Optimization of product of functions

Question

I have a function$\ f(x) $, which is not necessarily linear (actually it has some oscillations to it), this function is used to construct the function $\ g(\textbf{x})=f(x_1)*f(x_2)*...*f(x_N) $, where $\ \textbf{x} = [x_1, x_2, ..., x_N]$. How can I use my knowledge of $\ f(x)$ to determine the number of critical points of $\ g(\textbf{x})$ on some interval.

Answer 1

Having $\vec{x} \in \mathbb{R}^n$, one can compute the gradient of function:

$g(\vec{x})=\prod_{i=1}^n f(x_i)$

Obtaining:

$\frac{\partial g(\vec{x})}{\partial x_j} = \frac{d f(x_j)}{d x_j} \prod_{i=1, i \neq j}^n f(x_i)$

Setting this expression to zero for all $j=1 \dots n$ and solving for $(x_1, \dots x_n)$, we obtain stationary points of $g$:

$\frac{\partial g(\vec{x})}{\partial x_1} = \frac{d f(x_1)}{d x_1} \prod_{i=2}^n f(x_i) = 0$

$\frac{\partial g(\vec{x})}{\partial x_2} = \frac{d f(x_2)}{d x_2} \prod_{i=1,i\neq2}^n f(x_i) = 0$

$\dots \dots \dots\dots\dots\dots$

$\frac{\partial g(\vec{x})}{\partial x_n} = \frac{d f(x_j)}{d x_j} \prod_{i=1,}^{n-1} f(x_i) = 0$

• First remark, if $\vec{x}$ has at least two components which are roots of function $f$, it is a stationary point.
• Second remark, if $\vec{x}$ has one component which is root of function $f$, and this root is also an stationary point of $f$ (i.e. $df/dx = 0$), $\vec{x}$ is a stationary point of $g$.