MathB.in - Share Mathematics with LaTeX and Markdown

Define $f^{(k)}\left|[0,1]\rightarrow [0,1]\right., k=0,1,\ldots $ iteratively as follows,
$$
f^{(0)}(x)=2x(1-x)
$$
and for $k\geq 1$
$$
f^{(k)}(x)=f\circ f^{(k-1)}(x).
$$
The problem is to evaluate
$$
I_k:=\int_0^1f^{(k)}(x)\,\mathrm{d}x.
$$
To evaluate the integral the $\eta:=x-\frac{1}{2}$ is made.  With $g^{(k)}(\eta)\left|\left[-\frac{1}{2},\frac{1}{2}\right]\rightarrow \left[-\frac{1}{2},\frac{1}{2}\right]\right.$ defined by

$$
g^{(k)}(x)=f^{(k)}\left(\eta+\frac{1}{2}\right),
$$
one has
$$
I_k:=\int_{-\frac{1}{2}}^{\frac{1}{2}}g^{(k)}(\eta)\,\mathrm{d}\eta
$$
The advantage conferred by the substitution is that the functions $g{(k)}$ take on a rather simple form. Note that
$$
g^{(0)}(\eta)=2\left[\eta+\frac{1}{2}\right]\left[1-\left(\eta+\frac{1}{2}\right)\right]=2\left[\frac{1}{2}+\eta\right]\left[\frac{1}{2}-\eta\right]=\frac{1}{2}-2\eta^2
$$
and for $k\geq 1$
$$
g^{(k)}(\eta)=f^{(k)}\left(\eta+\frac{1}{2}\right)=f\left( f^{(k-1)}\left(\eta+\frac{1}{2}\right)\right)=f\circ g^{(k-1)}(\eta).
$$
It follows that 
$$
g^{(1)}(\eta)=2\left[\frac{1}{2}-2\eta^2\right]\left[\frac{1}{2}+2\eta^2\right]=\frac{1}{2}-2^3\eta^{2^2},
$$
and
$$
g^{(2)}(\eta)=2\left[\frac{1}{2}-2^3\eta^{2^2}\right]\left[\frac{1}{2}+2^3\eta^{2^2}\right]=\frac{1}{2}-2^7\eta^{2^3}.
$$
At this point the pattern is clear, and  induction can be used  to prove that for $k\geq 0$
$$
g^{(k)}(\eta)=\frac{1}{2}-2^{2^{k+1}-1}\eta^{2^{k+1}}.
$$
It follows that
$$
I_k=\frac{1}{2}\left[1-\frac{1}{2^{k+1}+1}\right]
$$

Sun, 16 Feb 2025 17:10 GMT