09-240/Classnotes for Thursday September 17 - Revision history

Drorbn at 00:15, 18 September 2009

2009-09-18T00:15:18Z

← Older revision		Revision as of 20:15, 17 September 2009
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	* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".		* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".


			{{09-240/Class Notes Warning}}
	==Class notes for today==		==Class notes for today==

Gdjones at 23:33, 17 September 2009

2009-09-17T23:33:41Z

← Older revision		Revision as of 19:33, 17 September 2009
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	• Convention for today: <math>x,y,a,b,c,d,...</math> will be real numbers; <math>z,w,u,v,...</math> will be complex numbers		• Convention for today: <math>x,y,a,b,c,d,...</math> will be real numbers; <math>z,w,u,v,...</math> will be complex numbers

	• Dream: Find a field <math>\mathbb C</math> that contains <math>\mathbb R</math> and also contains an element <math>i</math> such that <math>i^2=-1</math>		Dream: Find a field <math>\mathbb C</math> that contains <math>\mathbb R</math> and also contains an element <math>i</math> such that <math>i^2=-1</math>

	'''Implications:'''		'''Implications:'''
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	• <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>		• <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>

	• <math>=(a+c)+(bi+di)</math>		:<math>=(a+c)+(bi+di)</math>

	• <math>=(a+c)+(b+d)i</math>		:<math>=(a+c)+(b+d)i</math>

	• <math>=e+fi</math>		:<math>=e+fi</math>

	• <math>(a+bi)(c+di)=(a+c)+(b+d)i</math>		<math>(a+bi)(c+di)=(a+c)+(b+d)i</math>

	• <math>=a(c+di)+bi(c+di)</math>		:<math>=a(c+di)+bi(c+di)</math>

	• <math>=ac+adi+bic+bidi</math>		:<math>=ac+adi+bic+bidi</math>

	• <math>=ac+bdi^2 + adi+bci</math>		:<math>=ac+bdi^2 + adi+bci</math>

	• <math>=(ac-bd)+(ad+bc)i</math>		:<math>=(ac-bd)+(ad+bc)i</math>

	• <math>=e+fi</math>		:<math>=e+fi</math>

	• <math>0_C=0+0i</math>		:<math>0_C=0+0i</math>

	• <math>1_C=1+0i</math>		:<math>1_C=1+0i</math>

	• <math>(a+bi)+(c+di)=0+0i</math>		:<math>(a+bi)+(c+di)=0+0i</math>

	• <math>-(a+bi)=(-a)+(-b)i</math>		:<math>-(a+bi)=(-a)+(-b)i</math>

	• <math>a+bi \neq 0 \Rightarrow (a,b) \neq 0</math>		:<math>a+bi \neq 0 \Rightarrow (a,b) \neq 0</math>

	• Find another element of <math>\mathbb C</math>, <math>x+yi</math> such that <math>(a+bi)(x+yi)=(1+0i)</math>		• Find another element of <math>\mathbb C</math>, <math>x+yi</math> such that <math>(a+bi)(x+yi)=(1+0i)</math>

	• <math>(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1+0i</math>		:<math>(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1+0i</math>

	• <math>ax-by=1</math> (1)		:<math>ax-by=1</math> (1)

	• <math>bx+ay=0</math> (2)		:<math>bx+ay=0</math> (2)

	• <math>a,b</math> are given		:<math>a,b</math> are given

	• <math>x,y</math> unknowns		:<math>x,y</math> unknowns

	• <math>b \times (1)</math> <math>abx-b^2y=b</math>		• <math>b \times (1)</math> <math>abx-b^2y=b</math>
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	• <math>a \times (2)</math> <math>abx+a^2y=0</math>		• <math>a \times (2)</math> <math>abx+a^2y=0</math>

	• <math>\Rightarrow a^{2}y+b^{2}y=-b</math>		:<math>\Rightarrow a^{2}y+b^{2}y=-b</math>

	• <math>y=\frac{-b}{a^{2}+b^{2}}</math>		:<math>y=\frac{-b}{a^{2}+b^{2}}</math>

	• <math>x=\frac{a}{a^{2}+b^{2}}</math>		:<math>x=\frac{a}{a^{2}+b^{2}}</math>

	• (Note: We can divide since we assumed that <math>(a,b) \neq 0</math>		• (Note: We can divide since we assumed that <math>(a,b) \neq 0</math>

	• <math>(a+bi)^{-1}=\frac{a}{a^{2}+b^{2}}+\frac{-b}{a^{2}+b^{2}}i=\frac{a-bi}{a^{2}+b^{2}}=\frac{\overline{a+bi}}{\|a+bi\|^{2}}=\frac{\mbox{conjugate}}{\mbox{norm squared }}</math>		:<math>(a+bi)^{-1}=\frac{a}{a^{2}+b^{2}}+\frac{-b}{a^{2}+b^{2}}i=\frac{a-bi}{a^{2}+b^{2}}=\frac{\overline{a+bi}}{\|a+bi\|^{2}}=\frac{\mbox{conjugate}}{\mbox{norm squared }}</math>

	• Def: Let <math>\mathbb C</math> be the set of all pairs of real numbers <math>{(a,b)}={a+bi}</math>		Def: Let <math>\mathbb C</math> be the set of all pairs of real numbers <math>{(a,b)}={a+bi}</math>

	• with <math>+: (a,b)+(c,d)=(a+c,b+d)</math>		with <math>+: (a,b)+(c,d)=(a+c,b+d)</math>

