Vector weighting functions: Difference between revisions

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<math> \vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \vec \bold a_k \bullet \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \delta_{k,m} = \sum_{k=1}^3 v_k u_k </math>
<math> \vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \vec \bold a_k \bullet \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \delta_{k,m} = \sum_{k=1}^3 v_k u_k </math>


What if they aren't from a normalized system, so that
What if they aren't from a normalized system, so that

<math>\vec \bold a_k \bullet \vec \bold a_n = w_k \delta_{k,n} </math>

where the <math> w_k </math> is the square of the length of <math> \vec \bold a_k </math> and the symbol <math> \delta_{k,n} </math> is one when k = n and zero otherwise? Well the general inner product of <math> \vec \bold u </math> and <math> \vec \bold v </math> becomes

<math> \vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \vec \bold a_k \bullet \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m w_k\delta_{k,m} = \sum_{k=1}^3 w_k v_k u_k </math>.

Revision as of 10:34, 24 September 2004

Orthogonal but not Orthonormal Basis Sets

Suppose we have two vectors from an orthonormal system, and . Taking the inner product of these vectors, we get

What if they aren't from a normalized system, so that

where the is the square of the length of and the symbol is one when k = n and zero otherwise? Well the general inner product of and becomes

.