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Consider the function $$f(x_1, x_2) =2x_1^2-2x_1x_2+x_2^2-x_1+x_2-7$$

Perform Newton's method to minimize $f(x_1,x_2)$. The initial point is $x^0 = [2 -1]^T$

How man global minima does $f(x_1,x_2)$ have?

I know Newton's Method is $$x^{k+1} = x^k - \nabla^2f(x_1,x_2)^{-1}\nabla f(x_1,x_2)$$ and I found a minimum at $x = [0, \frac{1}{2}]^T$ using Newton's Method

How do I know for certain this is the only Minimum? (I used graphing software to comfirm this, but I would like to know intuitively from the function)

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  • $\begingroup$ Look at $f$. How many critical point does it have? $\endgroup$ – user251257 Nov 3 '16 at 2:49
  • $\begingroup$ An elementary approach would be to use a suitable rotation of coordinates to eliminate the $x_1\cdot x_2$ term,you will see that the graph is an ellipse. Otherwise, you might compute the Hessian and interpret. $\endgroup$ – Matematleta Nov 3 '16 at 2:49
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You might see whether the Hessian is positive definite (and it is). Then the function is convex, so the minimum is unique.

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