August 2019 - mathfun@mailman.xmission.com

[math-fun] Generalization
by françois mendzina essomba2 23 Aug '19

23 Aug '19

I summarize all identities into general expression sum(((2*n)!*(2*cosh(x)+2*cos(t))^(-n-1/2)*(abs(sin(t)))^(2*n+1)*cosh((n+1/2)*x))/((2*n+1)*2^(2*n)*n!^2),n=0..inf)=t/2;t in ]-Pi/2, Pi/2[ sum(((2*n)!*(2*cosh(x)+2*cos(t))^(-n-1/2)*(sin(t))^(2*n+1)*sinh((n+1/2)*x))/((2*n+1)*2^(2*n)*n!^2),n=0..inf)=Pi/2-t/2; t in ]Pi/2, 0[U]0 , 3*Pi/2[ Best regards

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[math-fun] The (in)famous windmill problem
by Fred Lunnon 23 Aug '19

23 Aug '19

Insightful discussion by 3Blue1Brown of an elegant problem I hadn't seen before, at https://www.youtube.com/watch?v=M64HUIJFTZM WFL

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[math-fun] Identities
by françois mendzina essomba2 22 Aug '19

22 Aug '19

hi, some identities for Pi, sum((3^(n+1/2)*(2*n)!*(2*cosh(2*x)+1)^(-n-1/2)*cosh((2*n+1)*x))/((4*n+2)*4^(2*n)*n!^2),n=0..inf)=Pi/6; sum(((2*n)!*(2*cosh(2*x)+sqrt(3))^(-n-1/2)*cosh((2*n+1)*x))/((4*n+2)*4^(2*n)*n!^2),n=0..inf)=Pi/12; sum((2^(1/4-(7*n)/2)*(2*n)!*(sqrt(2)*cosh(2*x)+1)^(-n-1/2)*cosh((2*n+1)*x))/((4*n+2)*n!^2),n=0..inf)=Pi/8; sum((3^(n+1/2)*(2*n)!*(2*cosh(2*x)-1)^(-n-1/2)*sinh((2*n+1)*x))/((4*n+2)*4^(2*n)*n!^2),n=0..inf)=Pi/6; sum(((2*n)!*(2*cosh(2*x)-sqrt(3))^(-n-1/2)*sinh((2*n+1)*x))/((4*n+2)*4^(2*n)*n!^2),n=0..inf)=Pi/12; sum((2^(1/4-(7*n)/2)*(2*n)!*(sqrt(2)*cosh(2*x)-1)^(-n-1/2)*sinh((2*n+1)*x))/((4*n+2)*n!^2),n=0..inf)=Pi/8; sum(((2*n)!*(2*cosh(t)+sqrt(2))^(n+1/2)*sech(2*t)^(n+1/2)*sinh((n+1/2)*t))/((2*n+1)*2^(4*n)*n!^2),n=0..inf)=Pi/4; Best regards

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[math-fun] Identities
by françois mendzina essomba2 22 Aug '19

22 Aug '19

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[math-fun] identity
by françois mendzina essomba2 21 Aug '19

21 Aug '19

Hello, This identity seems true for some values of x. But we can delete x. Pi/4=sum(((2*n)!*(cosh(5/(x/sqrt(x^17+1)+1))/(cosh(5/(x/sqrt(x^17+1)+1))^2-1/2)-1/(sqrt(2)*(cosh(5/(x/sqrt(x^17+1)+1))^2-1/2)))^((2*n+1)/2)*cosh(((5/2)*(2*n+1))/(x/sqrt(x^17+1)+1)))/((2*n+1)*2^(4*n)*n!^2),n=0..inf); best ragard

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[math-fun] Sometimes Simplify doesn't do much.
by Bill Gosper 16 Aug '19

