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Question Number 45356 by MrW3 last updated on 12/Oct/18

x=(1/(1+(1/(2+(1/(3+(1/(4+...))))))))=?

$${x}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}+...}}}}=? \\ $$

Answered by MJS last updated on 12/Oct/18

non−periodic infinite continued fractions are transcendental numbers, at least I seem to remember this. we can only approximate in this case the sequence is 1, (2/3), (7/(10)), ((30)/(43)), ((157)/(224)), ((972)/(1393)), ((6961)/(9976)), ((56660)/(81201)), ((516901)/(740785)), ((5225670)/(7489051)), ... the difference between the last two fractions is (1/(5547776645035))<10^(−12) I get x≈.697774657964

$$\mathrm{non}−\mathrm{periodic}\:\mathrm{infinite}\:\mathrm{continued}\:\mathrm{fractions} \\ $$$$\mathrm{are}\:\mathrm{transcendental}\:\mathrm{numbers},\:\mathrm{at}\:\mathrm{least}\:\mathrm{I}\:\mathrm{seem} \\ $$$$\mathrm{to}\:\mathrm{remember}\:\mathrm{this}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{is} \\ $$$$\mathrm{1},\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{7}}{\mathrm{10}},\:\frac{\mathrm{30}}{\mathrm{43}},\:\frac{\mathrm{157}}{\mathrm{224}},\:\frac{\mathrm{972}}{\mathrm{1393}},\:\frac{\mathrm{6961}}{\mathrm{9976}},\:\frac{\mathrm{56660}}{\mathrm{81201}},\:\frac{\mathrm{516901}}{\mathrm{740785}},\:\frac{\mathrm{5225670}}{\mathrm{7489051}},\:... \\ $$$$\mathrm{the}\:\mathrm{difference}\:\mathrm{between}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{fractions} \\ $$$$\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{5547776645035}}<\mathrm{10}^{−\mathrm{12}} \\ $$$$\mathrm{I}\:\mathrm{get}\:{x}\approx.\mathrm{697774657964} \\ $$

Commented by MrW3 last updated on 12/Oct/18

Thank you for this answer sir!

$${Thank}\:{you}\:{for}\:{this}\:{answer}\:{sir}! \\ $$

Commented by MJS last updated on 12/Oct/18

Sir, may I offer you these as a gift: these are by your special friend Lambert: (4/π)=1+(1^2 /(3+(2^2 /(5+(3^2 /(7+(4^2 /(9+(5^2 /⋱))))))))) tan x =(x/(1−(x^2 /(3−(x^2 /(5−(x^2 /(7−(x^2 /⋱))))))))) these are by Euler: e−2=(1/(1+(1/(2+(1/(1+(1/(1+(1/(4+(1/(1+(1/(1+(1/(6+(1/⋱)))))))))))))))))=(1/(1+(1/(2+(2/(3+(3/(4+(4/(5+(5/(6+(6/(7+(7/(8+(8/⋱)))))))))))))))))

$$\mathrm{Sir},\:\mathrm{may}\:\mathrm{I}\:\mathrm{offer}\:\mathrm{you}\:\mathrm{these}\:\mathrm{as}\:\mathrm{a}\:\mathrm{gift}: \\ $$$$ \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{by}\:\mathrm{your}\:\mathrm{special}\:\mathrm{friend}\:\mathrm{Lambert}: \\ $$$$\frac{\mathrm{4}}{\pi}=\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{3}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{5}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{7}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{9}+\frac{\mathrm{5}^{\mathrm{2}} }{\ddots}}}}} \\ $$$$\mathrm{tan}\:{x}\:=\frac{{x}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}−\frac{{x}^{\mathrm{2}} }{\mathrm{7}−\frac{{x}^{\mathrm{2}} }{\ddots}}}}} \\ $$$$ \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{by}\:\mathrm{Euler}: \\ $$$$\mathrm{e}−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{1}}{\ddots}}}}}}}}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}+\frac{\mathrm{3}}{\mathrm{4}+\frac{\mathrm{4}}{\mathrm{5}+\frac{\mathrm{5}}{\mathrm{6}+\frac{\mathrm{6}}{\mathrm{7}+\frac{\mathrm{7}}{\mathrm{8}+\frac{\mathrm{8}}{\ddots}}}}}}}}} \\ $$

Commented by MrW3 last updated on 13/Oct/18

very interesting, thank you sir.

$${very}\:{interesting},\:{thank}\:{you}\:{sir}. \\ $$

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