Solve the Global Minima Problem in Two Variable Functions

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Homework Statement
Let ##f(x,y)=\arctan(4\sin^2(y)+3\ln(x^2+1))## show that it has ##\infty## global minima
Relevant Equations
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I'm always struggling understand how to determine if a two variable function has global minima, I know that if I find a local minima and the function is convex than the local minima is also a global minima, in this case is really difficult to determine if the function is convex.

Sorry if I don't post any attempt but I got no clue how to do this.
 
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I'm afraid you will have to calculate the partial derivatives and consider where they vanish. Also look up the Hesse matrix.
 
Archimedess said:
no clue how to do this.
Not good enough per the PF guidelines !
Least you could do is find and discuss a few minima, remark that ##\sin^2## is periodical, etc ...
 
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No, it is not convex. A convex function has a UNIQUE global minimum.

Edit: OK, I'm going to back off on that statement. That's true for a strictly convex function. But you could imagine a convex function whose set of global minima was a finite flat region. For instance, a function with ##f(x,y) = 0## on the circle of radius 1, and positive elsewhere. That would qualify as a convex function with infinitely many global minima.

But that's not the reason for infinitely many minima here. @BvU has already identified the reason in comment #3.

To find the global minima, you're going to have to identify all the local minima. Then analyze them and find which subset of those have minimal values of ##f(x, y)##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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