I was experimenting with simple data points like squares, rectangles, and polygons to forecast my 0D and 1D persistent homology. I'm having trouble predicting persistent homology in the case of a circle. I can figure out 0D persistent homology for the case of the circle and the birth coordinate of H1, but I can't discover any logical explanation for H1's death coordinate. For example, in the picture below:
In the above example, radius of the circle is 1 and H1 is [0.15690251 1.78183484]. I am aware that the first coordinate reflects the time when the closed circle structure is first formed. I know the analytical formula when the birth of the structure will happen. This is given by $R=\frac{s}{2 sin(\pi/n)}$, where R is the radius of the circle, s is the side length of the regular polygon, n is the number of sides of the polygon (The circle inscribes any regular polygon). My goal now is to determine when this structure (circle) will expire, i.e., to develop an analytical formula to forecast the second coordinate of H1.
I have figured that it would be, of course, less than the diameter of the circle. But does there exist any exact analytical formula for this?
Yes, there exists a known exact analytical formula for this. As JHF says above, if you had the entire circle, then $H_1$ dies once the inscribed equilateral triangles appear in the Vietoris-Rips complex, which is at scale $\sqrt{3}$ with the Euclidean metric in the plane. Now, suppose we have instead $n$ evenly-spaced points around the circle. For convenience, label these points as $0,1,2,\ldots,n-1$ in circular order. If $n$ is a multiple of $3$, say $n=3k$, then $H_1$ dies again once such an equilateral triangle appears --- the side-length of such a triangle is still the distance from vertex $0$ to vertex $k$, which is still equal to $\sqrt{3}$. But if $n$ is not a multiple of $3$, say $n=3k+1$ or $n=3k+2$, then $H_1$ dies once the smallest-diameter triangle appears whose projection to the plane contains the origin. Such a triangle is no longer equilateral, but it is approximately equilateral. When $n=3k+1$, this triangle contains (for example) the vertices $0$, $k+1$, $2k+1$. When $n=3k+2$, this triangle contains (for example) the vertices $0$, $k+1$, $2k+2$. So when $n=3k+1$ or $n=3k+2$, the scale at which this triangle appears, and $H_1$ disappears, is equal to the Euclidean distance between vertex $0$ and vertex $k+1$. For any value of $n$, one can say that the scale at which $H_1$ disappears is equal to the Euclidean distance between vertex $0$ and vertex $\lceil n/3\rceil$. Using trigonometry, one can now solve for an analytical formula of this side-length, as a function of $n$ (and $\lceil n/3\rceil$).
For more on Vietoris--Rips complexes of the circle or of points on the circle, see the papers https://msp.org/pjm/2017/290-1/p01.xhtml and https://link.springer.com/article/10.1007/s00454-016-9803-5 and "Clique complexes of graph powers". Those papers don't use the Euclidean metric on the plane, but instead the path-length metric around the circle (sometimes of circumference 1). But you can use trigonometry to translate back and forth between scale parameters using one metric versus the other.