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Method of Proving Euler's Formula?

by Mandelbroth
Tags: euler, formula, method, proving
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Mandelbroth
#1
Dec21-12, 12:39 PM
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The other day, I was thinking about Fourier series. Because eix 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 eix, right?

Isn't this an easier way to prove Euler's formula than using Taylor expansion? Or...am I missing something?
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HallsofIvy
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Dec21-12, 01:06 PM
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How do you determine that eix has period [itex]2\pi[/itex] with using Euler's formula?

Sorry- I meant "without using Euler's formula".
Dickfore
#3
Dec21-12, 01:49 PM
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Because [itex]e^{z_1 + z_2} = e^{z_1} e^{z_2}[/itex], and [itex]e^{2 \pi i} = 1[/itex].

pasmith
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Dec21-12, 01:52 PM
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Method of Proving Euler's Formula?

Quote Quote by Dickfore View Post
Because [itex]e^{z_1 + z_2} = e^{z_1} e^{z_2}[/itex], and [itex]e^{2 \pi i} = 1[/itex].
So how do you know that [itex]e^{2\pi i} = 1[/itex] without using Euler's formula?
Mandelbroth
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Dec21-12, 02:07 PM
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Quote Quote by HallsofIvy View Post
How do you determine that eix has period [itex]2\pi[/itex] with using Euler's formula?
With Euler's formula? Sine and cosine have a period of 2 pi. Therefore, an addition of the two also has a period of 2 pi.

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

Quote Quote by pasmith View Post
So how do you know that [itex]e^{2\pi i} = 1[/itex] without using Euler's formula?
[itex]\displaystyle \forall k\in \mathbb{Z}, \ e^{2\pi i k} = \lim_{n \rightarrow \infty} \left(1+\frac{2\pi i k}{n}\right)^n = 1[/itex].

Though, you do make a valid point. I already knew that eix 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?
pasmith
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Dec21-12, 05:09 PM
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Quote Quote by Mandelbroth View Post

[itex]\displaystyle \forall k\in \mathbb{Z}, \ e^{2\pi i k} = \lim_{n \rightarrow \infty} \left(1+\frac{2\pi i k}{n}\right)^n = 1[/itex].
Why is that obvious (without assuming Euler's formula)? I think if you try to prove that rigorously, you will end up writing something like
[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].
Mandelbroth
#7
Dec21-12, 06:40 PM
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Quote Quote by pasmith View Post
Why is that obvious (without assuming Euler's formula)? I think if you try to prove that rigorously, you will end up writing something like
[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].
Obvious? (It's not...)

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]?
Mandelbroth
#8
Dec22-12, 10:04 AM
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Quote Quote by Mandelbroth View Post
Obvious? (It's not...)

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]?
I'm sorry. I was wrong here. The equation should be [itex]\displaystyle ln(e^{2\pi i k}) = \int_{1}^{e^{2\pi i k}} \frac{1}{t} \ dt = 0[/itex]. I forgot that not all properties of logarithms are consistent between real numbers and complex numbers.
Dickfore
#9
Dec22-12, 10:10 AM
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Quote Quote by pasmith View Post
So how do you know that [itex]e^{2\pi i} = 1[/itex] without using Euler's formula?
So, how do you know [itex]\cos 0 = 1[/itex]?
HassanEE
#10
Dec26-12, 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)=eix then eix=ei(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 eix, 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.
rbj
#11
Dec26-12, 09:51 PM
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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?
Boorglar
#12
Jan25-13, 10:27 PM
P: 177
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 n-th 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).
Boorglar
#13
Jan25-13, 10:29 PM
P: 177
Oops, I copy-pasted 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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