Hmm, but maybe I made a mistake — I got 2 sin(pi sqrt(2)), but Mma got 2 sin(pi 2 sqrt(2)). —Dan

On Mar 28, 2017, at 10:42 AM, Bill Gosper <billgosper@gmail.com> wrote:

Nice! Tnx! --rwg

Date: 2017-03-28 10:15 From: Dan Asimov <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com>

I took this as a personal challenge.

First note that

1/2 ((-I)^(1 + √2) + I^(1 + √2)) = cos(pi(1+sqrt(2))/2)

. Let

T = pi(1+sqrt(2))/2.

Also note that

1/4 ((-I)^(1 + √2) + I^(1 + √2))^2 = (cos T)^2

so that

sqrt(-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2) = i sin T

.

Then the whole expression is

(cos T - i sin T)^(1 + sqrt(2)) + (cos T + i sin T)^(1 + sqrt(2))

or

exp( -i T (1 + sqrt(2)) + exp( i T (1 + sqrt(2))

= 2 cos(T(1 + sqrt(2))) = 2 cos(pi(1 + sqrt(2))^2 / 2)

= 2 cos((pi/2) (3 + 2 sqrt(2))

= 2 cos(3 pi/2 + pi sqrt(2))

= 2 [cos(3 pi/2) cos(pi sqrt(2)) - sin(3 pi/2) sin(pi sqrt(2))]

= 2 sin(pi sqrt(2)) Q.E.D.

—Dan

Bill Gosper wrote (March 27, 2017 at 5:25 pm): ----- In[69]:= ((1/2 ((-I)^(1 + √2) + I^(1 + √2)) - Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2) + (1/2 ((-I)^(1 + √2) + I^(1 + √2)) + Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2)) // FullSimplify

Out[69]= 2 Sin[π 2√2]

(How??) ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun