Vector weighting functions - Revision history

Frohro at 00:36, 27 September 2004

2004-09-27T00:36:09Z

← Older revision		Revision as of 17:36, 26 September 2004
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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.		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.

	[[This idea is often extended to functions.]]		[[Orthogonal Functions\|This idea is often extended to functions.]]

	[[Orthogonal ~~Functions~~]]		[[Orthogonal functions]]

	Principle author of this page: [[User:Frohro\|Rob Frohne]]		Principle author of this page: [[User:Frohro\|Rob Frohne]]

Frohro at 00:32, 27 September 2004

2004-09-27T00:32:38Z

← Older revision		Revision as of 17:32, 26 September 2004
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	===Orthogonal but not ~~Orthonormal Basis Sets~~===		===Orthogonal but not orthonormal basis sets===

	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

Frohro at 21:24, 26 September 2004

2004-09-26T21:24:14Z

← Older revision		Revision as of 14:24, 26 September 2004
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	[[Orthogonal Functions]]		[[Orthogonal Functions]]

			Principle author of this page: [[User:Frohro\|Rob Frohne]]

172.17.69.1: /* Orthogonal but not Orthonormal Basis Sets */

2004-09-24T18:52:22Z

Orthogonal but not Orthonormal Basis Sets

← Older revision		Revision as of 11:52, 24 September 2004
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	[[This idea is often extended to functions.]]		[[This idea is often extended to functions.]]

			[[Orthogonal Functions]]

172.17.69.1: /* Orthogonal but not Orthonormal Basis Sets */

2004-09-24T18:49:58Z

Orthogonal but not Orthonormal Basis Sets

← Older revision		Revision as of 11:49, 24 September 2004
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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>.		<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>.

	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>.		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.

			[[This idea is often extended to functions.]]

172.17.69.1: /* Orthogonal but not Orthonormal Basis Sets */

2004-09-24T18:43:16Z

Orthogonal but not Orthonormal Basis Sets

← Older revision		Revision as of 11:43, 24 September 2004
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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>.		<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>.

			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>.

172.17.69.1 at 18:34, 24 September 2004

2004-09-24T18:34:35Z

← Older revision		Revision as of 11:34, 24 September 2004
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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>.

172.17.69.1 at 18:30, 24 September 2004

2004-09-24T18:30:26Z

New page

===Orthogonal but not Orthonormal Basis Sets===

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

<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