Cycle Groups and Boundary Groups

Leave a reply

Let us use $\langle\cdots\rangle$ for an unoriented simplex and $(\cdots)$ for an oriented simplex.

Examples. 1. $(p_0p_1)=-(p_1p_0)$.


2. \begin{eqnarray*}\sigma_2&=&(p_0p_1p_2)=(p_2p_0p_1)=(p_1p_2p_0)\\-(p_0p_2p_1)&=&-(p_2p_1p_0)=-(p_1p_0p_2).\end{eqnarray*}


Let $K=\{\sigma_\alpha\}$ be an $n$-dimensional simplicial complex of oriented simplexes.

Definition. The $r$-chain group $C_r(K)$ of a simplicial complex $K$ is a free abelian group generated by $r$-simplexes of $K$. If $r>\dim K$, $C_r(K):=0$. An element of $C_r(K)$ is called an $r$-chain.

Let there be $N_r$ $r$-simplexes in $K$. Denote them by $\sigma_{r,i}$ ($1\leq i\leq N$). Then $c\in C_r(K)$ is expressed as $$c=\sum_{i=1}^{N_r}c_i\sigma_{r,i},\ c_i\in\mathbb Z.$$ The integers $c_i$ are called the coefficients of $c$. The addition of two $r$-chains $\sum_ic_i\sigma_{r,i}$ and $c’=\sum_ic_i’\sigma_{r,i}$ is $$c+c’=\sum_i(c_i+c_i’)\sigma_{r,i}.$$ The unit element is $0=\sum_i0\cdot\sigma_{r,i}$. The inverse element of $c$ is $-c=\sum_i(-c_i)\sigma_{r,i}$. Hence we see that $C_r(K)$ is a free abelian group of rank $N_r$ $$C_r(K)\cong\stackrel{N_r}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}.$$

Denote the boundary of an $r$-simplex $\sigma_r$ by $\partial_r\sigma_r$. Since a 0-simplex has no boundary, $$\partial_0p_0=0.$$ For a 1-simplex $(p_0p_1)$, $$\partial_1(p_0p_1):=p_1-p_0.$$ Let $\sigma_r=(p_0\cdots p_r)$ ($r>0$) be an oriented $r$-simplex. The boundary $\partial_r\sigma_r$ of $\sigma_r$ is an $(r-1)$-chain defined by $$\partial_r\sigma_r:=\sum_{i=0}^r(-1)^i(p_0p_1\cdots\hat{p}_i\cdots p_r)$$ where the point $p_i$ under $\hat{}$ is omitted. For example, \begin{eqnarray*}\partial_2(p_0p_1p_2)&=&(p_1p_2)-(p_0p_2)+(p_0p_1),\\\partial_3(p_0p_1p_2p_3)&=&(p_1p_2p_3)-(p_0p_2p_3)+(p_0p_1p_3)-(p_0p_1p_2).\end{eqnarray*} The boundary $\sigma_r$ defines a homomorphism called the boundary operator $$\partial_r: C_r(K)\longrightarrow C_{r-1}(K);\ c=\sum_i c_i\sigma_{r,i}\longmapsto\partial_rc=\sum_ic_i\partial_r\sigma_{r,i}.$$

Let $K$ be an $n$-dimensional simplicial complex. Then there exists a sequence of free abelian groups and homomorphisms $$0\stackrel{i}{\hookrightarrow}C_n(K)\stackrel{\partial_n}{\longrightarrow}C_{n-1}(K)\stackrel{\partial_{n-1}}{\longrightarrow}\cdots\stackrel{\partial_2}{\longrightarrow}C_1(K)\stackrel{\partial_1}{\longrightarrow}C_0(K)\stackrel{\partial_0}{\longrightarrow}0.$$ This sequence is called the chain complex associated with $K$ and is denoted by $C(K)$.

Definition. $Z_r(K):=\ker\partial_r\subset C_r(K)$ is called the $r$-cycle group. The elements of $Z_r(K)$ are called $r$-cycles. If $c\in Z_r(K)$, i.e. if $c$ is an $r$-cycle, $\partial_rc=0$. If $r=0$, $\partial_rc=0$ for all $c\in C_0(K)$, so $C_0(K)=Z_0(K)$.

