Newsgroups: sci.math
Path: decwrl!sun!pitstop!sundc!seismo!uunet!attcan!utgpu!watmath!watcgl!watmum!gjfee
Subject: trig expression for sqrt of a prime
Posted: 16 Nov 88 08:02:13 GMT
Organization:
Richard Pavelle asks:
2) Is there some general scheme for generating these integer
identities?
I was able to generate a trig. expression for sqrt(13)
by starting with the "anomalous factorization" of x^(2*13) - 13^13 .
Then substitute the known root 13^(1/2)*exp(2*Pi*I/26) into the
appropriate factor
12 11 10 9 8 7 6 5
(x - 13 x + 91 x - 507 x + 2535 x - 10985 x + 41743 x - 142805 x
4 3 2
+ 428415 x - 1113879 x + 2599051 x - 4826809 x + 4826809)
of x^26 - 13^13 . Next solve for sqrt(13) in terms of cos functions
and rationalize the denominator to obtain
-1 + 4*cos(1/13*Pi) + 4*cos(3/13*Pi) - 4*cos(4/13*Pi) = sqrt(13)
A similar procedure (starting with x^14+7^7) leads to
2*sin(3/7*Pi) + 2*sin(2/7*Pi) - 2*sin(1/7*Pi) = sqrt(7)
and
2*sin(1/11*Pi) + 2*sin(2/11*Pi) + 2*sin(3/11*Pi)
-2*sin(4/11*Pi) + 2*sin(5/11*Pi) = sqrt(11)
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