Not, strictly speaking, a proof, Euler's relation e^{pi i} + 1 = 0 is a definition allowing one to take a complex power of e. It is, however, consistent with the analytical continuation of exp as defined on the real line. (In other words, it makes excellent sense, and behaves just as one would like).

One of the 2 formulae I scrawled on my helmet in the Army (the other was E=mc^2, not Born to Kill).

Euler's formula

ei*x = cis(x)

simply relates the transcendental functions of exponentiation and trigonometry. The statement given by inserting a value of pi for x demonstrates the uses of the natural exponentiation base e, pi, the negative unit, and the imaginary unit:

ei*pi = -1

Because it represents so many concepts in one small equality, it's part of the demonstration of intelligent life on Earth imprinted on the side of many NASA spacecraft. Of course, that assumes that the little green men will have the patience and luck enough to figure out that those little scrawlings represent mathematical truths and not a declaration of interstellar war.

Euler's formula is an example of the mathematical identity e^ix = cos(x)+i*sin(x) for all real values of x. In this case, x=pi, cos(pi)=-1 and sin(x)=0.

Deriving the identity itself involves some knowledge of calculus and combinatorics. You can short-cut it, if you'd like, by looking up Euler's formula in the sci.math FAQ at

Although the proof of Euler's formula through power series is sufficient, I personally like the proof that my calculus textbook uses, with good old differential equations. I'll try to explain it here, in all its creamy HTML goodness.

Suppose you "pull an equation out of thin air":

y = cos x + i sin x (also known as cis x)

where i = √-1. Note that this equation basically means that if x is an angle in radians, than y is the position of that angle's point on a unit circle in the complex plane. That's not important, though. We take the derivative of each side with respect to x:

d/dx(y) = d/dx(cos x + i sin x)
dy/dx = -sin x + i cos x
dy/dx = i(i sin x + cos x)
dy/dx = i(cos x + i sin x)

Therefore, because of the first equation,

dy/dx = iy

Freaky already. This is where the differential equation comes in. Manipulating things around:

1/y dy = i dx
1/y dy = ∫i dx
ln |y| = ix + C
|y| = Ceix
y = ±Ceix

And therefore by substitution:

±Ceix = cos x + i sin x

Setting x to 0 to find out C:

±Ce0i = cos 0 + i sin 0
±C = 1
C = 1

Since C has to equal 1 (and the negative C is unneccesary, as it's false for x = 0), we can forget about it.

eix = cos x + i sin x

Fun stuff. Substituting π for x gives the equation that everyone's talking about:

eiπ = -1

or, as most prefer to express it:

eiπ + 1 = 0.

Taylor Series Proof of Euler's Relation: eπi=-1

Step one: The Taylor Series

The Taylor Series is an infinite series that rewrites any function f(x) as

f(n)(x0) (x-x0)n
 Σ    ---------------
n=0         n!

If you don't know what that means, there's no way in hell I can explain it to you here. However, I will clarify that f(n)(x) means "the nth derivative of f(x)".

If none of this makes sense to you, you might want to look at:

Important Taylor Series #1: f(x)=ex

The function ex is its own derivative--this is an extremely unique property of ex that I will not prove here.

For this Taylor series, I will use x0=0. (¬e;: The x0=0 case of the Taylor Series is called the Maclurin Series.) Since when f(x)=ex, all f(n)(x)=f(x), and f(x0)=f(0)=e0=1, the term f(n) always cancels to 1 and drops out of the equation completely. Also, (x-x0)=(x-0)=(x). Therefore:

f(x)=ex=x0/0!+x1/1!+x2/2!+x3/3!...=  Σ  xn/n!

Important Taylor Series #2 and 3: f(x)=sin(x) and f(x)=cos(x)

Once again, x0=0. I will not derive this particular series here, but if you plug in a number for a few terms you can satisfy yourself that it works.

f(x)=sin(x)=x-x3/3!+x5/5!-x7/7!...= Σ  (-1)nx2n+1/(2n+1)!

And for f(x)=cos(x):

f(x)=sin(x)=1-x2/2!+x4/4!-x6/6!...= Σ  (-1)nx2n/(2n)!

Trust me, these work. You can derive them with only a little trouble, if you want--the key is that sin(x0)=0 and cos(x0)=1.

Step two: exi

If we want to find eπi, first we have to find a general expression for exi ∀ x.

Plug in xi for x in the ex Taylor Series.

f(xi)=exi=(xi)0/0!+(xi)1/1!+(xi)2/2!+(xi)3/3!+ (xi)4/4!...

Now, we'd like to get rid of as many i's as possible--remember, i=√(-1). Therefore, exponents of i are on a four-step cycle:

  i0                              =1
  i1                              =i
  i2=√-1)2                        =-1
  i3=√-1)2i=-1i                   =-i
( i4=√-1)22=-12                   =1 )

If this doesn't make any sense to you, read up on:

Using these identities, we can substitute into the Taylor series for f(x)=eix:

Separating out therealand imaginary terms:

Look familiar? You'll notice that the first series in parentheses is the same as the cosine Taylor series. The second one is the sine series. How handy!
            ∞                         ∞
         =( Σ  (-1)nx2n+1/(2n+1)!)+i( Σ  (-1)nx2n/(2n)!)
           n=0                       n=0


Step three: eπi

From here, it is a simple matter to substitute π in for x in our newly derived equation.
As we all know, cos(π)=-1 and sin(π)=0. If you didn't know that, you should read: Therefore:



Addendum: There was quite a horrible mistake in this node that was only corrected because of a helpful msg I got from ttsen483. Thanks, and shame on the rest of us for not noticing it earlier.