16 Aug '19

https://cs.uwaterloo.ca/~shallit/Papers/scf.ps (Much more: *https://tinyurl.com/y4kg4ng6 <https://tinyurl.com/y4kg4ng6>)* In[303]:= nxt[cf_] := Join[Drop[cf, -1], {cf[[-1]] + 1, cf[[-1]] - 1}, Reverse[cf[[2 ;; -2]]]] In[308]:= NestList[nxt, {0, u - 1, u + 1}, 4] // Column Out[308]= {0,-1+u,1+u} {0,-1+u,2+u,u,-1+u} {0,-1+u,2+u,u,u,-2+u,u,2+u,-1+u} {0,-1+u,2+u,u,u,-2+u,u,2+u,u,-2+u,2+u,u,-2+u,u,u,2+u,-1+u} {0,-1+u,2+u,u,u,-2+u,u,2+u,u,-2+u,2+u,u,-2+u,u,u,2+u,u,-2+u,2+u,u,u,-2+u,u,2+u,-2+u,u,2+u,u,-2+u,u,u,2+u,-1+u} In[312]:= Length /@ %308[[1]] Out[312]= {3, 5, 9, 17, 33} In[335]:= FromContinuedFraction@%308[[1, -1]] Out[335] CENSORED In[336]:= Simplify@% Out[336]= 1/u^32 + 1/u^16 + 1/u^8 + 1/u^4 + 1/u^2 + 1/u In[337]:= ByteCount@%% Out[337]= 931361224 A gigabyte! Note: Subtracting out u from %308 approaches a balanced ternary transcendental from which the continued fraction(u) is easily reconstructed. A090678 <http://oeis.org/A090678> —rwg

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[math-fun] Sum and difference of two Roth-type transcendentals are both Shallit-type
by Bill Gosper 15 Aug '19

