1-1-2-1-1-2-2-1-3-2-1-3-2-2-2-1-4-3-1-3-5-2-3-2-4-pi-Prove-the-above-Relation-

Question Number 128256 by Dwaipayan Shikari last updated on 05/Jan/21

1+(1/(2!1!))((1/2))^2 +(1/(3!2!))(((1.3)/2^2 ))^2 +(1/(4!3!))(((1.3.5)/2^3 ))^2 +....=(4/π) Prove the above Relation

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!\mathrm{1}!}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!\mathrm{2}!}\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}!\mathrm{3}!}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right)^{\mathrm{2}} +….=\frac{\mathrm{4}}{\pi} \\ $$$${Prove}\:{the}\:{above}\:{Relation} \\ $$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

1+Σ_(n≥1) ^∞ ((((1/2))_n ^2 )/((2)_n n!))= _2 F_1 ((1/2),(1/2);2;1)=((Γ(1)Γ(2−1))/(Γ^2 ((3/2))))=(4/π)

$$\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left(\mathrm{2}\right)_{{n}} {n}!}=\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{2};\mathrm{1}\right)=\frac{\Gamma\left(\mathrm{1}\right)\Gamma\left(\mathrm{2}−\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\mathrm{4}}{\pi} \\ $$

Commented by A8;15: last updated on 06/Jan/21

Mr. can we prove this relation with continued fraction ?

$$\mathrm{Mr}.\:\mathrm{can}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{relation}\:\mathrm{with} \\ $$$$\mathrm{continued}\:\mathrm{fraction}\:? \\ $$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

$${Hey}\:{sir}!\:{I}\:{am}\:{a}\:{student}.\:{Kindly}\:{don}'{t}\:{call}\:{me}\:{Sir}. \\ $$$${I}\:\:{also}\:{think}\:{this}\:{can}\:{be}\:{proved}\:{by}\:{Continued}\:{fractions} \\ $$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

tan^(−1) x=(x/(1+(x^2 /(3−x^2 +((9x^2 )/(5−3x^2 +((25x^2 )/(7−5x^2 +((49x^2 )/(..)))))))))) (π/4)=(1/(1+(1/(2+(3^2 /(2+(5^2 /(2+(7^2 /(2+...))))))))))⇒(4/π)=1+(1/(2+(3^2 /(2+(5^2 /(2+(7^2 /(2+(9^2 /(2+...))))))))))

$${tan}^{−\mathrm{1}} {x}=\frac{{x}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}−{x}^{\mathrm{2}} +\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{25}{x}^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}{x}^{\mathrm{2}} +\frac{\mathrm{49}{x}^{\mathrm{2}} }{..}}}}} \\ $$$$\frac{\pi}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{2}+…}}}}}\Rightarrow\frac{\mathrm{4}}{\pi}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{2}+…}}}}} \\ $$

Commented by A8;15: last updated on 06/Jan/21

I respect you. May I call you brother? (I think I am smaller than you, so you are my elder brother)

$$\mathrm{I}\:\mathrm{respect}\:\mathrm{you}.\:\mathrm{May}\:\mathrm{I}\:\mathrm{call}\:\mathrm{you}\:\mathrm{brother}? \\ $$$$\left(\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{am}\:\mathrm{smaller}\:\mathrm{than}\:\mathrm{you},\:\mathrm{so}\:\mathrm{you}\:\mathrm{are}\:\mathrm{my}\:\mathrm{elder}\:\mathrm{brother}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 23/Jan/21

$${My}\:{age}\:{e}^{\frac{\mathrm{34}}{\mathrm{12}}} \\ $$

Facebook Tweet Pin

1-1-2-1-1-2-2-1-3-2-1-3-2-2-2-1-4-3-1-3-5-2-3-2-4-pi-Prove-the-above-Relation-

Leave a Reply Cancel reply