Skip to main content

Any solution to system of equations?


I need ANY combination of 8 variables which satisfy:


a+b+c+d+e+f+g+h=0
a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2=1
a*b+b*c+c*d+d*e+e*f+f*g+g*h+a*h>2/3


Ideally these would be numbers having simple closed forms...


If I exchange third restriction with


a*b+b*c+c*d+d*e+e*f+f*g+g*h+a*h=2/3

I get a solution a=b=c=6^(-0.5); e=f=g=-6^(-0.5); d=h=0;



Answer



This is a special kind of problem which succumbs to a standard approach.


Analysis


The question can be solved by maximizing xAx and checking that this maximum exceeds 2/3. The restrictions are (1) that x=(a,b,,h) must have unit square length and (2) be orthogonal to e1=(1,1,,1), where A is the matrix


(0100000000100000000100000000100000000100000000100000000110000000)



Solution


Because A is obviously a rotation matrix (it represents a cyclic permutation of the eight basis vectors), e1 (by virtue of its constant coefficients) must be an eigenvector. Seven other complex eigenvectors can be found orthogonal to e1. Their eigenvalues fall into three groups of complex conjugates, representing rotations by primitive multiples of 2π/8, and one real eigenvalue of 1, representing a rotation by π. We seek the smallest rotation among these rotations, because that corresponds to the largest cosine of the angle, which geometrically is what xAx is computing.


The eigensystem can be obtained and displayed as


MatrixForm /@ ({e, ev} = Eigensystem[a = RotateRight[#, 1] & /@ IdentityMatrix[8]] // Simplify)

The cosines are the real parts of the eigenvalues e:


Re[e]


{12,12,12,12,1,0,0,1}




The two largest real parts, 1/20.7071>2/3, occur in the third and fourth places. The real parts of either the third or fourth normalized eigenvectors therefore afford values of x where the desired maximum is attained:


x = Normalize @ (Re /@ ev[[3]]) 


{122,0,122,12,122,0,122,12}



If any of this seems too abstract, let's check all the requirements:





  1. Confirm x is orthogonal to e1:


    x .ConstantArray[1, 8]


    0





  2. Confirm x has unit squared length:


    x . x



    1





  3. Compute xAx:


    x . a . x



    12





Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...