Definition. Let us consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)$ and let $c\in C_r(K)$. If there exists $d\in C_{r+1}(K)$ such that $c=\partial_{r+1}d$, then $c$ is called an $r$-boundary. The set of $r$-boundaries $B_r(K)$ ($=\partial_{r+1}C_{r+1}(K)={\rm Im}\partial_{r+1}$) is a subgroup of $C_r(K)$ called the $r$-boundary group. If $K$ is an $n$-dimensional simplicial complex, $B_n(K)=0$.

Consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)\stackrel{\partial_r}{\longrightarrow}C_{r-1}(K)$. Then the following lemma holds.

Lemma. The composite map $\partial_r\partial_{r+1}:C_{r+1}(K)\longrightarrow C_{r-1}(K)$ is a zero map.

Proof. Since $\partial_r$ is a linear operator on $C_r(K)$, it suffices to prove the identity $\partial_r\partial_{r+1}=0$ for the generators of $C_{r+1}(K)$. If $r=0$, $\partial_0\partial_1=0$ since $\partial_0$ is a zero operator. Let us assume that $r>0$. Take $\sigma=(p_0\cdots p_rp_{r+1})\in C_{r+1}(K)$. \begin{eqnarray*}\partial_r\partial_{r+1}\sigma&=&\partial_r\sum_{i=0}^{r+1}(-1)^i(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\partial_r(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\{\sum_{j=0}^{i-1}(-1)^j(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{j=i+1}^{r+1}(-1)^{j-1}(p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\}\\&=&\sum_{i=0}^{r+1}\sum_{j=0}^{i-1}(-1)^{i+j}(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{i=0}^{r+1}\sum_{j=i+1}^{r+1}(-1)^{i+j-1}p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\\&=&0.
\end{eqnarray*}

Theorem. $B_r(K)\subset Z_r(K)$ or equivalently $\mathrm{Im}\partial_{r+1}\subset\ker\partial_r$.

Proof. Let $c\in B_r(K)$ Then $c=\partial_{r+1}d$ for some $d\in C_{r+1}(K)$. By Lemma, $\partial_rc=\partial_r\partial_{r+1}d=0$. Hence, $c\in\ker\partial_r=Z_r(K)$.

A Physical Meaning of Derivative: Velocity and Acceleration

Leave a reply

Let us assume that a particle is moving along a straight line and that the function $s=f(t)$ describes the position of moving particle at the time $t$. In physics, such a function $s=f(t)$ is called a motion.

Suppose the particle passes the points $P$ and $Q$ at the times $t$ and $t+\Delta t$, respectively. If $s$ and $s+\Delta s$ are the respective distances from some fixed point $O$, then the average velocity of the particle during the time interval $\Delta t$ is $$\frac{\Delta s}{\Delta t}=\frac{f(t+\Delta t)-f(t)}{\Delta t}=\frac{\mbox{Distance Traveled}}{\mbox{Time Elapsed}}.$$ The instantaneous velocity $v$ of the particle at the time $t$ is then given by the derivative of motion $s=f(t)$ $$v=\frac{ds}{dt}=\lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}.$$ In physics, the intantaneous velocity is also denoted by $\dot{s}$ or $\dot{f}(t)$. This dot notation was introduced by Sir Issac Newton.

Similaryl, if $\Delta v$ is the change in the velocity of the particle as it moves from $P$ to $Q$ during the time interval $\Delta t$, then $$a=\frac{dv}{dt}=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t}$$ is the acceleration of the particle at the time $t$. Using dot notation, the acceleration is also denoted by $\dot{v}$, $\ddot{s}$, $\ddot{f}(t)$, or $\frac{d^2 s}{dt^2}$. The last notation $\frac{d^2 s}{dt^2}$ is due to Gottfried Leibniz.

If a body is thrown vertically upward with a certain initial velocity $v_0$, its distance $s$ from the starting point is given by the formula $$s(t)=v_0t-\frac{1}{2}gt^2,$$ where $g$ is the gravitational constant $g=9.8\mbox{m}/\mbox{sec}^2=32\mbox{ft}/\mbox{sec}^2.$

Example. From the top of a building 96 feet high, a ball is thrown directly upward with a velocity of 80 feet per second. Find (a) the time required to reach the highest point, (b) the maximum height attained, and (c) the velocity of the ball when it reaches the ground.

Solution. $v_0=80$ ft/sec and $g=32\mbox{ft}/\mbox{sec}^2$, so $$s=80t-16t^2$$ and $$v=\frac{ds}{dt}=80-32t.$$

(a) At the heighest point, $v=0$ that is $0=80-32t$. So, $t=\frac{5}{2}$.

