Euler's formula

There are two results known as Euler's formula, both named after the mathematician Leonhard Euler.

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In geometry and algebraic topology, there is a relationship called Euler's formula which relates the number of edges, vertices, and faces of a simply connected polyhedron. Given such a polyhedron, the sum of the vertices and the faces is always the number of edges plus two. i.e.: F - E + V = 2. The theorem also applies to any planar graph.

For nonplanar graphs, there is a generalization: If the graph can be embedded in a manifold M, then F - E + V = χ(M), where χ is the Euler characteristic of the manifold, a constant which is invariant under continuous deformations. The Euler characteristic of a simply-connected manifold such as a sphere or a plane is 2. A generalization of Euler's formula for arbitrary planar graphs exists: F - E + V - C = 1, where C is the number of components in the graph.

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In complex analysis, Euler's formula, attributed to the mathematician Leonhard Euler, states that

<math>e^{ix} = \cos x + i\;\sin x</math>

for any real number x. Here, e is the base of the natural logarithm, i is the imaginary unit and sin and cos are trigonometric functions.

This formula can be interpreted as saying that the function e^ix traces out the unit circle in the complex number plane as <math>x</math> ranges through the real numbers. Here, <math>x</math> is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians. The formula is valid only if sin and cos take their arguments in radians rather than in degrees.

The proof is based on the Taylor series expansions of the exponential function e^z (where z is a complex number) and of sin x and cos x for real numbers x (see below). In fact, the same proof shows that Euler's formula is even valid for all complex numbers x.

Euler's formula was proved (in an obscured form) for the first time by Roger Cotes in 1714, then rediscovered and popularized by Euler in 1748. It is interesting to note that neither of these men saw the geometrical interpretation alluded to above: the view of complex numbers as points in the plane arose only some 50 years later (see Caspar Wessel).

The formula provides a powerful connection between analysis and trigonometry. It is used to represent complex numbers in polar coordinates and allows the definition of the logarithm for complex arguments. By using the exponential laws

<math>e^{a + b} = e^a \cdot e^{b}</math>

and

<math>(e^a)^b = e^{a b}</math>

(which are valid for all complex numbers <math>a</math> and <math>b</math>), one can also readily derive several trigonometric identities as well as de Moivre's formula from it. Euler's formula also allows one to interpret the sine and cosine functions as mere variations of the exponential function:

<math>\cos x = {e^{ix} + e^{-ix} \over 2}</math>

<math>\sin x = {e^{ix} - e^{-ix} \over 2i}</math>

These formulas can even serve as the definition of the trigonometric functions for complex arguments <math>x</math>. You can derive the two equations above simply by adding Euler's formulas:

<math>e^{ix} = \cos x + i \sin x</math>

<math>e^{-ix} = \cos x - i \sin x</math>

and solving for either cosine or sine.

In differential equations, the function e^ix is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. Euler's identity is an easy consequence of Euler's formula.

In electrical engineering and other fields, signals that vary periodically over time are often described as a combination of sine and cosine functions (see Fourier analysis), and these are more conveniently expressed as the real part of exponential functions with imaginary exponents, using Euler's formula.

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Here is a derivation of Euler's formula using Taylor series expansions plus some basic facts about i:

The function e^x (assuming x is a real) can be written as:

<math> e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...

           = \sum_{n=0}^\infty \frac{x^n}{n!} </math>

and for complex x is defined by this series. Now if we throw i into the exponent, we get:

<math>

e^{ix} = \sum_{n=0}^\infty \frac{{(ix)}^n}{n!}

       = \sum_{n=0}^\infty \frac{i^n x^n}{n!}
       = \sum_{n=0}^\infty \frac{x^n}{n!} i^n

</math>

We can regroup its terms to produce this hairy-looking version:

<math>

e^{ix} = \sum_{n=0}^\infty \left(

           \frac{x^{4n}}  {(4n)!}   i^{\,4n}
         + \frac{x^{4n+1}}{(4n+1)!} i^{\,4n+1}
         + \frac{x^{4n+2}}{(4n+2)!} i^{\,4n+2}
         + \frac{x^{4n+3}}{(4n+3)!} i^{\,4n+3}
        \right)

</math>

To simplify this, we use the following basic facts about i:

<math>

  i^0 = 1, \qquad
  i^1 = i, \qquad
  i^2 = -1, \qquad
  i^3 = -i, \qquad
  i^4 = 1, \ldots

</math>

Or, to generalize this for all values of n:

<math>

  i^{\,4n} = 1, \qquad
  i^{\,4n+1} = i, \qquad
  i^{\,4n+2} = -1, \qquad
  i^{\,4n+3} = -i

</math>

So,

<math>

e^{ix} = \sum_{n=0}^\infty \left(

           \frac{x^{4n}}  {(4n)!}
         + \frac{x^{4n+1}}{(4n+1)!} i
         - \frac{x^{4n+2}}{(4n+2)!}
         - \frac{x^{4n+3}}{(4n+3)!} i
        \right)

</math>

or, rearranging terms and splitting the sum in two (allowed since the series is absolutely convergent):

<math>

e^{ix} = \sum_{n=0}^\infty \left(

           \frac{x^{4n}}  {(4n)!}
         - \frac{x^{4n+2}}{(4n+2)!}
        \right)
         +
         i\,\sum_{n=0}^\infty \left(
           \frac{x^{4n+1}}{(4n+1)!}
         - \frac{x^{4n+3}}{(4n+3)!}
        \right)

</math>

Taking it a step further, we use the following Taylor series expansions for cos(x) and sin(x):

<math> \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...

              = \sum_{n=0}^\infty \left(
                  \frac{x^{4n}}  {(4n)!}
                - \frac{x^{4n+2}}{(4n+2)!}
                \right)

</math>

<math> \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...

              = \sum_{n=0}^\infty \left(
                  \frac{x^{4n+1}}{(4n+1)!}
                - \frac{x^{4n+3}}{(4n+3)!}
                \right)

</math>

Which, when substituted into the previous formula for e^ix, gives:

<math>e^{ix} = \cos x + i\; \sin x</math>

as required.