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Question-137171




Question Number 137171 by mnjuly1970 last updated on 30/Mar/21
Answered by Dwaipayan Shikari last updated on 30/Mar/21
log(2)=1+1(−(1/2))+1(−(1/2))(−(2/3))+...  =(1/(1+((1/2)/(1−(1/2)+((2/3)/(1−(2/3)+((3/4)/(1−(3/4)+..))))))))=(1/(1+(1/(1+(2^2 /(1+(3^2 /(1+(4^2 /(1+...))))))))))  (1/(log(2)))=1+(1^2 /(1+(2^2 /(1+(3^2 /(1+..))))))
$${log}\left(\mathrm{2}\right)=\mathrm{1}+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)+… \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}+..}}}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{1}+…}}}}} \\ $$$$\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}=\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+..}}} \\ $$
Commented by Dwaipayan Shikari last updated on 30/Mar/21
Euler continued fractions  a_0 +a_0 a_1 +a_0 a_1 a_2 +...=(a_0 /(1−(a_1 /(1+a_1 +(a_2 /(1+a_2 −(a_3 /(1+a_3 −(a_4 /(1+a_4 −..))))))))))
$${Euler}\:{continued}\:{fractions} \\ $$$${a}_{\mathrm{0}} +{a}_{\mathrm{0}} {a}_{\mathrm{1}} +{a}_{\mathrm{0}} {a}_{\mathrm{1}} {a}_{\mathrm{2}} +…=\frac{{a}_{\mathrm{0}} }{\mathrm{1}−\frac{{a}_{\mathrm{1}} }{\mathrm{1}+{a}_{\mathrm{1}} +\frac{{a}_{\mathrm{2}} }{\mathrm{1}+{a}_{\mathrm{2}} −\frac{{a}_{\mathrm{3}} }{\mathrm{1}+{a}_{\mathrm{3}} −\frac{{a}_{\mathrm{4}} }{\mathrm{1}+{a}_{\mathrm{4}} −..}}}}} \\ $$
Commented by mnjuly1970 last updated on 30/Mar/21
grateful ... mr payan  thanks alot ....
$${grateful}\:…\:{mr}\:{payan} \\ $$$${thanks}\:{alot}\:…. \\ $$

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