"Distance" between summands in Taylor Series of Cosine The Taylor series of cosine is
$$\cos(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ ...$$
If we now plot the summands (ignoring the sign and the first summand for simplicity), one gets the following plot using GeoGebra:

In this picture it seems as if the distance between the graphs is the same for the different summands. Furthermore, the distance seems to be something around $\frac{\pi}{4}$.
Is this a fact? And if so, why?
I'd guess that it has something to do with the fact that we can write Pi using the Leibniz-series:
$$\frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\ ...$$
However I can't quite see the connection between those two series.
It seems that this has very little to do with the cosine, but rather with monomials themselves - but why do even monomials divided by the factorial of their exponent have a distance in $x$-direction of $\frac{\pi}{4}$?
 A: I don't know what you mean by distance between graphs, because they pairwise intersect and therefore have $0$ distance, but I will assume you mean the distance between the $y=1$ intercepts.
We have $x^{2n}/(2n)!=1$ exactly if $x=(2n)!^{1/(2n)}$, so let us study the sequence $n!^{1/n}$, which by your observation should grow about $\pi/8$ when $n$ increases by one. You can easily check that the growth is not exactly constant, but we can prove that $n!^{1/n}\sim n/e$, so $\lim_{n\to\infty}n!^{1/n}/n=1/e$. Furthermore, we can prove that $\lim_{n\to\infty}(n+1)!^{1/(n+1)}-n!^{1/n}=1/e$, which does not in fact follow from the previous limit. Now since $\pi/8\approx0.3927$ and $1/e\approx0.3679$ I guess you can say you were kinda close for someone eyeballing it.
A: The $n$-th curve crosses through the line $y=c$ at $x$ coordinate
$$
\frac{x^{2n}}{\left(2n\right)!}=c
$$
or, equivalently, $x=(k!c)^{1/k}$ where $k=2n$.
Therefore, the distance between two adjacent curves along this line is
$$
\left(\left(k+2\right)!c\right)^{1/(k+2)}-\left(k!c\right)^{1/k}
$$
which converges to $2/e\approx0.73576$ as $k \rightarrow \infty$.
To see this, let $F_{k}\equiv(\sqrt{2\pi c^{2}k})^{1/k}$ and use Stirling's approximation to get the asymptotic
$$
\frac{k+2}{e}F_{k+2}-\frac{k}{e}F_{k}
$$
and apply the identity $\log(ab)=\log a+\log b$ and L'Hopital's rule to get
$$
\begin{multline}
\lim_{k}F_{k}
=\lim_{k}\exp\left(\frac{\ln(2\pi c^{2}k)}{2k}\right)
=\exp\left(\lim_{k}\frac{\ln(2\pi c^{2}k)}{2k}\right)\\
=\exp\left(\lim_{k}\left\{ \frac{\ln(2\pi c^{2})}{2k}+\frac{\ln k}{2k}\right\} \right)
=1.
\end{multline}
$$
Note that the quantity $2/e$ is


*

*independent of $c$ and

*"relatively" close to $\pi/4\approx0.78540$.

A: Starting from @parsiad's answer and using 
$$d_k=\left(\left(k+2\right)!\,c\right)^{\frac 1{k+2}}-\left(k!\,c\right)^{\frac1k}$$ using one extra term of Stirling approximation we should end with
$$d_k \sim \frac{2 k+1}{e k}$$
