(2 intermediate revisions by the same user not shown)
Line 13: Line 13:
 
'''Supremum and infimum vs. maximum and minimum'''  
 
'''Supremum and infimum vs. maximum and minimum'''  
  
The concept of supremum, or least upper bound, is as follows: Let <math>S={a[n]}</math>, the sequence with terms <math>a[0],a[1],\cdots</math> over all the nonnegative integers. <math>S</math> has a supremum, called <math>\sup S</math> , if for every <math>n , a[n]\leq\sup S</math> (i.e. no a[n] exceeds <math>\sup S</math> ), and furthermore, <math>\sup S</math> is the least value with this property; that is, if <math>a[n]\leq b</math> for all <math>n</math>, then <math>\sup S\leq b</math> for all such <math>b</math> . This is why the supremum is also called the least upper bound, for a bound is a number which a function, sequence, or set, never exceeds. Similarly, one can define the infimum <math>\inf S</math> , or greatest lower bound.
+
The concept of supremum, or least upper bound, is as follows: Let <math class="inline">S={a[n]}</math>, the sequence with terms <math class="inline">a[0],a[1],\cdots</math> over all the nonnegative integers. <math class="inline">S</math> has a supremum, called <math class="inline">\sup S</math> , if for every <math class="inline">n , a[n]\leq\sup S</math> (i.e. no a[n] exceeds <math class="inline">\sup S</math> ), and furthermore, <math class="inline">\sup S</math> is the least value with this property; that is, if <math class="inline">a[n]\leq b</math> for all <math class="inline">n</math>, then <math class="inline">\sup S\leq b</math> for all such <math class="inline">b</math> . This is why the supremum is also called the least upper bound, for a bound is a number which a function, sequence, or set, never exceeds. Similarly, one can define the infimum <math class="inline">\inf S</math> , or greatest lower bound.
  
  
• Consider the set <math>\left\{ x:\;0<x<1\right\}</math>  . There is no maximum or minimum, however <math>0</math>  is the infimum and <math>1</math>  is the supremum.
+
• Consider the set <math class="inline">\left\{ x:\;0<x<1\right\}</math>  . There is no maximum or minimum, however <math class="inline">0</math>  is the infimum and <math class="inline">1</math>  is the supremum.
  
• Consider the set <math>S={a[n]},\; a[n]=1/n</math>  where <math>n</math>  is a positive integer.  
+
• Consider the set <math class="inline">S={a[n]},\; a[n]=1/n</math>  where <math class="inline">n</math>  is a positive integer.  
  
– <math>\sup S=1</math> , since <math>1/n>1/(n+1)</math>  for all such <math>n</math> , and so the largest term is the first. The maximum is also <math>1</math>.
+
– <math class="inline">\sup S=1</math> , since <math class="inline">1/n>1/(n+1)</math>  for all such <math class="inline">n</math> , and so the largest term is the first. The maximum is also <math class="inline">1</math>.
  
– <math>\inf S=0</math> . However, the minimum does not exist.
+
– <math class="inline">\inf S=0</math> . However, the minimum does not exist.
  
 
='''Well-known sets'''=
 
='''Well-known sets'''=
  
• <math>\mathbb{N}</math> : the set of natural numbers. It is countably infinite.
+
• <math class="inline">\mathbb{N}</math> : the set of natural numbers. It is countably infinite.
  
– <math>\mathbb{N}_{0}=\left\{ 0,1,\cdots\right\}</math>   
+
– <math class="inline">\mathbb{N}_{0}=\left\{ 0,1,\cdots\right\}</math>   
  
– <math>\mathbb{N}^{*}=\mathbb{N}_{1}=\left\{ 1,2,\cdots\right\}</math>   
+
– <math class="inline">\mathbb{N}^{*}=\mathbb{N}_{1}=\left\{ 1,2,\cdots\right\}</math>   
  
• <math>\mathbb{Z}_{n}</math> : the set of modulo <math>n</math>
+
• <math class="inline">\mathbb{Z}_{n}</math> : the set of modulo <math class="inline">n</math>
  
 
='''Logarithm'''=
 
='''Logarithm'''=
  
• <math>\log</math> : base 2
+
• <math class="inline">\log</math> : base 2
  
• <math>\ln</math> : base <math>e</math>  
+
• <math class="inline">\ln</math> : base <math class="inline">e</math>  
  