(b) $s\left(\frac{5}{2}\right)=80\left(\frac{5}{2}\right)-16\left(\frac{5}{2}\right)^2=100$ft. Hence the height of the ball above the ground is 196 feet.

(c) Since the ball will reach the ground when $s=-96$, it follows that $-96=80t-16t^2$ or $16(t-6)(t+1)=0$. Hence $t=6$ and the velocity is $v(6)=80-32\cdot 6=-112$ft/sec when the ball strikes the ground. The negative sign merely indicates that the velocity of the ball is directed downward.

Simplexes and Simplicial Complexes

Leave a reply

Definition. A 0-simplex $\langle p_0\rangle$ is a point or a vertex. A 1-simplex $\langle p_0p_1\rangle$ is a line or an edge. A 2-simplex $\langle p_0p_1p_2\rangle$ is a triangle with its interior included. A 3-simplex $\langle p_0p_1p_2p_3\rangle$ is a solid tetrahedron.

A 0-simplex $\langle p_0\rangle$ may be simply written as $p_0$.

Note that in order for an $r$-simplex to represent an $r$-dimensional object, the vertices $p_i$ must be geometrically independent, i.e. no $(r-1)$-dimensional hyperplane contains all the $r+1$ points. Let $p_0,\cdots,p_r$ be points geometrically independent in $\mathbb R^m$ ($m\geq r$). The $r$-simplex $$\sigma_r=\{x\in\mathbb R^m: x=\sum_{i=0}^r c_ip_i,\ c_i\geq 0,\ \sum_{i=0}^r c_i=1\}$$ has the points $p_0,\cdots,p_r$ as its vertices. The ordered $r+1$-tuple $(c_0,c_1,\cdots,c_r)$ is called the barycentric coordinate of $x$. The 3-simplex $\langle p_0p_1p_2p_3\rangle$ four 0-faces (vertices) $p_0,p_1,p_2,p_3$; six 1-faces (edges) $\langle p_0p_1\rangle$, $\langle p_0p_2\rangle$, $\langle p_0p_3\rangle$, $\langle p_1p_2\rangle$, $\langle p_1p_3\rangle$, $\langle p_2p_3\rangle$; four 2-faces (faces) $\langle p_0p_1p_2\rangle$, $\langle p_0p_2p_3\rangle$, $\langle p_0p_1p_3\rangle$, $\langle p_1p_2p_3\rangle$.

Let $K$ be a set of finite number of simplexes in $\mathbb R^m$. If these simplexes are nicely fitted together, $K$ is called a simplicial complex. By nicely fitted together we mean that:

  1. An arbitrary face of a simplex of $K$ belongs to $K$.
  2. If $\sigma$ and $\sigma’$ are two simplexes of $K$, $\sigma\cap\sigma’$ is either empty or a face of $\sigma$ and $\sigma’$.

The dimension of a simplicial complex is defined to be the maximum dimension of simplexes in $K$.

Let $\sigma_r$ be an $r$-simplex and $K$ be the set of faces of $\sigma_r$. Then $K$ is an $r$-dimensional simplicial complex.For example, take $\sigma_3=\langle p_0p_1p_2,p_3\rangle$. Then $$\begin{array}{c}K=\{p_0,p_1,p_2,p_3,\langle p_0p_1\rangle,\langle p_0p_2\rangle,\langle p_0p_3\rangle,\langle p_1p_2\rangle,\langle p_1p_3\rangle,\langle p_2p_3\rangle,\\\langle p_0p_1p_2\rangle,\langle p_0p_1p_3\rangle,\langle p_0p_2p_3\rangle,\langle p_1p_2p_3\rangle,\langle p_0p_1p_2p_3\rangle\}.\end{array}$$

Definition. Let $K$ be a simplicial complex of simplexes in $\mathbb R^m$. The union of all the simplexes of $K$ is a subset of $\mathbb R^m$ called the polyhedron $|K|$ of a simplicial complex $K$. Note that $\dim |K|=\dim K$.

Let $X$ be a topological space. If there is a simplicial complex $K$ and a homeomorphism $f:|K|\longrightarrow X$, $X$ is said to be triangulable and the pair $(K,f)$ is called a triangulation of $X$.

