- mathfun@mailman.xmission.com - mailman.xmission.com

[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."

1 0

[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

1 0

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 *****************************************

1 0

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 *****************************************

1 0

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 *****************************************

1 0

[math-fun] Golden Ratio
by françois mendzina essomba2 09 Oct '17

09 Oct '17

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+...))))));

1 0

Re: [math-fun] Random slice of a cube
by Bill Gosper 09 Oct '17

09 Oct '17

On 2017-10-08 21:16, Keith F. Lynch wrote: > James Propp <jamespropp(a)gmail.com> wrote: >> If I choose a random plane that intersects a cube, creating a >> polygon, how many sides does the polygon have on average? > >> (The most lowbrow way to define the probability measure I have in >> mind when I say "random plane" is to circumscribe a sphere of radius >> r around the cube with center O. Choose a point uniformly P at >> random on the sphere, and now choose a point Q uniformly at random >> on segment OP. ....) > > What about choosing three points independently and uniformly on a > sphere that encloses the cube, and using the plane defined by those > three points, skipping any planes that don't intersect the cube? > Would that get you the same answer? > > I am not a visual thinker. I'm not even sure what numbers of sides > the polygon could have. It can obviously have 3, 4, or 6. I doubt it > can have more than 6, but I'm not sure. I'm even less sure whether it > can have 5. gosper.org/pentagonal cubeslice.png --rwg > > I would ask, not what the average number of sides is, but the > proportion of each number of sides. That way I can quickly get any > of countless possible means, including the arithmetic mean, geometric > mean, harmonic mean, mode, median, root-mean-square, etc. Maybe > some of them are more interesting than others. Also, what's the > distribution of areas of the intersection? And what are the answers > to the same questions for the other four regular solids, and for the > sphere? Similarly in other dimensions. >

2 1

Re: [math-fun] Random slice of a cube
by Keith F. Lynch 08 Oct '17

08 Oct '17

James Propp <jamespropp(a)gmail.com> wrote: > If I choose a random plane that intersects a cube, creating a > polygon, how many sides does the polygon have on average? > (The most lowbrow way to define the probability measure I have in > mind when I say "random plane" is to circumscribe a sphere of radius > r around the cube with center O. Choose a point uniformly P at > random on the sphere, and now choose a point Q uniformly at random > on segment OP. ....) What about choosing three points independently and uniformly on a sphere that encloses the cube, and using the plane defined by those three points, skipping any planes that don't intersect the cube? Would that get you the same answer? I am not a visual thinker. I'm not even sure what numbers of sides the polygon could have. It can obviously have 3, 4, or 6. I doubt it can have more than 6, but I'm not sure. I'm even less sure whether it can have 5. I would ask, not what the average number of sides is, but the proportion of each number of sides. That way I can quickly get any of countless possible means, including the arithmetic mean, geometric mean, harmonic mean, mode, median, root-mean-square, etc. Maybe some of them are more interesting than others. Also, what's the distribution of areas of the intersection? And what are the answers to the same questions for the other four regular solids, and for the sphere? Similarly in other dimensions.

1 0

Re: [math-fun] Random slice of a cube
by Dan Asimov 08 Oct '17

08 Oct '17

Another pretty lowbrow way to define a probability measure on the planes meeting a cube is to first pick a random point in the cube, then pick a random direction through that point and take the perpendicular plane. Where random means uniformly distributed. Maybe like the Bertrand paradox (about a random chord of the circle) there are multiple "reasonable" probability measures to choose from, with inequivalent measures? —Dan Jim Propp wrote: ----- If I choose a random plane that intersects a cube, creating a polygon, how many sides does the polygon have on average? (The most lowbrow way to define the probability measure I have in mind when I say "random plane" is to circumscribe a sphere of radius r around the cube with center O. Choose a point uniformly P at random on the sphere, and now choose a point Q uniformly at random on segment OP. If Q is in the cube, erect a plane through Q perpendicular to OP. Otherwise, start over. The most highbrow way is to go via the standard measure on the affine Grassmannian of all planes in 3-space; if we restrict to the set of planes that pass through the cube, we get a finite measure, so we can rescale to get a probability measure.) -----

2 1

[math-fun] Random slice of a cube
by James Propp 08 Oct '17

08 Oct '17

If I choose a random plane that intersects a cube, creating a polygon, how many sides does the polygon have on average? (The most lowbrow way to define the probability measure I have in mind when I say "random plane" is to circumscribe a sphere of radius r around the cube with center O. Choose a point uniformly P at random on the sphere, and now choose a point Q uniformly at random on segment OP. If Q is in the cube, erect a plane through Q perpendicular to OP. Otherwise, start over. The most highbrow way is to go via the standard measure on the affine Grassmannian of all planes in 3-space; if we restrict to the set of planes that pass through the cube, we get a finite measure, so we can rescale to get a probability measure.) Jim Propp

2 3