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)?


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    $\begingroup$ 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 $\min f=0$". $\endgroup$ Jan 16 at 21:52

2 Answers 2


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.

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    $\begingroup$ I think a sentence needs to be added for this to be complete -- "we have proven the function is always zero or larger and we have also shown there's a point where it is zero thus the minimum is zero" because without tying it off your proof only shows the minimum is a non-negative number. (x^2+1 is always >= 0 but the minimum is 1 on the reals.) $\endgroup$
    – chx
    Jan 17 at 5:13

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