MATH 138

MATH 138 - Calculus 2
MWF 11:30AM - 12:20PM 
MC 4021
Katy Howell Escobar

1 | Integration

1.1 | The Definite Integral

Regular Partition: Consider an interval $[a,b]$ with $a<b$ and $n\in \mathbb{Z}$. If we define $\Delta x=b-\frac{a}{n}$, Then the points $x_i=a+i\Delta x,i\in \mathbb{N}$ separate $[a,b]$ into subintervals of equal width $\Delta x$. We call this a regular partition of $[a,b]$

Riemann Sum

$$\sum _{i=1}^{n}f(x_i)\Delta x=f(x_1)\Delta x+f(x_2)\Delta x+\cdots +f(x_n)\Delta x$$

Definite Integral

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}f(x_i)(\frac{b-a}{n})$$

Right and Left Endpoint Riemann Sums

$$R_n=f(x_1)\Delta x+f(x_2)\Delta x+\cdots +f(x_n)\Delta x=\sum _{i=1}^{n}f(x_i)(\frac{b-a}{n})$$ $$L_n=f(x_0)\Delta x+f(x_1)\Delta x+\cdots +f(x_{n-1})\Delta x=\sum _{i=1}^{n}f(x_{i-1})(\frac{b-a}{n})$$

Converting Function to Riemann Sum

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}f\left(a+\frac{i(b-a)}{n}\right)\left(\frac{b-a}{n}\right)$$

Example: Evaluate

$$\underset{n\to \infty }{\lim }\sum _{i=1}^n\sqrt{2-\frac{i3}{n}}(\frac{3}{n})$$

Right-handed Riemann sum: $\Delta x=\frac{3}{n}=\frac{b-a}{n}$ (IMPORTANT) $f(x)=\sqrt{2-x}$ $f(x_i)=\sqrt{2-(a+i(\frac{3}{n}))}$ $x\in [0,3]$ This is the same thing as

$${\int }_0^3\sqrt{2-x}$$

Summation Formulas

$$\begin{array}{rl}\sum _{i=1}^{n}1& =n\\ \sum _{i=1}^{n}i& =\frac{n(n+1)}{2}\\ \sum _{i=1}^n{i}^2& =\frac{n(n+1)(2n+1)}{6}\\ \sum _{i=1}^n{i}^3& =\frac{n^2(n+1{)}^2}{4}\end{array}$$

1.2 | Properties of the Definite Integral

Integral Properties

1. Constant Multiple

$${\int }_a^{b}kf(x)dx=k{\int }_a^{b}f(x)dx$$

2. Sum / Difference

$${\int }_a^b(f(x)\pm g(x))dx={\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx$$

3. Squeeze: $m\le f(x)\le M$ for all $x\in [a,b]$

$$m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$$

4. Positive: $f(x)\ge 0$ for all $x$

$${\int }_a^{b}f(x)dx\ge 0$$

5. Greater: $f(x)\le g(x)$ for all $x\in [a,b]$

$${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx$$

6. Absolute Value:

$$|{\int }_a^{b}f(x)dx|\le {\int }_a^b|f(x)|dx$$

Definite Integrals: Special Cases Zero

$${\int }_a^{a}f(x)dx=0$$

Inverted range

$${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$$

(Separating the Domain of Integration)

$${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$$

Even and Odd Functions: Let $f$ be a bounded integrable function defined on $[-a,a]$ If $f$ is odd $(f(-x)=-f(x))$, then

$${\int }_{-a}^{a}f(x)dx=0$$

If $f$ is even $(f(-x)=f(x))$, then

$${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$$

Function parity is just like number parity: odd / odd = even / even = even, else it’s odd

1.3 | Average Value of a Function

Definition: Average Value of a Continuous Function:
If $f$ is continuous on $[a,b]$, then the average value of $f$ on $[a,b]$ is

$$f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

Average Value Theorem (AVT) - similar to VT If $f$ is continuous on $[a,b]$, then there exists at least on $c\in [a,b]$ such that

$$f(x)=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

1.4 | FTC Part 1

Power Rule for Antiderivatives For all $n\in \mathbb{R},n\ne -1$

$$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$$

FTC I: If $f$ is continuous on an open interval $I$ containing $a$, then

$$\frac{d}{dx}{\int }_a^{x}f(t)dt=f(x)$$

Extended FTC I (Chain Rule): If $f$ is continuous and $g,h$ are differentiable, then

$$\frac{d}{dx}{\int }_{g(x)}^{h(x)}f(t)dt=f(h(x))h^{\prime }(x)-f(g(x))g^{\prime }(x)$$

