# “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 $$\frac{\pi}{4}$$?

• 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 $10^7$) – Sudix Jul 7 at 14:46

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 $$\pi/4\approx0.78540$$.
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.
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 $$d_k \sim \frac{2 k+1}{e k}$$