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Sum of cosines

  1. Sep 29, 2014 #1
  2. jcsd
  3. Sep 29, 2014 #2

    mathman

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    Are you interested in absolute value (easy = 0) or most negative (harder)?
     
  4. Sep 30, 2014 #3
    Thank You Mathman! Actually I am interested to find minimum value not absolute . What I am thinking Geometrically is that the possibility seems by imagining unit vectors inclined at 120 degrees to each other in 3D space.Am I correct?It gives thevalue -3/2.
     
  5. Sep 30, 2014 #4

    mathman

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    You can put 3 vectors at 120 deg. apart in a plane. I haven't looked at the problem in any detail, but I would guess you could do better in 3d.
     
  6. Sep 30, 2014 #5
    In the general case, it's easiest to do the problem in vector fashion. The angle cosines are easy: ## \cos \theta_{12} = n_1 \cdot n_2 ##. For unit vector n relative to unit vectors n1, n2, n3, we get
    $$ S = \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = n \cdot n_1 + n \cdot n_2 + n \cdot n_3 = n \cdot (n_1 + n_2 + n_3) $$
    We now want to maximize ## S = n \cdot n_{total} ## and we wish to do so with the constraint that ## |n| = 1 ##. This is easiest to do with Lagrange multipliers:
    $$ S' = S - \frac12 \lambda (n^2 - 1) $$
    Taking derivatives with respect to all components of n, we get ## \lambda n = n_1 + n_2 + n_3 = n_{total} ##, meaning that n is proportional to ntotal.
     
  7. Oct 2, 2014 #6
    It seems to me that -3/2 is indeed the minimum. Denote three unit vectors in 3-space by a,b,c with components a(i), b(i), c(i) where i=1,2,3. Then we wish to minimize a.b + b.c + c.a subject to a^2=b^2=c^2=1. Using Lagrange multipliers L,M,N we wish to minimize:

    a(i)b(i) + b(i)c(i) + c(i)a(i) - L(a^2-1) - M(b^2-1) - N(c^2-1) where I have used summation convention for repeated indices.

    Differentiating by a(i), b(i), c(i) respectively we get

    b(i) + c(i) =2La(i)
    a(i) + c(i) =2Mb(i)
    b(i) + a(i) =2Nc(i)

    it is immediately apparent from this that all the vectors are coplanar. furthermore, subtracting the first two of these gives
    b(i) (1+2M) = a(i) (1+2L)

    But by inspection the minimum is not achieved by having a(i) and b(i) be the same vector, since the cosine is then maximized, so we must have L=M=N= -1/2

    Therefore b(i) + c(i) = -a(i), or a + b + c = 0. Hence at the extremum, a,b,c are at the vertices of an equilateral triangle, as conjectured.
     
  8. Oct 5, 2014 #7
    Thank You Friends!
     
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