Example: Evaluate

\frac{d}{dx}\int_{5x}^\sqrt{x}\cos(t^2)dt=\cos(x)\cdot\frac{1}{2\sqrt{x}}-\cos(25x^2)\cdot5

1.5 | FTC Part 2

The Antiderivative Theorem: If $F,G$ are antiderivatives of a function $f$ on an open interval $I$, there exists a constant $C\in \mathbb{R}$ such that

$$G(x)=F(x)+C$$

FTC II:

$${\int }_a^{b}f(t)dt=f^{\prime }(b)-f^{\prime }(a)$$

1.6 | The Substitution Rule

Substitution Rule / Change of Variables If $f$ and $g$ are functions such that $g^{\prime }$ is continuous on an interval $[a,b]$ and $f$ is continuous on the range of $g$

$$\int f(g(x))g^{\prime }(x)dx=\int f(u)du{|}_{u=g(x)}$$

Example: Evaluate

$$\int x^2\sqrt{x^3+4}dx$$ $$u=x^3+4⟹du=3x^{2}dx⟹dx=\frac{du}{3x^2}$$ $$\begin{array}{rl}\int x^2\sqrt{x^3+4}dx& =\int x^2\sqrt{u}\frac{du}{3x^2}\\ & =\frac{1}{3}\int u^{1/2}du\\ & =\frac{1}{3}\frac{u^{3/2}}{3/2}+C\\ & =\frac{2}{9}{u}^{3/2}+C\\ & =\frac{2}{9}(x^3+4{)}^{3/2}+C\end{array}$$

Example: Evaluate

$$\int \frac{\ln (x)}{x}dx=\int \frac{1}{x}\ln (x)dx$$ $$u=\ln (x)⟹\frac{du}{dx}=\frac{1}{x}⟹dx=xdu$$ $$\int \frac{1}{x}u(xdu)=\int udu=\frac{1}{2}{u}^2+C=\frac{1}{2}(\ln (x){)}^2+C$$

Example: Evaluate

$$\int \frac{\cos (\sqrt{x})}{x}dx$$ $$u=\sqrt{x}⟹\frac{du}{dx}=\frac{1}{2\sqrt{x}}⟹dx=2\sqrt{x}du=2udu$$ $$\int \frac{\cos (u)}{u}\cdot 2udu=\int 2\cos (u)du=2\sin (u)+C=2\sin (\sqrt{x})+C$$

Example: Evaluate

$$\int \frac{x^2}{\sqrt{x+1}}dx$$ $$u=x+1⟹\frac{du}{dx}=1⟹dx=du$$ $$\int \frac{(u-1{)}^2}{\sqrt{u}}du=\int \frac{u^2-2u+1}{u^{1/2}}=\int u^{3/2}-2u^{1/2}+u^{-1/2}=\frac{2}{5}{u}^{5/2}-\frac{4}{3}{u}^{3/2}+2u^{1/2}+C$$ $$=\frac{2}{5}(x+1{)}^{5/2}-\frac{4}{3}(x+1{)}^{3/2}+2(x+1{)}^{1/2}+C$$

Example: Evaluate

$$\int {\sin }^{6}x\cos xdx$$ $$u=\sin (x)⟹\frac{du}{dx}=\cos (x)⟹du=\cos (x)dx$$ $$=\int u^{6}du=\frac{1}{7}{u}^7+C=\frac{1}{7}{\sin }^7(x)+C$$

Example: Evaluate

$$\int x{e}^{5x^2}dx$$ $$u=5x^2⟹\frac{du}{dx}=10x⟹dx=\frac{du}{10x}⟹xdx=\frac{du}{10}$$ $$\frac{1}{10}\int e^{u}du=\frac{1}{10}\int e^u+C=\frac{1}{10}\int e^5{x}^2+C$$

