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'''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
Contents
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) $