	• <math>(a+bi)+(c+di)=(a+c)+(b+d)i</math>		:<math>(a+bi)+(c+di)=(a+c)+(b+d)i</math>

	• <math>x:(a+bi)(c+di)=</math>...you know what		<math>\times :(a+bi)(c+di)=</math>...you know what

	• 0 = you know what		• 0 = you know what
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	• 1 = you know what		• 1 = you know what

			'''Theorem:'''
	• Thm:

	~~• 1.~~ <math>\mathbb C</math> is a field		#:<math>\mathbb C</math> is a field

	~~• 2.~~ <math>(0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0)</math>		#:<math>(0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0)</math>

	~~• 3.~~ <math>\mathbb R \rightarrow \mathbb C</math> by <math>a \rightarrow a+0i</math>		#:<math>\mathbb R \rightarrow \mathbb C</math> by <math>a \rightarrow a+0i</math>

	• Proof: <math>F_{1},F_{2},F_{3},...</math>		Proof: <math>F_{1},F_{2},F_{3},...</math>

	• Example: <math>F_{5}</math> (distributivity)		'''Example:''' <math>F_{5}</math> (distributivity)

	• Show that <math>z(u+v)=zu+zv</math>		• Show that <math>z(u+v)=zu+zv</math>

	• Let <math>z=(a+bi)</math>		Let <math>z=(a+bi)</math>

	• <math>u=(c+di)</math>		:<math>u=(c+di)</math>

	• <math>v=(e+fi)</math>		:<math>v=(e+fi)</math>

	• When <math>a,b,c,d,e,f \in \mathbb R</math>		When <math>a,b,c,d,e,f \in \mathbb R</math>

	• <math>(a+bi)[(c+di)+(e+fi)]=(a+bi)(c+di)+(a+bi)(e+fi)=(ac-bd)+\ldots</math>		:<math>(a+bi)[(c+di)+(e+fi)]=(a+bi)(c+di)+(a+bi)(e+fi)=(ac-bd)+\ldots</math>

	• NEXT WEEK: Complex numbers have geometric meaning, geometric interpretation (waves)		• NEXT WEEK: Complex numbers have geometric meaning, geometric interpretation (waves)

Gdjones at 23:18, 17 September 2009

2009-09-17T23:18:42Z

@@ Line 35: / Line 35: @@
 • <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>
-• <math>= (a+c)+(bi+di)</math>
+• <math>=(a+c)+(bi+di)</math>
-• <math>= (a+c)+(b+d)i</math>
+• <math>=(a+c)+(b+d)i</math>
 • <math>=e+fi</math>
@@ Line 43: / Line 43: @@
 • <math>(a+bi)(c+di)=(a+c)+(b+d)i</math>
-• <math>= a(c+di)+bi(c+di)</math>
+• <math>=a(c+di)+bi(c+di)</math>
 • <math>=ac+adi+bic+bidi</math>
@@ Line 59: / Line 59: @@
 • <math>(a+bi)+(c+di)=0+0i</math>

Gdjones at 21:59, 17 September 2009

2009-09-17T21:59:00Z

← Older revision		Revision as of 17:59, 17 September 2009
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	* Dori Eldar's work on "mechanical computations": {{Home Link\|People/Eldar/thesis/linkfunc.htm\|Machines as Calculating Devices}} and {{Home Link\|People/Eldar/thesis/squaring.htm\|Computing the function <math>W=Z^2</math> the hard way}}.		* Dori Eldar's work on "mechanical computations": {{Home Link\|People/Eldar/thesis/linkfunc.htm\|Machines as Calculating Devices}} and {{Home Link\|People/Eldar/thesis/squaring.htm\|Computing the function <math>W=Z^2</math> the hard way}}.
	* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".		* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

			==Class notes for today==

			• Convention for today: <math>x,y,a,b,c,d,...</math> will be real numbers; <math>z,w,u,v,...</math> will be complex numbers

			• Dream: Find a field <math>\mathbb C</math> that contains <math>\mathbb R</math> and also contains an element <math>i</math> such that <math>i^2=-1</math>

			'''Implications:'''

			• <math>b \in \mathbb R \Rightarrow bi \in \mathbb C</math>

			• <math>a \in \mathbb R \Rightarrow a+bi \in \mathbb C</math>

			• <math>c,d \in \mathbb R \Rightarrow c+di \in \mathbb C</math>

			• <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>

			• <math>= (a+c)+(bi+di)</math>

			• <math>= (a+c)+(b+d)i</math>

			• <math>=e+fi</math>

			• <math>(a+bi)(c+di)=(a+c)+(b+d)i</math>

			• <math>= a(c+di)+bi(c+di)</math>

			• <math>=ac+adi+bic+bidi</math>

			• <math>=ac+bdi^2 + adi+bci</math>

			• <math>=(ac-bd)+(ad+bc)i</math>

			• <math>=e+fi</math>

			• <math>0_C=0+0i</math>

			• <math>1_C=1+0i</math>

			• <math>(a+bi)+(c+di)=0+0i</math>

			(rest of notes will be added in 1/2-hour)

Drorbn at 16:01, 17 September 2009

2009-09-17T16:01:13Z

@@ Line 1: / Line 1: @@
 {{09-240/Navigation}}
 * Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
 * The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

Drorbn at 15:57, 17 September 2009

2009-09-17T15:57:07Z

New page

{{09-240/Navigation}}

==Some links==

* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".