Gradient of polynomial and apply points

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Bryan Ambrósio
Bryan Ambrósio 2021-4-28
评论： Bryan Ambrósio 2021-4-28
Hi everyone,
First of all, I have this polynomial (symbolic):
PF = 0.56057*x1^4 - 2.3567*x1^3*x3 + 7.7971*x1^3*x4 - 10.544*x1^3 - 1.5171*x1^2*x3^2 - 4.5825*x1^2*x3*x4 + 5.1824*x1^2*x3 + 11.583*x1^2*x4^2 - 30.391*x1^2*x4 + 28.869*x1^2 + 0.87025*x1*x3^3 - 3.9352*x1*x3^2*x4 + 5.6176*x1*x3^2 - 1.2696*x1*x3*x4^2 + 1.3133*x1*x3*x4 + 0.23756*x1*x3 + 5.2212*x1*x4^3 - 19.615*x1*x4^2 + 31.103*x1*x4 - 18.59*x1 + 0.37366*x3^4 + 0.33446*x3^3 - 1.0902*x3^2*x4^2 + 2.6402*x3^2*x4 - 1.7305*x3^2 - 0.48794*x3*x4^2 + 1.1816*x3*x4 - 0.80798*x3 + 0.72325*x4^4 - 3.503*x4^3 + 8.1821*x4^2 - 9.5425*x4 + 2.6176
I want to generate the gradient of this polynomial, and then apply some points on this gradient, for example, the points are:
x1 x2 x3 x4
1 0.638 1.710422667 2.00712864
1 0.55225 1.396263402 1.710422667
1 0.44475 1.151917306 1.483529864
1 0.40205 0.951204442 1.3543755
1 0.341825 0.865683309 1.258382391
1 0.33075 0.942477796 1.291543646
Where the first column corresponds to variable x1, the second column to x2, and so on… Then I want to store the results in a vector.
I have tried using the function gradient(PF), but it is returning a 3 line matrix, so I presumed something is wrong, as PF is a 4 variable function.
Can anyone help me doing this gradient and then apply those points? If the application of points can be generic it will be better to me, because I have more points than I showed here.
Eventually I will have to do this same process for much larger polynomials and apply more points. So this computational processing should be as light as it can be.
  1 个评论
Bryan Ambrósio
Bryan Ambrósio 2021-4-28
Just now i noticed that there is not variable x2 on this particular PF. So i have to apply only the points x1, x3 and x4 on the gradient.

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