The Number “e”: Taylor Series and Euler’s Formula

Let us play for a moment with Euler’s famous formula:

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

R.P. Feynman, in his Lectures, referred to it as “the most remarkable formula in mathematics.” Many important results follow from it. For $x=\pi$, it yields the famous identity $e^{i\pi}+1=0$, linking five fundamental constants of mathematics.

This post is an adapted chapter from my book: M. Szewczyk, Analytical Methods in the Calculation of Switching Processes in Power Systems, OWPW 2024. At a deeper source level, it is also based on the beautiful textbook: G. Fichtenholz, Differential and Integral Calculus, Vol. II, PWN, Warsaw 1965. Original title: Г. М. Фихтенгольц, Course of Differential and Integral Calculus, Moscow–Leningrad 1958.

The Taylor Series

Euler’s formula $e^{ix}=\cos x + i\sin x$ can be obtained by examining the Taylor expansions of the functions $e^x$, $\cos x$, and $\sin x$ about $x_0=0$:

\[f(x)=\sum_{n=0}^{\infty}{f^{(n)}(0)\over n!}x^n\:\:.\]

For elementary functions, successive derivatives $f^{(n)}(0)$ in Taylor series are computed according to the familiar formulas:

\[\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x,\:\:\: \frac{\mathrm{d}}{\mathrm{d}x}\cos{x} =-\sin{x},\:\:\: \frac{\mathrm{d}}{\mathrm{d}x}\sin{x} = \cos{x}.\]

Since the function $e^x$ reproduces itself under differentiation, $(e^x)’ = e^x$, its Taylor expansion follows immediately:

\[e^x=\sum_{n=0}^{\infty}{x^n\over n!}.\]

In the case of $\cos x$, only the even terms are nonzero, while for $\sin x$, only the odd terms remain. This happens because the remaining derivatives are proportional to $\sin x$, which vanishes at $x=0$. Moreover, the signs alternate, starting with a positive term.

Thus, for $\cos x$:

\[\cos x = {x^0\over0!}+0-{x^2\over2!}+0+{x^4\over4!}+0-{x^6\over6!}+0+{x^8\over8!}+0-\ldots \approx\] \[\approx 1-{x^2\over2!}+{x^4\over4!}-{x^6\over6!}+{x^8\over8!}-\ldots +{(-1)^Nx^{2N}\over (2N)!} =\sum_{n=0}^{N}{(-1)^nx^{2n}\over (2n)!}\]

and similarly for $\sin x$:

\[\sin x ={x^1\over1!}+0-{x^3\over3!}+0+{x^5\over5!}+0-{x^7\over7!}+0+{x^9\over9!}+0-\ldots\approx\] \[\approx x-{x^3\over3!}+{x^5\over5!}-{x^7\over7!}+{x^9\over9!}-\ldots +{(-1)^Nx^{2N+1}\over (2N+1)!} =\sum_{n=0}^{N}{(-1)^nx^{2n+1}\over (2n+1)!}.\]

Therefore, for all three functions, $e^x$, $\cos x$ i $\sin x$, we have:

\[e^x=\sum_{n=0}^{\infty}{x^n\over n!}\:\:, \:\:\: \cos x=\sum_{m=0}^{\infty}{(-1)^nx^{2n}\over (2n)!}\:\:, \:\:\: \sin x=\sum_{n=0}^{\infty}{(-1)^nx^{2n+1}\over (2n+1)!}\:\:.\]

Rearranging the Series

The next step is to appropriately rearrange and group the terms in the expansion

\[e^x=\sum_{n=0}^{\infty}{x^n\over n!},\]

so that within the series for $e^x$ we can recognize the series for $\cos x$ and $\sin x$.

There are interesting theorems concerning permutations (rearrangements) of series terms which justify that such regrouping is legitimate in this case.

Euler’s Formula

Since the Taylor series may be rearranged, we now substitute $ix$ for $x$, where $i^2=-1$, and obtain:

\[e^{ix}=\sum\limits_{n=0}^\infty{(ix)^n\over n!} =\sum\limits_{n=0}^\infty{\left[ {(ix)^{2n}\over (2n)!} + {(ix)^{2n+1}\over (2n+1)!} \right]} = \sum\limits_{n=0}^\infty{ {(-1)^n\over (2n)!}x^{2n}} + i\sum\limits_{n=0}^\infty{ {(-1)^n\over (2n+1)!}x^{2n+1}},\]

where we use the identities: $(ix)^n=i^nx^n$,
$i^{2n}=(i^2)^n = (-1)^n$,
$i^{2n+1}=(-1)^ni$.

Recognizing in these expressions the Taylor series of $\cos x$ and $\sin x$, we obtain the formula known as Euler’s formula:

\[T_{e^{ix}}=T_{\cos x} + i~T_{\sin x}\:\:\Rightarrow\:\: \boxed{\:e^{ix}=\cos x + i\sin x\:}\:.\]