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This article is about Euler's formula in complex analysis. For Euler's formula in graph theory see planar graph. See also topics named after Euler.

Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that shows a deep relationship between the trigonometric functions and the complex exponential function. (Euler's identity is a special case of the Euler formula.)

Euler's formula states that, for any real number x,

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

where

<math>e</math> is the base of the natural logarithm

<math>i</math> is the imaginary unit

<math>\sin</math> and <math>\cos</math> are trigonometric functions.

Contents 1 History 2 Applications in complex number theory 3 Relationship to trigonometry 4 Other applications 5 Proofs 5.1 Using Taylor series 5.2 Using calculus 5.3 Using ordinary differential equations 5.4 A heuristic argument 6 See also 7 References 8 External links

[edit] History

Euler's formula was proven for the first time by Roger Cotes in 1714 in the form <math>\log(\cos x+i\sin x)=ix</math><ref>John Stillwell (2002). Mathematics and Its History. Springer.</ref>. It was Euler who published the equation in its current form in 1748, basing his proof on the infinite series of both sides being equal. Neither of these men saw the geometrical interpretation of the formula: the view of complex numbers as points in the plane arose only some 50 years later (see Caspar Wessel).

[edit] Applications in complex number theory

This formula can be interpreted as saying that the function e^ix traces out the unit circle in the complex number plane as x ranges through the real numbers. Here, x 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 can be used to represent complex numbers in polar coordinates. Any complex number z=x+iy can be written as

<math> z = x + iy = |z| (\cos \phi + i\sin \phi ) = |z| e^{i \phi} \,</math>

where

<math> x = \mathrm{Re}\{z\} \,</math>

<math> y = \mathrm{Im}\{z\} \,</math>

<math> |z| </math> is the magnitude of z

and <math>\phi</math> is the argument of z— the angle between the x axis and the vector z measured counterclockwise and in radians — which is defined up to addition of 2π.

Now, taking this derived formula, we can use Euler's formula to define the logarithm of a complex number. To do this, we also use the facts that

<math>a = e^{ln (a)}\,</math>

and

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

both valid for any complex numbers a and b.

Therefore, one can write:

<math>

z=|z| e^{i \phi} = e^{\ln |z|} e^{i \phi} = e^{\ln |z| + i \phi}\, </math>

for any <math>z\ne 0</math>. Taking the logarithm of both sides shows that:

<math>\ln z= \ln |z| + i \phi.\,</math>

and in fact this can be used as the definition for the complex logarithm. The logarithm of a complex number is thus a multi-valued function, due to the fact that <math>\phi</math> is multi-valued.

Finally, the other exponential law

<math>(e^a)^k = e^{a k}, \,</math>

which can be seen to hold for all integers k, together with Euler's formula, implies several trigonometric identities as well as de Moivre's formula.

[edit] Relationship to trigonometry

Euler's formula provides a powerful connection between analysis and trigonometry, and provides an interpretation of the sine and cosine functions as weighted sums of the exponential function:

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

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

The two equations above can be derived by adding or subtracting 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.

These formulas can even serve as the definition of the trigonometric functions for complex arguments x. For example, letting x = iy, we have:

<math> \cos(iy) = {e^{-y} + e^{y} \over 2} = \cosh(y) </math>

<math> \sin(iy) = {e^{-y} - e^{y} \over 2i} = i \sinh(y). </math>

[edit] Other applications

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.

[edit] Proofs

[edit] Using Taylor series

Here is a proof of Euler's formula using Taylor series expansions as well as basic facts about the powers of i:

<math>\begin{align}

i^0 &{}= & 1 \\ i^1 &{}= & i \\ i^2 &{}= & -1 \\ i^3 &{}= & -i \\ i^4 &={} & 1 \\ i^5 &={} & i \\ & \vdots & \vdots \end{align}</math>

and so on. The functions e^x, cos(x) and sin(x) (assuming x is real) can be expressed using their Taylor expansions around 0:

<math> e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots </math>

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

</math>

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

</math>

and for complex z we define each of these function by the above series, replacing x with z. This is possible because the radius of convergence of each series is infinite. We then find that

<math>e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots</math>

<math>= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots</math>

<math>= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} - \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right) </math>

<math>= \cos (z) + i\sin (z) \,</math>

The rearrangement of terms is justified because each series is absolutely convergent. Taking z = x to be a real number gives the original identity as Euler discovered it.

[edit] Using calculus

Define the function <math>f \ </math> by

<math>f(x) = \frac{\cos x+i\sin x}{e^{ix}}. \ </math>

This is allowed since the equation

<math>e^{ix}\cdot e^{-ix}=e^0=1 \ </math>

implies that <math>e^{ix} \ </math> is never zero.

