Explain Euler's Theorem/Identityby BloodyFrozen Tags: calculus, euler, explanation, theorem, trigonometry 

#19
Apr1911, 08:30 AM

P: 352

I personally think that you could not replace i because the modulus would not be the same in every case. Correct me if I'm mistaken?




#20
Apr1911, 06:16 PM

P: 91

But he probably new about those expansions since he was contemporary of both, if he hadn't figured it out himself. 



#22
Apr2811, 01:15 PM

P: 91

Took a while to find the book and the time. Here it is.




#23
Apr2811, 01:35 PM

P: 498

Let f(x) = cos x + i sin x. df/dx = sin x + i cos x = i f(x) Hence f(x) is its own derivative except for a factor of i. This is a clue that f(x) might be something like e^ix. You can check this by "guessing" that f(x) = e^u(x) for some unknown u(x). Substituting this form for f(x) into the above equation: df/dx = i f(x) d/dx (e^u(x)) = i e^u(x) e^u(x) du/dx = i e^u(x) du/dx = i u = ix + C Hence f(x) = e^(ix + C) = (e^C) (e^ix) = A e^ix, where A = 1 since for x = 0, e^ix = 1 = cos(0) + i sin(0). 



#24
Apr2811, 01:59 PM

P: 91

Yeah, that's another way to do it. You show that the magnitude of the of the derivative is unchanged. You can find it in Needham's Visual Complex Analysis, a couple of other places too.




#26
Apr3011, 09:35 AM

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#27
Apr3011, 10:31 AM

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P: 16,554

Well, if I put 1=unicorn and 2=smurf, then unicorn+unicorn=smurf. That's perfectly correct, so I don't really understand Bob's point. Care to explain, Bob??




#28
Apr3011, 02:33 PM

P: 12

If you indeed replace i on the right side of the equation with a unicorn, it will be zeroed out by sin(pi). However, on the left side of the equation, that unicorn must still be sitting precisely at the pi/2 position of the unit circle (where i is currently at). Otherwise, there is no equivalence to the right side. 



#30
Jun2011, 08:49 AM

P: 8

and my original point is that i itself is completely irrelevant to the identity in that the sin of pi=0, therefor there is no complex component and the equation resolves as a real number. But, further, Euler's formula works perfectly well, no matter what you replace i with, as long as you keep it on the complex plane with the imaginary coefficient being defined by that term. All I am saying is that the use of the term i in Euler's identity is perfuntory. I realize you have to have it to put it on the complex plane and therefore define it as e^ipi = cos (pi) + i sin (pi) , but it's value in the equation itself is ZERO. It's only function is to allow the use of the complex plane. 



#31
Jun2011, 08:52 AM

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#32
Jun2011, 08:56 AM

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[tex]e^{2i\pi}=1[/tex] which is similar to Euler's identity, did you mean that. Or take 3+i, then you'd get [tex]e^{(3+i)\pi}=e^{3\pi}[/tex] The value of i in the equation is not 0 as you can see, because I can change it with something else and get a different equation... 



#33
Jun2011, 08:57 AM

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#34
Jun2011, 01:41 PM

P: 12

'' rotates a value 180 degrees around the origin of the complex plane just as i rotates it 90 degrees. If you want to think of them only as operators, then it makes more sense to express i as 1i. 



#35
Jun2111, 12:42 PM

P: 8

I have a difficult time seeing how e^(2ipi) = 1.
Are you saying e^((2pi)i)? I think that would be a bit different from simply replacing i. That is changing the coefficient of the angle. If you are truly replacing i with 2i then that would be equal to cos(pi) + 2i(sin(pi)), would that not also be 1? I don't see how e^(3+i)pi equals e ^pi/3, to my thinking that would be equal to cos(pi) + (i+3)sin(pi), or 1. Maybe you are going with e^3+ipi, but that would be very different from replacing i, now wouldn't it? In fact any multiple of i should work out just the same, but I am talking about replacing the square root of a negative one with some, other imaginary factor. Not just multiplying i or adding a real number to it. That is a change to the identity and the general equation. But, in short; My basic point is that, on the complex plan, at pi the nonreal coefficient is always zero, because the sin of pi is zero. Define your imaginary number however you wish, at pi it's not relevant. It's more an artifact of how the imaginary plane is constructed than any magical property of the equation involving e, i or pi. As to the notion that what I am saying is equivalent to saying "the value of 2 in the equation is zero", well, as a matter of fact, if you wish to take 2 times the sin of pi and plot it on the "2 plane", it has no value and doesn't move the equation off of the real number line. This is a very interesting equation, no doubt. But I find it troubling that at the exact point of the identity, the equation doesn't really exist in the complex dimension, or that the complex dimension has no particular affinity with the (square root of a negative one) for Euler's equation at all. Very troubling. Just my humble observations. 


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