Euler’s formula,

\[e^{i \theta} = \cos \theta + i \sin \theta,\]

is both a beautiful and eminently useful result about the complex numbers. It leads directly to the more widely known Euler’s identity,

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

which shows a somewhat suprising connection between five of the most significant numbers in mathematics. There are many proofs of Euler’s formula, but as someone who has taught Calculus II for many years, the proof using Taylor series is rather close to my heart. In this post, we’ll explore this proof using `sympy`

, a Python library for symbolic mathematics, in order to avoid using calculus ourselves.

The basic idea of Taylor series is that many (though not all) functions on the real line may be represented by polynomials of infinite degree. The most well- known example of a Taylor series is the sum the geometric series,

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

for \(|x| < 1\).

We can illustrate this sum in `sympy`

as follows.

```
from sympy.interactive import printing
printing.init_printing()
import sympy as sym
```

```
x = sym.Symbol('x')
sym.series(1 / (1 - x))
```

\[1 + x + x^{2} + x^{3} + x^{4} + x^{5} + \mathcal{O}\left(x^{6}\right)\]

The term

`sym.Order(x**6)`

\[\mathcal{O}\left(x^{6}\right)\]

here indicates that all of the omitted terms have degree greater than or equal to six. Naturally, `sympy`

can only ever calculate finitely many terms of a functions Taylor series. (We’ll discuss the impact of this limitation on our “proof” later.) We may control the number of terms of the Taylor series calculated by `sympy`

by passing the optional argument `n`

to `sympy.series`

. For example, we may calculate the first ten terms of the geometric series as follows.

`sym.series(1 / (1 - x), n=10)`

\[1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} + x^{7} + x^{8} + x^{9} + \mathcal{O}\left(x^{10}\right)\]

We now turn our attentation to Euler’s formula by first defining the variable `theta`

.

```
theta = sym.Symbol('\\theta', real=True)
theta
```

\[\theta\]

We also define functions to calculate the Taylor series of \(\sin \theta\) and \(\cos \theta\) for any degree \(n\).

```
def sin_series(n):
return sym.series(sym.sin(theta), n=n)
def cos_series(n):
return sym.series(sym.cos(theta), n=n)
```

The first few terms for the series of \(\sin \theta\) are

```
n = 10
sin_series(n)
```

\[\theta - \frac{\theta^{3}}{6} + \frac{\theta^{5}}{120} - \frac{\theta^{7}}{5040} + \frac{\theta^{9}}{362880} + \mathcal{O}\left(\theta^{10}\right)\]

and the first few for \(\cos \theta\) are

`cos_series(n)`

\[1 - \frac{\theta^{2}}{2} + \frac{\theta^{4}}{24} - \frac{\theta^{6}}{720} + \frac{\theta^{8}}{40320} + \mathcal{O}\left(\theta^{10}\right)\]

Let’s compare these two series to that for \(e^{i \theta}\).

```
def exp_series(n):
return sym.series(sym.exp(sym.I * theta), n=n)
exp_series(n)
```

\[1 + i \theta - \frac{\theta^{2}}{2} - \frac{i \theta^{3}}{6} + \frac{\theta^{4}}{24} + \frac{i \theta^{5}}{120} - \frac{\theta^{6}}{720} - \frac{i \theta^{7}}{5040} + \frac{\theta^{8}}{40320} + \frac{i \theta^{9}}{362880} + \mathcal{O}\left(\theta^{10}\right)\]

The key idea behind this proof of Euler’s formula is that the terms of this sum containing \(i\) are identical (apart form the \(i\)) to the terms of the Taylor series for \(\sin \theta\). Similary, the terms not containing \(i\) are identical to the terms from the Taylor series of \(\cos \theta\).

We can see this more clearly by asking `sympy`

to collect the terms containing \(i\).

`sym.collect(exp_series(n), sym.I)`

\[1 + i \left(\theta - \frac{\theta^{3}}{6} + \frac{\theta^{5}}{120} - \frac{\theta^{7}}{5040} + \frac{\theta^{9}}{362880} + \mathcal{O}\left(\theta^{10}\right)\right) - \frac{\theta^{2}}{2} + \frac{\theta^{4}}{24} - \frac{\theta^{6}}{720} + \frac{\theta^{8}}{40320} + \mathcal{O}\left(\theta^{10}\right)\]

Odd placement of the leading one aside, we immediately recognize this expression as \(\sin \theta + i \cos \theta\).

In addition to this visual inspection, `sympy`

makes it fairly easy to verify that the Taylor series agree to a fairly large number of terms.

```
N = 100
sym.simplify(exp_series(N) - (cos_series(N) + sym.I * sin_series(N)))
```

\[\mathcal{O}\left(\theta^{100}\right)\]

So we see a bit more formally that Euler’s formula is quite likely to be true. The weasel work “likely” is necessary here because we really only checked that finitely many terms of the infinite Taylor series for \(e^{i \theta}\) and \(\cos \theta + i \sin \theta\) agree. Fortunately, armed with a bit of calculus knowledge, a pen, and some paper, we could verify by hand that all of the terms coincide, if we were so inclided.

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