Explain Euler's Theorem/Identity

[Bob Kutz]

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]

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

e^{2i\pi}=1

which is similar to Euler's identity, did you mean that.

Or take 3+i, then you'd get

e^{(3+i)\pi}=-e^{3\pi}

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]

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]

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]

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]

In general:

e^{i\theta}=\cos(\theta)+i\sin(\theta)

So

e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1

And

e^{(3+i)\pi}=e^{3\pi}e^{i\pi}=e^{3\pi}(\cos(\pi)+i\sin(\pi))=-e^{3\pi}

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

 

[HallsofIvy]

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.

?? "2ipi", 2i\pi, is exactly the same as "(2pi)i", 2\pi i. I have no idea what you mean by "changing the coefficient of the angle'. There is no angle and no coefficient here.

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?

No, that's not what it means.: e^{ix}= cos(x)+ i sin(x). The "i" in cos(x)+ i sin(x) is a fixed number, not a variable. If you have some multiple of i that has to be incorporated in the "x", not the i. e^{2\pi i}= cos(2\pi)+ i sin(2\pi)= 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?

You are once again misunderstanding "e^{ix}= cos(x)+ i sin(x)". The "i" in that is NOT a variable. e^{(3+ i)\pi= e^{3\pi+ i\pi}= e^{3\pi}e^{i\pi}. Since e^{i\pi}= cos(\pi)+ i sin(\pi)= -1, e^{(3+i)\pi}= -e^{3\pi}.

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.

All that says is that you have no idea what complex numbers are or what everyone is trying to tell you.

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.

I have no idea what you could mean by "the equation doesn't really exist in the complex dimension" or why you think that "the complex dimension has no particular affinity with the (square root of a negative one) for Euler's equation at all."

If you mean you are troubled by the fact that you do not comprehend complex numbers, I can understand that but surely that can be easily remedied.

 

[Bob Kutz]

I am sorry.

I had looked at Euler's as a trig function where i doesn't seem to have much of a role.

Had I looked at it as literally the limit of (1+zi/n)^n, nothing else besides i works at all.

Again, sorry for my narrow view of the topic.

 

[Bob Kutz]

Hootanany; I did mistate when I said the 'value of i is zero', it probably should've read 'the coefficient of i is zero', or 'the imaginary part is zero' or something. That would've been clearer to my point. Point of fact; the imaginary part of the solution to e^iPi is in fact zero. It is not complex.

In Re; HallsofIvy; I am pretty sure I understand the complex plane. Your assertion that I do not is a bit insulting.

But, since you didn't insult me ad-hominem, but did question my understanding of the basic concept of complex numbers, I will likewise take you to task;

You stated; "There is no angle and no coefficient here."

Not sure how you defend that. By denying that there is in fact an angle, and that the coefficient of that angle is i, you demonstrate a rather thin understanding of Euler's identity. I don't know what place you have acting as an authority on this topic at all.

Whereas I simply failed to evaluate Euler's identity as an infinite polynomial (limit), rather than a trig. function, you have failed to recognize that it springs from Euler's formula which inherently and explicitly involves angles and a coefficient of i. It is hard to take the sine or cosine of something that isn't an angle. Euler's formula converts cartesian coordinates to polar coordinates in the imaginary plain. Can you elaborate how this is possible without angles?

You can derive Eulers Identity without using limits. You cannot derive it without the formula though.

The complex plane is simple enough to be understood by most 9th graders. I am pretty sure everyone visiting this message board gets that. To state that I do not is a thinly veiled insult.

In the end; I feel confident in my comprehension of the issues at hand, but I think you've demonstrated yours is lacking.

 

[SteveL27]

Hootanany; I did mistate when I said the 'value of i is zero', it probably should've read 'the coefficient of i is zero', or 'the imaginary part is zero' or something.

Just to be precise: the imaginary part of i is 1. In general, the imaginary part of a + bi is b. In other words the imaginary part of a complex number is defined as a particular real number.

That would've been clearer to my point. Point of fact; the imaginary part of the solution to e^iPi is in fact zero. It is not complex.

Zero is a complex number. Getting the terminology right is a big part of discussing math.

 

[alexfloo]

Bob Kutz, what you said is not not strictly speaking true. You said that if "i" is relaced with anything in the formula, then it follows that e^{anything*x}=-1. However, it is not a property of "anything" (even any mathematical object) that anything*0=0. This is a property of real and complex numbers. The operation of "multiplying a real number by a unicorn" is not well defined. The operation of "multiplying a real number by a complex number," however, is.

 

[Bob Kutz]

Steve,

"The imaginary part of i" is an odd construct and a mis-use of what I said. In Euler's Identity, there is no imaginary part at all in the solution. e^ipi + 1= 0. It certainly does not equal 0+i. So you've misread what I said or misunderstand the discussion. Again and for clarity;

e^ipi +1 = 0

It does not equal 0 + i, so the imaginary part is in fact not 1. In fact, to your precise statement "the imaginary part of i is 1" does not even make sense.

i is imaginary b/c i^2 = -1 and nobody had been able to define a real number that satisfies i.

1 is not imaginary, and in Euler's Identity the imaginary component is in fact zero, not one.As to the notion that zero is a complex number; do you mean in the sense that all real numbers are a subset of the complex numbers?

I always thought of complex numbers as an extention to the real numbers, but what ever. The imaginary component in Euler's identity is still zero. If you wish to invoke the complex plane to solve cos (pi) + i sin(pi) then God bless, but you don't need to. Sin of pi is zero, the solution rests on the real number line.Alexfloo; As to the notion you cannot multiply a real number by a unicorn; would you object, then if I multiplied a real number (say 5) by cow and had five cows? That is the underlying math, is it not?

