Vector weighting functions: Difference between revisions
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===Orthogonal but not |
===Orthogonal but not orthonormal basis sets=== |
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Suppose we have two vectors from an orthonormal system, <math> \vec \bold u </math> and <math> \vec \bold v </math>. Taking the inner product of these vectors, we get |
Suppose we have two vectors from an orthonormal system, <math> \vec \bold u </math> and <math> \vec \bold v </math>. Taking the inner product of these vectors, we get |
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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>. |
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You can interpret the <math>w_k</math> as a weighting factor between the different directions so that different directions all end up in the units you would like. For example, suppose that the x and y directions were measured in meters, and the z direction was measured in centimeters, and you would like to use meters as your base unit. You could either convert the z dimensions to meters (probably simpler) or use a weighting function <math> w_x = 1</math>, <math>w_y = 1</math> and <math> w_z = 10^{-6} </math>. In this sense, the system could be considered orthonormal with these units and this weighting arrangement. |
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[[Orthogonal Functions|This idea is often extended to functions.]] |
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[[Orthogonal functions]] |
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Principle author of this page: [[User:Frohro|Rob Frohne]] |
Latest revision as of 16:36, 26 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
.
You can interpret the as a weighting factor between the different directions so that different directions all end up in the units you would like. For example, suppose that the x and y directions were measured in meters, and the z direction was measured in centimeters, and you would like to use meters as your base unit. You could either convert the z dimensions to meters (probably simpler) or use a weighting function , and . In this sense, the system could be considered orthonormal with these units and this weighting arrangement.
This idea is often extended to functions.
Principle author of this page: Rob Frohne