Example. The following picture shows a triangulation of $S^1\times [0,1]$.

Example. The following example is not a triangulation of $S^1\times [0,1]$.

Let $\sigma_2=\langle p_0p_1p_2\rangle$ and $\sigma_2’=\langle p_2p_3p_0\rangle$. Then $\sigma_2\cap\sigma_2’=\langle p_0\rangle\cup\langle p_2\rangle$. This is neither $\emptyset$ nor a simplex.

Free Abelian Groups

Leave a reply

Before we discuss homology groups, we review some basics of abelian group theory.

The group operation for an abelian group is denoted by $+$. The unit element is denoted by $0$.

Let $G_1$ and $G_2$ be abalian groups. A map $f: G_1\longrightarrow G_2$ is said to be a homomorphism if $$f(x+y)=f(x)+f(y),\ x,y\in G_1.$$ If $f$ is also a bijection (i.e one-to-one and onto), $f$ is called an isomorphism. If there is an isomorphism $f: G_1\longrightarrow G_2$, $G_1$ is said to be isomorphic to $G_2$ and we write $G_1\stackrel{f}{\cong} G_2$ or simply $G_1\cong G_2$.

Example. Define a map $f: \mathbb Z\longrightarrow\mathbb Z_2=\{0,1\}$ by $$f(2n)=0\ \mbox{and}\ f(2n+1)=1.$$ Then $f$ is a homomorphism.

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$.

Example. For any $k\in\mathbb N$, $k\mathbb Z=\{kn: n\in\mathbb Z\}$ is a subgroup of $\mathbb Z$.

Example. $\mathbb Z_2=\{0,1\}$ is not a subgroup of $\mathbb Z$.

Let $H$ be a subgroup of $G$. Define a relation on $G$ by $$\forall x,y\in G,\ x\sim y\ \mbox{if}\ x-y\in H.$$ Then $\sim$ is an equivalence relation on $G$. The equivalence class of $x\in G$ is denoted by $[x]$, i.e. \begin{eqnarray*}[x]&=&\{y\in G: y\sim x\}\\&=&\{y\in G: y-x\in H\}.\end{eqnarray*} Let $G/H$ be the quotient set $$G/H=\{[x]: x\in G\}.$$ Define an operation $+$on $G/H$ by $$[x]+[y]=[x+y],\ \forall [x],[y]\in G/H.$$ Then $G/H$ becomes an abelian group with this operation.

Example. $\mathbb Z/2\mathbb Z=\{[0],[1]\}$. Define $\varphi: \mathbb Z/2\mathbb Z\longrightarrow\mathbb Z_2$ by $$\varphi([0])=0\ \mbox{and}\ \varphi([1])=1.$$ Then $\mathbb Z/2\mathbb Z\cong\mathbb Z_2$. In general, for every $k\in\mathbb N$, $\mathbb Z/k\mathbb Z\cong\mathbb Z_k$.

Lemma 1. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then

(a) $\ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0)$ is a subgroup of $G_1$.

(b) ${\mathrm im}f=\{f(x): x\in G_1\}$ is a subgroup of $G_2$.

Theorem 2 [Fundamental Theorem of Homomorphism]. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then $$G_1/\ker f\cong{\mathrm im}f.$$

Example. Let $f: \mathbb Z\longrightarrow\mathbb Z_2$ be defined by $$f(2n)=0,\ f(2n+1)=1.$$ Then $\ker f=2\mathbb Z$ and ${\mathrm im}f=\mathbb Z_2$. By Fundamental Theorem of Homomorphism, $$\mathbb Z/2\mathbb Z\cong\mathbb Z_2.$$

Take $r$ elements $x_1,x_2,\cdots,x_r$ of $G$. The elements of $G$ of the form $$n_1x_1+n_2x_2+\cdots+n_rx_r\ (n_i\in\mathbb Z,\ 1\leq i\leq r)$$ form a subgroup of $G$, which we denote $\langle x_1,\cdots,x_r\rangle$. $\langle x_1,\cdots,x_r\rangle$ is called a subgroup of $G$ generated by the generators $x_1,\cdots,x_r$. If $G$ itself is generated by finite lelements, $G$ is said to be finitely generated. If $n_1x_1+\cdots+n_rx_r=0$ is satisfied only when $n_1=\cdots=n_r=0$, $x_1,\cdots,x_r$ are said to be linearly independent.

