perplexus.info :: Calculus : Integrable Nest

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)), ....
h_n+1(x) = h_n(h(x))

Determine:

1
∫ h_n(x) dx in terms of n.
0

No Solution Yet Submitted by K Sengupta

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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, h_n(x), join (0, 0) to (1, 1) using

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

   (2^1-n/3, 2/3), (2^2-n/3, 5/6), (2^3-n/3, 11/12), ..... ,(1/3, 1 – 2^1-n/3).

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

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

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

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

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

So the total area required for the integral is:

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

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

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

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

=          1 - (3 + n)/(3*2ⁿ⁺¹)

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

explains the simplicity of the total area under the 1^st 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*2^n-1)

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

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

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