Wednesday, February 22, 2006

Euler's Identity

One of the simplest and most elegant equations is Euler's Identity:
eπi + 1 = 0
.

It states that Euler's number e (which is equal to roughly 2.718...) to the power of i*π (taken at roughly 3.14...) is equal to -1. This equation is used to derive cyclotomic integers which are used in Kummer's proof for Fermat's Last Theorem for regular primes.

To be fair, for many people, xi does not have a clear value. What does it mean to put an exponent to an imaginary power? In mathematics, it is ok if the details are not intuitive as long as they are logically consistent. The Maclaurin Series, for example, can be used to define exponents to a complex power (see here). Newton did something similar when he generalized the Binomial Theorem to include complex powers (see here for details)

Theorem: Euler's Identity is the equation: eπi = -1

This equation derives directly from Euler's Formula:
eix = cos x - isin x [See here for proof]

We get:
eπi = cos(π) + isin(π) = -1 + i(0) = -1. [Review of e (see here), sin and cos (see here), i (see here), and π (see here)]

QED

Corollary: eπi + 1 = 0

This directly follows from above.

QED

10 comments:

Avo said...

You don't really mean to say e = 0.552... do you??

Larry Freeman said...

Ouch. That's a bad typo. I just changed it.

Thanks for posting! :-)

-Larry

Veefessional said...

Well, what happen if you substitute x by 2π, 4π, 6π, or in general, 2kπ, where k=1,2,3,... ?

You will actually get:

2πi=0, 4πi=0 and so on, which, by dividing the constants 2 or 4 (and so on) on both sides, gives πi=o, which is inanity. What is this?

I have been thinking about it myself, you know. Haha. We share a similar wavelength range.

Do reply to me personally at Veefessional@yahoo.com.


With regards,

Vee-Liem.

Zachary said...

If you ask me, i itself is insanity. That it should behave so well in this identity is just amazing.

PsychoKia said...

hmmm. i think that veefessional has a point. could you please post the reply on the blog? So that more people could benefit. Thanks

Larry Freeman said...

Hi Veefessional,

You make the assumption that if e^(2kπ) = 1 that 2kπ = 0.

But this is not the case as you demonstrate in the argument. In fact, the solutions to this type of equation where 2kπ ≠ 0 are called roots of unity.

A root of unity is a number that solves the following solution:
x^n = 1 where n ≠ 0.

I discuss this issue in another blog. See here

Please let me know if you have any additional questions.

Cheers,

-Larry

llouk said...

I dont get it...How e^x=-1
e is a positiv number ^x always will give positiv...
..??

Larry Freeman said...

Hi llouk,

Everything changes when the exponent is an imaginary number.

You can see the details on reasoning with imaginary exponents here

EULER PIPHI said...

its just a notation....dont take it in its face value....e^i(theta)=cos(theta)+isin(theta)...thats it...its just a shorthand notation

Vijay said...

This is one of my favorite blogs!

Just wanted to let you know that there's a typo in the post:

e^(ix) = cos x - isin x [See here for proof]

should read:

e^(ix) = cos x + isin x [See here for proof]