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Thread: Circles enclosed in regular polygons

  1. #1 Circles enclosed in regular polygons 
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    I found this is a puzzle book once and it intrigued me!

    It starts off with a circle of radius 1m. That circle is enclosed with a regular triangle. This triangle is enclosed in another circle which is enclosed by a square... then a circle and then a pentagon, if you get the picture i.e. a circles are increasingly enclosed with regular polygons which increase their number of sides by 1.

    The question is... as the number of sides of the regular polygons approach infinity, will the radius of the outer circle approach a definite limit or go on for ever.

    I am unable to work out this sum on paper. I figured that the radius of each circle is equal to the radius of the previous circle divided by the sin of 1/2 the angle of the regular polygon.

    As the angle of a regular polygon with sides n, is ((n-2)*180)/n)

    Now let t = ((n-2)*180)/n)

    Now using simple trigonometry, we see that Sin (t/2) = R(x)/R(x+1)

    => R(n+1) = R(n) / sin[t(1)/2]

    => R(n+2) = R(n+1) / sin[t(2)/2]

    => R(n+2) = [R(n) / sin[t(1)/2)] / sin[t(2)/2]

    => R(n+2) =R(n) / sin[t(1)/2)*sin(t(2)/2)]

    I follows that as n approaches infinity the radius of the outermost circle will be the infinite product:-

    R(inf) = R(1) / sin[t(1)/2)*sin(t(2)/2)*sin(t3)/2*............]

    I hope this is clear, if you draw it out it will help...

    Now my question is this:- Is there an explicit way to find this limit by hand, I did it by computer and it seems to approach 8.657m after the 1000 sided polygon is added, but it seems to increase by ever smaller amounts.

    My second question is:- Have I made any mistakes with my reasoning, I have lost the puzzle book I read it in but I remember reading the answer which was different from my result of about 8.657m.


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  3. #2  
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    I realise that I didn't explain it properly in the last post, I have put the problem up on a web site check it out here....

    http://www.geocities.com/mobiusmaths/infpoly.htm


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  4. #3  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    I dont see how you can do this without an expression for t(n). Anyway I have come up with a different expression without t(n).
    R(n+1)=R(n)/cos(PI/2n)
    This is probably from using the other angle of the same triangle which you used. Any way this might be easier to work with. This leads to a similar infinite product.
    R(inf) = R(1)/[cos(PI/6)cos(PI/8)cos(PI/10)cos(PI/12)...]

    But I have not made any progress with this yet.

    You cannot just plug in a high n, you need to demonstrate convergence. The start of the series is
    1.1547005383792515
    1.24983885641513
    1.3141583439056648
    1.3605168307326971
    1.3955050560486582
    1.4228446163822157
    1.444794288052866
    1.462803830591233
    1.4778461737619097
    1.4905984501388307
    1.5015464140915882
    1.5110475640154442
    So if it is up to 8.657 at n=1000 as you say, I would say that prospect of convergence is not very promising.
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

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  5. #4  
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    Interesting....

    How did you come up with this different function for the radius of the infinite sided polygon.

    Also my series gives the answers:-
    r1=1
    r2=2
    2.828427125
    3.496128196
    4.03698111
    4.480710992
    4.849886629
    5.16114155
    5.426745374
    5.655846914
    5.85536359
    6.030601922
    6.185690087
    6.323882083
    6.447774257
    6.559461555
    6.660651823
    6.752750539
    6.836924424
    6.914149764
    6.985249532
    7.050922202 which is r22.

    this series does indeed seem to converge.

    I posted this up on a different forum if you want to check out what came from it. (I was getting no replies from this forum, thank you for yours :-D

    http://www.scienceforums.net/forums/...ad.php?t=14440
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  6. #5  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    I found an error in my derivation. The correct recursion is
    R(n+1)=R(n)/cos(PI/n)
    no 2 in the denominator, which yields the same starting sequence that you have
    1.9999999999999996
    2.8284271247461894
    3.4961281955905674
    4.036981109691309
    4.480710991812583
    4.849886629302337
    5.161141549931671
    5.426745373699692
    5.6558469138722804
    5.85536358997785
    6.0306019216478965
    6.185690086718116
    This certainly is in a lot better agreement with a convergence 8.657 at n=1000 but ....
    Convergence cannot be established just by looking at the sequence but convergence would be much more believable if you have a couple of significant figures match between n=100 and n=1000 otherwise you need to see how much change you get in your answer if you go to n=10000. If you dont have two significant figures match between n=1000 and n=10000, I would begin to seriously doubt that you have a convertent sequence. Of course three or four significant figures matching for a 10 times difference in n would be a lot more reassuring.
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

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  7. #6  
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    My series does indeed converge....

    The general term in the series is

    1/{sin[(n-2)*180/2n]}

    Multiply out:-

    1/{sin[(180n - 360)/2n]}

    Divide by n top and bottom (of the sin term that is)

    1/{sin[(180 - 360/n)/2]}

    Now as n approaches infinity the 360/n term approaches 0.

