Recent content by Luckyroad21

Okay, thanks a lot for your help. It was a fun problem!

Is this enough to be able to apply the Lagrangian method?

Here is an example of building infinite solutions; with m being a positive number:

Using the method, it's easy to find all the other solutions, for example:

This proves that there are infinite solutions

For any Numbers x1,x2…xn with x1+x2…+xn=0 and x1^2*x2^2…+xn^2= u, get xi/sqrt(u)=xi’. We Will have: Sum(xi’)=0 and Sum(xi’^2)=1

How can I check this?

Am I right?

Obviously, the set of points that maximizes our sum is equal to the one that minimizes with the signs of the numbers swapped (our sum is an odd function), so: and

With : (1) (2) (3) Equating the partial derivatives to 0, we have: or (4) So I will assume that lambda is negative and that there are b numbers of xi's that are negative and (n-b) numbers of xi's that are positive. (5) Applying (5) and (4) to (3), we get: (6) Applying (5) and...

Maybe it's easier to work with that.

So what do you mean is that we only retrieve the solutions for n multiple 3? Do the others not exist?

Your result for n=3 is positive for my conjecture.For n=4, the solutions I obtained from the partial derivatives do not follow the imposed restrictions. For forcing such restrictions on the obtained system I arrived at xi=(2/n)^(1/2) or xi= - (1/2n)^(1/2); it's easy to see that this only adds up...

I applied the partial derivatives you suggested and arrived at x1=0, the problem is that for n=3, the optimal solution does not contain null terms. I apologize if this is too obvious, I'm just a humble medical student who likes numbers.

To solve the elegant asymmetry, I took ai = 1/n + xi. (n^2-n)^(-1/2), such a transformation led me to non-negative real numbers such that: a1+a2+a3...+an=1 It is a1^2+a2^2+a3^2...+an^2= (n-1)^(-1). The problem is to get the maximum value of a1^3+a2^3+a3^3...+an^3.