Euler's Identity. So What?

[Anjruu]

e^(i*pi)+1=0. What is it used for? Why is it special? Ok, it relates pi,e, i, 1, and 0, but does it have a practical application?

[□h=-16πT]

No, not really.

[matt grime]

Define practical. Why should it be practical? Maths is more than its mere applcitaions just as English is more than a mere language to communicate the fact that we want food.

 

But the point was that it in one simple line links the most important mathematical objects there are. it doesn't need a practical application anymore than the Mona Lisa. You are entitled not to find it interesting or beautiful, but it is certainly succinct not to say elegant.

[Yggdrasil]

Euler's identity is used a lot in solving differential equations. It is also useful because it allows one to express the trigonometric functions in terms of exponential functions.

[Dave]

I think you're talking about the equation that the Euler identity is grabbed from:

 

[math]e^{i\theta} = \cos\theta + i \sin\theta[/math]

[Yggdrasil]

My bad.

[Dave]

Not a problem. Just in case it isn't obvious for anyone, put [imath]\theta = \pi[/imath] into that equation, and Euler's identity will drop out.

[ydoaPs]

it also is an easy way to find the natural log of -1...[imath]\ln{-1}=i\pi[/imath]

[Tom Mattson]

e^(i*pi)+1=0. What is it used for? Why is it special? Ok, it relates pi,e, i, 1, and 0, but does it have a practical application?

 

Take the more general version cited by dave, and yes it has numerous practical applications, by virtue of the fact that exponential functions are easier to calculate with than trigonometric functions.

 

One very important application of complex exponentials is in electric circuit analysis in the frequency domain.

[Dave]

it also is an easy way to find the natural log of -1...[imath']\ln{-1}=i\pi[/imath]

 

Whilst this is a bit offtopic, I should point out that it's not quite as simple as this. The above statement is certainly true; however, it can be expanded upon. You should be able to see that [imath]e^{(2n+1)i\pi} = -1[/imath] for [imath]n \in \mathbb{Z}[/imath], since [imath](-1)^{2n+1} = -1[/imath]. So you can quite easily say that [imath]\log(-1) = (2n+1)i\pi[/imath].

 

There's quite a nice little article about this at Dr. Math that I found from a quick google. I suggest people check it out, since it's rather informative

[Ollie]

e^(i*pi)+1=0. What is it used for?

 

Historically, the best use of that line is to get annoying philosophers to shut up.

[DV8 2XL]

It's a stunningly beautiful relationship; what else does it need to be?

[Anjruu]

Nothing. I have only known it has a beautiful relationship, I was curious if there was more to it.