This provides us with the ability to cut right to the chase: Maximize[Īnd that's a quick and easy distillation of the discussion above. So the maximum value of $|x^3-3x+m|$ is either a maximum value of $x^3-3x+m$ or a minimum value of $x^3-3x+m$ (which we then negate to get a relatively large positive value). The latter would be the same as $-\min ( x^3-3x+m )$ in the case where $\min f(x)$ is so low that in the absolute value, it is higher than other potential maxima of $f(x)$. That means either $\max f(x) = x^3-3x+m$ or else $\max f(x) = -(x^3-3x+m)$, for the particular value of $x$. One way to use built-in functions more clearly is to consider that you have a very simple function - the absolute value of a low degree polynomial. in the comments, but it is certainly one way to do it. MaxValue is typically used to find the largest. Maxdata (3.22) Running through rosetta2, I got a max result of 2.7.there is some.
Your "work around" is not necessary, as noted by J.M. MaxValue finds the global maximum of f subject to the constraints given. With a newly reset computer, and a newly reset kernel (completely fresh started mathematica) I wrote a little script to force the benchmark to be done 100 times in a row, in the hopes that the cpu would ramp up some more and give a better result. Coefficient expr, form, n gives the coefficient of form n in expr. A common approach is taken to both elementary complete. Coefficient expr, form gives the coefficient of form in the polynomial expr. Let's talk a little bit about how this works in the code first. Data and outline Mathematica code are given for several examples of maximum likelihood estima- tion. To confirm, the two values of $m$ that you want are indeed $m=-2$ and $m=-14$, which is what your code produces.
#MAX MATHEMATICA HOW TO#
I'm afraid part of the problem might be that you are focusing too much on how to reformulate the maximum in a way that can be used with Solve (and also that you have presupposed knowing that $x=1$ is relevant).
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