B Differentiating Euler formula vs. multiplying by i

ke7ijo
Messages
2
Reaction score
1
TL;DR Summary
I ran into an apparent contradiction when working with Euler's formula and I can't find the mistake.
I differentiated both sides of Euler's formula with respect to x :
e^ix = sin x + i cos x => ie^ix = cos x - i sin x

Then for comparison I multiplied both sides of Euler's formula by i:
e^ix = sin x + i cos x => ie^ix = i sin x - cos x

Each of these two procedures seems to yield the additive inverse of the other, and I can't seem to figure out why even after a couple of hours of going back over it.
 
Physics news on Phys.org
ke7ijo said:
I differentiated both sides of Euler's formula with respect to x :
e^ix = sin x + i cos x
The mistake is that you have misremembered Euler's formula. The correct version is ##e^{ix} = \cos(x) + i\sin(x)##, which differs from what you wrote.

You can think of it this way. On the unit circle, with ##x## being the angle a ray makes with the horizontal axis, ##e^{ix}## represents the point on the unit circle. The coordinates of the point are ##(\cos(x), \sin(x))##. As a complex number, this point is ##\cos(x) + i\sin(x)##.
 
Thank you! That's it.
 
if you are mathematically curious, look at the different derivations of Euler's formula. My favorite is the one that uses Taylor Series.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top