In continuous-time differential equations, if I have:
$\quad \frac{d}{dt}x_1(t) = A_1\cdot x_1(t) + B_1$
$\quad \frac{d}{dt}x_2(t) = A_2\cdot x_2(t) + B_2$.
the square of $x_1(t)$ and $x_2(t)$ can propagate as:
$\quad \frac{d}{dt}\{x_1^2(t)\} = 2 x_1(t) \cdot \frac{d}{dt}{x}_1(t) = 2x_1(t) \cdot\{A_1\cdot x_1(t) + B_1\} = 2A_1 \cdot x_1^2(t) + 2B_1 \cdot x_1(t)$
$\quad \frac{d}{dt}\{x_2^2(t)\} = 2 x_2(t) \cdot \frac{d}{dt}{x}_2(t) = 2x_2(t) \cdot\{A_2\cdot x_2(t) + B_2\} = 2A_2 \cdot x_2^2(t) + 2B_2 \cdot x_2(t)$,
the squares of $x_1(t)$ and $x_2(t)$ are of squares and the original $x_1(t)$ and $x_2(t)$.
I am curious about the propagation of powers of discrete-time maps:
$\quad x_1(k+1) = a_1\cdot x_1(k) + b_1$
$\quad x_2(k+1) = a_2\cdot x_2(k) + b_2$,
where $k = 1, 2, 3, \dots \in \mathbb{N}$.
For example, I am interested in the trajectories of $x_{1}^2(k)$ and $x_{2}^2(k)$.
My guess would be to do the following:
$\quad x_{1}^2(k+1) = (x_1(k+1))^2 = (a_1\cdot x_1(k) + b_1)^2 = a_1^2 \cdot x_1^2(k) + 2a_1\cdot x_1(k)\cdot b_1 + b_1^2$
$\quad x_{2}^2(k+1) = (x_2(k+1))^2 = (a_2\cdot x_2(k) + b_2)^2 = a_2^2 \cdot x_2^2(k) + 2a_2\cdot x_2(k)\cdot b_2 + b_2^2$.
Is this correct?
That is correct. This solid green curve in this image shows $x_1x_2$ for a two-dimensional system calculated as the product of the two solutions. Directly below the solid curve is another curve that gives the simulated trajectory of $x_1x_2$. This is no proof, but rather empirical evidence that what you think is correct is what is seen in practice.