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 Integrable Nest (Posted on 2013-08-16)
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 No Rating

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 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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