I think real numbers multiplied by ordinary (or even imaginary) objects is quite well defined. That is where math came from, you know.

You can certainly multiply real numbers by imaginary things, or this whole topic is a waste.

As I said in the earlier post; my mistake was in evaluating Eulers based on it's derived formula rather than as a function of the natural log; the limit of (1+zi/n)^n. THAT is where i comes into it's own; if you don't have an i component, it doesn't make any sense at all. THAT is where Euler's goes from being a mathematical oddity to being truly transcendental. When evaluating it as a trigonometric function, i plays no important role other than as the scale of the y axis. As I said; could be anything, really, equation works just the same.

And I am not sure you are correct that 'it is not a property of "anything" (even any mathematical object) that anything*0=0.' Isn't it? Is it not actually a mathematical property that for any x, x(0) = 0?From some unimportant website;

Zero element
Any number multiplied by zero is zero. This is known as the zero property of multiplication: x(0) = 0

Let x be a number. Or let it be boxes, ballots, cows or unicorns. If you have zero of them then you don't have any. If you have one and multilply it by 5, then you have 5.

Say you have 5 boxes, and they're empty, then you multiply those boxes by zero; well then, you no longer have any boxes, do you?

That is pretty rudimentary stuff right there.

I thought we were well beyond all that, talking about Euler's formula and equation and all.

 

[micromass]

I think real numbers multiplied by ordinary (or even imaginary) objects is quite well defined.

Please, educate me, what is the definition??
What is the definition of

e^{\pi *\text{cow}}

?

 

[alexfloo]

"And I am not sure you are correct that 'it is not a property of "anything" (even any mathematical object) that anything*0=0.' Isn't it? Is it not actually a mathematical property that for any x, x(0) = 0?"

This is not true. 0 times any two-by-two matrix is not zero, it's a matrix of zeros. The rule you're discussing is a property of NUMBERS as the page you cited states.

"If you wish to invoke the complex plane to solve cos (pi) + i sin(pi) then God bless, but you don't need to. Sin of pi is zero, the solution rests on the real number line."

This on the other hand is not wrong, it just completely misses the point. Euler's formuls isn't there for "solving cos (pi) + i sin(pi)," it's a much more profound and general statement about a relationship between two important functions on the complex plane.

 

[alexfloo]

To reiterate, the point is not that:

cos (pi) + i sin(pi) = -1

This, as you've correctly stated many times, is trivial. The point is that:

e^{i\pi}=-1

which is *not* trivial, and is in fact a very interesting fact.

 

[alanling]

I came across Lakoff's book "where math comes from" in which there's a section devoted to explain the meaning of the Identy instead of using a "tautology" that is generally given by many a professors, textbooks, intimidators, etc. They almost succeeded in making me understand, but at the last moment they also fall victim to using such vague qualities such as"periodicity", "rate of change" mumbo/jumbo that I feel an intuitive understanding is still lacking. Most people will tell U that it is because it equals cos x + i sinx or because e^xi happens to equal taylor expansion when sin and cos terms are added with a complex number twist lol, yeh but why?

The best way to understand it comes first, from khalid's "betterexplained" plus page of songho ahn's where a unit helix in 3d is shown with an img, real, and an X axis respectively(where x is just a linear axis representing the angle truncated at 2PI.

Khalid's explained http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/#comment-53478 e^x as growth(as in compound int) ---> lim (1+1/n)^n and x represents rate/duration mix because e is the no. when u set n to infin of instantaneous 100% compounding.

Similar with the unit circle, e^0 first grows to 1x the base e ( the radius), then because of the part i in the exp and where the rate of e^x change is just e^x as a def. Then the radius R is like a vector of unit length driven by a const i vector perpen to it (i.R is vel or accel if u will,moving counterclockwise at unit increments). And, note ln(x) means at what rate/period is needed to reach x time growth in e base.

In Anh's page, http://www.songho.ca/math/euler/euler.html he sets ln(cosx+isinx) = xi, then cosx+isinx = e^xi so xi is the rate/period needed to reach position cosx+isinx.

Now although with e^xi ,x as an angle is a very convenient indication of the position of the radius, it is hard to see how this exp resolves to something like .5678+ .8790i proportion as an example. Evidently, it is a result of the trig function and xi has a different ratio of the vertical vs horizontal depending on where the radius is pointed at along the circle, meaning different slope/tangent. So at any pos there is a changing rate of growth with regards both axes, but only in angles, not in length.

On Anh's page , the 3D diag would dispel a lot of confusion generated by using imaginary no.
Actually, with 2 extra dimensions, imag no. is just a math convenience to make sure U don't add the two nos as though they are both real and to deal with the sign when phase changes.

So if U notice, the projection on the x-Real plane is the Cos function tracing how the real shrinks and grow as the Img-x plane shadows a sin function showing how it grows and shrink on a complementary rate tracing a helix, but a 2D circle projection on the Img-Real plane.

 

[alan2]

"Labeling i as imaginary is just nonsense, because it does exist."

That was a hoot.

 

[alan2]

Historically, much of abstract mathematics is motivated by a desire to generalize familiar concepts and Euler's formula can be justified the same way. Suppose we seek a generalization of the exponential function to the complex plane. That is we seek a function f such that df/dz=f for all z, f is entire, and f=e^x on the real line, i.e. when y=0. Clearly f=(e^x)cosy+i(e^x)siny satisfies these requirements so we may define f=e^z. Now when z=iy is purely imaginary the result is Euler's formula. Proof? Maybe, maybe not, but it is an interesting historical justification.