Sum of cosines

1. Sep 29, 2014

gianeshwar

Dear Friends! How can I find the minimum value of

2. Sep 29, 2014

mathman

Are you interested in absolute value (easy = 0) or most negative (harder)?

3. Sep 30, 2014

gianeshwar

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.

4. Sep 30, 2014

mathman

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.

5. Sep 30, 2014

lpetrich

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.

6. Oct 2, 2014

davidmoore63@y

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.

7. Oct 5, 2014

gianeshwar

Thank You Friends!