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BloodyFrozen
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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
HallsofIvy said: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).
snipez90 said: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).
mathwonk said: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 said: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.
snipez90 said: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).
Bob Kutz said: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.
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!Bob Kutz said: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.
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 \(\displaystyle e^{\pi}= 24.14\), 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.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?
Bob Kutz said: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?
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.
dimitri151 said: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.
Here's another way to understand the "differential equations" approach:BloodyFrozen said: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 = ex
olivermsun said: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).
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.Bob Kutz said: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.
Bob Kutz said: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.
baric said: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.
HallsofIvy said: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.
Bob Kutz said:But, further, Euler's formula works perfectly well, no matter what you replace i with
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?Bob Kutz said: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 Kutz said: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.