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Explain Euler's Theorem/Identity

by BloodyFrozen
Tags: calculus, euler, explanation, theorem, trigonometry
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BloodyFrozen
#1
Apr11-11, 07:38 PM
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Please explain Euler's theorem. I don't get how he got this formula and how it can be used instead of trigonometry. Thanks
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Stephen Tashi
#2
Apr11-11, 10:29 PM
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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
#3
Apr12-11, 05:01 PM
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What Stephen Tashi said!

Specifically, the MacLaurin series for [itex]e^x[/itex] is
[tex]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[/tex]

If you replace "x" with "ix" that becomes
[tex]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[/tex]
[tex]= 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[/tex]

But it is easy to see that [itex]i^2= -1[/itex], [itex]i^3= i(-1)= -i[/itex], [itex]i^4= (i^2)(i^2)= (-1)(-1)= 1[/itex], [itex]i^5= (1)(i)= i[/itex], 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:
[tex]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)[/tex]

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
#4
Apr12-11, 09:18 PM
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Explain Euler's Theorem/Identity

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
#5
Apr13-11, 02:49 PM
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Quote Quote by HallsofIvy View Post
What Stephen Tashi said!

Specifically, the MacLaurin series for [itex]e^x[/itex] is
[tex]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[/tex]

If you replace "x" with "ix" that becomes
[tex]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[/tex]
[tex]= 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[/tex]

But it is easy to see that [itex]i^2= -1[/itex], [itex]i^3= i(-1)= -i[/itex], [itex]i^4= (i^2)(i^2)= (-1)(-1)= 1[/itex], [itex]i^5= (1)(i)= i[/itex], 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:
[tex]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]

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).

Is there anyway to "understand" this without the knowledge of Calculus beyond the fundamentals (limits, derivatives, integrals)? As I'm not very familiar with Taylor series, MacLaurin Series, and Series in general I suppose. I don't understand how this is applicable for trigonometry.

Thanks to answerers so far.
snipez90
#6
Apr13-11, 03:24 PM
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The associated wikipedia article 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
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Apr13-11, 03:35 PM
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Quote Quote by snipez90 View Post
The associated wikipedia article 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
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Apr13-11, 03:38 PM
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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. thats probably how it arose.
BloodyFrozen
#9
Apr13-11, 03:51 PM
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Quote Quote by mathwonk View Post
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. thats 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 dont see how this has a relation to cos x + i sin x = ex
snipez90
#10
Apr13-11, 04:05 PM
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Quote Quote by BloodyFrozen View Post
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
#11
Apr13-11, 04:11 PM
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Quote Quote by snipez90 View Post
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
#12
Apr15-11, 03:10 PM
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Anyone got a trigonometric/complex way to express this?
Bob Kutz
#13
Apr18-11, 02:18 PM
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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
#14
Apr18-11, 03:25 PM
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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.

Hmm.. I see your point, but I'm not really sure what to say... Anyone Else?
Nebuchadnezza
#15
Apr18-11, 03:50 PM
P: 76
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 alot 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... therfore taking the pi angle, will give you -1.

Anyho, just read that book.
Bob Kutz
#16
Apr18-11, 04:23 PM
P: 8
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
#17
Apr18-11, 06:21 PM
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Quote Quote by Bob Kutz View Post
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
#18
Apr18-11, 09:30 PM
P: 150
Quote Quote by Bob Kutz View Post
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.


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