mathrm{2}{k}−\mathrm{1}\right){x}\right)}{\pi\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{3}} }\: \\ $$$$\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \left(\pi{x}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:{dx}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{8}}{\pi\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{3}} }\right)^{\mathrm{2}} \\ $$$${simplifying}\:{this}\:{yields} \\ $$$$\frac{\pi^{\mathrm{4}} }{\mathrm{15}}\:=\:\frac{\mathrm{64}}{\pi^{\mathrm{2}} }\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right)^{\mathrm{6}} \:{and}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{6}} }\:=\:\frac{\pi^{\mathrm{6}} }{\mathrm{960}} \\ $$$${however}\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{6}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{6}} }\:−\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}}\:=\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{6}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{63}}{\mathrm{64}}\:\zeta\left(\mathrm{6}\right) \\ $$$$\zeta\left(\mathrm{6}\right)\:=\:\frac{\mathrm{64}}{\mathrm{63}}.\frac{\pi^{\mathrm{6}} }{\mathrm{960}}\:=\:\frac{\pi^{\mathrm{6}} }{\mathrm{945}}. \\ $$
Commented by mnjuly1970 last updated on 26/Dec/20
very nice bravo mr liberty excellent...
$${very}\:{nice}\:{bravo}\:{mr}\:\:\:{liberty} \\ $$$$\:{excellent}… \\ $$

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