[seqfan] Re: Fun little identity (to me anyway)
Paul D Hanna
pauldhanna at juno.com
Fri Jan 23 08:33:18 CET 2009
Seqfans,
This is directly related to an identity of Jacobi Elliptic functions:
(a^2 + k^2*b^2)*(c^2 + k^2*d^2) = (a*c + k^2*b*d)^2 + k^2*(a*d - b*c)^2
where
a = dn(u) ,
b = sn(u)*cn(v) ,
c = dn(v) ,
d = sn(v)*cn(u) ,
and k is the elliptic modulus:
cn(u)^2 + sn(u)^2 = 1,
dn(u)^2 + k^2*sn(u)^2 = 1.
This identity involves the Addition Theorem since:
a*c + k^2*b*d = dn(u-v)*(1 - k^2*sn(u)^2*sn(v)^2),
a*d - b*c = -sn(u-v)*(1 - k^2*sn(u)^2*sn(v)^2)
and
a^2 + k^2*b^2 = c^2 + k^2*d^2 = (1 - k^2*sn(u)^2*sn(v)^2) ,
dn(u-v)^2 + k^2*sn(u-v)^2 = 1.
So David's fun little identity is simple but not trivial.
Paul
-- avik roy <avik_3.1416 at yahoo.co.in> wrote:
This identity can be more generalized:
[(n+p)^2 + k][n^2 + k] = [n(n+p) + k]^2 + kp^2;
This identity can contribute to expressing the product of sums of two squares as a sum of two squares [if k is a square].
As an illustration, check:
(3^2 + 2^2)*(10^2 + 2^2)=34^2 + 14^2
More information about the SeqFan
mailing list