Euler’s formula is \(e^{ix} = \cos(x) + i\sin(x)\) where \(i\) is the imaginary unit representing “\(\sqrt{-1}\)”. Euler’s formula is what is used to produce cute and popular results like $$e^{i\pi} + 1 = 0$$ and $$i^{i} = e^{-\frac{\pi}{2}}$$
Many students, especially introductory engineering students, are taught Euler’s formula as something to know as fact since its derivation requires a little bit of Calculus that they will see later in their math study. However, by the time they take a Calculus course that teaches them about Taylor series, this little nugget slips through the cracks. It’s not a purposeful omission, but there is sometimes just a disconnect between what a math instructor will teach and what, say, an engineering instructor will think is being taught. The math instructor may assume that mathematics specific to an engineering student’s discipline will be covered in an engineering course while the engineering instructor may assume that mathematics, even though it may be related to engineering, would be taught in a mathematics course. Such miscommunication is inadvertent.
In any case, this post will show how to derive Euler’s formula through Taylor series (there are other ways, however). As such, some knowledge of Calculus is required. Additionally, I do not intend to write an entire chapter on this in this post, so some specific technicalities are not going to be proven. The interested reader can reference any reasonably good undergraduate Calculus text to fill in any gaps.
The Taylor series expansion of a function \(f\) about some point \(a\) for which \(f(a)\) is known is given as $$f(x) = \sum_{k=0}^{\infty}f^{(k)}(a)\frac{(x-a)^{k}}{k!}$$ where \(f^{k}(a)\) represents the \(k\)th derivative of \(f\) at \(a\) (so \(f^{(0)}(a)\) is zeroth derivative of \(f\) at \(a\) and the zeroth derivative is the original function). The Maclaurin series is just the Taylor series for when \(a = 0\). So, to generate the Maclaurin series for \(f(x) = e^{x}\) it is just a matter of taking derivatives of \(f\) and evaluating them at \(a = 0\). Since \(e^{x}\) is its own derivative and since \(f(0) = e^{0} = 1\), the Maclaurin series of \(e^{x}\) is $$e^{x} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$$
This is a nice series to know (not memorize by rote! but gain a learned familiarity with!) since it shows up in many places (e.g., exponential generating functions in Combinatorics) and has some nice properties (e.g., absolutely convergent which is pretty much a requirement to allow what happens in the next few paragraphs).
Consequently, we have $$e^{ix} = \sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!}$$
and recognizing the cyclic nature of powers of \(i\), we have
$$\sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!} = 1 + ix – \frac{x^{2}}{2!} – i\frac{x^{3}}{3!} + \frac{x^{4}}{4!} + i\frac{x^{5}}{5!} – \cdots$$
which after regrouping gives
$$\sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!} = (1 – \frac{x^{2}}{2!} + \frac{x^{4}}{4!} – \cdots) + i(x – \frac{x^{3}}{3!} + \frac{x^{5}}{5!} – \cdots)$$
So how in the world does this relate to Euler’s formula? Well, since Euler’s formula seems to involve sines, cosines, and imaginary numbers, we may as well see what can be done with those things.
Recall the first two derivatives of \(\sin(x)\). If \(g(x) = \sin(x)\) then \(g^{(1)}(x) = \cos(x)\) and \(g^{(2)}(x) = -\sin(x)\). As such, the Maclaurin series of \(\sin(x)\) contains only odd powers of \(x\) since \(\cos(0) = 1\) and \(\sin(0) = 0\). Therefore $$\sin(x) = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$$
and writing this out for visual purposes, we have
$$\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!} = x – \frac{x^{3}}{3!} + \frac{x^{5}}{5!} – \cdots$$
See something familiar? The imaginary portion of \(\sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!}\) is just the Maclaurin series for \(\sin(x)\)!
Similarly, the Maclaurin series for \(\cos(x)\) is
\[\begin{aligned}
\cos(x) & = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!} \\
& = 1 – \frac{x^{2}}{2!} + \frac{x^{4}}{4!} – \cdots
\end{aligned}\]
See something else familiar? The real portion of \(\sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!}\) is just the Maclaurin series of \(\cos(x)\)!
So, combining things together we have \(e^{ix} = \cos(x) + i\sin(x)\).
This is a fun, little exercise to work through if for no other reason than to see how cleanly things can fall into place.
Incidentally, if you are curious: since \(i = e^{i\frac{\pi}{2}}\) then \(i^{i} = e^{(i\frac{\pi}{2})^{i}} = e^{i^{2}\frac{\pi}{2}} = e^{-\frac{\pi}{2}}\). But ye be warned: there are boundary conditions. This can easily lead to mumbo jumbo like \(e^{-\frac{\pi}{2}} = e^{-2\pi – \frac{\pi}{2}}\) which is clearly false. It’s all about how we traverse the unit circle.