Show that ∏1∞ (1 + 2 cos(2z/3n)/3 = (sin z)/z for all complex z.
For z = 0, we may take (sin z)/z = 1 by continuity. In this case, each term on the lhs is 1, so the relation holds. Assume now that z is non-zero.
We have sin z = sin(2z/3 + z/3) = sin 2z/3 cos z/3 + cos 2z/3 sin z/3, and sin z/3 = sin(2z/3 - z/3) = sin 2z/3 cos z/3 - cos 2z/3 sin z/3. Hence sin z = (1 + cos 2z/3) sin z/3. Hence (sin z)/z = (sin z/3)/(z/3) (1 + cos 2z/3)/3. Iterating: (sin z)/z = (sin z/3N)/(z/3N) ∏1N (1 + 2 cos(2z/3n) )/3. But as N tends to infinity z/3N tends to zero and hence (sin z/3N)/(z/3N) tends to 1. Hence the result.
9th Putnam 1949
© John Scholes
11 Mar 2002