 
='''Partition'''=
 
='''Partition'''=
  
• A set <math>\mathcal{A}=\left\{ A_{1},A_{2},\cdots,A_{k}\right\}</math>  of disjoint subsets of a set <math>A</math>  is a partition of <math>A</math>  if <math>A=\bigcup_{i=1}^{k}A_{i}</math>  and <math>A_{i}\neq\varnothing</math>  for every <math>i</math> .
+
• A set <math class="inline">\mathcal{A}=\left\{ A_{1},A_{2},\cdots,A_{k}\right\}</math>  of disjoint subsets of a set <math class="inline">A</math>  is a partition of <math class="inline">A</math>  if <math class="inline">A=\bigcup_{i=1}^{k}A_{i}</math>  and <math class="inline">A_{i}\neq\varnothing</math>  for every <math class="inline">i</math> .
  
• Another partition <math>\left\{ A'_{1},A'_{2},\cdots,A'_{l}\right\}</math>  of <math>A</math>  refines the partition <math>\mathcal{A}</math>  if <math>A'_{i}</math>  is contained in some <math>A_{j}</math> .
+
• Another partition <math class="inline">\left\{ A'_{1},A'_{2},\cdots,A'_{l}\right\}</math>  of <math class="inline">A</math>  refines the partition <math class="inline">\mathcal{A}</math>  if <math class="inline">A'_{i}</math>  is contained in some <math class="inline">A_{j}</math> .
  
'''<math>k</math> -sets and <math>k</math> -subsets'''
+
'''<math class="inline">k</math> -sets and <math class="inline">k</math> -subsets'''
  
• <math>\left[A\right]^{k}</math>  is the set of all <math>k</math> -element subsets of <math>A</math> .
+
• <math class="inline">\left[A\right]^{k}</math>  is the set of all <math class="inline">k</math> -element subsets of <math class="inline">A</math> .
  
• Sets with <math>k</math>  elements will be called <math>k</math> -sets.
+
• Sets with <math class="inline">k</math>  elements will be called <math class="inline">k</math> -sets.
  
• Similarly, subsets with <math>k</math>  elements are <math>k</math> -subsets.
+
• Similarly, subsets with <math class="inline">k</math>  elements are <math class="inline">k</math> -subsets.
  
 
='''1.1.2 Taylor series'''=
 
='''1.1.2 Taylor series'''=
  
<math>e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}</math>.  
+
<math class="inline">e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}</math>.  
  
<math>\frac{x}{\left(1-x\right)^{2}}=\sum_{n=1}^{\infty}nx^{n}=\sum_{n=0}^{\infty}nx^{n}</math> for <math>\left|x\right|<1</math>.  
+
<math class="inline">\frac{x}{\left(1-x\right)^{2}}=\sum_{n=1}^{\infty}nx^{n}=\sum_{n=0}^{\infty}nx^{n}</math> for <math class="inline">\left|x\right|<1</math>.  
  
<math>\log\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}</math> for <math>-1\leq x<1</math>.  
+
<math class="inline">\log\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}</math> for <math class="inline">-1\leq x<1</math>.  
  
 
='''1.1.3 Differentiation and Integration'''=
 
='''1.1.3 Differentiation and Integration'''=
  
<math>\left(\frac{b}{a}\right)^{\prime}=\frac{b^{\prime}a-ba^{\prime}}{a^{2}}</math>
+
<math class="inline">\left(\frac{b}{a}\right)^{\prime}=\frac{b^{\prime}a-ba^{\prime}}{a^{2}}</math>
  
<math>\left(f\left(g\left(x\right)\right)\right)^{\prime}=f^{\prime}\left(g\left(x\right)\right)\cdot g^{\prime}\left(x\right)</math>  
+
<math class="inline">\left(f\left(g\left(x\right)\right)\right)^{\prime}=f^{\prime}\left(g\left(x\right)\right)\cdot g^{\prime}\left(x\right)</math>  
  
<math>\int uv^{\prime}=uv-\int u^{\prime}v</math>
+
<math class="inline">\int uv^{\prime}=uv-\int u^{\prime}v</math>
  
<math>\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\longleftrightarrow\left(x^{n}\right)^{\prime}=n\cdot x^{n-1}</math>  
+
<math class="inline">\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\longleftrightarrow\left(x^{n}\right)^{\prime}=n\cdot x^{n-1}</math>  
  
<math>\int e^{x}dx=e^{x}+C\longleftrightarrow\left(e^{x}\right)^{\prime}=e^{x}</math>  
+
<math class="inline">\int e^{x}dx=e^{x}+C\longleftrightarrow\left(e^{x}\right)^{\prime}=e^{x}</math>  
  
