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n-dimensional spheres (Posted on 2014-09-15) Difficulty: 3 of 5

  
An n-dimensional sphere of radius one is intersected with another
n-dimensional sphere of radius one and whose center lies on the
first sphere. The intersection is a (n-1)-dimensional sphere.

The (n-1)-dimensional sphere is intersected with another
n-dimensional sphere of radius one and whose center lies on the
(n-1)-dimensional sphere. The intersection is a (n-2)-dimensional sphere.

This procedure is continued until we have a 1-dimensional sphere.
What is its radius?

This all takes place in an n-dimensional space.
  

  Submitted by Bractals    
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Solution: (Hide)

  
We can assume that the first n-dimensional sphere and the subsequent
(n-k)-dimensional spheres are centered at the origin of the n-space by
translation.
   x12 + x22 + ... + xk-12 + xk2 = rk2                     (1)
and the n-dimensional sphere centered on it
   x12 + x22 + ... + xk-12 + (xk - rk)2 = 1                (2)

   where rn = 1.
Solving (1) and (2) gives
   xk = rk - 1/(2*rk)                                    (3)
Plugging xk into (1) or (2) gives
   x12 + x22 + ... + xk-12 = 1 - 1/(4*rk2)                (4)
Equation (4) implies the recurrence relation
   rk-12 = 1 - 1/(4*rk2)                                 (5)
I don't know if there is a procedure for finding a closed form solution
to a recurrence relation, but I came up with
   rk2 = (n - k + 2)/[2*(n - k + 1)]                    (6)                                                                        

   rn2 = (n - n + 2)/[2*(n - n + 1)] = 1                                    

   r12 = (n - 1 + 2)/[2*(n - 1 + 1)] = (n + 1)/(2*n)                                                                           
Therefore,
   r1 = √[ (n + 1)/(2*n) ]                               (7)                                                      
QED

Check it out for n = 1, 2, and 3 and Charlie's post.
  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: The sequence, if my trig is rightBractals2014-09-15 09:57:27
Some ThoughtsThe sequence, if my trig is rightCharlie2014-09-15 08:26:49
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