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Topic:
Spiral of Theodorus  unwound with inverse of integers
Replies:
3
Last Post:
Sep 26, 2017 12:07 PM




Re: Spiral of Theodorus  unwound with inverse of integers
Posted:
Sep 26, 2017 12:59 AM


On 26/09/2017 04:50, joshipura@gmail.com wrote: > I am not a mathematician. > > Earlier in someone in this group pointed me to the spiral of Theodorus in response of some question. > > Today I thought of constructing it "the inverse way". That is, start with a right angle triangle of sides with lengths 1 and 1. On its hypotenuse of it (length sqrt(2)), make a right angle with a side with length 1/2. On the hypotenuse of it (length 3/2), make a right angle triangle with the side 1/3 and so on. > > Obviously, as added sides get smaller and smaller, growth of hypotenuse grows slower and slower. Over 35,000+ steps it seems to grow very slowly closer to 1.626316648. > > Questions are simple  if we let this spiral unwind indefinitely, will it converge to a final value? If so, which? What is the name of such a spiral? > > It must be a wellknown/obvious result for you, but I am no mathematician so I can't decide. > > Thanks in advance, > Bhushit >
Yes it will converge to sqrt( (Pi^2/6) + 1) = 1.626325326...
If you work out the SQUARES of lengths of successive hypotenuses, using Pythagorus's theorem, you will see they are:
1 + 1/1^2 = 1 + 1/1 1 + 1/1^2 + 1/2^2 = 1 + 1/1 + 1/4 1 + 1/1^2 + 1/2^2 + 1/3^2 = 1 + 1/1 + 1/4 + 1/9 ...etc.
Apart from the extra first term (1), this is a famous series:
1/1^2 + 1/2^2 + 1/3^2 ...+ 1/n^2 ...
(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), although this is not easy to prove (or it wouldn't have taken Euler to do it! :)).
The answer for your problem is based on this, but has to be adjusted, because you have an extra 1 as the first term in your series, and because the series is giving the SQUAREs of successive hypotenuses (hence the final squareroot operation in the formula I gave).
Regards, Mike.



