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Explain Euler's Theorem/Identity

by BloodyFrozen
Tags: calculus, euler, explanation, theorem, trigonometry
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BloodyFrozen
#19
Apr19-11, 08:30 AM
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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?
dimitri151
#20
Apr19-11, 06:16 PM
P: 99
Also, seeing as how Euler managed to manipulate series with ease, it seems pretty unlikely that he was unaware of how to obtain the identity from series expansion.
It depends on what you mean by series expansion. In his Analysis of the infinite he gets the expansion for ex by really fancy use of the binomial theorem rather than Taylor or Mclaurin series. I wish I had my copy to confirm definitely, but I believe he didn't use those types of expansion, just your plain old binomial theorem and substituting variables then taking limits.

But he probably new about those expansions since he was contemporary of both, if he hadn't figured it out himself.
BloodyFrozen
#21
Apr19-11, 09:00 PM
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Quote Quote by dimitri151 View Post
It depends on what you mean by series expansion. In his Analysis of the infinite he gets the expansion for ex by really fancy use of the binomial theorem rather than Taylor or Mclaurin series. I wish I had my copy to confirm definitely, but I believe he didn't use those types of expansion, just your plain old binomial theorem and substituting variables then taking limits.

But he probably new about those expansions since he was contemporary of both, if he hadn't figured it out himself.
Can you post how he got it?
dimitri151
#22
Apr28-11, 01:15 PM
P: 99
Took a while to find the book and the time. Here it is.
Attached Files
File Type: pdf notes_euler_identity.pdf (53.6 KB, 39 views)
olivermsun
#23
Apr28-11, 01:35 PM
P: 789
Quote Quote by BloodyFrozen View Post
I get what you are saying about the first and second derivative, but then I lost ya. Yet, so far I dont see how this has a relation to cos x + i sin x = ex
Here's another way to understand the "differential equations" approach:

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).
dimitri151
#24
Apr28-11, 01:59 PM
P: 99
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.
BloodyFrozen
#25
Apr29-11, 10:23 PM
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Quote Quote by olivermsun View Post
Here's another way to understand the "differential equations" approach:

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).


Perfect explanation. Thanks to everyone who has helped!
HallsofIvy
#26
Apr30-11, 09:35 AM
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Quote Quote by Bob Kutz View Post
Has anyone else noticed that if you replace i with (insert whatever imaginary meme you chose here), the equation works just the same;

Try smurf for example; e^(smurf*Pi) = -1

Or Pterodactyl; e^(Pt*Pi) = -1

As long as e^((Really Imaginary Constant)*Pi) = cos (Pi) + RIC sin (Pi), then e^((whatever you want to put here) * Pi) = - 1 because the sine of Pi is Zero.

I don't mean to poke fun at 'our little jewel', but I cannot help ponder the implications of this alarming corollary. Am I missing something? Could it be that e^(unicorn*pi) = -1?

Doesn't that seem inherently wrong and evil? What happened to the unicorn? Did the Pi EAT THE UNICORN?

This concept seems to belie a complete lack of meaning for Euler's, but I don't seem to be able to prove it wrong.

In case you're wondering, it works with theta too, if you just change out your y-axis label for your imaginary friend of choice.

Just a random thought from someone who's knowledge of math occasionally plays hell with their sanity. I think I hear Nietzsche calling my name, perhaps I should go now.
The only sense I can make of this is that if you replace "i" with a different symbol meaning the same thing the equation is still true. Well that's true of any statement and any symbol! I can't imagine why anyone would think such a thing is worthy of saying.
micromass
#27
Apr30-11, 10:31 AM
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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??
baric
#28
Apr30-11, 02:33 PM
P: 12
Quote Quote by Bob Kutz View Post
Has anyone else noticed that if you replace i with (insert whatever imaginary meme you chose here), the equation works just the same;

Try smurf for example; e^(smurf*Pi) = -1

Or Pterodactyl; e^(Pt*Pi) = -1

As long as e^((Really Imaginary Constant)*Pi) = cos (Pi) + RIC sin (Pi), then e^((whatever you want to put here) * Pi) = - 1 because the sine of Pi is Zero.

I don't mean to poke fun at 'our little jewel', but I cannot help ponder the implications of this alarming corollary. Am I missing something? Could it be that e^(unicorn*pi) = -1?

Doesn't that seem inherently wrong and evil? What happened to the unicorn? Did the Pi EAT THE UNICORN?

This concept seems to belie a complete lack of meaning for Euler's, but I don't seem to be able to prove it wrong.

In case you're wondering, it works with theta too, if you just change out your y-axis label for your imaginary friend of choice.

Just a random thought from someone who's knowledge of math occasionally plays hell with their sanity. I think I hear Nietzsche calling my name, perhaps I should go now.
haha, unfortunately Bob, i has a very specific mathematical value and, as a result, e ^ipi = -1.

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.
BloodyFrozen
#29
Apr30-11, 02:57 PM
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I see your point
Bob Kutz
#30
Jun20-11, 08:49 AM
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Quote Quote by baric View Post
haha, unfortunately Bob, i has a very specific mathematical value and, as a result, e ^ipi = -1.

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.
Well, that's technically true; but by defining your y-axis to match, it still works perfectly;

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.
Bob Kutz
#31
Jun20-11, 08:52 AM
P: 8
Quote Quote by HallsofIvy View Post
The only sense I can make of this is that if you replace "i" with a different symbol meaning the same thing the equation is still true. Well that's true of any statement and any symbol! I can't imagine why anyone would think such a thing is worthy of saying.
No; my original point is that the value of i in the equation itself is Zero, since it is multiplied by the sin of pi, which is zero; so you can replace i with anything you want to without effect. There is no i coefficient to the number, so it is no longer complex and resolves to a real number. The only purpose of i in the identity is to put it on the complex plane.
micromass
#32
Jun20-11, 08:56 AM
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Quote Quote by Bob Kutz View Post
But, further, Euler's formula works perfectly well, no matter what you replace i with
I have a hard time seeing why you say that. If you take 2i, then Euler's Identity isn't true anymore, instead we would have

[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...
Hootenanny
#33
Jun20-11, 08:57 AM
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Quote Quote by Bob Kutz View Post
No; my original point is that the value of i in the equation itself is Zero, since it is multiplied by the sin of pi, which is zero; so you can replace i with anything you want to without effect. There is no i coefficient to the number, so it is no longer complex and resolves to a real number. The only purpose of i in the identity is to put it on the complex plane.
Bob, I think you are either confused or using confusing terminolgy. The number "i" is a constant. To say that the "value of i" is misleading at best. You wouldn't say "the value of 2 in the equation is zero", now would you?
baric
#34
Jun20-11, 01:41 PM
P: 12
Quote Quote by Bob Kutz View Post
All I am saying is that the use of the term i in Euler's identity is perfunctory. 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.
Yes, but the complex plane is a necessary part of mathematics. Describing i as a plane-accessing constant is misleading, imo, since that plane exists whether we choose to acknowledge it. To me, that's analogous to saying the only purpose '-' symbol is to allow the use of the negative side of the real number line.

'-' 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.
Bob Kutz
#35
Jun21-11, 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 non-real 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.
micromass
#36
Jun21-11, 12:47 PM
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In general:

[tex]e^{i\theta}=\cos(\theta)+i\sin(\theta)[/tex]

So

[tex]e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1[/tex]

And

[tex]e^{(3+i)\pi}=e^{3\pi}e^{i\pi}=e^{3\pi}(\cos(\pi)+i\sin(\pi))=-e^{3\pi}[/tex]

I think this might be the root of your misunderstanding...


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