15 Aug '19

Shallit must know this, but I'm a poor scholar. Explosive "partial floors", even case: In[278]:= ContinuedFraction[Sum[ 2^-2^(2 n), {n, 0, 11}], 32] Out[278]= {0, 1, 1, 3, 2, 255, 1, 1, 3, 1, 1, 4294967295, 2, 3, 1, 1, 255, 2, 3, 1, 1, 340282366920938463463374607431768211455, 2, 3, 2, 255, 1, 1, 3, 2, 4294967295, 1} Odd case: In[279]:= ContinuedFraction[Sum[2^-2^(2 n + 1), {n, 0, 11}], 32] Out[279]= {0, 3, 1, 15, 4, 65535, 1, 3, 15, 1, 3, 18446744073709551615, 1, 2, 1, 15, 3, 1, 65535, 4, 15, 1, 3, 115792089237316195423570985008687907853269984665640564039457584007913129639935, 1, 2, 1, 15, 4, 65535, 1, 3} Difference and sum have bounded, predictable partial floors: In[280]:= ContinuedFraction /@ {#1 - #2, #1 + #2} & @@FromContinuedFraction /@ {%%, %} Out[280]= { {0, 3, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 5, 1, 1, 2, 2, 33, 1, 2, 1, 1, 1, 4, 12, 292, 1, 1, 86, 1, 53, 5, 1, 2, 4, 5, 1, 6, 1, 1, 26, 23, 1, 3, 1, 16, 1, 2, 5, 3, 1, 4, 1, 2, 1, 2, 2, 1, 2, 3, 1, 17, 4, 2, 7, 6, 1, 5, 1, 29, 1, 8, 1, 1, 2, 1, 1, 2, 5, 1, 1, 3, 3, 1, 3, 2, 1, 21, 1, 1, 2, 1, 1, 1, 19, 3, 6, 1, 3, 2, 2, 2, 2, 62, 3, 1, 13, 1, 49, 1, 1, 3, 2, 5, 5, 6, 7, 4, 7, 1, 6, 8, 2, 1, 1, 1, 1, 1, 1, 196,3, 1, 2, 2, 1, 3, 1, 74, 832, 1, 29, 1, 5, 3472, 8, 1, 3, 3, 25, 1, 14, 1, 13, 4, 4, 6, 12, 1, 1, 6, 1, 2, 1, 2, 6, 1, 4, 10, 1, 1, 6, 3, 1, 4, 13, 1, 3, 1, 1, 1, 2, 2, 1, 5, 1, 1, 65, 6, 63, 2, 2, 1, 1, 3, 1, 2, 2, 1, 1, 4, 2, 37, 1, 1, 15, 1, 1, 3, 1, 1, 3, 4, 1, 3, 1,1, 2, 4, 7, 2, 4, 7, 1, 3, 1, 1, 2, 2, 2, 2, 2, 1, 15, 3, 5, 16, 1, 1, 1, 2, 1, 2, 1, 13}, {0, 1, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 5, 1, 1, 1, 2, 6, 1, 1, 14, 2, 1, 3, 7, 2, 1, 291, 1, 8, 1, 8, 1, 8, 1, 11, 3, 20, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 227, 1, 6, 1, 2, 1, 77, 6, 2, 27, 3, 9, 1, 2, 1, 54, 11, 4, 4, 1, 2, 15, 1, 2, 1, 3, 1, 2, 6, 1, 3, 1, 12, 2, 1, 5, 1, 5, 1, 2, 1, 12, 2, 2, 1, 3, 8, 1, 5, 1, 37, 9, 1, 1, 3, 156, 2, 26, 3, 1, 4, 2, 3, 1, 3, 116, 1, 14, 1, 5, 1, 5, 4, 3, 1, 5, 1, 1, 3, 5, 8, 1, 2, 1, 1, 1, 7, 1, 1, 1, 9, 4, 3, 2, 2, 2, 2, 7, 1, 4, 4, 1, 1, 1, 1, 4, 2, 1, 1, 1, 2, 121, 2, 3, 4, 5, 2, 1, 1, 1, 1, 2, 11, 7, 1, 3, 134, 15, 3, 7, 1, 1, 2, 13, 1, 1, 2, 4, 2, 2, 1, 3, 1, 2, 1, 5, 2, 2, 2, 6, 1, 3, 16, 2, 1, 1, 24, 3, 7, 8, 1, 1, 6, 2, 1, 1, 11, 4, 2, 1, 8756, 1, 1, 50, 1, 1, 1, 4, 1, 1, 25, 1, 5, 5,1, 2, 5, 4, 1, 52, 1, 55, 2, 15}} (Presumably the eventual noise is an artifact of series truncation.) For an (almost) explanation of the sum pattern in ContinuedFraction[Sum[ u^-2^n, {n, 0, ∞}], ...], https://cs.uwaterloo.ca/~shallit/Papers/scf.ps, except it only treats integer u>2, presumably because the u=2 pattern is haired up by vanishing partial floors. Also: In[288]:= ContinuedFraction[Sum[2^n 4^-2^n, {n, 0, 13}], 64] Out[288]= {0, 2, 1, 1, 3, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 3, 1, 1, 2, 7, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 1, 2, 1, 3, 1, 1, 2, 15} In[289]:= ContinuedFraction[Sum[(-2)^n 4^-2^n, {n, 0, 13}], 64] Out[289]= {0, 7, 8, 1, 1, 9, 7, 3, 1, 6, 9, 1, 1, 8, 6, 1, 7, 7, 8, 1, 1, 9, 6, 1, 3, 7, 9, 1, 1, 8, 7, 15, 1, 6, 8, 1, 1, 9, 7, 3, 1, 6, 9, 1, 1, 8, 7, 7, 1, 6, 8, 1, 1, 9, 6, 1, 3, 7, 9, 1, 1, 8, 6, 1} (Probably unbounded but predictable.) Likewise: In[286]:= ContinuedFraction[Sum[(-2)^n 2^-2^n, {n, 0, 13}], 64] Out[286]= {0, 4, 1, 1, 3, 3, 1, 2, 1, 1, 3, 1, 7, 4, 1, 1, 2, 1, 3, 3, 1, 1, 4, 15, 1, 3, 1, 1, 3, 3, 1, 2, 1, 1, 4, 7, 1, 3, 1, 1, 2, 1, 3, 3, 1, 1, 3, 1, 31, 4, 1, 1, 3, 3, 1, 2, 1, 1, 3, 1, 7, 4, 1, 1} In[287]:= ContinuedFraction[Sum[2^n 3^-2^n, {n, 0, 13}], 64] Out[287]= {0, 1, 1, 1, 1, 5, 1, 8, 1, 1, 2, 1, 1, 15, 1, 1, 1, 2, 6, 1, 1, 1, 10, 31, 1, 10, 4, 3, 1, 9, 1, 2, 1, 2, 1, 1, 3, 5, 1, 14, 1, 1, 1, 63, 2, 1, 14, 1, 5, 6, 2, 3, 1, 7, 4, 2, 1, 1, 3, 1, 3, 10, 4, 3} —rwg

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[math-fun] Ocellus Homology
by Brad Klee 15 Aug '19

