Explain Euler's Theorem/Identity

[BloodyFrozen]

Please explain Euler's theorem. I don't get how he got this formula and how it can be used instead of trigonometry. Thanks

 

[Stephen Tashi]

I don't know how Euler got it, but one way of looking at is to compare the McLaurin series for e^x, cos(x) and sin(x). The terms from the cos(x) and sin(x) series appear in the series for e^x but with some differences in their signs. You might start wondering if there is some way to make the series for e^x or e^(-x) equal to something like the series for sin(x) + cos(x) or cos(x) - sin(x) etc. That might lead to the idea of comparing the terms in the series for e^x with x = i theta to the terms in the series for cos(theta) and i times the terms in the series for sin(theta).

 

[HallsofIvy]

What Stephen Tashi said!

Specifically, the MacLaurin series for e^x is
e^x= 1+ x+ \frac{1}{2}x^2+ \frac{1}{3!}x^3+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot

If you replace "x" with "ix" that becomes
e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \frac{1}{3!}(ix)^3+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot
= 1+ ix+ \frac{1}{2}i^2x^2+ \frac{1}{3!}i^3x^3+ \cdot\cdot\cdot+ \frac{1}{n!}i^nx^n+ \cdot\cdot\cdot

But it is easy to see that i^2= -1, i^3= i(-1)= -i, i^4= (i^2)(i^2)= (-1)(-1)= 1, i^5= (1)(i)= i, etc. In particular, all odd powers of i are imaginary and all even powers are real- and their signs alternate. We can separate the series above into "imaginary" and "real" parts:
e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4- \cdot\cdot\cdot+ \frac{(-1)^n}{(2n)!}x^{2n}+ \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5- \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}x^{2n+1}+ \cdot\cdot\cdot)

Now, if you take the MacLaurin series for cosine and sine you will see that they are just the two series above (cosine is an even function and so will have only even powers, sine is an odd function and so will have only odd powers).

 

[dimitri151]

You can read how he got it in his Analysis of the Infinite.

Here's a good synopsis.

http://www.maa.org/editorial/euler/How Euler Did It 46 e pi and i.pdf
http://www.maa.org/news/howeulerdidit.html
August 2007

 

[BloodyFrozen]

What Stephen Tashi said!

Specifically, the MacLaurin series for e^x is
e^x= 1+ x+ \frac{1}{2}x^2+ \frac{1}{3!}x^3+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot

If you replace "x" with "ix" that becomes
e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \frac{1}{3!}(ix)^3+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot
= 1+ ix+ \frac{1}{2}i^2x^2+ \frac{1}{3!}i^3x^3+ \cdot\cdot\cdot+ \frac{1}{n!}i^nx^n+ \cdot\cdot\cdot

But it is easy to see that i^2= -1, i^3= i(-1)= -i, i^4= (i^2)(i^2)= (-1)(-1)= 1, i^5= (1)(i)= i, etc. In particular, all odd powers of i are imaginary and all even powers are real- and their signs alternate. We can separate the series above into "imaginary" and "real" parts:
e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4- \cdot\cdot\cdot+ \frac{(-1)^n}{(2n)!}x^{2n}+ \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5- \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}x^{2n+1}+ \cdot\cdot\cdot)[/itex]<br /> <br /> Now, if you take the MacLaurin series for cosine and sine you will see that they are just the two series above (cosine is an even function and so will have only even powers, sine is an odd function and so will have only odd powers).

<br /> <br /> <br /> Is there anyway to &quot;understand&quot; this without the knowledge of Calculus beyond the fundamentals (limits, derivatives, integrals<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" />)? As I&#039;m not very familiar with Taylor series, MacLaurin Series , and Series in general I suppose. I don&#039;t understand how this is applicable for trigonometry.<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> Thanks to answerers so far.<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" />

 

[snipez90]

http://en.wikipedia.org/wiki/Power_series" is always a good place to start.

Maclaurin series are just taylor series centered at 0 (explained in the article). The applications to trigonometry are partly due to the nice properties of the exponential function. They are abundant in problem solving texts, but you might want to start by googling "applications of de Moivre's theorem".

One approach to the identity is to consider the power series solution to f'(z) = f(z) and from that derive all the properties of the complex exponential, including Euler's identity.