<math>\int a^{x}dx=\frac{a^{x}}{\ln a}+C\longleftrightarrow\left(a^{x}\right)^{\prime}=a^{x}\cdot\ln a</math>  
+
<math class="inline">\int a^{x}dx=\frac{a^{x}}{\ln a}+C\longleftrightarrow\left(a^{x}\right)^{\prime}=a^{x}\cdot\ln a</math>  
  
<math>\int e^{kx}dx=\frac{e^{kx}}{k}+C\longleftrightarrow\left(e^{kx}\right)^{\prime}=k\cdot e^{kx}</math>  
+
<math class="inline">\int e^{kx}dx=\frac{e^{kx}}{k}+C\longleftrightarrow\left(e^{kx}\right)^{\prime}=k\cdot e^{kx}</math>  
  
<math>\int\cos xdx=\sin x+C\longleftrightarrow\left(\sin x\right)^{\prime}=\cos x</math>
+
<math class="inline">\int\cos xdx=\sin x+C\longleftrightarrow\left(\sin x\right)^{\prime}=\cos x</math>
  
<math>\int\sin xdx=-\cos x+C\longleftrightarrow\left(\cos x\right)^{\prime}=-\sin x</math>  
+
<math class="inline">\int\sin xdx=-\cos x+C\longleftrightarrow\left(\cos x\right)^{\prime}=-\sin x</math>  
  
<math>\int\tan xdx=-\ln\left|\left(\cos x\right)^{-1}\right|+C=\ln\left|\sec x\right|+C\longleftrightarrow\left(\tan x\right)'=\sec^{2}x</math>  
+
<math class="inline">\int\tan xdx=-\ln\left|\left(\cos x\right)^{-1}\right|+C=\ln\left|\sec x\right|+C\longleftrightarrow\left(\tan x\right)'=\sec^{2}x</math>  
  
<math>\int\frac{\sin x}{x}dx=?</math>  
+
<math class="inline">\int\frac{\sin x}{x}dx=?</math>  
  
 
='''Integration using substitution'''=
 
='''Integration using substitution'''=
  
1. The problem is <math>\int\frac{2t}{1+t^{2}}dt</math>.  
+
1. The problem is <math class="inline">\int\frac{2t}{1+t^{2}}dt</math>.  
  
2. We substitute <math>1+t^{2}</math>  as r . We can get <math>1tdt=dr</math>  because <math>1+t^{2}=r</math> .
+
2. We substitute <math class="inline">1+t^{2}</math>  as r . We can get <math class="inline">1tdt=dr</math>  because <math class="inline">1+t^{2}=r</math> .
  
3. Now, we can get the solution using sutitution as <math>\int\frac{1}{r}dr=\ln r=\ln\left(1+t^{2}\right)</math>.
+
3. Now, we can get the solution using sutitution as <math class="inline">\int\frac{1}{r}dr=\ln r=\ln\left(1+t^{2}\right)</math>.
  
 
='''1.1.4 Relations between cosine, sine and exponential functions'''=
 
='''1.1.4 Relations between cosine, sine and exponential functions'''=
  
<math>e^{\pm i\theta}=\cos\theta\pm i\sin\theta</math>{ (Euler's formula)}.  
+
<math class="inline">e^{\pm i\theta}=\cos\theta\pm i\sin\theta</math>{ (Euler's formula)}.  
  
<math>\sin\theta=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)=\frac{1}{2i}\left(\left(\cos\theta+i\sin\theta\right)-\left(\cos\theta-i\sin\theta\right)\right)=\frac{2i\sin\theta}{2i}=\sin\theta</math>.  
+
<math class="inline">\sin\theta=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)=\frac{1}{2i}\left(\left(\cos\theta+i\sin\theta\right)-\left(\cos\theta-i\sin\theta\right)\right)=\frac{2i\sin\theta}{2i}=\sin\theta</math>.  
  
<math>\cos\theta=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)=\frac{1}{2}\left(\left(\cos\theta+\sin\theta\right)+\left(\cos\theta-\sin\theta\right)\right)=\frac{2\cos\theta}{2}=\cos\theta</math>.  
+
<math class="inline">\cos\theta=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)=\frac{1}{2}\left(\left(\cos\theta+\sin\theta\right)+\left(\cos\theta-\sin\theta\right)\right)=\frac{2\cos\theta}{2}=\cos\theta</math>.  
  