15 Aug '19

Thanks Again Scott Vorthmann, the Nijhout book was a GREAT SUGGESTION!! According to chapter 2, "Pattern Elements and Homologies", a proper and history-respective interpretation of colour patterns on butterfly wings should be stated in terms of the standard //Nymphalid Ground Plan//, a template invented by scientists who sought out the notion of a "general butterfly" (LOL) between 1920 and 1930. Near the distal end of the wing, in the "Occelenreihe", it is possible--depending on species--to find a line of ocelli (Latin for "eyespots"). We have recently seen such features in data sets for the pipevine swallowtail: https://www.butterfliesandmoths.org/species/Battus-philenor Except for spot number five, it appears that each ocellus is usually a union of three topological disks. One black and one white disk together form an annulus, which entirely surrounds a third orange disk. Orange connects to black along one edge, and to white along one edge. Black and white connect on two edges, separated by an orange-white edge. The color field surrounding is usually iridescent blue. Sadly, a sole black Enso is usually too simple to impress anyone, so I went ahead and designed an orange and white attachment, that fits together to form a model homologous to the above: https://0x0.st/zVxq.JPG Hmm... What is that orange marble doing in there? Well, I think it is a symbol of //unbounded curiosity//. Could it be that the orange protrusions of the pipevine caterpillar somehow metamorphose to become the orange spots of the pipevine butterfly? I hope that we can find an answer to this question soon!! Cheers, Brad

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[math-fun] equality
by françois mendzina essomba2 14 Aug '19

14 Aug '19

Hello I still did not notice this equality between sinus functions, I find it harmonious sin(3*x)/(1-cos(3*x))=(cos(3*x)+cos(2*x)+cos(x)+1)/(sin(3*x)+sin(2*x)+sin(x)); sin(5*x)/(1-cos(5*x))=(cos(5*x)+cos(4*x)+cos(3*x)+cos(2*x)+cos(x)+1)/(sin(5*x)+sin(4*x)+sin(3*x)+sin(2*x)+sin(x)); sin(11*x)/(1-cos(11*x))=(cos(11*x)+cos(10*x)+cos(9*x)+cos(8*x)+cos(7*x)+cos(6*x)+cos(5*x)+cos(4*x)+cos(3*x)+cos(2*x)+cos(x)+1)/(sin(11*x)+sin(10*x)+sin(9*x)+sin(8*x)+sin(7*x)+sin(6*x)+sin(5*x)+sin(4*x)+sin(3*x)+sin(2*x)+sin(x)); general sin(k*x)/(1-cos(k*x))=(sum(cos(n*x),n=0..k))*(sum(sin(n*x),n=0..k))^(-1);

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Re: [math-fun] Solving some logarithmic equations exactly
by Bill Gosper 14 Aug '19

14 Aug '19

We need a similar function to solve Kepler's equation! *M* = *E* - e sin *E .* http://www.tweedledum.com/rwg/pizza.htm Will this suffice to describe motion in an elliptical orbit? E.g., {A cos(u(t)), B sin(u(t))}. What is u(t)? Is it the same, modulo scaling t, for all orbits of the shape "A by B"? Does scaling the "sun"'s gravity just scale t somehow? —rwg Date: 2019-08-12 13:11 From: Andres Valloud <avalloud(a)smalltalk.comcastbiz.net> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: "andres.valloud(a)gmail.com" <andres.valloud(a)gmail.com>, math-fun < math-fun(a)mailman.xmission.com> Thanks! The Wikipedia page notes W cannot be expressed in terms of elementary functions. No wonder trying to clear out x from the original equation was rather difficult! On 8/12/19 2:37 , Neil Bickford wrote: > I think Example 1 in > https://en.wikipedia.org/wiki/Lambert_W_function#Example_1 might be > applicable to this! > > For instance, for a x + b = ln x, set y = ln x - b to get (a e^b) e^y = > y. Then -a e^b = -y e^(-y), and so y = -W(-a e^b). Substituting back > then gives x = e^(b-W(-a e^b)), assuming I’ve done my math right! > > Mathematica informs me that since a x + b = ln x, this can also be > written as x = -W(-a e^b)/a. > > --Neil Bickford > > On Mon, Aug 12, 2019 at 2:11 AM Andres Valloud > <avalloud(a)smalltalk.comcastbiz.net > <mailto:avalloud@smalltalk.comcastbiz.net>> wrote: > > What's a neat way to solve equations such as the below exactly? > > ax + b = ln x > > In particular, what to do when b = 0? > > Andres.

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