Consider the unit disk $\mathcal{R}=\{(x,y)|x^2+y^2=1\}$. Let's say we partition it into $p$ non-intersecting segments using $p-1$ vertical lines such that the area of each segment is $\dfrac{\pi}{p}$.

Specifically, we're partitioning the interval $[-1,1]$ using a sequence $x_{p,k}$ indexed by $k\in\{0,1,2,\ldots,p\}$ so that
$$-1=x_{p,0}

Question

Consider the unit disk $\mathcal{R}=\{(x,y)|x^2+y^2=1\}$. Let's say we partition it into $p$ non-intersecting segments using $p-1$ vertical lines such that the area of each segment is $\dfrac{\pi}{p}$.

Specifically, we're partitioning the interval $[-1,1]$ using a sequence $x_{p,k}$ indexed by $k\in\{0,1,2,\ldots,p\}$ so that
$$-1=x_{p,0}<\cdots<x_{p,k}<\cdots<x_{p,p}=1$$I'll post an example shortly for further clarification.

holsteremission · Answer

In the case when $p=3$, the partition is made up of the points
$$\begin{cases}x_{3,0}=-1\
x_{3,1}\approx -0.2649\
x_{3,2}\approx  0.2649\
x_{3,3}=1\end{cases}$$
When $p=4$, we have
$$\begin{cases}x_{4,0}=-1\
x_{4,1}\approx -0.404\
x_{4,2}=0\
x_{4,3}\approx  0.404\
x_{4,4}=1\end{cases}$$
And so on.

The task is to find out whether or not the sum $\displaystyle\sum_{n=1}^\infty x_{n,n-1}$ converges. If this diverges, what about $\displaystyle\sum_{n=1}^\infty (-1)^{n-1}x_{n,n-1}$? Bonus points for exact values if either case converges.

holsteremission · Answer

Traditional calc and numerical methods get me so far as determining each of the intermediate $x_{p,k}$ values for a given. We only care about the rightmost one that's not $1$, which is given by the solution to
$$\arccos x_{p,k-1}-x_{p,k-1}\sqrt{1-{x_{p,k-1}}^2}=\frac{\pi}{p}$$

holsteremission · Answer

*for a given $p$*

holsteremission · Answer

I'm pretty sure the non-alternating series diverges, since it seems $x_{n,n-1}\to1$, but haven't proved anything along these lines.

kainui · Answer

OK I think I finally understand what it is you're trying to do, now I'm actually gonna start thinking about how to start solving it; seems like what you're doing is reasonable. My knee jerk reaction is to like use a comparison test to sorta turn the square root into some other nice exponent or compare it to an integral which might be easier to evaluate. I really have no clue yet though what'll work but seems like a fun problem.

holsteremission · Answer

I'm glad it's understandable! I had a hell of a time formulating the problem with all these indices.

kainui · Answer

You could put an upper bound and lower bound on the circle by squeezing it between two trapezoids, it might somehow make it easier to prove that it converges/diverges that way too.

kainui · Answer

Err not areas, I guess you're doing like little arc lengths or something... Haha maybe I haven't quite understood it after all, it's somehow that second to last bit of all the circles with $\pi/p$ area slices #_# wow this is confusing haha

kainui · Answer

I see now, it's the lengths of the projection of all the first cuts on the x-axis added up.

holsteremission · Answer

Right, the summand refers to the $x$ coordinates of the dots in this quick sketch:
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