    1/ sin (180/2)

    =1/sin(90)

    =1

    Therefore the 'final' term to be multiplied by the series is 1.
    Each progessive term gets closer and closer to 1.

    Hence the series must converge....

    Also the 100th term is 8.28314283
    The 1000th term is 8.657230754
    The 10,000th term is 8.695744603
    The 65,500th term is 8.69938119

    I would imagine that the limit is 8.7m.
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  8. #7  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Mobius
    Also the 100th term is 8.28314283
    The 1000th term is 8.657230754
    The 10,000th term is 8.695744603
    The 65,500th term is 8.69938119
    I would imagine that the limit is 8.7m.
    This is encouraging although not 100% conclusive. Your argument here (copied below) is most definitely flawed which I will prove below.
    Quote Originally Posted by Mobius
    My series does indeed converge....
    The general term in the series is
    1/{sin[(n-2)*180/2n]}
    Multiply out:-
    1/{sin[(180n - 360)/2n]}
    Divide by n top and bottom (of the sin term that is)
    1/{sin[(180 - 360/n)/2]}
    Now as n approaches infinity the 360/n term approaches 0.
    1/ sin (180/2)
    =1/sin(90)
    =1

    Therefore the 'final' term to be multiplied by the series is 1.
    Each progessive term gets closer and closer to 1.
    Hence the series must converge....
    Consider a series represented by the following recursion formula
    R(n+1) = R(n) (1+1/n)
    Just like your arguement you can argue that since 1/n goes to zero
    then as n goes to infinity (1+1/n) goes to one and therefore the series converges, right?

    But this series does not converge. If R(0)>=1 then every term in the series above is greater than the series given by
    R(n+1) = R(n) [1+1/(n R(n))]
    This means that if this new series diverges then so does the previous one.
    But this is the famous harmonic series:
    R(inf) = R(0) (1 + 1/2 + 1/3 + 1/4 +1/5 + ....)
    which is well known and fairly easy to show that it diverges. Look up harmonic series in google. So the series given by R(n+1) = R(n) (1+1/n) also diverges. If R(0)<1 then multiply the whole series by 1/R(0) and by the same argument R(inf)/R(0) diverges, which means R(inf) diverges.

    Your series has a lot in common with the harmonic series so its convergence is seriously in doubt. If you can get the series in an additive form like a0+a1+a2+a2+..... , then there are a lot of tools you can use to prove its convergence. I have tried but have not found a way to do this yet.

    You may also want to look at http://en.wikipedia.org/wiki/Infinite_product where it looks at simpler infinite product. They show how you can change the infinite product A = 1/[cos(pi/3)cos(pi/4)cos(pi/5).....] to
    log A = -log(cos(pi/3))-log(cos(pi/4))-log(cos(pi/5))-..... which is an additve series. if log A converges then so does A. One test is to show that a(n+1)/a(n) goes to a number less than 1 as n goes to infinity. This fails for the hamonic series since a(n+1)/a(n) = n/n+1 but in the limit as n goes to infinity n/n+1 goes to 1. For your series we have
    a(n+1)/a(n) = log(cos(pi/(n+1))/log(cos(pi/n))
    And now I haven't the foggiest what to do with this. I suspect the limit as n goes to inf is probably 1 in which case we still don't know if it converges. It only proves divergence if the limit is greater than 1.
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
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  9. #8  
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    I was told on another forum that the product

    (1+x_1)(1+x_2)(1+x_3)...

    converges if and only if the series sum x_i converges.

    or in this case the product (y_1)(y_2)(y_3)...

    converges if and only if the series sum(y_i - 1) converges

    I tried this and the sum does indeed converge....
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  10. #9  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Mobius
    I was told on another forum that the product

    (1+x_1)(1+x_2)(1+x_3)...

    converges if and only if the series sum x_i converges.

    or in this case the product (y_1)(y_2)(y_3)...

    converges if and only if the series sum(y_i - 1) converges

    I tried this and the sum does indeed converge....
    This is great progress. But hold on a minute, you cannot prove convergence just by looking at the series. What is your expression for (y_i - 1)?

    But wait, I think I have something. I believe that for small x, 1/cos(x) ~ 2-cos(x) in which case we would have that 1/cos(x) - 1 ~ 1 - cos(x) for small x. But using the series expansion,
    1-cos(x) = x^2/2 - x^4/24 + x^6/720 - .....
    which means that for large n
    1/cos(pi/n) - 1 ~ pi^2/(2 n^2) - pi^4/(24 n^4) + pi^6/(720 n^6) - ....
    and this not only goes to zero but a sum of such terms would coverge as well ( in other words (1/cos(pi/3) -1) + (1/cos(pi/4) -1) + (1/cos(pi/6) -1) + .... is a sum that converges).

    Therefore if what you have been told is correct and I understand it correctly then indeed your series does converge!
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
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