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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: Graphs 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 $\frac{\pi}{4}$?

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  • $\begingroup$ Be aware though, that the picture you're seeing there, breaks down if you choose the scale of the y-axis high enough (and since we've got high exponents here, I mean really high, for your example e.g. around $10^7$ (so that the scale in vision goes form $-10^7 $ to $10^7$) $\endgroup$ – Sudix Jul 7 at 14:46
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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 $$ \lim_{k}F_{k} =\lim_{k}\exp\left(\frac{\ln(2\pi c^{2}k)}{2k}\right) =\exp\left(\lim_{k}\frac{\ln(2\pi c^{2}k)}{2k}\right) =\exp\left(\lim_{k}\left\{ \frac{\ln(2\pi c^{2})}{2k}+\frac{\ln k}{2k}\right\} \right)=1. $$ Note that the quantity $2/e$ is

  • independent of $c$ and
  • "relatively" close to $\pi/4\approx0.78540$.
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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.

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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 $$d_k \sim \frac{2 k+1}{e k}$$

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