[math-fun] Golden Ratio
Hello,thank you foradding me in the Math-fun mailing list .I take thisopportunity to convey to you four formulas of the golden ratio.Thank you foryour timeand bestregards,formula 1:periodicity of order 4 a:=(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+e^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(sqrt(5)*sqrt(exp^(2*Pi)+e^Pi+1)); b:=(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)-1)/(2*(exp^(2*Pi)+exp^Pi+1))-(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(2*sqrt(5)*sqrt(exp^(2*Pi)+exp^Pi+1)); phi:=a/(b+1/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+…)))))))))); formula 2: r:=(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(sqrt(p)*sqrt(exp^(2*Pi)+exp^Pi+1));a:= ((p-1)/4)*r;b:=(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)-1)/(2*(exp^(2*Pi)+exp^Pi+1))-(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(2*sqrt(p)*sqrt(exp^(2*Pi)+exp^Pi+1));j:=a/(b+1/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+…))))))))));j :=(sqrt(p)+1)/2 ; formula3;periodicity of order 4 ;a:=sqrt(7132129815550981725745577127418524500105436241812778561)/(4983267051713428905499820187*sqrt(5));b:=-(sqrt(7132129815550981725745577127418524500105436241812778561)+2670593656810572179193662447*sqrt(5))/(9966534103426857810999640374*sqrt(5));phi:=a/(b+1/(1+1^1/(1+2^2/(1+3^3/(1+4^4/(1+5^5/(1+6^6/(1+7^7/(1+8^8/(1+9^9/(1+10^10/(1+11^11/(1+1^1/(1+2^2/(1+3^3/(1+4^4/(1+5^5/(1+6^6/(1+7^7/(1+8^8/(1+9^9/(1+10^10/(1+11^11/(1+… )))))))))))))))))))))))); formula4 ; 165580141*phi^(-1)-102334155=1/(sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+...))))));
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françois mendzina essomba2