10/01 - Vectors & Functions - Revision history

Fonggr: 10/1 - Vectors & Functions moved to 10/01 - Vectors & Functions

2008-11-12T02:15:15Z

10/1 - Vectors & Functions moved to 10/01 - Vectors & Functions

← Older revision	Revision as of 19:15, 11 November 2008
(No difference)

Fonggr: 1/10/08 - Vectors & Functions moved to 10/1 - Vectors & Functions

2008-11-12T02:14:39Z

1/10/08 - Vectors & Functions moved to 10/1 - Vectors & Functions

← Older revision	Revision as of 19:14, 11 November 2008
(No difference)

Fonggr: /* Questions */

2008-11-10T23:38:43Z

Questions

← Older revision		Revision as of 16:38, 10 November 2008
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	**<math>\vec A \cdot \vec B = \left \| \vec A \right \| \left \| \vec B \right \| \cos \theta </math>		**<math>\vec A \cdot \vec B = \left \| \vec A \right \| \left \| \vec B \right \| \cos \theta </math>
	*Then is the picture assuming something is a unit vector?		*Then is the picture assuming something is a unit vector?
			**You will have to choose whether you're interested in projecting A onto B or B onto A. The lengths will be the same, but the direction will be different.
	*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?		*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
	**Completely arbitrary, the end result is the same either way		**Completely arbitrary, the end result is the same either way
	*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector		*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector

Fonggr: /* Questions */

2008-11-10T23:15:29Z

Questions

← Older revision		Revision as of 16:15, 10 November 2008
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	**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this		**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
	**<math> \hat b_j </math> is a unit vector for one direction in our new basis set		**<math> \hat b_j </math> is a unit vector for one direction in our new basis set
	*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?		*When you do the dot product of say <math>\vec A \cdot \vec B</math>, is it always the projection of <math>\vec A</math> onto <math>\vec B</math> and not the opposite way around?
			**<math>\vec A \cdot \vec B = \left \| \vec A \right \| \left \| \vec B \right \| \cos \theta </math>
			*Then is the picture assuming something is a unit vector?
	*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?		*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
	**Completely arbitrary, the end result is the same either way		**Completely arbitrary, the end result is the same either way
	*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector		*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector

Fonggr: /* Questions */

2008-11-10T23:11:40Z

Questions

← Older revision		Revision as of 16:11, 10 November 2008
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	*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?		*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?
	**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this		**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
	**<math> \hat b_j </math> is		**<math> \hat b_j </math> is a unit vector for one direction in our new basis set
	*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?		*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
	*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?		*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
	**Completely arbitrary, the end result is the same either way		**Completely arbitrary, the end result is the same either way
	*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector		*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector

Fonggr: /* Questions */

2008-11-10T23:10:44Z

Questions

← Older revision		Revision as of 16:10, 10 November 2008
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	==Questions==		==Questions==
	*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?		*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?
	**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set		**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
			**<math> \hat b_j </math> is
	*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?		*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
	*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?		*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
	**Completely arbitrary, the end result is the same either way		**Completely arbitrary, the end result is the same either way
	*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector		*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector

Fonggr: /* Questions */

2008-11-10T23:10:01Z

Questions

← Older revision		Revision as of 16:10, 10 November 2008
Line 32:		Line 32:
	==Questions==		==Questions==
	*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?		*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?
			**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set
	*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?		*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
	*Why did you decide to make it k_m instead of k_j?		*Why did you decide to make it <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
			**Completely arbitrary, the end result is the same either way
			*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector

Fonggr: /* Questions */

2008-11-10T22:51:03Z

Questions

← Older revision		Revision as of 15:51, 10 November 2008
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	==Questions==		==Questions==
	*How did you get the last two lines of the last page?
	*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?		*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?
	*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?		*When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
	*Why did you decide to make it k_m instead of k_j?		*Why did you decide to make it k_m instead of k_j?

Fonggr: /* Defining k_m \,\! */

2008-11-10T22:50:15Z

Defining k_m \,\!

← Older revision		Revision as of 15:50, 10 November 2008
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	==Defining <math> k_m \,\! </math>==		==Defining <math> k_m \,\! </math>==
	Taking <math> k_m \,\! </math> from the previous section:		Taking <math> k_m \,\! </math> from the previous section:
	:<math> \hat b_j \cdot \hat b_m = k_m \delta mj \Longrightarrow \hat b_m \cdot \hat b_m = k_m \Longrightarrow \left \| b_m \right \|^2 = k_m </math>		:<math> \hat b_j \cdot \hat b_m = k_m \delta mj \Longrightarrow \hat b_m \cdot \hat b_m = k_m \Longrightarrow \left \| \hat b_m \right \|^2 = k_m </math>

			Thus <math> k_m \,\! </math> is the length of <math>\hat b_m</math> squared

	==Questions==		==Questions==

Fonggr: /* Defining k_m \,\! */

2008-11-10T22:49:09Z

Defining k_m \,\!

← Older revision		Revision as of 15:49, 10 November 2008
Line 26:		Line 26:
	==Defining <math> k_m \,\! </math>==		==Defining <math> k_m \,\! </math>==
	Taking <math> k_m \,\! </math> from the previous section:		Taking <math> k_m \,\! </math> from the previous section:
	:<math> \hat b_j \cdot \hat b_m = k_m \delta mj \Longrightarrow \hat b_m \cdot \hat b_m = k_m ~~\delta mj~~ \Longrightarrow \left \| b_m \right \|^2 = k_m </math>		:<math> \hat b_j \cdot \hat b_m = k_m \delta mj \Longrightarrow \hat b_m \cdot \hat b_m = k_m \Longrightarrow \left \| b_m \right \|^2 = k_m </math>

	==Questions==		==Questions==