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> |
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What if they aren't from a normalized system, so that |
What if they aren't from a normalized system, so that |
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<math>\vec \bold a_k \bullet \vec \bold a_n = w_k \delta_{k,n} </math> |
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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 |
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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 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
.