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I am new to math. How to approach the following problem?

$\min_{a,b} \sum_{t=1}^N (-4aX_t\sin(Z_tb) -4aY_t\cos(Z_tb)+a^2Z_t^2 + X_t^2 + Y_t^2)$

where $X_t,Y_t,Z_t$ for $t\in \{1,...N\}$ are given. Say $N$ is around 800.

Do I need software? Which? I am not having luck with simple optim() in R, but maybe I am using wrong parameters.

Would this be a problem that could be solved in Mathematica? (I don't have access to Mathematica, and am also not familiar with it. I just heard that it is powerful.)

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With a little bit of work you can convert your two-dimensional minimization problem into a one-dimensional root-finding problem.

Call the quantity to be minimised $L(a,b)$. Then computing the partial derivatives and setting to zero gets you

$$\frac{\partial L}{\partial a} = 0 \;\Rightarrow\; \frac{a}{2} \sum_{t=1}^n Z_t^2 - \sum_{t=1}^N X_t\sin(Z_tb) + Y_t\cos(Z_tb) = 0$$

$$\frac{\partial L}{\partial b} =0 \;\Rightarrow\; a \sum_{t=1}^N X_t\cos(Z_tb) - Y_t\sin(Z_tb) = 0$$

and hence either $a=0$ or the sum in the second equation is zero. If you can find $b$ such that the sum is zero (using some numerical root finder) then $a$ is determined by the first equation. On the other hand, if $a$ is zero then you determine $b$ from the first equation using a numerical root finder.

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    Thanks for your reply. It is still not clear for me: How do I find the global minimum?2011-05-27
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    Global minimization is a difficult problem in general. The extremal points of your equation are given by the solutions to the two equations in my answer. This will find all local maxima, minima and saddle points. To find the _global_ minimum you should take those solutions and put them back into your original expression to find out which one gives the smallest result.2011-05-27
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    Thanks, had hoped there was some way to avoid this.2011-05-27