Example: Integrate

$${\int }_0^1{e}^x\cos (e^x)dx$$ $$u=e^x⟹\frac{du}{dx}=e^x⟹dx=\frac{du}{u}⟹e^0=1,e^1=e$$ $${\int }_1^{e}u\cos (u)\frac{du}{u}=\int \cos (u)du=\sin (u){\mid }_1^e=\sin (e)-\sin (1)$$

Theorem: Integrating $f(ax)$ Let $a\in \mathbb{R},a\ne 0$. If $\int f(x)dx=F(x)+C$, then

$$\int f(ax)dx=\frac{1}{a}F(ax)+C$$

Theorem: Substitution Rule for Definite Integrals If $f,g$ are functions such that $g^{\prime }$ is continuous on an interval $[a,b]$ and $f$ is continuous on the range of $g$, then

$${\int }_{x=a}^{x=b}f(g(x))g^{\prime }(x)dx={\int }_{u=g(a)}^{u=g(b)}f(u)du$$

Example: Integrate

$${\int }_0^1\frac{x^3}{1+x^4}dx$$ $$u=1+x^4⟹\frac{du}{dx}=4x^3⟹dx=\frac{du}{4x^3}⟹\frac{du}{4}=x^{3}dx$$ $${\int }_{x=1}^{x=0}=\frac{1}{4u}du=\frac{1}{4}\ln (u){\mid }_{x=0}^{x=1}=\frac{1}{4}(\ln 2-\ln 1)=\frac{\ln 2}{4}$$

1.7 | Trig Substitution

Theorem: Antiderivatives of Trig functions

$$\int \sec (x)dx=\ln |\sec (x)+\tan (x)|+C$$ $$\int \csc (x)dx=-\ln |\csc (x)+\cot (x)|+C$$ $$\int \tan (x)dx=-\ln |\cos (x)|+C$$ $$\int \cot (x)dx=\ln |\sin (x)|+C$$

Example 1: Consider

$$\int \frac{\sqrt{9-4x^2}}{x^2}dx$$

Let $x=\frac{3}{2}\sin \theta ,-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ $\frac{dx}{d\theta }=\frac{3}{2}\cos \theta$

$$\begin{array}{rl}& =\int \frac{\sqrt{9-4{\left(\frac{3}{2}\sin \theta \right)}^2}}{\frac{3}{2}\sin {\theta }^2}\cdot \frac{3}{2}\cos \theta d\theta \\ & =\int \frac{\sqrt{9-9{\sin }^2\theta }}{\frac{9}{4}{\sin }^2\theta }\cdot \frac{3}{2}\cos \theta d\theta \\ & =\int \frac{3\cos \theta }{\frac{9}{4}{\sin }^2\theta }\cdot \frac{3}{2}\cos \theta d\theta \\ & =\int 2\frac{{\cos }^2\theta }{{\sin }^2\theta }d\theta \\ & =2{\cot }^2\theta d\theta \\ & =2\int cs{c}^2\theta -1d\theta \\ & =-2\cot \theta -2\theta +C\\ & =-2(\frac{\sqrt{9-4x^2}}{2x})-2\arcsin (\frac{2x}{3}+C))\end{array}$$

Motivation for picking $x=\frac{3}{2}\sin \theta$:

$$9-9{\sin }^2\theta =9(1-{\sin }^2\theta )=9{\cos }^2\theta =(3\cos \theta {)}^2$$

Writing in terms of $x$:

$$\frac{2x}{3}=\sin \theta ⟹\theta =\arcsin (\frac{2x}{3}+C)$$

Example 2: Consider

$$\int \frac{1}{\sqrt{x^2+4}}dx$$

Let $x=2\tan \theta$ $\frac{dx}{d\theta }=2{\sec }^2\theta d\theta ,\theta \in (\frac{-\pi }{2},\frac{\pi }{2})$

$$\begin{array}{rl}& =\int \frac{1}{\sqrt{4{\tan }^2\theta +4}}\cdot 2{\sec }^2\theta d\theta \\ & =\int \frac{2{\sec }^2\theta }{\sqrt{{\tan }^2\theta +1}}d\theta \\ & =\int \frac{{\sec }^2\theta }{\sqrt{{\sec }^2\theta }}d\theta \\ & =\int \frac{se{c}^2\theta }{|\sec \theta |}d\theta \\ & =\int \sec \theta d\theta \\ & =\ln |\sec \theta +\tan \theta |+C\\ & =\ln |\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}|+C\end{array}$$

