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Problem 93 Solution

e^x = $\Sigma$ (n=0 to infinity) (xⁿ/n!).

sin(x) = $\Sigma$ (n=0 to infinity) (-1)ⁿ*x²ⁿ⁺¹/(2n+1)!.

cos(x) = $\Sigma$ (n=0 to infinity) (-1)ⁿ*x²ⁿ/(2n)!.

So e^i$\pi$ = $\Sigma$ (n=0 to infinity) (i$\pi$)ⁿ/n! =

(i*$\pi$)¹/1! + (i*$\pi$)²/2! + (i*$\pi$)³/3! + ... + (i*$\pi$)⁴/4! + (i*$\pi$)⁵/5! + (i*$\pi$)⁶/6! + ... =

(i*$\pi$)¹/1! + (i*$\pi$)³/3! + (i*$\pi$)⁵/5! + ... +

(i*$\pi$)²/2! + (i*$\pi$)⁴/4! + (i*$\pi$)⁶/6! + ... =

i * [$\pi$¹/1! -$\pi$³/3! + $\pi$⁵/5! - ... ]

-1 * [$\pi$²/2! - $\pi$⁴/4! + $\pi$⁶/6! - ... ]=

= i*sin($\pi$) - cos($\pi$) = i*0 - 1 = -1

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