# Find a suitable zero equation to solve the optimization problem $\min_{x \in \mathbb{R}^N} f(x)$

Suppose we have an optimization problem for this general form of $$f: \mathbb{R}^N \rightarrow \mathbb{R}$$

$$\min_{x \in \mathbb{R}^N} f(x)$$

and this problem is solvable. How could I construct a suitable zero equation to solve the optimization problem?

I think if $$x\in \mathbb{R}$$, then we can construct the zero equation as $$f'(x) = 0$$. With the zero equation in hand then the solution can be found with iterations of Newton's method. However I am not sure how to proceed when $$x \in \mathbb{R}^N$$.

Any ideas would be appreciated!

• What is a "zero equation" and how can you claim that the equation $f'(x)=0$ leads to a solution via some method? There is no method to solve a general optimization problem such as you give. – Michael Jun 12 at 3:23

If $$f$$ is differentiable, the minimiser will satisfy the equation $$\nabla f = 0$$, where $$\nabla f = (f'_{x_1}, \cdots, f'_{x_n})$$. Newton's method is also quite similar: if you want to numerically solve an equation $$g(x) = 0$$, where $$g:\mathbb{R}^n\to \mathbb{R}^n$$, you take an initial approximation $$x^{(0)}$$ and proceed using the iteration $$x^{(k+1)}= x^{(k)} - [J_g(x^{(k)})]^{-1} g(x^{(k)}),$$
where $$J_g$$ is the Jacobian matrix of $$g$$. Please note that the convergence of Newton's method is by no means guaranteed, and even if it converges it can converge to a local minimum (which may not be the global minimiser), a local/global maximum or a saddle point.