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You can easily derive e^ix = i cos x + sin x. Just look at the Taylor series for e^x and compare with the Taylor series for sin x + cos x. You'd see that they
by username90 7y ago
You can easily derive e^ix = i cos x + sin x.
Just look at the Taylor series for e^x and compare with the Taylor series for sin x + cos x. You'd see that they are basically the same, just that the trigonometric functions alternate their signs every 2 steps. How do we alternate the signs in e^x? We just add a factor of i, that way we get alternate signs every two steps but get an additional i on every odd step. Odd steps are sin, so just add i to sin and we are done: e^ix = i sin x + cos x.
Now we also see that you remembered the wrong formula, it the sine part is imaginary since it corresponds to the odd factors.
- plopilop 7y ago(it's the other way round, e^(ix) = cos(x) + i sin(x))
- contravariant 7y agoI have no problems with this route to show that e^ix = cos(x) + i sin(x) but to do it properly does require showing that this is the unique analytic continuation of the exponential function. The alternative method is to note that exp() is defined as the unique function satisfying exp(0) = 1 and dexp(ax)/dx = a exp(ax). So exp(ix) describes a point moving orthogonal to it's current position, which means it's a circle, and since it's speed is proportional to it's distance from the origin it is moving at a constant speed. So e^ix is just a point x radions around the unit circle.