Feynman gives I think gives a numerical approach to the identity in his lectures on physics (volume 1).

 

[BloodyFrozen]

http://en.wikipedia.org/wiki/Power_series" is always a good place to start.

Maclaurin series are just taylor series centered at 0 (explained in the article). The applications to trigonometry are partly due to the nice properties of the exponential function. They are abundant in problem solving texts, but you might want to start by googling "applications of de Moivre's theorem".

One approach to the identity is to consider the power series solution to f'(z) = f(z) and from that derive all the properties of the complex exponential, including Euler's identity.

Feynman gives I think gives a numerical approach to the identity in his lectures on physics (volume 1).

As I said before I'm not familiar with calculus series, the only thing I do know that you mentioned is De'Moivres theorem. Thanks though, I'll check if I can understand this.

 

[mathwonk]

look at the differential equations. if the derivative of e^cx is c.e^cx, then the second derivative of e^ix is -e^ix, so it satisfies the same equation y'' + y = 0 as sin and cosine. so e^ix must be a linear combination of sin and cosine. that's probably how it arose.

 

[BloodyFrozen]

look at the differential equations. if the derivative of e^cx is c.e^cx, then the second derivative of e^ix is -e^ix, so it satisfies the same equation y'' + y = 0 as sin and cosine. so e^ix must be a linear combination of sin and cosine. that's probably how it arose.

I get what you are saying about the first and second derivative, but then I lost ya. Yet, so far I don't see how this has a relation to cos x + i sin x = e^x

 

[snipez90]

As I said before I'm not familiar with calculus series, the only thing I do know that you mentioned is De'Moivres theorem. Thanks though, I'll check if I can understand this.

Well actually you asked if it was possible to understand what the above posters did with basic knowledge of limits, derivatives, and integrals. Series are defined in terms of a convergent sequence, which is a certain type of limit.

If you had no calculus background and were only interested in the trig applications, then a more hand-wavy approach is to the identity is probably fine. Also note this.

But if you do have the basic calculus background, everything mentioned so far should be accessible to you. I remember learning power series after integration (I think this starts Calc II for some people, but it's more of the same basic stuff).

*EDIT* If you're some innocent high schooler who just started learning calculus, you're not expected to fully understand what mathwonk said. There is some basic linear algebra involved, and unsurprisingly you would understand this after studying some basic linear ODE theory.

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.

 

[BloodyFrozen]

Well actually you asked if it was possible to understand what the above posters did with basic knowledge of limits, derivatives, and integrals. Series are defined in terms of a convergent sequence, which is a certain type of limit.

If you had no calculus background and were only interested in the trig applications, then a more hand-wavy approach is to the identity is probably fine. Also note this.

But if you do have the basic calculus background, everything mentioned so far should be accessible to you. I remember learning power series after integration (I think this starts Calc II for some people, but it's more of the same basic stuff).

I'm not that far into calc yet, but I think i got it. The animation on wiki helped alot. Anyone think of a more trigonometric way to approach this?

 

[BloodyFrozen]

Anyone got a trigonometric/complex way to express this?

 

[Bob Kutz]

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.

 

[BloodyFrozen]

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.

Hmm.. I see your point, but I'm not really sure what to say... Anyone Else?

 

[Nebuchadnezza]

Bob Kutz should be banned from the forum for giving ambigous and bogus information...
Labeling i as imaginary is just nonsense, because it does exist. Just not in the space we are normaly working in. And It does have practical use!

From the top of my head electrical engineers works a lot with complex numbers when trying to measure the capacity of electrical circuits i think. (Correct me on this one.)

What I am sorry about is that you can not fully appreciate this equation, because you do not have the proper background knowledge for it.

If i were you I would run down to a local bookstore and buy the book. "Visual Complex Analysis"
It explains everything that has to do with complex analysis in a simple matter with many pictures that illustrates the points made. I really liked the way it explains the question you are asking as well.

What does it mean to raise a number to the i? Short line is, it is not the same as raising a normal number to a power. It denotes the angle between the real and the complex axis... therefore taking the pi angle, will give you -1.

Anyho, just read that book.

 

[Bob Kutz]

Uhm, excuse me, Mr. Nebuchanezza, but the term i is defined as the imaginary unit. So labeling it as imaginary isn't really nonsense at all.

