## Monday, January 24, 2011

### "Our Jewel" - Euler's Formula and Euler's Identity

Richard Feynman called Euler's formula "our jewel"[2] and "one of the most remarkable, almost astounding, formulas in all of mathematics."[3]
 One equation to bring them all and in the darkness bind them. In the land of Euler (that would be Switzerland, Biatch!), where the Shadows (of the Alps) lie.
We will be discussing two equations. The ring above is Euler's Identity, which we will discuss second. Feynman's quote refers to Euler's formula, which we will discuss first. From Wikipedia:

Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the deep relationship between the trigonometric functions and the complex exponential function. Euler's formula states that, for any real number x,
$e^{ix} = \cos x + i\sin x \$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions cosine and sine respectively, with the argument x given in radians. This complex exponential function is sometimes called cis(x). The formula is still valid if x is a complex number, and so some authors refer to the more general complex version as Euler's formula.[1]

## History

It was Bernoulli [1702] who noted that
$\frac{1}{1+x^2}=\frac{1}{2} \left(\frac{1}{1-ix}+\frac{1}{1+ix} \right) \ .$
And since
$\int \frac{dx}{1+ax}=\frac{1}{a}\ln(1+ax)+C \ ,$
the above equation tells us something about complex logarithms. Bernoulli, however, did not evaluate the integral. His correspondence with Euler (who also knew the above equation) shows that he didn't fully understand logarithms. Euler also suggested that the complex logarithms can have infinitely many values.

Meanwhile, Roger Cotes, in 1714, discovered
$\ln(\cos x + i\sin x)=ix \$
(where "ln" means natural logarithm, i.e. log with base e).[4] We now know that the above equation is only true modulo integer multiples of i, but Cotes missed the fact that a complex logarithm can have infinitely many values which owes to the periodicity of the trigonometric functions.

It was Euler (presumably around 1740) who turned his attention to the exponential function instead of logarithms, and obtained the correct formula now coined after his name. It was published in 1748, and his proof was based 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 complex plane arose only some 50 years later (see Caspar Wessel).

## Applications in complex number theory

This formula can be interpreted as saying that the function eix 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 original proof is based on the Taylor series expansions of the exponential function ez (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 z.

A point in the complex plane can be represented by a complex number written in cartesian coordinates. Euler's formula provides a means of conversion between cartesian coordinates and polar coordinates. The polar form reduces the number of terms from two to one, which simplifies the mathematics when used in multiplication or powers of complex numbers. Any complex number z = x + iy can be written as
$z = x + iy = |z| (\cos \phi + i\sin \phi ) = r e^{i \phi} \$
$\bar{z} = x - iy = |z| (\cos \phi - i\sin \phi ) = r e^{-i \phi} \$
where
$x = \mathrm{Re}\{z\} \,$ the real part
$y = \mathrm{Im}\{z\} \,$ the imaginary part
$r = |z| = \sqrt{x^2+y^2}$ the magnitude of z
$\phi = \arg z = \,$ atan2(y, x) .
$\phi \,$ is the argument of z—i.e., the angle between the x axis and the vector z measured counterclockwise and in radians—which is defined up to addition of 2π. Many texts write tan-1(y/x) instead of atan2(y,x) but this needs adjustment when x ≤ 0.

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 definition of the logarithm (as the inverse operator of exponentiation) that
$a = e^{\ln (a)} \$
and that
$e^a e^b = e^{a + b} \$
both valid for any complex numbers a and b.

Therefore, one can write:
$z = |z| e^{i \phi} = e^{\ln |z|} e^{i \phi} = e^{\ln |z| + i \phi} \$
for any z ≠ 0. Taking the logarithm of both sides shows that:
$\ln z= \ln |z| + i \phi \ .$
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, because φ is multi-valued.

Finally, the other exponential law
$(e^a)^k = e^{a k} \ ,$
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.

## 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:
$\cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2}$
$\sin x = \mathrm{Im}\{e^{ix}\} ={e^{ix} - e^{-ix} \over 2i} \ .$
The two equations above can be derived by adding or subtracting Euler's formulas:
$e^{ix} = \cos x + i \sin x \;$
$e^{-ix} = \cos(- x) + i \sin(- x) = \cos x - i \sin x \;$
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:
$\cos(iy) = {e^{-y} + e^{y} \over 2} = \cosh(y)$
$\sin(iy) = {e^{-y} - e^{y} \over 2i} = -{1 \over i} {e^{y} - e^{-y} \over 2} = i\sinh(y) \ .$
Complex exponentials can simplify trigonometry, because they are easier to manipulate than their sinusoidal components. One technique is simply to convert sinusoids into equivalent expressions in terms of exponentials.