Definition. If $G$ is fintely generated by $r$ linearly independent elements, $G$ is called a free abelian group of rank $r$.

Example. $\mathbb Z$ is a free abelian group of rank 1 generated by 1 (or $-1$).

Example. Let $\mathbb Z\oplus\mathbb Z=\{(m,n):m,n\in\mathbb Z\}$. The $\mathbb Z\oplus\mathbb Z$ is a free abelian group of rank 2 generated by $(1,0)$ and $0,1)$. More generally, $$\stackrel{r\ \mbox{copies}}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}$$ is a free abelian group of rank $r$.

Example. $\mathbb Z_2=\{0,1\}$ is fintely generated by 1 but is not free. $1+1=0$ so 1 is not linearly independent.

If $G=\langle x\rangle=\{0,\pm x,\pm 2x,\cdots\}$, $G$ is called a cyclic group. If $nx\ne 0$ $\forall n\in\mathbb Z\setminus\{0\}$, it is an infinite cyclic group. If $nx=0$ for some $n\in\mathbb Z\setminus\{0\}$, it is a finite cyclic group. Let $G=\langle x\rangle$ and let $f:\mathbb Z\longrightarrow G$ be a homomorphism defined by $f(k)=kx$, $k\in\mathbb Z$. $f$ is an epimorphism (i.e. onto homomorphism), so by Fundamental Theorem of Homomorphism, $$G\cong\mathbb Z/\ker f.$$ If $G$ is a finite group, then there exists the smallest positive integer $N$ such that $Nx=0$. Thus $$\ker f=\{0,\pm N,\pm 2N,\cdots\}=N\mathbb Z.$$ Hence $$G\cong\mathbb Z/N\mathbb Z\cong\mathbb Z_N.$$ If $G$ is an infinite cyclic group, $\ker f=\{0\}$. Hence, $$G\cong\mathbb Z/\{0\}\cong\mathbb Z.$$

Lemma 3. Let $G$ be a free abelian group of rank $r$, and let $H$ be a subgroup of $G$. Then one may always choose $p$ generators $x_1,\cdots,x_p$ out of $r$ generators of $G$ so that $k_1x_1,\cdots,k_px_p$ generate $H$. Hence, $$H\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z$$ and $H$ is of rank $p$.

Theorem 4 [Fundamental Theorem of Finitely Generated Abelian Groups] Let $G$ be a finitely generated abelian group with $m$ generators. Then $$G\cong\stackrel{r}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus \mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}$$ where $m=r+p$. The number $r$ is called the rank of $G$.

Proof. Let $G=\langle x_1, \cdots,x_m\rangle$ and let $f: \mathbb Z\oplus\cdots\oplus\mathbb Z\longrightarrow G$ be the surjective homomorphism $$f(n_1,\cdots,n_m)=n_1x_1+\cdots +n_mx_m.$$ Then by Fundamental Theorem of Homomorphism $$\mathbb Z\oplus\cdots\oplus\mathbb Z/\ker f\cong G.$$ $\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}$ is a free abelian group of rank $m$ and $\ker f$ is a subgroup of $\mathbb Z\oplus\cdots\oplus\mathbb Z$, so by Lemma 3 $$\ker f\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z.$$ Define $\varphi:\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus \cdots\oplus k_p\mathbb Z\longrightarrow\mathbb  Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z$ by $$\varphi((n_1,\cdots,n_p)+k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z)=(n_1+k_1\mathbb Z,\cdots,n_p+k_p\mathbb Z).$$ Then $$\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\stackrel{\varphi}{\cong}\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z.$$ Hence, \begin{eqnarray*}G&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/\ker f\\&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus Z/k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}.\end{eqnarray*}

Derivatives

4 Replies

In this lecture, I am going to introduce you a new idea, which was discovered by Sir Issac Newton and Gottfried Leibiz, to find the slope of a tangent line. This is in fact a quite ingenious idea as you will see. Let a function $y=f(x)$ be given. We want to find the slope of a line tangent to the graph of $y=f(x)$ at a point $x=a$. First consider another point on the $x$-axis that is away from $x=a$. If the distance from $x=a$ to this point is $h$, then the point can be written as $x=a+h$. Let $P(a,f(a))$ and $Q(a+h,f(a+h))$. Then the slope of line segment $\overline{PQ}$ is given by $$\frac{f(a+h)-f(a)}{h}.$$

Now we continuously change $h$ so that it gets smaller and smaller close to $0$, consequently the point $a+h$ gets closer to $a$. We want to see how the rate $\frac{f(a+h)-f(a)}{h}$ changes as $h\to 0$. To illustrate the situation better, I will use a specific example, say $f(x)=x^2$ with $a=2$. First we take $h=1$. The following picture shows you the graph of $f(x)=x^2$ (in black), where $1.5\leq x\leq 3$ and the line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ (in blue), and the line tangent to the graph $f(x)=x^2$ at $x=2$ (in red).

