How many Eulers identity proofs are there?

[vco96]

Euler's identity proof

If you recall the famous Euler's identity e(xi) = cos(x) + i sin(x) there is one a proof using infinite series expansion.

My question is: Are there any other proofs of this identity.

Thanks
Art.

 

[waterman]

Re: Euler's identity proof

I remember that one could prove it using the defferential equation.
d(e(xi))dx = i*e(xi)

Sorry I forget the detail. I will try to find out. (I believe it could be found in some
textbooks)

 

[jorgito]

Euler's identity proof

Hello,
Look at
h**p://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/eixproof.htm

Best regards

 

[eltonjohn]

Re: Euler's identity proof

Hey .. Somebody posted the proof and he must erased ..
It was the clasic expansions series proof as i recall it

 

[greek]

Euler's identity proof

Sure there is a proof else we would talk about Euler's Last Theorem

 

[sifeddin]

Euler's identity proof

what is Euler's Last Theorem?

 

[snaider]

Re: Euler's identity proof

is the following:
In number theory, Euler's theorem (also known as the Fermat-Euler theorem) states that if n is a positive integer and a is relatively prime to n, then

aφ = 1 (mod n)
where φ denotes Euler's totient function.

The theorem is a generalization of Fermat's little theorem, and is further generalized by Carmichael's theorem.

The theorem may be used to easily reduce large powers modulo n. For example, consider finding the last decimal digit of 7222, i.e. 7222 mod 10. Note that 7 and 10 are coprime, and φ(10) = 4. So Euler's theorem yields 74 = 1 (mod 10), and we get 7222 = 74·55 + 2 = (74)55·72 = 155·72 = 49 = 9 (mod 10).

In general, when reducing a power of a modulo n (where a and n are coprime), one needs to work modulo φ in the exponent of a:

if x = y (mod φ), then ax = ay (mod n).

 

[Element_115]

Re: Euler's identity proof

It's not hard just clever.

Go to any Calculas book and find the series expantion for "e^x"
Then find the series expantion for Sin(x) and cos(x).

1) Change "e^x" to (x=jω) "e^jω"

2) Do all the j^2=-1 , j^3 =-j , j^4 = 1

3) Now separate the "e^jω" so one line looks ~like cos(ω) and the other
line looks ~like jsin(ω); (Factor out the"j"!)

Now you should be able to see that e^jω = cos(ω) +jsin(ω)

Cheers

 

[mayyan]

Re: Euler's identity proof

its not hard to prove ( without a series ) for: e^(ax) = b*sin(x)+c*cos(x) just "play" with it a little.
if someone manage to prove for e^(ax) = b*f(x)+c*g(x) i would be interested.

regards
mayyan

 

[mmatica]

Re: Euler's identity proof

> if someone manage to prove for e^(ax) = b*f(x)+c*g(x) i would be interested.

You probably mean b=b(a), c=c(a) and f,g depend of x only.
It is not difficult to show that such function of 1 variable (b,c,f,g) do not exist!

Mate.

 

[sp]

Re: Euler's identity proof

https://ccrma.stanford.edu/~jos/mdft/e_j_theta.html

the above will giv u the prove using the series....taylor and maclaurin seies.

 

[tantoun2004]

Re: Euler's identity proof

one of the easiest proof is to expand exp(i*x) in series (Macclauen for example)
you'll get : 1+(i*x)+(i*x)^2/2!+......
you'll find that it's the expansion of cos(x)+i*sin(x)
Regards,