Euler’s Identity

Behold the most beautiful formula in the whole of mathematics,

\[e^{i\pi}+1=0\]

Or, using Hama beads (made together with my my seven year old daughter):

Euler’s identity using Hama beadsEuler’s identity using Hama beads

This seemingly simple expression is named Euler’s identity (also Euler’s formula or Euler’s equation), after Leonard Euler, an important Swiss mathematician and physicist. The expression establishes an amazing relation between the five numbers \(0\), \(1\), \(e\), \(\pi\), and \(i\). \(0\) is the neutral element for addition; if you add \(0\) to something, nothing changes: \(x+0=x\). \(1\) is the neutral element for multiplication; if you multiply something by \(1\), nothing changes: \(1\,x=x\). \(e\) is the base number of the natural logarithm. It is also the number for which the derivative of the exponential function is equal to the function itself, so

\[\frac{d}{dx}e^x=e^x\]

is only true for \(e\). The value of \(e\) is about 2.718281828459045… \(\pi\) is of course the ratio of a circle’s circumference to its diameter, but it pops up everywhere else in science and engineering. The value of \(\pi\) is about 3.141592653589793… And, finally, \(i\) is the imaginary unit, equal to \(\sqrt{-1}\).

Note that, starting from the second decimal, \(e\) contains the rather peculiar sequence 1828 1828 45 90 45. This doesn’t really mean anything, as both \(e\) and \(\pi\) are expected to contain any sequence of numbers somewhere in their decimal expansion. Starting from the 762nd decimal, for example, \(\pi\) contains the sequence 999999. This doesn’t really mean anything either. Look up your phone number in \(\pi\) or \(e\) using the pi search engine!

Derivation

Euler’s identity is a special case (set \(x=\pi\)) of Euler’s formula, which is

\[e^{ix}=\cos x+i\sin x\]

This equation is just as stunning as Euler’s identity itself. The meaning of the right-hand side becomes clear if we combine trigonometry with complex numbers. Take the complex number \(c=a+bi\). If you set \(a\) to \(\cos{x}\) and \(b\) to \(\sin{x}\), for the same \(x\), you get points on a circle with radius one. But that is not the stunning part. The stunning part is how you get to the left-hand side. We have the following three Taylor series.

\[e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots\]

\[\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots\]

\[\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots\]

This looks like there must be some nifty trick lurking around somewhere, doesn’t it? Well, there is. Since the Taylor series is valid for all \(x\), it is also valid for \(ix\). Since \(i=\sqrt{-1}\), we have \(i^2=-1\), \(i^3=-i\), \(i^4=1\), \(i^5=i\), etc. So, following the expression for \(e^x\) above, the expression for \(e^{ix}\) is

\[e^{ix}=1+i\frac{x}{1!}-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\frac{x^6}{6!}-i\frac{x^7}{7!}+\frac{x^8}{8!}+i\frac{x^9}{9!}-\cdots\]

We can then rearrange the terms (but only because the series is absolutely convergent, otherwise this would change the result) as follows.

\[e^{ix}=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots\right)+i\left(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots\right)\]

The expressions in brackets are exactly the Taylor series of \(\cos{x}\) and \(\sin{x}\), which proves Euler’s formula.

Submitted by Tom Roelandts on 11 March 2012

Add new comment