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pi-2-1-pi-2-3-pi-2-9pi-2-5-3pi-2-25pi-2-7-5pi-2-49pi-2-9-7pi-2-81pi-2-11-9pi-2-121pi-2-




Question Number 127682 by Dwaipayan Shikari last updated on 31/Dec/20
(π^2 /(1+(π^2 /(3−π^2 +((9π^2 )/(5−3π^2 +((25π^2 )/(7−5π^2 +((49π^( 2) )/(9−7π^2 +((81π^2 )/(11−9π^2 +((121π^2 )/(.....))))))))))))))
π21+π23π2+9π253π2+25π275π2+49π297π2+81π2119π2+121π2..
Commented by Dwaipayan Shikari last updated on 01/Jan/21
πtan^(−1) (π)
πtan1(π)
Commented by A8;15: last updated on 03/Jan/21
It will be nice if you show us a proof,sir!
Itwillbeniceifyoushowusaproof,sir!
Answered by Dwaipayan Shikari last updated on 03/Jan/21
tan^(−1) x=x−(x^3 /3)+(x^5 /5)−(x^7 /7)+....           =  x+x(−(x^2 /3))+x(−(x^2 /3))(−((3x^2 )/5))+x(−(x^2 /3))(−((3x^2 )/5))(−((5x^2 )/7))+..         =  (x/(1+((x^2 /3)/(1−(x^2 /3)+((x^2 /5)/(1−(x^2 /5)+((x^2 /7)/(1−(x^2 /7)+...))))))))=(x/(1+(x^2 /(3−x^2 +((3^2 x^2 )/(5−3x^2 +((5^2 x^2 )/(7−5x^2 +..))))))))  Take x=π
tan1x=xx33+x55x77+.=x+x(x23)+x(x23)(3x25)+x(x23)(3x25)(5x27)+..=x1+x231x23+x251x25+x271x27+=x1+x23x2+32x253x2+52x275x2+..Takex=π
Commented by Dwaipayan Shikari last updated on 03/Jan/21
https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula
Commented by A8;15: last updated on 03/Jan/21
  Thanks sir. My respect for you!
Thankssir.Myrespectforyou!

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