<math>\cos\left(2x\right)=\cos^{2}\left(x\right)-\sin^{2}\left(x\right)=2\cos^{2}\left(x\right)-1=1-2\sin^{2}\left(x\right)</math>.  
+
<math class="inline">\cos\left(2x\right)=\cos^{2}\left(x\right)-\sin^{2}\left(x\right)=2\cos^{2}\left(x\right)-1=1-2\sin^{2}\left(x\right)</math>.  
  
<math>\cos^{2}\left(x\right)+\sin^{2}\left(x\right)=1</math>.  
+
<math class="inline">\cos^{2}\left(x\right)+\sin^{2}\left(x\right)=1</math>.  
  
<math>\sec x=\frac{1}{\cos x}</math>.  
+
<math class="inline">\sec x=\frac{1}{\cos x}</math>.  
  
<math>\csc x=\frac{1}{\sin x}</math>.  
+
<math class="inline">\csc x=\frac{1}{\sin x}</math>.  
  
<math>\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}</math>.  
+
<math class="inline">\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}</math>.  
  
 
='''1.1.5 Set operation'''=
 
='''1.1.5 Set operation'''=
  
• <math>A-B=A\cap\bar{B}</math>  
+
• <math class="inline">A-B=A\cap\bar{B}</math>  
  
• <math>A\triangle B=A\cup B-A\cap B=\left(A\cup B\right)\cap\overline{\left(A\cap B\right)}</math>  
+
• <math class="inline">A\triangle B=A\cup B-A\cap B=\left(A\cup B\right)\cap\overline{\left(A\cap B\right)}</math>  
  
• <math>A=B\Longleftrightarrow A\subset B\textrm{ and }A\supset B</math>  
+
• <math class="inline">A=B\Longleftrightarrow A\subset B\textrm{ and }A\supset B</math>  
  
• <math>A=A\cap S=A\cap\left(B\cup\bar{B}\right)=\left(A\cap B\right)\cup\left(A\cap\bar{B}\right)</math>
+
• <math class="inline">A=A\cap S=A\cap\left(B\cup\bar{B}\right)=\left(A\cap B\right)\cup\left(A\cap\bar{B}\right)</math>
 
----
 
----
 
[[ECE600|Back to ECE600]]
 
[[ECE600|Back to ECE600]]
  
 
[[ECE 600 Prerequisites|Back to ECE 600 Prerequisites]]
 
[[ECE 600 Prerequisites|Back to ECE 600 Prerequisites]]

Latest revision as of 11:20, 30 November 2010


1.1 Basic Mathematics

From the ECE600 Pre-requisites notes of Sangchun Han, ECE PhD student.


1.1.1 Mathematical notation

 : approximately equal

~ : CST ·

Supremum and infimum vs. maximum and minimum

The concept of supremum, or least upper bound, is as follows: Let $ S={a[n]} $, the sequence with terms $ a[0],a[1],\cdots $ over all the nonnegative integers. $ S $ has a supremum, called $ \sup S $ , if for every $ n , a[n]\leq\sup S $ (i.e. no a[n] exceeds $ \sup S $ ), and furthermore, $ \sup S $ is the least value with this property; that is, if $ a[n]\leq b $ for all $ n $, then $ \sup S\leq b $ for all such $ b $ . This is why the supremum is also called the least upper bound, for a bound is a number which a function, sequence, or set, never exceeds. Similarly, one can define the infimum $ \inf S $ , or greatest lower bound.


• Consider the set $ \left\{ x:\;0<x<1\right\} $ . There is no maximum or minimum, however $ 0 $ is the infimum and $ 1 $ is the supremum.

• Consider the set $ S={a[n]},\; a[n]=1/n $ where $ n $ is a positive integer.

$ \sup S=1 $ , since $ 1/n>1/(n+1) $ for all such $ n $ , and so the largest term is the first. The maximum is also $ 1 $.

$ \inf S=0 $ . However, the minimum does not exist.

Well-known sets

$ \mathbb{N} $ : the set of natural numbers. It is countably infinite.

$ \mathbb{N}_{0}=\left\{ 0,1,\cdots\right\} $

$ \mathbb{N}^{*}=\mathbb{N}_{1}=\left\{ 1,2,\cdots\right\} $

$ \mathbb{Z}_{n} $ : the set of modulo $ n $

Logarithm

$ \log $ : base 2

$ \ln $ : base $ e $

Partition

• A set $ \mathcal{A}=\left\{ A_{1},A_{2},\cdots,A_{k}\right\} $ of disjoint subsets of a set $ A $ is a partition of $ A $ if $ A=\bigcup_{i=1}^{k}A_{i} $ and $ A_{i}\neq\varnothing $ for every $ i $ .

