[math-fun] Are you a Gammaturge?
Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == GoldenRatio In[345]:= N[%, 11111] Out[345]= True --rwg
There is Euler's formula: Gamma(z) Gamma(1-z) = pi/sin(pi*z) and Gauss's formula: sum{k=0 to n-1} Gamma(z + k/n) = (2*pi)^((n-1)/2) * n^(1/2 - n*z) * Gamma(n*z) for n a positive integer Apply the latter to z=1/10, z=3/10 to get Gamma(1/10) = sqrt(2) * sqrt(pi) * 2^(3/10) * Gamma(1/5)/ Gamma(3/5) and Gamma(3/10) = sqrt(2) * sqrt(pi) * 2^(-1/10) * Gamma(3/5) / Gamma(4/5) to get Gamma(1/10) * Gamma(3/10) = 2 * pi * 2^(1/10) * Gamma(1/5)/ Gamma(4/5) Plug that into the left hand side of Bill's formula to get (Gamma(1/5) * Gamma(4/5))^2 * sqrt(5)/(4*pi^2) Using Euler's formula, this becomes sqrt(5)/(4 * (sin(pi/5))^2). Finishing it off is now easy. Victor On Sun, Nov 5, 2017 at 7:12 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == GoldenRatio
In[345]:= N[%, 11111]
Out[345]= True --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Whoops, I meant Gauss's formula to be: prod_{k=0 to n-1} Gamma(z + k/n) = (2*pi)^((n-1)/2) * n^(1/2 - n*z) * Gamma(n*z) for n a positive integer On Mon, Nov 6, 2017 at 10:14 AM, Victor Miller <victorsmiller@gmail.com> wrote:
There is Euler's formula: Gamma(z) Gamma(1-z) = pi/sin(pi*z)
and Gauss's formula: sum{k=0 to n-1} Gamma(z + k/n) = (2*pi)^((n-1)/2) * n^(1/2 - n*z) * Gamma(n*z) for n a positive integer
Apply the latter to z=1/10, z=3/10 to get
Gamma(1/10) = sqrt(2) * sqrt(pi) * 2^(3/10) * Gamma(1/5)/ Gamma(3/5)
and
Gamma(3/10) = sqrt(2) * sqrt(pi) * 2^(-1/10) * Gamma(3/5) / Gamma(4/5) to get
Gamma(1/10) * Gamma(3/10) = 2 * pi * 2^(1/10) * Gamma(1/5)/ Gamma(4/5)
Plug that into the left hand side of Bill's formula to get
(Gamma(1/5) * Gamma(4/5))^2 * sqrt(5)/(4*pi^2)
Using Euler's formula, this becomes
sqrt(5)/(4 * (sin(pi/5))^2).
Finishing it off is now easy.
Victor
On Sun, Nov 5, 2017 at 7:12 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == GoldenRatio
In[345]:= N[%, 11111]
Out[345]= True --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
An interesting is that if there were an identity of the form Gamma(n) = f(n) . Gamma(n/2)^2 where (for integer n) f(n) mod y can be computed efficiently, i.e., in time polynomial in the number of digits of n and y, then we could factor integers in polynomial time. - Cris
On Nov 6, 2017, at 9:36 AM, Victor Miller <victorsmiller@gmail.com> wrote:
Whoops, I meant Gauss's formula to be:
prod_{k=0 to n-1} Gamma(z + k/n) = (2*pi)^((n-1)/2) * n^(1/2 - n*z) * Gamma(n*z) for n a positive integer
On Mon, Nov 6, 2017 at 10:14 AM, Victor Miller <victorsmiller@gmail.com> wrote:
There is Euler's formula: Gamma(z) Gamma(1-z) = pi/sin(pi*z)
and Gauss's formula: sum{k=0 to n-1} Gamma(z + k/n) = (2*pi)^((n-1)/2) * n^(1/2 - n*z) * Gamma(n*z) for n a positive integer
Apply the latter to z=1/10, z=3/10 to get
Gamma(1/10) = sqrt(2) * sqrt(pi) * 2^(3/10) * Gamma(1/5)/ Gamma(3/5)
and
Gamma(3/10) = sqrt(2) * sqrt(pi) * 2^(-1/10) * Gamma(3/5) / Gamma(4/5) to get
Gamma(1/10) * Gamma(3/10) = 2 * pi * 2^(1/10) * Gamma(1/5)/ Gamma(4/5)
Plug that into the left hand side of Bill's formula to get
(Gamma(1/5) * Gamma(4/5))^2 * sqrt(5)/(4*pi^2)
Using Euler's formula, this becomes
sqrt(5)/(4 * (sin(pi/5))^2).
Finishing it off is now easy.
Victor
On Sun, Nov 5, 2017 at 7:12 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == GoldenRatio
In[345]:= N[%, 11111]
Out[345]= True --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Cristopher Moore Professor, Santa Fe Institute Why should we be deported? This is very, very hard for a family. What will our fellow citizens think if honest subjects are faced with such a decree — not to mention the great material losses it would incur. I would like to become a Bavarian citizen again. Your most humble and obedient, Friedrich Trump (1905)
participants (3)
-
Bill Gosper -
Cris Moore -
Victor Miller