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Calculus
Integrable Nest (
Posted on 20130816
)
h(x) =2*x, for 0 ≤ x ≤ 1/3
= x/2 + 1/2, for 1/3 ≤ x ≤ 1
and, h
_{2}
(x) = h(h(x)), h
_{3}
(x) = h
_{2}
(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
No Rating
Comments: (
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)
Solution
Comment 1 of 1
The graph of h
_{1}
(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
^{1n}
/3, 2/3), (2
^{2n}
/3, 5/6), (2
^{3n}
/3, 11/12), ..... ,(1/3, 1 2
^{1n}
/3).
The triangular areas below the first and last segments together yield:
(1/2)(2
^{1n}
/3)2/3
+
(1/2)(1 1/3)(1 2
^{1n}
/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
^{1nr}
/3 2
^{rn}
/3)(2 2
^{1r}
/3 2
^{r}
/3)
which simplifies to
A
_{r}
= (2
^{r+1}
1)/(3*2
^{n+1}
)
So the total area required for the integral is:
2/3
+
Sum(r from 1 to n1) of A
_{r}
=
2/3
+
(1/(3*2
^{n+1}
))*[Sum(r from 1 to n1) of (2
^{r+1}
1)]
=
2/3
+
(1/(3*2
^{n+1}
))*[4(2
^{n1}
1) (n 1)]
=
2/3
+
[2
^{n+1}
n 3] / (3*2
^{n+1}
)
=
1

(3 + n)/(3*2
^{n+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 theres an easier approach! Also, all the
intermediate vertices lie on the hyperbola
x(1 y) = 1/(9*2
^{n1}
)
and the integral > 1 as n > infinity as expected.)
Posted by
Harry
on 20130822 11:15:11
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