• Another partition $ \left\{ A'_{1},A'_{2},\cdots,A'_{l}\right\} $ of $ A $ refines the partition $ \mathcal{A} $ if $ A'_{i} $ is contained in some $ A_{j} $ .

$ k $ -sets and $ k $ -subsets

$ \left[A\right]^{k} $ is the set of all $ k $ -element subsets of $ A $ .

• Sets with $ k $ elements will be called $ k $ -sets.

• Similarly, subsets with $ k $ elements are $ k $ -subsets.

1.1.2 Taylor series

$ e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!} $.

$ \frac{x}{\left(1-x\right)^{2}}=\sum_{n=1}^{\infty}nx^{n}=\sum_{n=0}^{\infty}nx^{n} $ for $ \left|x\right|<1 $.

$ \log\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n} $ for $ -1\leq x<1 $.

1.1.3 Differentiation and Integration

$ \left(\frac{b}{a}\right)^{\prime}=\frac{b^{\prime}a-ba^{\prime}}{a^{2}} $

$ \left(f\left(g\left(x\right)\right)\right)^{\prime}=f^{\prime}\left(g\left(x\right)\right)\cdot g^{\prime}\left(x\right) $

$ \int uv^{\prime}=uv-\int u^{\prime}v $

$ \int x^{n}dx=\frac{x^{n+1}}{n+1}+C\longleftrightarrow\left(x^{n}\right)^{\prime}=n\cdot x^{n-1} $

$ \int e^{x}dx=e^{x}+C\longleftrightarrow\left(e^{x}\right)^{\prime}=e^{x} $

$ \int a^{x}dx=\frac{a^{x}}{\ln a}+C\longleftrightarrow\left(a^{x}\right)^{\prime}=a^{x}\cdot\ln a $

$ \int e^{kx}dx=\frac{e^{kx}}{k}+C\longleftrightarrow\left(e^{kx}\right)^{\prime}=k\cdot e^{kx} $

$ \int\cos xdx=\sin x+C\longleftrightarrow\left(\sin x\right)^{\prime}=\cos x $

$ \int\sin xdx=-\cos x+C\longleftrightarrow\left(\cos x\right)^{\prime}=-\sin x $

$ \int\tan xdx=-\ln\left|\left(\cos x\right)^{-1}\right|+C=\ln\left|\sec x\right|+C\longleftrightarrow\left(\tan x\right)'=\sec^{2}x $

$ \int\frac{\sin x}{x}dx=? $

Integration using substitution

1. The problem is $ \int\frac{2t}{1+t^{2}}dt $.

2. We substitute $ 1+t^{2} $ as r . We can get $ 1tdt=dr $ because $ 1+t^{2}=r $ .

3. Now, we can get the solution using sutitution as $ \int\frac{1}{r}dr=\ln r=\ln\left(1+t^{2}\right) $.

1.1.4 Relations between cosine, sine and exponential functions

$ e^{\pm i\theta}=\cos\theta\pm i\sin\theta ${ (Euler's formula)}.

$ \sin\theta=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)=\frac{1}{2i}\left(\left(\cos\theta+i\sin\theta\right)-\left(\cos\theta-i\sin\theta\right)\right)=\frac{2i\sin\theta}{2i}=\sin\theta $.

$ \cos\theta=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)=\frac{1}{2}\left(\left(\cos\theta+\sin\theta\right)+\left(\cos\theta-\sin\theta\right)\right)=\frac{2\cos\theta}{2}=\cos\theta $.

$ \cos\left(2x\right)=\cos^{2}\left(x\right)-\sin^{2}\left(x\right)=2\cos^{2}\left(x\right)-1=1-2\sin^{2}\left(x\right) $.

$ \cos^{2}\left(x\right)+\sin^{2}\left(x\right)=1 $.

$ \sec x=\frac{1}{\cos x} $.

$ \csc x=\frac{1}{\sin x} $.

$ \cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x} $.

1.1.5 Set operation

$ A-B=A\cap\bar{B} $

$ A\triangle B=A\cup B-A\cap B=\left(A\cup B\right)\cap\overline{\left(A\cap B\right)} $

$ A=B\Longleftrightarrow A\subset B\textrm{ and }A\supset B $

$ A=A\cap S=A\cap\left(B\cup\bar{B}\right)=\left(A\cap B\right)\cup\left(A\cap\bar{B}\right) $


Back to ECE600

Back to ECE 600 Prerequisites

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett