How to minimize the following polynomial over a four-dimensional sphere: $(c_0x_0^2+c_1x_1^2+c_2x_2^2+c_3x_3^2)^2+sum_nmc_nm x_nx_m $

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I need to find the global minimum of the following polynomial over a four-dimensional sphere.



$$f(x_0, x_1, x_2, x_3) = left( sumlimits_j=0^3 c_j x_j^2 right)^2 + sumlimits_n,m=0^3 c_nm x_n x_m $$



Here $x_0,x_1,x_2,x_3$ are the variables and are constrained to be on the four-dimensional sphere, i.e.,



$$sum_n=0^3 x_n^2=1$$



and $c_j (j = 0, 1, 2, 3)$ and $c_nm ( n, m = 0,1,2,3)$ are problem constants that can be positive or negative or zero.



Well, I am aware of the general method; find the derivatives calculate the roots etc... however as you can see the problem has some sort of simple structure (quartic part can be diagonalized) so I am wondering if there is a simpler solution. It is also possible to diagonalize the quadratic part and leave the quartic part complicated.







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  • Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
    – David G. Stork
    yesterday











  • Does the range of summation include $0,1,2,3$?
    – copper.hat
    yesterday










  • Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
    – Buddha_the_Scientist
    yesterday










  • Are the $c_0,...,c_3$ positive?
    – copper.hat
    yesterday










  • @copper.hat No actually, they can be negative.
    – Buddha_the_Scientist
    yesterday














up vote
0
down vote

favorite












I need to find the global minimum of the following polynomial over a four-dimensional sphere.



$$f(x_0, x_1, x_2, x_3) = left( sumlimits_j=0^3 c_j x_j^2 right)^2 + sumlimits_n,m=0^3 c_nm x_n x_m $$



Here $x_0,x_1,x_2,x_3$ are the variables and are constrained to be on the four-dimensional sphere, i.e.,



$$sum_n=0^3 x_n^2=1$$



and $c_j (j = 0, 1, 2, 3)$ and $c_nm ( n, m = 0,1,2,3)$ are problem constants that can be positive or negative or zero.



Well, I am aware of the general method; find the derivatives calculate the roots etc... however as you can see the problem has some sort of simple structure (quartic part can be diagonalized) so I am wondering if there is a simpler solution. It is also possible to diagonalize the quadratic part and leave the quartic part complicated.







share|cite|improve this question





















  • Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
    – David G. Stork
    yesterday











  • Does the range of summation include $0,1,2,3$?
    – copper.hat
    yesterday










  • Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
    – Buddha_the_Scientist
    yesterday










  • Are the $c_0,...,c_3$ positive?
    – copper.hat
    yesterday










  • @copper.hat No actually, they can be negative.
    – Buddha_the_Scientist
    yesterday












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to find the global minimum of the following polynomial over a four-dimensional sphere.



$$f(x_0, x_1, x_2, x_3) = left( sumlimits_j=0^3 c_j x_j^2 right)^2 + sumlimits_n,m=0^3 c_nm x_n x_m $$



Here $x_0,x_1,x_2,x_3$ are the variables and are constrained to be on the four-dimensional sphere, i.e.,



$$sum_n=0^3 x_n^2=1$$



and $c_j (j = 0, 1, 2, 3)$ and $c_nm ( n, m = 0,1,2,3)$ are problem constants that can be positive or negative or zero.



Well, I am aware of the general method; find the derivatives calculate the roots etc... however as you can see the problem has some sort of simple structure (quartic part can be diagonalized) so I am wondering if there is a simpler solution. It is also possible to diagonalize the quadratic part and leave the quartic part complicated.







share|cite|improve this question













I need to find the global minimum of the following polynomial over a four-dimensional sphere.



$$f(x_0, x_1, x_2, x_3) = left( sumlimits_j=0^3 c_j x_j^2 right)^2 + sumlimits_n,m=0^3 c_nm x_n x_m $$



Here $x_0,x_1,x_2,x_3$ are the variables and are constrained to be on the four-dimensional sphere, i.e.,



$$sum_n=0^3 x_n^2=1$$



and $c_j (j = 0, 1, 2, 3)$ and $c_nm ( n, m = 0,1,2,3)$ are problem constants that can be positive or negative or zero.



Well, I am aware of the general method; find the derivatives calculate the roots etc... however as you can see the problem has some sort of simple structure (quartic part can be diagonalized) so I am wondering if there is a simpler solution. It is also possible to diagonalize the quadratic part and leave the quartic part complicated.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









David G. Stork

7,3102728




7,3102728









asked yesterday









Buddha_the_Scientist

115




115











  • Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
    – David G. Stork
    yesterday











  • Does the range of summation include $0,1,2,3$?
    – copper.hat
    yesterday










  • Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
    – Buddha_the_Scientist
    yesterday










  • Are the $c_0,...,c_3$ positive?
    – copper.hat
    yesterday










  • @copper.hat No actually, they can be negative.
    – Buddha_the_Scientist
    yesterday
















  • Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
    – David G. Stork
    yesterday











  • Does the range of summation include $0,1,2,3$?
    – copper.hat
    yesterday










  • Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
    – Buddha_the_Scientist
    yesterday










  • Are the $c_0,...,c_3$ positive?
    – copper.hat
    yesterday










  • @copper.hat No actually, they can be negative.
    – Buddha_the_Scientist
    yesterday















Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
– David G. Stork
yesterday





Try Lagrange optimization (i.e., a mento for optimization given a constraint). But your problem is unclear: the unit sphere has just $x_1$, $x_2$ and $x_3$. What is $x_0$? Can it actually be a variable? What does it have to do with a sphere?
– David G. Stork
yesterday













Does the range of summation include $0,1,2,3$?
– copper.hat
yesterday




Does the range of summation include $0,1,2,3$?
– copper.hat
yesterday












Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
– Buddha_the_Scientist
yesterday




Yes I have edited now. @David G. Stork Yes, thanks. Does the fact that the quartic part can be diagonalized leads to some simplification here?
– Buddha_the_Scientist
yesterday












Are the $c_0,...,c_3$ positive?
– copper.hat
yesterday




Are the $c_0,...,c_3$ positive?
– copper.hat
yesterday












@copper.hat No actually, they can be negative.
– Buddha_the_Scientist
yesterday




@copper.hat No actually, they can be negative.
– Buddha_the_Scientist
yesterday















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