eit by arithmetic

e^it by arithmetic

Let us start with small values of t. We know that when t is small, we can write e^it = 1 + it. 1/2048 = 0.00048828125 is small enough for our purpose. We successively square exp(i/2048) and make the following table:

e^it = x + yi

e^i/2048 = 1.00000000 + 0.00048828125 i

e^i/1024 = 0.99999976 + 0.00097656250 i

e^i/512 = 0.99999856 + 0.0019531245 i

e^i/256 = 0.99999330 + 0.0039062433 i

e^i/128 = 0.99997134 + 0.0078124344 i

e^i/64 = 0.99988164 + 0.015624421 i

e^i/32 = 0.99951917 + 0.031245143 i

e^i/16 = 0.99806231 + 0.062460239 i

e^i/8 = 0.99222709 + 0.12467842 i

e^i/4 = 0.96896988 + 0.24741861 i

e^i/2 = 0.87768665 + 0.47948237 i

eⁱ = 0.54043051 + 0.84167055 i

e²ⁱ = -0.41634418 + 0.90972889 i

It is clear that somewhere between eⁱ and e²ⁱ, we have the pure imaginary number i. It can be verified by multiplication that,

eⁱ e^i/2 eⁱ/16 eⁱ/128 eⁱ/2048 = -0.0000043665471 + 1.0003725 i

This is the closest we can get to i with our table (due to the finite precision and accumulated round off errors). The exponent upon simplification gives 1.5708 which is the value of p/2, correct to five significant digits. Thus, we have, exp(ip/2) = i.

Since, we have e^it = x + iy and e^-it = x – iy, the amplitude of e^it is unity for any t. Further, since x² + y² = 1, we could take x and y to be the sine and cosine of an angle, let us say q. From the above arithmetic, it is apparent that q is same as t. In other words, e^it= cos t + i sin t. (When t = p/2, it rightly gives i, and when t is small it approximates as 1 + it). Feynman called this formula the "jewel" (The Feynman Lectures on Physics, Vol 1, ch. 22.)