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Integrable Nest (Posted on 2013-08-16) Difficulty: 4 of 5
h(x) =2*x, for 0 ≤ x ≤ 1/3
= x/2 + 1/2, for 1/3 ≤ x ≤ 1

and, h2(x) = h(h(x)), h3(x) = h2(h(x)), ....
hn+1(x) = hn(h(x))

Determine:
1
∫ hn(x) dx in terms of n.
0

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
The graph of h1(x) joins (0, 0) to (1, 1), using 2 straight segments via
a single intermediate ‘vertex’ at (1/3, 2/3).

As n increases, successive functions, hn(x), join (0, 0) to (1, 1) using

n+1 segments, via n intermediate vertices with coordinates as follows:

   (21-n/3, 2/3), (22-n/3, 5/6), (23-n/3, 11/12), ..... ,(1/3, 1 – 21-n/3).

The triangular areas below the first and last segments together yield:

  (1/2)(21-n/3)2/3  +  (1/2)(1 – 1/3)(1 – 21-n/3)  =  2/3

The areas below other segment are trapezia, and the r th trapezium

has an area given by:     Ar = (1/2)(21-n-r/3 – 2r-n/3)(2 – 21-r/3 – 2-r/3)

which simplifies to          Ar = (2r+1 – 1)/(3*2n+1)

So the total area required for the integral is:

            2/3  +  Sum(r from 1 to n-1) of Ar

=          2/3  +  (1/(3*2n+1))*[Sum(r from 1 to n-1) of (2r+1 – 1)]

=          2/3  +  (1/(3*2n+1))*[4(2n-1 – 1) – (n – 1)]

=          2/3  +  [2n+1 – n – 3] / (3*2n+1)

=          1  -  (3 + n)/(3*2n+1)

(..Interesting that all the hn are symmetric about x + y = 1, which

explains the simplicity of the total area under the 1st and last segments

and makes you wonder if there’s an easier approach! Also, all the

intermediate vertices lie on the hyperbola  x(1 – y) = 1/(9*2n-1)

and the integral -> 1 as n -> infinity as expected.)



  Posted by Harry on 2013-08-22 11:15:11
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