Next we take $h=0.5$. Then the picture becomes

For $h=0.1$, the picture becomes

As one can clearly see, the line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ gets closer to the tangent line as $h$ gets smaller close to $0$. We can still do better. For $h=0.001$, the picture becomes

The line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ and the tangent line now appear to be overlapping. From this observation, we can see that the rate $\frac{f(a+h)-f(a)}{h}$ gets closer and closer to the slope of tangent line as $h$ gets smaller and smaller close to $0$. In fact, the slope would be exactly the limit of $\frac{f(a+h)-f(a)}{h}$ as $h$ approaches $0$. Denote the limit by $f'(a)$. Then $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$ $f'(a)$ is called the derivative of $f(x)$ at $x=a$. One may wonder why we need another name for the slope of a tangent line. The reason is that as we will see later the slope of a tangent line can mean something else in different contexts. Let $x=a+h$. Then $x\to a$ as $h\to 0$. So $f'(a)$ can be also written as $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$ The equation of tangent line to $y=f(x)$ at $x=a$ is then given by $$y-f(a)=f'(a)(x-a).$$

Example. Find the equation of tangent line to the graph of $f(x)=x^2$ at $x=2$.

Solution. First we need to find $f'(2)$, i.e. the slop of the tangent line. \begin{eqnarray*}f'(2)&=&\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}\\&=&\lim_{h\to 0}\frac{(2+h)^2-4}{h}\\&=&\lim_{h\to 0}\frac{4+4h+h^2-4}{h}\\&=&\lim_{h\to 0}(4+h)\\&=&4.\end{eqnarray*}

Of course, we can also use the alternative definition of $f'(a)$ to calculate the slope:\begin{eqnarray*}f'(2)&=&\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}\\&=&\lim_{x\to 2}\frac{x^2-4}{x-2}\\&=&\lim_{x\to 2}\frac{(x+2)(x-2)}{x-2}\\&=&\lim_{x\to 2}(x+2)\\&=&4.\end{eqnarray*}

The equation of tangent line is then $y-4=4(x-2)$ or $y=4x-4$.

Remark. One may wonder which definition of $f'(a)$ to use. I would say that is the matter of a personal taste. For a polynomial function, one notable difference between the two definitions is that if you use the first definition, you will end up expanding a polynomial, while you will have to factorize a polynomial with the second definition. Since the expansion of a polynomial is easier than the factorization, you may want to use the first definition if you are not confident with factorizing polynomials.

Example. Find the equation of tangent line to the graph of $f(x)=x^5$ at $x=1$.

Solution. As we discussed in the previous lecture, this is an extremely difficult problem to solve by using only algebra if not impossible. But surprise! With the new method, this is more or less a piece of cake. First we calculate the slope $f'(1)$. \begin{eqnarray*}f'(1)&=&\lim_{h\to 0}\frac{(1+h)^5-1}{h}\\&=&\lim_{h\to 0}\frac{(1+h)^5-1}{h}\\&=&\lim_{h\to 0}\frac{1+5h+10h^2+10h^3+5h^4+h^5-1}{h}\\&=&\lim_{h\to 0}(5+10h+10h^2+5h^3+h^4)\\&=&5.\end{eqnarray*} Or by the second definition, \begin{eqnarray*}f'(1)&=&\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}\\&=&\lim_{x\to 1}\frac{x^5-1}{x-1}\\&=&\lim_{x\to 1}\frac{(x-1)(x^4+x^3+x^2+x+1)}{x-1}\\&=&\lim_{x\to 1}(x^4+x^3+x^2+x+1)\\&=&5.\end{eqnarray*}Therefore the equation of the tangent line is given by $y-1=5(x-1)$ or $y=5x-4$. The following picture shows the graph of $y=x^5$ (in blue) and the graph of tangent line $y=5x-4$.