The derivative of <math>f \ </math> is, according to the quotient rule:

<math>f'(x)\,</math>	<math>= \displaystyle\frac{(-\sin x+i\cos x)\cdot e^{ix} - (\cos x+i\sin x)\cdot i\cdot e^{ix}}{(e^{ix})^2} \ </math>
	<math>= \displaystyle\frac{-\sin x\cdot e^{ix}-i^2\sin x\cdot e^{ix}}{(e^{ix})^2} \ </math>
	<math>= \displaystyle\frac{-\sin x-i^2\sin x}{e^{ix}} \ </math>
	<math>= \displaystyle\frac{-\sin x-(-1)\sin x}{e^{ix}} \ </math>
	<math>= \displaystyle\frac{-\sin x+\sin x}{e^{ix}} \ </math>
	<math>= 0 \ </math>

Therefore, <math>f \ </math> must be a constant function. Thus,

<math>f(x)=f(0)=\frac{\cos 0 + i \sin 0}{e^0}=1</math>

<math>\frac{\cos x + i \sin x}{e^{ix}}=1</math>

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

Q.E.D.

[edit] Using ordinary differential equations

Define the function g(x) by

<math>g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} .\ </math>

Considering that i is constant, the first and second derivatives of g(x) are

<math>g'(x) = i e^{ix} \ </math>

<math>g(x) = i^2 e^{ix} = -e^{ix} \ </math>

because i ² = -1 by definition. From this the following 2^nd order linear ordinary differential equation is constructed:

<math>g(x) = -g(x) \ </math>

or

<math>g(x) + g(x) = 0. \ </math>

Being a 2^nd order differential equation, there are two linearly independent solutions that satisfy it:

<math>g_1(x) = \cos(x) \ </math>

<math>g_2(x) = \sin(x). \ </math>

Both cos(x) and sin(x) are real functions in which the 2^nd derivative is identical to the negative of that function. Any linear combination of solutions to a homogeneous differential equation is also a solution. Then, in general, the solution to the differential equation is

<math>g(x)\,</math>	<math>= A g_1(x) + B g_2(x) \ </math>
	<math>= A \cos(x) + B \sin(x) \ </math>

for any constants A and B. But not all values of these two constants satisfy the known initial conditions for g(x):

<math>g(0) = e^{i0} = 1 \ </math>

<math>g'(0) = i e^{i0} = i \ </math>.

However these same initial conditions (applied to the general solution) are

<math>g(0) = A \cos(0) + B \sin(0) = A \ </math>

<math>g'(0) = -A \sin(0) + B \cos(0) = B \ </math>

resulting in

<math>g(0) = A = 1 \ </math>

<math>g'(0) = B = i \ </math>

and, finally,

<math>g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} = \cos(x) + i \sin(x). \ </math>

Q.E.D.

[edit] A heuristic argument

In doing trigonometry without complex numbers, one may prove the two identities

<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y) = c_1 c_2 - s_1 s_2,\,</math>

<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) = s_1 c_2 + c_1 s_2.\,</math>

Similarly in treating multiplication of complex numbers (with no involvement of trigonmetry), one may observe that the real and imaginary parts of the product of c₁ + is₁ and c₂ + is₂ are respectively

<math>c_1 c_2 - s_1 s_2,\,</math>

<math>s_1 c_2 + c_1 s_2. \,</math>

Thus one sees this same pattern arising in two disparate contexts:

• trigonometry without complex numbers, and
• complex numbers without trigonometry.

This coincidence can serve as a motivation for conjoining the two contexts and thereby discovering the trigonometric identity

<math>g(x+y) = g(x) g(y),\,</math>

where

<math>g(x) \ \stackrel{\mathrm{def}}{=}\ \cos(x) + i\sin(x).\,</math>

This identity for g(x) of a sum is simpler than the identities for sin and cos of a sum. Having proved this identity, recall which familiar sort of function satisfy this same functional equation

<math>f(x+y) = f(x)f(y).\,</math>

The answer is exponential functions. That suggests that g(x) may be an exponential function

<math>g(x) = b^x.\,</math>

Then the question is: what is the base b? The definition of g(x) and the local behavior of sin and cos near zero suggest that

<math>g(0+dx) = g(0) + i\,dx,</math>

(where dx is an infinitesimal increment of x). Thus the rate of change at 0 is i, so the base should be eⁱ. Thus we have

<math>g(x) = e^{ix}\,</math>

and g(x) is indeed an exponential function.

[edit] See also

• Leonhard Euler
• Euler's identity
• Complex number
• Exponential function
• Trigonometry

[edit] References

<references/>

• Feynman, Richard P., The Feynman Lectures on Physics, vol. I Addison-Wesley (1977), ISBN 0-201-02010-6, ISBN 0-201-02115-3 (set).

[edit] External links

• Euler and his beautiful and extraordinary formula by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.
• Euler's Formula - Puzzle: 55 pieces in a six star style of piece by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.
• Detailed Proof of Euler's Relation by Craig Lewis.
• Proof of Euler's Formula by Julius O. Smith III
• Euler's Formula and Fermat's Last Theorembg:Формула на Ойлер

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