The sequence \(u_0, u_1, \ldots\) is said to be a constant sequence if \(u_n = u_{n+1}\) for \(n = 0, 1, 2, \ldots\).
The sequence is said to be a sequence of period 2 if \(u_n = u_{n+2}\) for \(n = 0, 1, 2, \ldots\) and the sequence is not constant.
A sequence of real numbers is defined by \(u_0 = a\) and \(u_{n+1} = f(u_n)\) for \(n = 0, 1, 2, \ldots\), where
$$f(x) = p + (x - p)x,$$
and \(p\) is a given real number.
Find the values of \(a\) for which the sequence is constant.
Show that the sequence has period 2 for some value of \(a\) if and only if \(p > 3\) or \(p < -1\).
A sequence of real numbers is defined by \(u_0 = a\) and \(u_{n+1} = f(u_n)\) for \(n = 0, 1, 2, \ldots\), where
$$f(x) = q + (x - p)x,$$
and \(p\) and \(q\) are given real numbers.
Show that there is no value of \(a\) for which the sequence is constant if and only if \(f(x) > x\) for all \(x\).
Deduce that, if there is no value of \(a\) for which the sequence is constant, then there is no value of \(a\) for which the sequence has period 2.
Is it true that, if there is no value of \(a\) for which the sequence has period 2, then there is no value of \(a\) for which the sequence is constant?
Solution:
If \(f(a) = a\) then the sequence is constant, ie \(a = p+a^2-pa \Rightarrow 0 = (a-p)(a-1)\). Therefore \(a = 1, p\)
If there sequence has period \(2\) then there must be a solution to \(f(f(x)) = x\), ie \begin{align*}
&& x &= p+(f(x)-p)f(x) \\
&&&= p+(p+(x-p)x-p)(p+(x-p)x) \\
&&&= p + (x-p)x(p+(x-p)x) \\
&&&= p+(x^2-px)(x^2-px+p) \\
\Rightarrow && 0 &= x^4-2px^3+(p+p^2)x^2-(p^2+1)x+p \\
&&&= (x-1)(x-p)(x^2-(p-1)x+1)
\end{align*}
The first two roots (\(x = 1, p\)) are constant sequences, so we need the second quadratic to have a root, ie \((p-1)^2-4 \geq 0 \Rightarrow p \geq 3 , p \leq -1\). We also need this root not to be \(1\) or \(p\), ie \(1-(p-1)+1 = 3-p \neq 0\) and \(p^2-(p-1)p + 1 = 1+p \neq 0\) so \(p \neq -1, 3\). Therefore \(p > 3\) or \(p < -1\).
There exists a constant sequence iff there is a solution to \(f(x) = x\), ie
\begin{align*}
&& x &= f(x) \\
&&&= q + (x-p)x \\
\Leftrightarrow && 0 &= x^2-(p+1)x + q \tag{has a solution} \\
\end{align*}
But if it doesn't have a solution, clearly the RHS is always larger, and if it does have a solution then there is some point where the inequality doesn't hold.
Suppose \(f(x) > x\) for all \(x\) then \(f(f(x)) > f(x) > x\) therefore there is no value where \(f(f(x)) = x\) which is required for any sequence of period 2.
No, consider \(p = q = 0\) so \(f(x) = x^2\) then there cannot be a period \(2\) sequence by the first part, but also clearly \(u_n = 1\) is a valid constant sequence.