Question-121875

Question Number 121875 by rs4089 last updated on 12/Nov/20

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

((sinπx)/(πx))=Π_(n=1) ^∞ (1−(x^2 /n^2 )) ((sinu)/u)=Π_(n=1) ^∞ (1−(u^2 /(π^2 n^2 ))) 1−(u^2 /6)+(u^4 /(120))−...=(1−(u^2 /π^2 ))(1−(u^2 /(4π^2 )))(1−(u^2 /(9π^2 )))(1−(u^2 /(16π^2 )))... 1−(u^2 /6)+(u^4 /(120))−...=1−((u^2 /π^2 )+(u^2 /(4π^2 ))+...)+(3/4)((u^4 /π^4 )+(u^4 /(16π^4 ))+...)+.... (u^2 /6)=u^2 ((1/π^2 )+(1/(4π^2 ))+...) 1+(1/4)+(1/9)+..=(π^2 /6) (u^4 /(120))=(3/4)((u^4 /π^4 )+(u^4 /(16π^4 ))+.....) 1+(1/(16))+(1/(81))+...=(π^4 /(90))

$$\frac{{sin}\pi{x}}{\pi{x}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$\frac{{sinu}}{{u}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{6}}+\frac{{u}^{\mathrm{4}} }{\mathrm{120}}−…=\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{9}\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{16}\pi^{\mathrm{2}} }\right)… \\ $$$$\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{6}}+\frac{{u}^{\mathrm{4}} }{\mathrm{120}}−…=\mathrm{1}−\left(\frac{{u}^{\mathrm{2}} }{\pi^{\mathrm{2}} }+\frac{{u}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }+…\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{{u}^{\mathrm{4}} }{\pi^{\mathrm{4}} }+\frac{{u}^{\mathrm{4}} }{\mathrm{16}\pi^{\mathrm{4}} }+…\right)+…. \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{6}}={u}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} }+…\right) \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}+..=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\frac{{u}^{\mathrm{4}} }{\mathrm{120}}=\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{{u}^{\mathrm{4}} }{\pi^{\mathrm{4}} }+\frac{{u}^{\mathrm{4}} }{\mathrm{16}\pi^{\mathrm{4}} }+…..\right) \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{81}}+…=\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$

Commented by mnjuly1970 last updated on 12/Nov/20

$${nice}\:\:{very}\:{nice}\: \\ $$$${without}\:{using}\:{the}\:{fourier}\:{series}\: \\ $$$${thank}\:{you}… \\ $$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

�� with pleasure sir!

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