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

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Mike Terry

Posts: 763
Registered: 12/6/04
Re: Spiral of Theodorus - unwound with inverse of integers
Posted: Sep 26, 2017 12:59 AM
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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 well-known/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 square-root operation in the formula I gave).

Regards,
Mike.





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