(The equation is still true and will be for the forseeable future. And that, folks, is why math is better than fizziks.)

e^i(theta)=cis(theta) (idea)

This has been reffered to the most beautiful relation in mathematics. For those who have trouble reading it is e to the power of i times theta(where theta is a variable for the angle of the argument (angle)of a complex number) is equal to cos(theta)+isin(theta) or cis(theta). And cis(theta) is a very convenient way of expressing a complex number.

Now onto the proof: A long While back some irish mathematician named something MacLauren started analysing a function equal to a series A0 + A1x + A2x2 + A3x3 + .... He started finding the derivatives of such a function.

f'(x)= A1 + 2A2x + 3A3x2 + 4A3x3 + ...

f''(x)= 2!A2 + 3•2A3x + 4•3A4x2 + 5•4A5x3 + ...

f'''(x)= 3!A3 + 4•3•2A4x + 5•4•3A5x2 + 6•5•4A6x3 + ...

f''''(x)= 4!A4 + 5•4•3•2A5x + 6•5•4•3A6x2 + 7•6•5•4A7x3 + ...


Then he observed that if he knew f(0) and f'(0) and f''(0) and so on, he could solve the equation for f(x)

f(0)=A0 + A1x + A2x2 + A3x3 + ....

f'(0)= A1 + 2A2(0) + 3A3(0)2 + 4A3(0)3 + ...

f''(0)= 2!A2 + 3•2A3(0) + 4•3A4(0)2 + 5•4A5(0)3 + ...

f'''(0)= 3!A3 + 4•3•2A4(0) + 5•4•3A5(0)2 + 6•5•4A6(0)3 + ...

f''''(0)= 4!A4 + 5•4•3•2A5(0) + 6•5•4•3A6(0)2 + 7•6•5•4A7(0)3 + ...






Then he figured that A1=f'(0)/1! , A2=f''(0)/2!, A(3)=f'''/3! and so on, that An=f(0)n/n! where ! denotes the factorial notation. sp...

f(x)= f(0) + f'(0)x + f''(0)x2/2! + f'''(0)x3/3! + ... + f n(0)xn/n! + ...

This is function let's us figure out any function for which we know f(0) and all the derivatives of zero. Three functions for which we know this are sinx, cosx and e^x.

Briefly about the calculus of trigonometric functions

The derivative of f(x)=sinx is equal to cosx that is f'(x)=cosx, and the derivative of cosx is -sinx so f''(x)=-sinx and then the derivative of this; f'''(x)=-cosx and f''''(x)=sinx and then it repeats for further derivatives. (There is a rather convenient proof for this but it requires the aid of graphical representation to accurately convey the idea, and is difficult to prove in HTML.) So now sin 0=0 and cos 0=1 now inserting this into our above identity sinx= 0 + 1*x + 0*x^2/2! + (-1)x^3/3! + (-0)x^4/4! + this can be rewritten as sinx= x - x^3/3! + x^5/5! - x^7/7! + x^9/9! + ... and cos x can be written as 1-x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...

Now assuming that e is a number such that the derivative of e^x=e^x then we can find and any number to the power of zero equals one, then we can safely say that all derivatives of e^x for x=0 are 1. Again taking MacLauren's series.

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

now taking a number i such that i = ?-1 and putting it into the equation for e^x we get:

e^i = 1 + i +i^2/2! + i^3/3! + i^4/4! + i^5/5! + ...

well first off we know that for all even values of n in i^n/n! that i^n/n! will be real(will not utilize i). And seeing as i^2=-1 then i^4=1 and i^6=-1 we can conclude that all values of n that are only divisible by 2 and not 4 will be negative, and all values of n divisible by four will be positive. so looking at only even values of n in e^i, e^i= 1 + i^2/2! + i^4/4! + i^6/6! + i^8/8! + ...

and simplifying with our two theories e^i = 1- 1/2! + 1/4! - 1/6! + 1/8! + ... which looks startlingly familiar to our equation of cosx and indeed if we had ^ix then the real part would be equal to cosx.

As for the nonreal part; for all values of n that aren't divisible by two we know that the i's won't multiply out, and that this part is imaginary, and agaain using theories similar to those above we can state that for all values of n-1 divisble by 2 and not four, i^n/n! is negative, and for all values of n-1 divisible by 4, i^n/n! is positive.

e^i = 1 + i - 1/2! - i/3! + 1/4! + i/5! - 1/6! - i/7! + 1/8! + i/9! - ...

rearranged e^i = 1 - 1/2! + 1/4!- 1/6! +1/8! -... + i - i/3! + i/5! - i/7! + i/9! - ...

which looks suspiciously like cos(1) + isin(1) and indeed we could substitute x or any variable(such as theta) in and it will come out to cos(x) + isin(x) or cis(x).

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