October 2017 - mathfun@mailman.xmission.com

[math-fun] Winning number
by James Propp 14 Oct '17

14 Oct '17

The number 2 (\[Pi] - ArcTan[2 \[Pi]]) turned up in a problem I'm working on, and I asked Mathematica to compute it numerically; it reported the answer as 3.45726 My instinctive reaction was that 3,4,5,7,2,6 would be a really good hand in some version of poker. Jim Propp

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[math-fun] Golden ratio and Pi
by François Mendzina Essomba 13 Oct '17

13 Oct '17

Bonjour à tous : Two small formulas for the day. formula 7 Golden ratio in an infinite nested root (Phi)^(-1)=(exp(-Pi)*sqrt(5*exp(2*Pi)-4)-1)/2+exp(-Pi)/sqrt(1+sqrt((5*exp(2*Pi)-4)+(5*exp(2*Pi)-4)*sqrt((5*exp(2*Pi)-4)+(5*exp(2*Pi)-4)*sqrt((5*exp(2*Pi)-4)+(5*exp(2*Pi)-4)*sqrt((5*exp(2*Pi)-4)+(5*exp(2*Pi)-4)*sqrt((5*exp(2*Pi)-4)+(5*exp(2*Pi)-4)*sqrt((5*exp(2*Pi)-4)+...))))))); formula 8 a small algorithm for Pi ^ 4 u_(0)=-1; u_(n+1)=(1/2)*sqrt(2+2*u_(n)); a_(n)=2^(4*n+5)*(3-2*sqrt(2+2*u_(n))+u_(n)); n → infinity ; a_(n) → Pi^4;

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[math-fun] Largest sum-free subset of {0,1,2,...,2n}
by James Propp 13 Oct '17

13 Oct '17

Prove that if S is a subset of {0,1,2,...,2n} containing more than n elements, then there must exist x,y,z in S such that x+y=z. (This smells like an Olympiad problem to me. In fact, I may well have solved this very problem forty years ago, but I don't see how to solve it now. Some sort of pigeonhole argument, sure, but how to set it up?) The bound is tight; both 1,3,5,...,2n-1 and n+1,...,2n are sum-free. Jim Propp

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[math-fun] Unsolved puzzle from Le Monde
by Don Reble 13 Oct '17

13 Oct '17

> First problem: in Alice's small city, 10 council members had to be > elected. Each voter could vote for 10 candidates, out of a list of 25. > What is the largest possible number of voters... > [such that] every voter has at least one of the candidates he voted > for elected in the council, no matter what... > First problem: 1358 One can show that 1358 voters can be satisfied, using a fairly obvious algorithm to choose the elected from the ballots. The problem is to rig 1359 ballots so that (at least) one voter must be left unsatisfied. Here's an easier (?) derived problem: there are 17 candidates; 2 will be chosen; and 8 voters, each voting for 10. (7 voters can be satisfied.) Can you rig 8 ballots? -- Don Reble djr(a)nk.ca

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[math-fun] Formulas (5, 6) ... number pi
by françois mendzina essomba2 12 Oct '17

12 Oct '17

Hello,Two smallapproximation formulas of pi. Formula 5:A formula that weighs,the number of terms explodes with m. Pi=Limit(1/2^(m-1)*(1+2*Sum(sqrt(1-((n^2)*2^(-2*m))),n=1..2^(m)-1)),m=infinity); Formula 6:if Lambda istaken in the interval ] 1; 1 [ and different from 0, -1 / 2,1 / 2, -1 / 4,1 / 4. Pi=(1/lambda)*Sum(((-1)^(n+1)/(2*n-1))*((tan(lambda*Pi))^(2*n-1))*cos((4*n-2)*lambda*Pi),n=1..infinity)+(1/lambda)*Sum(((-1)^(n+1)/(2*n))*((tan(lambda*Pi))^(2*n))*sin((4*n)*lambda*Pi),n=1..infinity);

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[math-fun] infinity minus one
by Henry Baker 11 Oct '17

11 Oct '17

FYI -- Mary Bono must have had a physicist on her staff: https://en.wikipedia.org/wiki/Copyright_Term_Extension_Act "Actually, Sonny wanted the term of copyright protection to last forever. I am informed by staff that such a change would violate the Constitution. ... As you know, there is also [then-MPAA president] Jack Valenti's proposal for term to last *forever less one day*. Perhaps the Committee may look at that next Congress." -- Mary Bono, speaking on the floor of the House https://en.wikipedia.org/wiki/Renormalization "Renormalization is a collection of techniques in quantum field theory, the statistical mechanics of fields, and the theory of self-similar geometric structures, that are used to treat *infinities* arising in calculated quantities by altering values of quantities to compensate for effects of their self-interactions." "Renormalization replaces the originally postulated mass and charge with new numbers such that the observed mass and charge matches those originally postulated."

