e, i, pi - linked?

Question

How is "e^(i*pi)=-1" or whatever it is derived? Why is it true? Does it have any use?

Answer 1

e^x can be represented as the infinite sum

1 + x/1! + (x^2)/2! + ... + (x^n)/n!

If x is angular measurement in radians, then it turns out
that cos(x) can be represented by the infinite sum

1 - (x^2)/2! + (x^4)/4! + ... + (-1)^(2n) (x^(2n))/(2n)!

and sin(x) by the infinite sum

x - (x^3)/3! + (x^5)/5! +...+ (-1)^(2n)(x^(2n+1))/(2n+1)!


If we extend the exponential function just by plugging in the number ix where i is the "imaginary" number that we imagine so that -1 can have a square root, then (keeping in mind that
i^2 = -1 (and so i^3 = -i and i^4 = 1), we have that

e^(i * x) = cos x + i sin x

and so e^(pi * i) = cos(pi) + i sin(pi) = -1 + (i * 0) = -1

The easiest explanations for why cos and sin can be defined by those series above involve some calculus.

To generalize the fact you are asking about, a point in the plane (or, equivalently, a complex number c) can be represented by rectangular coordinates (x,y) or by polar coordinates (r, theta).

where r > 0 and 0 <= theta < 2*pi.

In the first case, the complex number c = x + iy and x and y simply denote the horizontal and vertical displacements from the origin of your coordinates; in the second, theta gives the counterclockwise radian measurement at the origin from the positive horizontal axis to the ray from the orign leading to the point of interest in the plane, and r gives the distance along this ray from the origin. theta tells which infinitely thin spoke of concentric circles the point is on, and r tells which circle centered at the origin the point is on.

This other representation of a complex number is called the polar representation or the polar coordinates of the number, and it is sometimes much easier to work with than the rectangular representation of the number.

Answer 2

You need to know how Taylor series work. ( http://en.wikipedia.org/wiki/Taylor_series ). Basically, you can engineer an polynomial function that has the same derivatives as any other function. For example, the derivative of e^x with respect to x is e^x. So you could create a function which has the same derivative as e^x at the origin of the coordinate system. Thus e^x is about equal to 1 + x + x^2/2 + x^3/3, and so on at x=0. e^x | x = 0 is 1, which is correct by that formula, but we lose accuracy as we move away from 0 (actually, if you were to create an infinitely long taylor series, it would perfectly match e^x, although this is not true of all functions), so if we use x = 1 in the above formula, we get e = 2.83333, which is close to the actual value of e = 2.718. So there's that. The scoop is that if you find taylor series for sin(x) and cos(x), you find that cos(x) + i*sin(x) = e^(i*x) (all of those functions really being their taylor series equivalents). So, by this e^(i*pi) = cos(pi) + i*sin(pi) = -1. You should read the wikipedia article for a full derivation.
This formula is used to define coordinates on the complex number plane, and also comes in when looking at stuff like oscillating functions and Schrodinger's equation, which has an imaginary part.

Answer 3

Here's one use for the e-pi identity. In 19th century, mathematicians thought that e and pi were transcendental numbers, but didn't have the proof. When a proof for e was found, pi was automatically proved to be transcendental, because only then could a function of the two, as in this identity, be a rational number.

Answer 4

What fascinated me about this was that e and pi are irrational and i has no Real value. Yet combining the three (e^(i*pi)) gives -1.

Answer 5

Let f(x)=e*(ix). Then f^(n)(x)=(i^n)•f(x) where f^(n) is the nth derivative of f.

Since f(0)=1, f^(n)(0)=i^n=(-1)^(n+1) for n even and (-1)*(n+1)•i for n odd.

Therefore consider the Taylor expansion of f(x) around 0.
f(x)=1+ix-x^2/2-ix^3/3!+ x^4/4!+ix^5/5!+ . . .=(1-x^2/2+x^4/4!-x^6/6!+ . . .)+ i•(x-x^3/3!+x^5/5!-x^7/7!+ . . .) = cos(x)+i•sin(x).

therefore e^(iπ)=f(π) =cos(π)+i•sin(π)=-1+i•0=-1.

Answer 6

e^(ix) = cosx + i*sinx ( Moivre's law)
Choose x=pi and you find what you want:
e^(i*pi) = -1 + 0
The relation e^(i*pi) = -1 connects four important 'units' in math.
Peculiar!

Answer 7

From Euler relation :
e^(i*x)=cos(x)+i*sin(x)

if u substitute x for pi , you get :

cos(pi)+i*sin(pi)

cos(pi)=-1
sin(pi)=0
--> cos(pi)+i*sin(pi)=-1

Answer 8

i actually learned this earlier this year. cosine and sine were involved in the formula

cos(x) + i sin(x) = e^(i*x)
cos(pi) + i sin(pi) = e^(i*pi)
cos(pi) + i sin(pi) = -1