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Question Number 184969 by SEKRET last updated on 14/Jan/23

3 + (1/(6+(3^2 /(6+(5^2 /(6+(7^2 /(6+(9^2 /(6+......)) )) )) )) )) = 𝛑 provet that.

$$\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\:+\:\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{6}+......}\:\:\:}\:\:}\:\:}\:\:\:\:}\:\:=\:\boldsymbol{\pi} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{provet}}\:\:\boldsymbol{\mathrm{that}}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Frix last updated on 15/Jan/23

I know the proof for this one π=(4/(1+(1^2 /(2+(3^2 /(2+(5^2 /(2+(7^2 /(2+...)))))))))) which was found by John Wallis in 1665.

$$\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{for}\:\mathrm{this}\:\mathrm{one} \\ $$$$\pi=\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{2}+...}}}}} \\ $$$$\mathrm{which}\:\mathrm{was}\:\mathrm{found}\:\mathrm{by}\:{John}\:{Wallis}\:\mathrm{in}\:\mathrm{1665}. \\ $$

Commented by SEKRET last updated on 15/Jan/23

??

$$?? \\ $$

Answered by aba last updated on 14/Jan/23

[3]=3 [3;6]=3+(1/6)=((19)/6)=3,166... [3;6;6]=3+(1/(6+(9/6)))=((141)/(45))=3,1333... [3;6;6;6]=3+(1/(6+(9/(6+((25)/6)))))=((1321)/(420))=3,145... [3;6;6;6;6]=3+(1/(6+(9/(6+((25)/(6+((49)/6)))))))=((989)/(315))=3,13968... [3;6;6;6;6;6]=3+(1/(6+(9/(6+((25)/(6+((49)/(6+((121)/6)))))))))=((248851)/(79170))=3,143... [3;6;6;6;6..........∞]=π

$$\left[\mathrm{3}\right]=\mathrm{3} \\ $$$$\left[\mathrm{3};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{19}}{\mathrm{6}}=\mathrm{3},\mathrm{166}... \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}}}=\frac{\mathrm{141}}{\mathrm{45}}=\mathrm{3},\mathrm{1333}... \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}}}}=\frac{\mathrm{1321}}{\mathrm{420}}=\mathrm{3},\mathrm{145}... \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}+\frac{\mathrm{49}}{\mathrm{6}}}}}=\frac{\mathrm{989}}{\mathrm{315}}=\mathrm{3},\mathrm{13968}... \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}+\frac{\mathrm{49}}{\mathrm{6}+\frac{\mathrm{121}}{\mathrm{6}}}}}}=\frac{\mathrm{248851}}{\mathrm{79170}}=\mathrm{3},\mathrm{143}... \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}..........\infty\right]=\pi \\ $$

Commented by Frix last updated on 15/Jan/23

This is not a proof.

$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{proof}. \\ $$

Commented by aba last updated on 15/Jan/23

this is a proposal

$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{proposal} \\ $$

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