2/17/96 3:47:32 PM

Please explain the following mathematical equation. (How do you obtain -1) (1+(1/1!)+(1/2!)+(1/3!)+...)^pi*i=-1

Matt:

There is a result called De Moivre's Theorem (Abraham De Moivre, 1667-1754) which says that (cos t + i*sin t)^n = cos n*t + i sin n*t. (You can verify this for n=2 by squaring the left hand side, using i^2 = -1 and the double angle formulas for sine and cosine.)

The series that you mention, 1+(1/1!)+(1/2!)+(1/3!)+..., is usually designated by the letter e, called Euler's constant, and the quantity e^(i*t) is cos t + i sin t. The equation you have given then follows from De Moivre's Theorem with t = pi since cos pi = -1 and sin pi = 0.

Harley

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