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[math-fun] Unsolved puzzle from Le Monde
by Seb Perez-D 11 Oct '17

11 Oct '17

Dear all, the French newspaper Le Monde has a weekly math-oriented puzzle. In number 1021 the authors (Elisabeth Busser, Gilles Cohen, Jean-Louis Legrand) have not been able to convince readers of their solution to their puzzle, and seem not to be too sure that their solution is correct. The authors have invited everyone to provide a convincing solution - which I have not been able to do. I thus turn to the collective intelligence of the math-fun list! The puzzle reads as follows (my translation, simplifying a bit from http://www.affairedelogique.com/espace_probleme.php?corps=probleme&num=1021 ): In a democratically advanced country, elections to the municipal council are run in a particular way. The council has a pre-determined size. Once the number of candidates is known, it is decided for how many candidates each voter has to vote [all votes have to be cast; it is not allowed to vote twice for the same candidate]. The number of votes is set so that the composition of the elected council (to be selected by an algorithm based on all the votes) is such that every voter has at least one of the candidates he voted for elected in the council, no matter what the votes where. First problem: in Alice's small city, 10 council members had to be elected. Each voter could vote for 10 candidates, out of a list of 25. What is the largest possible number of voters? Second problem: in Bob's medium-sized city, there were 55,555 voters and 33 candidates. By some coincidence, as in Alice's case, the number of votes that each voter can cast is equal to the size of the council. What is, at the minimum, the size of the council? Cheers Sébastien ---- solutions of the puzzle authors below, without the justification (which I can make available in French if needed) ----- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ---- First problem: 1358 Second problem: 14

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Re: [math-fun] math-fun Digest, Vol 176, Issue 12
by françois mendzina essomba2 10 Oct '17

10 Oct '17

formula 3 periodicity of order 11 ; \\ 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+… )))))))))))))))))))))))); formula 4 ; \\ 165580141*phi^(-1)-102334155=1/(sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+...)))))); Le Mardi 10 octobre 2017 20h07, françois mendzina essomba2 <m_essob(a)yahoo.fr> a écrit : formulas 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 ; Le Mardi 10 octobre 2017 20h02, françois mendzina essomba2 <m_essob(a)yahoo.fr> a écrit : formulas 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 ; Le Mardi 10 octobre 2017 19h57, françois mendzina essomba2 <m_essob(a)yahoo.fr> a écrit : Formula1 a:=(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(5)*sqrt(exp(2*Pi)+exp(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+…)))))))))) ; Le Mardi 10 octobre 2017 19h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. Golden Ratio (fran?ois mendzina essomba2) ---------------------------------------------------------------------- Message: 1 Date: Tue, 10 Oct 2017 03:40:08 +0000 (UTC) From: fran?ois mendzina essomba2 <m_essob(a)yahoo.fr> To: "math-fun(a)mailman.xmission.com" <math-fun(a)mailman.xmission.com> Subject: [math-fun] Golden Ratio Message-ID: <201334138.7974682.1507606808311(a)mail.yahoo.com> Content-Type: text/plain; charset=UTF-8 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+...))))));? ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 12 *****************************************

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Re: [math-fun] math-fun Digest, Vol 176, Issue 12
by françois mendzina essomba2 10 Oct '17

10 Oct '17

formulas 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 ; Le Mardi 10 octobre 2017 20h02, françois mendzina essomba2 <m_essob(a)yahoo.fr> a écrit : formulas 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 ; Le Mardi 10 octobre 2017 19h57, françois mendzina essomba2 <m_essob(a)yahoo.fr> a écrit : Formula1 a:=(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(5)*sqrt(exp(2*Pi)+exp(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+…)))))))))) ; Le Mardi 10 octobre 2017 19h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. Golden Ratio (fran?ois mendzina essomba2) ---------------------------------------------------------------------- Message: 1 Date: Tue, 10 Oct 2017 03:40:08 +0000 (UTC) From: fran?ois mendzina essomba2 <m_essob(a)yahoo.fr> To: "math-fun(a)mailman.xmission.com" <math-fun(a)mailman.xmission.com> Subject: [math-fun] Golden Ratio Message-ID: <201334138.7974682.1507606808311(a)mail.yahoo.com> Content-Type: text/plain; charset=UTF-8 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+...))))));? ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 12 *****************************************

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Re: [math-fun] math-fun Digest, Vol 176, Issue 12
by françois mendzina essomba2 10 Oct '17

10 Oct '17

Formula1 a:=(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(5)*sqrt(exp(2*Pi)+exp(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+…)))))))))) ; Le Mardi 10 octobre 2017 19h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. Golden Ratio (fran?ois mendzina essomba2) ---------------------------------------------------------------------- Message: 1 Date: Tue, 10 Oct 2017 03:40:08 +0000 (UTC) From: fran?ois mendzina essomba2 <m_essob(a)yahoo.fr> To: "math-fun(a)mailman.xmission.com" <math-fun(a)mailman.xmission.com> Subject: [math-fun] Golden Ratio Message-ID: <201334138.7974682.1507606808311(a)mail.yahoo.com> Content-Type: text/plain; charset=UTF-8 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+...))))));? ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 12 *****************************************

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