# Can this Minimization Problem be Proved through Induction?

Consider the following function (this function has $$n$$ dimensions):

$$f(x_1,...,x_n)=10n+\sum_{i=1}^n(x_i^2-10\cos(2\pi x_i));\quad -5.12\leq x_i\leq 5.12$$, $$\text{minimum at }f(0,...,0)=0$$.

Using basic trigonometry, we can see that no matter the value of $$n$$, this function will always have a minimum when the inputs of this function are at $$0$$. This is my logic:

• $$2\pi\cdot 0 = 0$$
• $$\cos(0) = 1$$
• $$10\cos(0) = 10$$
• $$0^2 - 10 = - 10$$
• $$-10$$ summed from $$i = 1$$ to $$n$$ is $$-10n$$
• $$10n - 10n = 0$$

My Question: Is this a sufficient mathematical proof to show that the above function will always have a minimum at $$f(0,...,0) = 0$$ , regardless of the value of $$n$$?

Or like most things in math, does this require a formal mathematical proof (e.g. proof by induction)?

Thanks!

• Writing a proof is telling a story, You have to explain how one thing leads to another. A sequence of apparently unrelated assertions ending in $10n-10n=0$ is not a properly written proof that should end with "Therefore $\min f=0$". Jan 16 at 21:52

You've demonstrated that the value $$f(0,0,\dots,0)$$ is in fact $$0$$. You have not shown that it's the minimum value of $$f$$ yet.
This does not require induction, necessarily. However, crucially, you must at some point compare the value $$f(0,0,\dots,0)$$ to other values $$f(x_1, x_2, \dots, x_n)$$. That's what minimum means: it means "lower than any other value".
It may help you to notice that the different $$x_i$$'s don't interact with each other in the definition of $$f$$. You can separately optimize $$x_i^2 - 10 \cos(2\pi x_i)$$ for each $$i$$. (And the best value of $$x_i$$ is in fact $$0$$, but you need to justify this, by comparing it to other values of $$x_i$$.)
Example of a formal proof: (1). $$x^2\ge 0$$ for every $$x\in\Bbb R.$$ (2). $$-10\cos 2\pi x\ge -10$$ for every $$x\in\Bbb R.$$ Therefore for any $$n\in\Bbb N$$ and any $$(x_1,...,x_n)\in\Bbb R^n$$ we have $$f(x_1,...,x_n)=10n+\sum_{j=1}^n(x_j^2-10\cos 2\pi x_j)\ge$$ $$\ge 10n+\sum_{j=1}^n(x_j^2-10) \ge\quad \text {....by (2)}$$ $$\ge 10n+\sum_{j=1}^n(0-10)=\quad \text {....by (1)}$$ $$=10n-10n=0.$$ On the Q of whether a formal proof is needed, every assertion in math needs one, otherwise you don't know that it's true.