@Bill: Thanks for the comment! I consider the coefficient on i as how “long” you go around the unit circle, assuming unit speed. So, pi “seconds” takes you halfway around. Another way is to think about it in terms of distance, i.e. halfway around directly (speed and time are “equivalent” here since it’s a unit speed… but I like thinking e^rate*time where the “rate” is i and the time is pi).
We’re using radians because we want measurements from the perspective of the mover (vs. degrees, which are an arbitrary measure from the perspective of the observer).
Can I ask you why exactly is imaginary exponential growth at 90 degree?
I mean seen one way it seems very intuitive: since there is no real growth, the length (i.e. the radius of the circle) cannot change. Therefore the e frustratingly goes in circles, unable to really grow!The more it is told that its growth is imaginary, the faster it spins around in circles.
But if I look at what’s happening to it in the Re-Im plane, it seems to be oscillating between being completely real and being completely imaginary! Even though growth is imaginary, it is possible for it to become real again, only with a minus sign! Why this oscillation, given that it is only imaginary growth?
So to satisfy my curiosity, I request you to forward me any resource you know regarding the reason imaginary growth happens at 90 degrees
Hi Rajesh, sine & cosine are very related so there’s several ways you can look at it. On a circle, however, you have two dimensions, and sine only wiggles in one dimension (between 1 and -1), so you need both components if you want to describe a circular path. But, because circles are so symmetrical, it’s indeed the case that cosine is a “leading indicator” of what sine will do :).
Hi Charles, thanks so much for the comment and support. Once I first experienced what a math insight could feel like, I had to keep looking for them to share!
@Christian: Good question! When talking about sine, I think about it more like this:
i^i is equal to the computation e^(-pi/2) which is approximately 0.207
The fact that e^(-pi/2) = e^(-5pi/2) is a bit like 4/4 = 2/2 = 1 – different “computations” that approximate the same real number. So, i^i still only has one spot on the number line because e^(5pi/2) = e^(pi/2).
You claim i^i = e^(-π/2). Is it possible you might be forgetting an infinite amount of possible values? I could be wrong, but hear me out:
In general, sin([4n+1]π/2) = 1 and cos([4n+1]π/2) = 0 for any integer n
So e^(i*[4n+1]π/2) = i
Exponentiating, i^i = {e^(i*[4n+1]π/2)}^i = e^(–[4n+1]π/2)
For example, taking n = 1 yields your answer, whereas taking n = 2 (for which it seems this equation is still valid), yields i^i = e^(-5π/2). So I would conclude i^i doesn’t map to only one place on the real line.
Thanks for this Kalid, it helped me to visualize the maths involved which is the only way that I ever end up actually learning anything. Formulas are a road-block, I need pictures.
So I ended up here because I’m trying to wrap my head around the Fast Fourier Transformation in two dimensions (image processing) and understanding Euler’s Formula is an essential stepping stone along the way. I eventually figured that out after watching this: http://www.youtube.com/watch?v=ObklYbQaX24
Now there’s a challenge for you, give the FFT the Better Explained treatment so that even a calculus flunker like me can have an “aha” moment. Massive challenge that
Thank you so much for such an information stuffed website and sharing with everybody.
The article about Euler’s theorem is very helpful. I need few clarifications regarding this. Can I say SINE theta as component which defines my position and COS Theta as the one which gives the rate of change of the position component, because if I differentiate SINE I’ll get COS theta.
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