Informal Derivation of Euler’s Identity

“Who would have thought that \pi which enters as the ratio of circumference to diameter, e, as the natural base for logarithms, i, as the fundamental imaginary unit and 0 and 1 (which we know all about from infancy) would all be tied together in any way, not to mention such a simple and compact way? I hope I never stumble into anything like this formula, for nothing I do after that in life would have any significance.” – Physicist Ramamurti Shankar 1

 

(1)   \begin{equation*} e^{i \pi} + 1 = 0 \end{equation*}

The Nobel Prize winning physicist Richard Feynman called equation (1), known as Euler’s identity, “one of the most remarkable, almost astounding, formulas in all of mathematics” 2.

Euler’s identity can be derived easily from Taylor series…

(2)   \begin{equation*} f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \end{equation*}

…as well as defining e^{ax} as

(3)   \begin{equation*} {e^{ax}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {ax} \right)}^n}}}{{n!}}} \end{equation*}

In equation (2), f^{(n)} represents the n-th derivative of the function f, with the zeroth derivative of f being equal to f.

The Taylor series of \cos{x} can be easily derived from equation (2).

    \[{f^0}\left( x \right) = f\left( x \right) = \cos x\]

    \[{f^0}\left( 0 \right) = f\left( x \right) = \cos 0 = 1\]

    \[{f^1}\left( x \right) = \frac{{df\left( x \right)}}{{dx}} = \frac{d}{{dx}}\cos x = - \sin x\]

    \[{f^1}\left( 0 \right) = - \sin 0 = 0\]

    \[{f^2}\left( x \right) = \frac{{{d^2}f\left( x \right)}}{{dx}} = - \frac{d}{{dx}}\sin x = - \cos x\]

    \[{f^2}\left( 0 \right) = - \cos 0 = - 1\]

    \[{f^3}\left( x \right) = - \frac{d}{{dx}}\cos x = \sin x\]

    \[{f^3}\left( 0 \right) = \sin 0 = 0\]

    \[{f^4}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^4}\left( 0 \right) = \cos 0 = 1\]

From the equations above, as well as equation (2), it follows that:

(4)   \begin{equation*} \cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - ...\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{x^{2n}}} \end{equation*}

The Taylor series of \sin{x} can also be derived easily from equation (2).

    \[{f^0}(x) = f\left( x \right) = \sin x\]

    \[{f^0}\left( 0 \right) = \sin 0 = 0\]

    \[{f^1}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^1}\left( 0 \right) = \cos 0 = 1\]

    \[{f^2}\left( x \right) = \frac{d}{{dx}}\cos x = - \sin x\]

    \[{f^2}\left( 0 \right) = - \sin 0 = 0\]

    \[{f^3}\left( x \right) = - \frac{d}{{dx}}\sin x = - \cos x\]

    \[f^3\left( 0 \right) = - \cos 0 = - 1\]

    \[{f^4}\left( x \right) = - \frac{d}{{dx}}\cos x = \sin x\]

    \[{f^5}\left( 0 \right) = \sin 0 = 0\]

    \[{f^5}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^5}\left( 0 \right) = \cos 0 = 1\]

From the equations above and equation (2), it follows that:

(5)   \begin{equation*} \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \cdots = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}{x^{2n + 1}}} \end{equation*}

From equations (3), (4), and (5), if \theta is a real number:

    \[{e^{i\theta }} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {i\theta } \right)}^n}}}{{n!}}} = 1 + i\theta + \underbrace {{i^2}}_{ - 1}\frac{{{\theta ^2}}}{{2!}} + \underbrace {{i^3}}_{ - i}\frac{{{\theta ^3}}}{{3!}} + \underbrace {{i^4}}_1\frac{{{\theta ^4}}}{{4!}} + \underbrace {{i^5}}_i\frac{{{\theta ^5}}}{!} - ...\]

    \[{e^{i\theta }} = 1 + i\theta - \frac{{{\theta ^2}}}{{2!}} - \frac{{i{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + \frac{{i{\theta ^5}}}{{5!}}\]

(6)   \begin{equation*} {e^{i\theta }} = \underbrace {\left( {1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - ...} \right)}_{\cos \theta } + i\underbrace {\left( {\theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}} - ...} \right)}_{\sin \theta } = \cos \theta + i\sin \theta \end{equation*}

Finally, let \theta = \pi. Equation (6) becomes:

    \[{e^{i\pi }} = \cos \pi + i\sin \pi = - 1\]

And:

    \[{e^{i\pi }} + 1 = 0\]

Notes:

  1. “Basic Training in Mathematics: a Fitness Program for Science Students.” Basic Training in Mathematics: a Fitness Program for Science Students, by Ramamurti Shankar, Springer, 2006, p. 95.
  2. Feynman, Richard. “Algebra.” The Feynman Lectures on Physics, www.feynmanlectures.caltech.edu/I_22.html.