Example 3:

$$\int \frac{dx}{x^2\sqrt{x^2-4}}$$

Let $x=2\sec \theta$ $\frac{dx}{d\theta }=2\sec \theta \tan \theta d\theta$

$$\begin{array}{rl}& =\int \frac{2\sec \theta \tan \theta }{4{\sec }^2\theta \sqrt{(2\sec \theta {)}^2-4}}d\theta \\ & =\int \frac{\tan \theta }{2\sec \theta \sqrt{4{\sec }^2\theta -4}}d\theta \\ & =\frac{1}{4}\int \frac{\tan \theta }{2\sqrt{{\sec }^2\theta -1}}d\theta \\ & =\frac{1}{4}\int \frac{\tan \theta }{\sec \theta \tan \theta }d\theta \\ & =\frac{1}{4}\int \frac{1}{\sec \theta }d\theta \\ & =\frac{1}{4}\int \cos \theta d\theta \\ & =\frac{\sin \theta }{4}+C\\ & =\frac{\sqrt{x^2-4}}{4x}+C\end{array}$$

hyp = x adj = 2 opp = $\sqrt{x^2-4}$

Example 5:

\int_0^\sqrt{3}\frac{x}{(1+x^2)}dx

Let $x=\tan \theta$ $dx={\sec }^2\theta d\theta$ $0=\tan \theta$ $\theta =0$ $\sqrt{3}=\tan \theta$ $\theta =\frac{\pi }{3}$

$$\begin{array}{rl}& ={\int }_0^{\frac{\pi }{3}}\frac{\tan \theta {\sec }^2\theta }{(1+\tan \theta {)}^2}d\theta \\ & ={\int }_0^{\frac{\pi }{3}}\frac{{\tan }^{\theta }{\sec }^2\theta }{{\sec }^4\theta }d\theta \\ & ={\int }_0^{\frac{\pi }{3}}\frac{\tan \theta }{{\sec }^2\theta }d\theta \\ & ={\int }_0^{\frac{\pi }{3}}\frac{\sin \theta }{\cos \theta }{\cos }^2\theta d\theta \\ & ={\int }_0^{\frac{\pi }{3}}\sin \theta \cos \theta d\theta \\ & ={\int }_0^{\frac{\sqrt{3}}{2}}udu\\ & =\frac{u^2}{2}{\mid }_0^{\frac{\sqrt{3}}{2}}\\ & =\frac{3}{8}\end{array}$$

Example 6:

$$\int \frac{x}{(3-2x-x^2{)}^{3/2}}dx$$

Aside:

$$\begin{array}{rl}3-2x-x^2& =-x^2-3x+3\\ & =-(x^2+2x)+3\\ & =-(x^2+2x+1)+3\\ & =-(x+1{)}^2+4\end{array}$$

Let $x+1=2\sin \theta ⟹x=2\sin \theta -1$ $\frac{dx}{d\theta }=2\cos \theta$

$$\begin{array}{rl}& =\int \frac{x}{[4-(x+1{)}^2{]}^{3/2}}dx\\ & =\int \frac{(2\sin \theta -1)2\cos \theta }{(4-4{\sin }^2\theta {)}^{3/2}}d\theta \\ & =\frac{1}{4}\int \frac{(2\sin \theta -1)\cos \theta }{({\cos }^2\theta {)}^{3/2}}d\theta \\ & =\frac{1}{4}\int \frac{2\sin \theta -1}{{\cos }^2\theta }d\theta \\ & =\frac{1}{4}\int \frac{2\sin \theta }{{\cos }^2\theta }-\frac{1}{{\cos }^2\theta }d\theta \\ & =\frac{1}{4}\int 2\tan \theta \sec \theta -{\sec }^2\theta d\theta \\ & -\frac{1}{4}(2\sec \theta -\tan \theta )+C\\ & =\frac{1}{4}(\frac{4}{\sqrt{4-(x+1{)}^2}}-\frac{x-1}{\sqrt{4-(x+1{)}^2}})+C\end{array}$$

