1
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ 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

4
$\begingroup$

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 $x_i$.)

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.