So maybe it isn't me who should be banned.

I never said that Euler's identity was wrong or didn't have practical value, I just pointed out that it works with deliberately anachronistic values. Values that don't appear to have any meaning. And yet the equation works, just the same.

Since you don't even understand the true nature of the term "i" used in the equation, I don't think you ought to run around telling others that they don't appreciate Euler's formula because they don't have the background. I think it takes a fair amount of understanding to look at it, realize why it works, and why it would work no matter how the y-axis is labeled.

What does it mean to raise a number to the i? Therefore taking the pi angle will give you -1? Well, yes that's true, but only because the coefficient of of the i component is zero, at that point, because sine of pi is zero. There is no angle, and the complex axis is perfectly irrelevant at pi.

Regardless; replace i with whatever term you want and the underlying equation works exactly as it did.

What part of that don't you understand?

 

[HallsofIvy]

Uhm, excuse me, Mr. Nebuchanezza, but the term i is defined as the imaginary unit. So labeling it as imaginary isn't really nonsense at all.

You did say that it could be replaced with other 'imaginary' things with no change in the formula. I thought you were joking. If not, then maybe Mr. Nebuchanezza was right!

So maybe it isn't me who should be banned.

I never said that Euler's identity was wrong or didn't have practical value, I just pointed out that it works with deliberately anachronistic values. Values that don't appear to have any meaning. And yet the equation works, just the same.

Since you don't even understand the true nature of the term "i" used in the equation, I don't think you ought to run around telling others that they don't appreciate Euler's formula because they don't have the background. I think it takes a fair amount of understanding to look at it, realize why it works, and why it would work no matter how the y-axis is labeled.

What does it mean to raise a number to the i? Therefore taking the pi angle will give you -1? Well, yes that's true, but only because the coefficient of of the i component is zero, at that point, because sine of pi is zero. There is no angle, and the complex axis is perfectly irrelevant at pi.

Regardless; replace i with whatever term you want and the underlying equation works exactly as it did.

What part of that don't you understand?

Well, I don't understand "replace i with whatever term you want and the underlying equation works exactly as it did". If I replace "i" with the number 1, I get [math]e^{\pi}= 24.14[/math], approximately, not even closed to -1. I would get the "underlying equation" working exactly as it did only if I replace "i" with a number that had exactly the same properties- and that is the same as saying "replace "i" with "i" apparently renamed. If that is what you meant, then it is very trivial and not at all what you said.

 

[camilus]

Uhm, excuse me, Mr. Nebuchanezza, but the term i is defined as the imaginary unit. So labeling it as imaginary isn't really nonsense at all.

So maybe it isn't me who should be banned.

I never said that Euler's identity was wrong or didn't have practical value, I just pointed out that it works with deliberately anachronistic values. Values that don't appear to have any meaning. And yet the equation works, just the same.

Since you don't even understand the true nature of the term "i" used in the equation, I don't think you ought to run around telling others that they don't appreciate Euler's formula because they don't have the background. I think it takes a fair amount of understanding to look at it, realize why it works, and why it would work no matter how the y-axis is labeled.

What does it mean to raise a number to the i? Therefore taking the pi angle will give you -1? Well, yes that's true, but only because the coefficient of of the i component is zero, at that point, because sine of pi is zero. There is no angle, and the complex axis is perfectly irrelevant at pi.

Regardless; replace i with whatever term you want and the underlying equation works exactly as it did.
What part of that don't you understand?

this is absolutely wrong in every sense of the word. 'i' has a very precise definition, as does everything mathematics. if you replace 'i' by 'unicorn' that equation WILL NOT work, unless you define " unicorn² = -1 ". In essence, you are creating an entirely new axis apart from the real axis, seeing as how any number in the real number line squared is greater than or equal to zero, or in other words (any Real number)²cannot be negative.

 

[BloodyFrozen]

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]

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 e^x 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]

It depends on what you mean by series expansion. In his Analysis of the infinite he gets the expansion for e^x 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]

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

 

[olivermsun]

I get what you are saying about the first and second derivative, but then I lost ya. Yet, so far I don't see how this has a relation to cos x + i sin x = e^x

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]

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]

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]

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]

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]

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]

I see your point

 

[Bob Kutz]

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]

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.