The geometric interpretation of the common powerseries for \(cos\) and \( sin \)
$$
\begin{align}
sin \theta &= \sum^\infty_{n=0} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!} \\
cos \theta &= \sum^\infty_{n=0} (-1)^n \frac{\theta^{2n}}{(2n)!} \\
\end{align}
$$
These are the commonly taught power series for the
trigonometric functions \(sin \) and \(cos \).
While learning about these power series and the role they posess in mathematics (for example,
to prove Euler's identity: \( e^{i\pi} = -1)\)
I learned to appreciate these formulas.
One thing was very dissatisfying to me though. These formulas seemed to be quite arbitrary and no geometric interpretation was there
to be found. All this was made even worse, knowing how intuitive \( sin \) and \( cos \) are geometrically, while these series
just seemed like random numbers added and subtracted together without any coherence.
As a result I began to dig deeper and found this gem of geometry which aims to provide an interactive experience to
teach the geometric interpretation of the power series for \(sin \) and \(cos \).
Legend
light blue is the \( sin \theta\)
dark blue is the \( cos \theta\)
red lines are the lengths of the involutes which go vertically
(\( \forall 2k+1, \, k \in \mathbb{N}_0 \))
green lines are the lengths of the involutes which go horizontally
(\( \forall 2k, \, k \in \mathbb{N}_0 \))
When adding all the red lines together
(each going up is added, each going down is subtracted), we end up with a length equal to
\(sin \theta\).
When adding all the green lines together
(each going right is added, each going left is subtracted), we end up with a length equal to
\(cos \theta\)
These red and green lines can be represented
as a series \( I_k \) of lines that alternates between red and
green lines. We take \( | I_k | \) to mean the length of the \( k^{th}\) line.
In order to achieve our "left means subtract for green, right means ..." rules, we need to construct the following:
$$
\begin{align}
Red_k &= (-1)^k |I_{2k+1}| \\
Green_k &= (-1)^k |I_{2k}| \\
& k \in \mathbb{N}_0
\end{align}
\\
\Rightarrow \sum^\infty_{n=0} Red_k = sin \theta
\\
\Rightarrow \sum^\infty_{n=0} Green = cos \theta
$$
Now all we still need to prove is that \( |I_k| = \frac{\theta^k}{k!}\)
But with that assumption we already proved the power series of \(sin \) and \(cos \)
\( n = \)
Number of subdivisions of the angle \( \theta \). Determines how many triangles are used to approximate the circle.
\( \theta = \)
Angle of which to calculate \(cos \theta , \, sin \theta \)
\( k \)
Number of involutes to draw. Each involute will evolve
to the line fromed by the endpoints of the involute before it, as that line evolved to the line before it.
We know that exponentiating \( i = \sqrt{-1} \) with a number \( z \in \mathbb{Z} \), the following pattern will emerge:
$$
\begin{align}
\dots \quad
i^{-4} = 1 \quad
i^{-3} = i \quad
i^{-2} = -1 \quad
i^{-1} = -i \quad
i^0 = 1 \quad
i^1 = i \quad
i^2 = -1 \quad
i^3 = -i \quad
i^4 = 1 \quad
\dots
\end{align}
$$
This can be visualized in the following way
We also know that \(e\) is defined as \( \sum^\infty_{n=0}\frac{x^n}{n!}\) because:
$$
\begin{equation}
\frac{d}{dx} e^x = e^x \\
e^x \stackrel{!}{=} \sum^\infty_{n=0} \frac{x^n}{n!} = 1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\
\frac{d}{dx} \bigg( 1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \bigg) = 0+1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\
\frac{d}{dx} \sum^\infty_{n=0} \frac{x^n}{n!} = \sum^\infty_{n=0} \frac{x^n}{n!} \\
\end{equation}
$$
We can now also see why there is the connection between \(e\) and \(cos, \, sin\) called
Euler's identity: \( e^{i\theta} = cos \theta + i sin\theta\)
With each term, the exponent in the power series for \(e\) increases by \(1\). Therefor, if we plug in \(i\) as a basis, it will
be rotated by \(90°\) on every term. We also know, that each term in the sum has an absolute value (ie. "line length") of
\( \frac{\theta^n}{n!} \). This corresponds to the observations about \( |I_k| \) we made in the section
Introduction.
By leaving out every even term, we can therefor derive \(sin\theta \), or by leaving out the odd terms \(cos \theta \).
This however still leaves the question, why \( \frac{d}{dx} e^x = e^x \) based on the knowledge of
what we learned about e