X1 belongs to (0, 1) xn+1=xn^3-3xn^2+3xn where n + 1, n are indices * Study the monotony. As detailed as pissible please.

Question

anonymous · Answer

Since $x_1$ belongs to (presumably) the interval $(0,1)$, then $x_1$ is some positive rational number between 0 and 1.

This means you have
$$x_2={x_1}^3-3{x_1}^2+3x_1$$
To describe the monotony of a recursive sequence, you must compare successive terms. For the particular recursion, you want to be able to compare the effect of higher powers of the $n$-th term.

For example, suppose $x_1=\dfrac{1}{2}$. Higher powers of this paticular $x_1$ will be smaller than the first power, i.e. ${x_1}^2=\dfrac{1}{4}$ and ${x_1}^3=\dfrac{1}{8}$, so $x_1>{x_1}^2>{x_1}^3>\cdots$. This is true for any number $x_1$ in $(0,1)$.

Keeping with the example of $x_1=\dfrac{1}{2}$, we get
$$x_2=\frac{1}{8}-\frac{3}{4}+\frac{3}{2}=\frac{7}{8}>\frac{1}{2}$$
You should be able to see that the $3x_n$ term will always overcome the size of the ${x_n}^3-3{x_n}^2$ terms, so the sequence is increasing (and actually caps at 1).