After the manipulations, the simplified result is still real-valued. For example:
\begin{align} \cos x\cdot \cos y & = \frac{(e^{ix}+e^{-ix})}{2} \cdot \frac{(e^{iy}+e^{-iy})}{2} \\ & = \frac{1}{2}\cdot \frac{e^{i(x+y)}+e^{i(x-y)}+e^{i(-x+y)}+e^{i(-x-y)}}{2} \\ & = \frac{1}{2} \left[ \underbrace{ \frac{e^{i(x+y)} + e^{-i(x+y)}}{2} }_{\cos(x+y)} + \underbrace{ \frac{e^{i(x-y)} + e^{-i(x-y)}}{2} }_{\cos(x-y)} \right] \ . \end{align}
Another technique is to represent the sinusoids in terms of the real part of a more complex expression, and perform the manipulations on the complex expression. For example:
\begin{align} \cos(nx) & = \mathrm{Re} \{\ e^{inx}\ \} = \mathrm{Re} \{\ e^{i(n-1)x}\cdot e^{ix}\ \} \\ & = \mathrm{Re} \{\ e^{i(n-1)x}\cdot (e^{ix} + e^{-ix} - e^{-ix})\ \} \\ & = \mathrm{Re} \{\ e^{i(n-1)x}\cdot \underbrace{(e^{ix} + e^{-ix})}_{2\cos(x)} - e^{i(n-2)x}\ \} \\ & = \cos[(n-1)x]\cdot 2 \cos(x) - \cos[(n-2)x] \ . \end{align}
This formula is used for recursive generation of cos(nx) for integer values of n and arbitrary x (in radians).

## Other applications

In differential equations, the function eix is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. The reason for this is that the complex exponential is the eigenfunction of differentiation. 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. Also, phasor analysis of circuits can include Euler's formula to represent the impedance of a capacitor or an inductor.

## Definitions of complex exponentiation

The exponential function ex for real values of x may be defined in a few different equivalent ways (see Characterizations of the exponential function). Several of these methods may be directly extended to give definitions of ez for complex values of z simply by substituting z in place of x and using the complex algebraic operations. In particular we may use either of the two following definitions which are equivalent. From a more advanced perspective, each of these definitions may be interpreted as giving the unique analytic continuation of ex to the complex plane.

### Power series definition

For complex z
$e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{z^n}{n!} ~.$
Using the ratio test it is possible to show that this power series has an infinite radius of convergence, and so defines ez for all complex z.

### Limit definition

For complex z
$e^z = \lim_{n \rightarrow \infty} \left(1+\frac{z}{n}\right)^n ~.$

## Proofs

Various proofs of the formula are possible.

### Using power series

Here is a proof of Euler's formula using power series expansions as well as basic facts about the powers of i:
\begin{align} i^0 &{}= 1, \quad & i^1 &{}= i, \quad & i^2 &{}= -1, \quad & i^3 &{}= -i, \\ i^4 &={} 1, \quad & i^5 &={} i, \quad & i^6 &{}= -1, \quad & i^7 &{}= -i, \end{align}
and so on. Using now the power series definition from above we see that for real values of x
\begin{align} e^{ix} &{}= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots \\[8pt] &{}= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\[8pt] &{}= \cos x + i\sin x \ . \end{align}
In the last step we have simply recognized the Taylor series for sin(x) and cos(x). The rearrangement of terms is justified because each series is absolutely convergent.

