Euler

# Euler - cos x + i sin x = 1 + ix + 1 2! (ix) 2 + 1 3! (ix)...

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Complex Waveforms-Euler Identity We can prove that e ix = cos x + i sin x by finding the Taylor expansions for e ix , sin x, and cos x. Remember that : e x = 1 + x + 1 2! x 2 + 1 3! x 3 + 1 4! x 4 +..... + 1 n! x n or e ix = 1 + ix + 1 2! (ix) 2 + 1 3! (ix) 3 + 1 4! (ix) 4 +...... + 1 n! (ix) n Then expanding the trig functions: sin x = x - 1 3! x 3 + 1 5! x 5 -...... + (-1) n 1 (2n+1)! x 2n+1 cos x = 1 - 1 2! x 2 + 1 4! x 4 -...... + (-1) n 1 (2n)! x 2n Multiplying the sin expansion by i gives: i sin x = ix - 1 3! ix 3 + 1 5! ix 5 -...... + (-1) n 1 (2n+1)! ix 2n+1 Now noting that i . i = -1 , i sin x = ix + 1 3! (ix) 3 + 1 5! (ix) 5 +...... + 1 (2n+1)! (ix) 2n+1 cos x = 1 + 1 2! (ix) 2 + 1 4! (ix) 4 +...... + 1 (2n)! (ix) 2n Now combine the cos x and i sin x expansions:
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Unformatted text preview: cos x + i sin x = 1 + ix + 1 2! (ix) 2 + 1 3! (ix) 3 + 1 4! (ix) 4 + 1 5! (ix) 5 + . .... and you see that the result is the same expansion as for e ix . From trigonometric identities remember that sin(-x) = - sin(x) and cos(-x) = cos(x). This is handy for finding the real and imaginary parts of wavefunctions. To find the real part of a wavefunction note that you can take for the real part: cos x = 1 2 ( e ix + e-ix ) and the imaginary part sin x = 1 2i ( e ix- e-ix )...
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## This document was uploaded on 02/22/2012.

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