On 26/09/2017 12:11, firstname.lastname@example.org wrote: > >> Yes it will converge to sqrt( (Pi^2/6) + 1) = 1.626325326... > > Oh wow! We can draw a circle with that radius and entire spiral will fit in! Nifty!
Yes, I'd not thought of it like that, but you're right. Also, looking at the length of the spiral we get the series
1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ...
This is another well known series called the harmonic series, and can be shown fairly easily to diverge (i.e. sums become arbitrarily large as we go on).
So... this implies the spiral circles the origin infinitely often, but still fitting in the bounding circle.
I wonder if the spiral already has a name. (If not, we can call it the Joshi spiral! :))
> >> (i.e. the sum of inverses of square numbers) whose limit was first >> evaluated by the famous mathematician Euler in 1734. It's called the >> Basel problem, and you can search/read all about it in Wikipedia. The >> infinite series sums to (Pi^2/6), > > Thanks. Will look up. >