### Using calculus

Treating i as a constant, albeit an imaginary constant, note that
$\frac{d}{dx} e^{ix} = i e^{ix} \ .$
Then define the function
$f(x) = (\cos x - i \sin x) \cdot e^{ix} \ .$
Because the product rule holds for complex valued functions of a real variable for the same reason as in the real case, the derivative of ƒ(x) according to the product rule is:
\begin{align} \frac{d}{dx}f(x) &= (\cos x - i\sin x)\cdot\frac{d}{dx}e^{ix} + \frac{d}{dx}(\cos x - i\sin x)\cdot e^{ix} \\ &= (\cos x - i\sin x)(i e^{ix}) + (-\sin x - i\cos x)\cdot e^{ix} \\ &= (i\cos x + \sin x - \sin x - i\cos x)\cdot e^{ix} \\ &= 0 \ . \end{align}
Therefore, ƒ(x) must be a constant function in x. Because ƒ(0) = 1 by inspection, ƒ(x) = 1, giving
$1 = (\cos x - i \sin x) \cdot e^{ix} \ .$
Multiplying both sides by cos x + i sin x, we obtain
\begin{align} \cos x + i \sin x &= (\cos x + i \sin x)(\cos x - i \sin x) \cdot e^{ix} \\ &= (\cos^2 x -(i \sin x)^2) \cdot e^{ix} = (\cos^2 x + \sin^2 x) \cdot e^{ix} = e^{ix} \ . \end{align}

### Using differential equations

Here is another proof that follows from the differential identity above. Define a new function ƒ(x) of the real variable x as
$f(x) = \cos x + i \sin x \ .$
Then we may check that
\begin{align} \frac{d}{dx}f(x) &= -\sin x + i \cos x \\ &= i f(x) \ . \end{align}
Thus ƒ(x) and eix satisfy the same first-order ordinary differential equation (here the complex values are considered as points in the plane ℝ2). Note also that both functions are equal to 1 at x = 0, then by the uniqueness of solutions to ordinary differential equations they must be equal everywhere (see Picard–Lindelöf theorem and note the comments concerning global uniqueness in the proof section there).

## References

1. ^ Moskowitz, Martin A. (2002). A Course in Complex Analysis in One Variable. World Scientific Publishing Co.. pp. 7. ISBN 981-02-4780-X.
2. ^ Feynman, Richard P. (1977). The Feynman Lectures on Physics, vol. I. Addison-Wesley. pp. 22–10. ISBN 0-201-02010-6.
3. ^ Feynman, Richard P. (1977). The Feynman Lectures on Physics, vol. I. Addison-Wesley. pp. 22–1. ISBN 0-201-02010-6.
4. ^ John Stillwell (2002). Mathematics and Its History. Springer.

Steven Colyer said...

Some of my regular readers may be offended I make fun of Switzerland in my comments under the picture of the Titanium Euler's Identity ring caption above.

Let me assure them that first, I consider it to be gentle fun. I certainly feel no malice to the lovely Swiss.

Living in America where I live, I have met many people from the Germanic counties of central Europe, meaning Germany, Austria, and Switzerland. Each and every one of them to the last man and women are simply outstanding people, and in my humble opinion add greatly to the ever-changing melting pot that is the horribly-named "United States of America."

Welcome aboard "shotzes" (whatever that means ... we speak but one language here, heh), glad to have you here.

Having said that ....

I was very distressed to read a few years that Switzerland wasn't as "neutral" as they pride themselves as being during WWII, when it was proven they held various pieces of pricey Jewish artwork confiscated by The Third Reich. That in turn makes me wonder how "neutral" these Swiss Banks are today with their "numbered" accounts.

Julian Assange says he will blow the cover off many when he (eventually) releases his hacked "bank records" of the wealthy. I won't be surprised whatever it is, and I hope that Switzerland weathers the storm, whatever it is. But I must say ...

Leonhard Euler! My GOD, what a GREAT Swiss! What a great human, and Switzerland owns him, so thanks for that.

Also, Switzerland is the most beautiful country in Europe, based on the account of one person I know who has visited all of them.

Aoife said...

Would it be inappropriate for one to want that ring as a wedding ring?! Can you actually buy them? I'd love to get one. I got euler's formula henna tattooed onto my wrist over the summer. If I ever did get a tattoo (which will never happen) it would definitely be this equation though.
It really is just perfection.

Steven Colyer said...

If it's not inappropriate to you AND Dan then it's not inappropriate at all. :-)

Yes, give me a bit, I'll find where to buy it, I wish a few myself.

Steven Colyer said...

Well there's this, but it's not the same, bear with me ....

Steven Colyer said...

Sheesh, \$215 ?? Um, OK.

Chris & Sandy Boothe, Exotica Jewelry said...

You can buy that ring here: http://www.titaniumringsforever.com/blog/2011/titanium-rings-formulas-equations

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