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Method of Proving Euler's Formula? 
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#1
Dec2112, 12:39 PM

P: 615

The other day, I was thinking about Fourier series. Because e^{ix} is periodic, with period of 2 pi, we can use the Fourier series...
[itex]\displaystyle \frac{1}{2\pi} \int_{\pi}^{\pi} e^{ix} \ dx + \frac{1}{\pi}\sum_{n=1}^{\infty}\left[\left(\int_{\pi}^{\pi} e^{ix} cos(nx) \ dx \right) cos(nx) + \left(\int_{\pi}^{\pi} e^{ix} sin(nx) \ dx\right) sin(nx)\right] = \frac{0}{2\pi} + \frac{1}{\pi}(\pi cos{x} + i\pi sin{x}) = cos{x} + i sin{x}[/itex] ...to describe e^{ix}, right? Isn't this an easier way to prove Euler's formula than using Taylor expansion? Or...am I missing something? 


#2
Dec2112, 01:06 PM

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How do you determine that e^{ix} has period [itex]2\pi[/itex] with using Euler's formula?
Sorry I meant "without using Euler's formula". 


#3
Dec2112, 01:49 PM

P: 3,014

Because [itex]e^{z_1 + z_2} = e^{z_1} e^{z_2}[/itex], and [itex]e^{2 \pi i} = 1[/itex].



#4
Dec2112, 01:52 PM

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Method of Proving Euler's Formula?



#5
Dec2112, 02:07 PM

P: 615

Without, [itex]\displaystyle e^{ix} = \lim_{n \rightarrow \infty}\left(1 + \frac{ix}{n}\right)^n[/itex]. Observing the graph of [itex]f(x) = \left(1+\frac{ix}{n}\right)^n[/itex], it can be seen that the graph looks more and more sinusoidal as n grows larger. Using a large n, it would be fairly easy to approximate the period of the function's limit as n approaches infinity. Using WolframAlpha (an essential tool for deriving Euler's formula in the 1700s ) to graph f(x) with n=100, we get this (link goes to WolframAlpha). It can be seen that the imaginary part looks like it is approaching the form of the sine function, and the real part looks like cosine. If we look at x ≈ 3.14, the imaginary part is around 0 and the real part is around 1. This can also be used to answer Though, you do make a valid point. I already knew that e^{ix} was periodic based on Euler's formula. Thus, there is a minor element of circular logic to my "proof". However, assuming that we say that the function f from above is asymptotically periodic, is there any reason that this is not correct? 


#6
Dec2112, 05:09 PM

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[tex]\lim_{n \to \infty} \left(1 + \frac{2\pi k \mathrm{i}}{n}\right)^n = \sum_{n=0}^{\infty} \frac{(2\pi k\mathrm{i})^n}{n!} = \sum_{n=0}^{\infty} \frac{(1)^n(2\pi k)^{2n}}{(2n)!} + \mathrm{i} \sum_{n=0}^{\infty} \frac{(1)^n(2\pi k)^{2n+1}}{(2n+1)!}\\ = \cos(2\pi k) + \mathrm{i}\sin(2\pi k) = 1 [/tex] which is just the Taylor series proof of Euler's formula in the special case [itex]x = 2\pi k[/itex]. 


#7
Dec2112, 06:40 PM

P: 615

Perhaps, if someone was ridiculously intent on not using Euler's formula, they might use [itex]\displaystyle 2\pi i k = ln(\frac{1}{2}) + \int_{\frac{1}{2}}^{e^{2\pi i k}} \frac{1}{t} \ dt[/itex]? 


#8
Dec2212, 10:04 AM

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#9
Dec2212, 10:10 AM

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#10
Dec2612, 05:18 AM

P: 11

I see where HallsofIvy is getting at. Before you go ahead and find the Fourier series, you have to show periodicity. For f(x) to be periodic it must satisfy f(x+T)=f(x) for all t. If f(x)=e^{ix} then e^{ix}=e^{i(x+T)}. At this point you will not be able to proceed unless you used Euler's formula (or Taylor series expansion, but you are trying to avoid that altogether).
You mentioned graphs of e^{ix}, but to plot these graphs you need to use Euler's formula! If you were to find a way to show periodicity without using Euler's formula, then you can probably go ahead and use the Fourier series to prove Euler. 


#11
Dec2612, 09:51 PM

P: 2,251

the easiest way i know of to prove Euler's formula was in wikipedia for a while, but they deleted it (for no good reason).
Consider the function [tex] f(x) = \frac{e^{ix}}{\cos(x) + i \sin(x)} [/tex] where [tex] i^2 = 1 [/tex]. so, treating [itex]i[/itex] as a constant with the above property, compute the first derivative of [itex]f(x)[/itex] and also what [itex]f(0)[/itex] is. what does that tell you? 


#12
Jan2513, 10:27 PM

P: 173

I use the limit definition of the exponential:
[tex] e^{i\pi}=\lim_{n→∞}(1+\frac{ i\pi }{ n }) [/tex] Then I rewrite [itex] 1 + \frac{ i\pi }{ n } [/itex] as: [tex] \sqrt{1+\frac{\pi^2}{n^2}}[\cos(\arctan(\frac{\pi}{n})) + i\sin(\arctan(\frac{\pi}{n}))] [/tex] i.e. in the trigonometric form (the module is the stuff with the square root, and the argument is arctan(pi/n) ). (Not assuming Euler's formula here) Then using de Moivre's formula on this rewritten number (which again can be proven by induction for integer n, and then for rational n, without Euler's formula), [tex] (1+\frac{i\pi}{n})^n = \sqrt{1+\frac{\pi^2}{n^2}}^n[\cos(\arctan(\frac{\pi}{n})) + i\sin(\arctan(\frac{\pi}{n}))]^n = \sqrt{1+\frac{\pi^2}{n^2}}^n[\cos(n\arctan(\frac{\pi}{n})) + i\sin(n\arctan(\frac{\pi}{n}))] [/tex] The limit as n approaches infinity of the module raised to the nth power approaches 1, and the limit for n*arctan(pi/n) will be pi (both can be proven with l'Hopital's rule or some other theorem about limits, which do not use Euler's formula). And plugging pi into cos and sin will finally give the answer of 1. You could argue that this is just a special case of a proof of Euler's formula, but at least I didn't use it anywhere in the proof. Oh and a technical detail: when I took the limit, I actually did it only for rational n, but I think that the continuity of (1+i*pi/n)^n implies the limit for any real n, because Q is dense (but I haven't checked into that). 


#13
Jan2513, 10:29 PM

P: 173

Oops, I copypasted my previous post from a previous post I wrote, but the Latex apparently didn't...
EDIT: Ok, I rewrote my above post in a more readable format. Also, I just realised this thread asks for a proof of Euler's formula in general, not the special case for pi. But I think my argument is exactly the same, by replacing pi with x. (My original answer was to a person asking for proof of Euler's identity without using Euler's formula) 


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