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I am working on the following exercise:

Consider the function $$f(x,y) = \sin(x)+\sin(y)+\sin(x+y)$$ for $0\le x,y \le 2\pi$.

I calculated the gradient vector:

$$\nabla f(x,y) = \begin{bmatrix} \cos(x)+\cos(x+y)x\\ \cos(y)+\cos(x+y)y \end{bmatrix}$$

But how can I determine the zeroes of $\nabla f(x,y)$ ?

Bernard
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3nondatur
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2 Answers2

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We get $$f_x=\cos(x)+\cos(x+y)=0$$ and $$f_y=\cos(y)+\cos(x+y)=0$$ Now use $$\cos(x)+\cos(y)=2 \cos \left(\frac{y}{2}\right) \cos \left(x+\frac{y}{2}\right)$$ It is $$\cos(x+\frac{y}{2})\cos(y/2)=0$$ etc. and $$2 \cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}+y\right)=0$$

Dr. Sonnhard Graubner
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Your gradient vector is not correct.$$\nabla f(x,y) = \begin{bmatrix} \cos(x)+\cos(x+y)x\\ \cos(y)+\cos(x+y)y \end{bmatrix}$$

should have been $$\nabla f(x,y) = \begin{bmatrix} \cos(x)+\cos(x+y)\\ \cos(y)+\cos(x+y) \end{bmatrix}$$

You need to solve $$\cos(x)+\cos(x+y)=0, \cos(y)+\cos(x+y)=0$$

Note that $$\cos (\pi \pm x)= - \cos(x)$$ which makes your solution simple to find.

Mohammad Riazi-Kermani
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