1.8 | Integration by Parts

Derivation

$$\begin{array}{rl}[f(x)g(x){]}^{\prime }& =f^{\prime }(x)g(x)+f(x)g^{\prime }(x)\\ f(x)g^{\prime }(x)& =[f(x)g(x){]}^{\prime }-f^{\prime }g(x)\\ \int f(x)g^{\prime }(x)dx& =\int [f(x)g(x){]}^{\prime }dx-\int f^{\prime }(x)g(x)dx\\ \int f(x)g^{\prime }(x)dx& =f(x)g(x)-\int f^{\prime }(x)g(x)dx\end{array}$$

Theorem: Integration by parts

$$\int udv=uv-\int vdu$$

Example 1:

$$f(x)=ln(x)⟹f^{\prime }(x)=\frac{1}{x}$$ $$g^{\prime }(x)=x^2⟹g(x)=\frac{1}{3}{x}^2$$ $$\begin{array}{rl}\int x^2\ln xdx& =\frac{x^3\ln x}{3}-\int \frac{1}{x}\frac{x^3}{3}dx\\ & =\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx\\ & =\frac{x^3\ln x}{3}-\frac{x^3}{9}+C\end{array}$$

Example 2: $f(x)=x⟹f^{\prime }(x)=1$ $g^{\prime }(x)=e^x⟹g(x)=e^x$

$$\begin{array}{rl}\int x{e}^{x}dx& =x{e}^x-\int e^x(1)\\ & =x{e}^x-e^x+C\\ & =e^x(x-1)+C\end{array}$$

1.9 | Partial Fractions

Robert Garbary

Goal: Find a derivative of

$$\int \frac{a(x)}{b(x)}dx$$

If $\mathrm{deg}(a)\ge deg(b)$, write $a(x)=b(x)q(x)+r(x)$

$$\int \frac{a(x)}{b(x)}=\int \frac{b(x)q(x)+r(x)}{b(x)}dx=q(x)+\frac{r(x)}{b(x)}dx$$

Ex:

$$\int \frac{x+1}{(x+3)(x+4)}dx$$

Plan: Find $A,B\in \mathbb{R}$ such that

$$\frac{x+1}{(x+3)(x+4)}=\frac{A}{x+3}+\frac{B}{x+4}$$

Spoiler: It turns out a unique $A,B$ always exist as long as $\mathrm{deg}(P(x))\le 5$

$$\frac{A(x+4)+B(x+3)}{(x+3)(x+4)}$$ $$x+1=A(x+4)+B(x+3)$$

Method 1: Comparing coefficients

$$x+1=(A+B)x+(4A+3B)$$ $$A+B=1$$ $$4A+3B=1$$ $$(A,B)=(-2,3)$$

Method 2:

$$x=-4,-4=B(-1)⟹B=3$$ $$x=-3,-2=A(1)⟹A=-2$$ $$\int \frac{x+1}{(x+3)(x+4)}dx=\int \frac{-2}{x+3}+\frac{-3}{x+4}dx=-2\ln |x+3|+3\ln |x+4|+C=\ln |\frac{(x+4{)}^3}{(x+3{)}^2}|+C$$

Theorem: Cool stuff

$$\begin{array}{rlr}& ax+b& \frac{A}{ax+b}\\ & (ax+b{)}^n& \frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+\cdots +\frac{A_n}{(ax+b{)}^n}\\ & \underset{\text{irreducible}}{\underbrace{a{x}^2+bx+c}}& \frac{Ax+B}{a{x}^2+bx+c}\\ & (a{x}^2+bx+c{)}^n& \frac{A_{1}x+B_1}{a{x}^2+bx+c}+\frac{A_{2}x+B_2}{(a{x}^2+bx+c{)}^2}+\cdots +\frac{A_{n}x+B_n}{(a{x}^2+bx+c{)}^n}\end{array}$$

Ex:

$$\frac{x^2+x-3}{(x-3{)}^2(x+3)}=\frac{A}{x-3}+\frac{B}{(x-3{)}^2}+\frac{C}{x+3}$$ $$x^3+x-1=(A+C)x^3+(B+D)x^2+(SA)x+(SB)$$

$A+C=1⟹C=\frac{4}{5}$ $B+D=0⟹D=\frac{1}{5}$ $SA=1⟹A=\frac{1}{5}$ $SB=-1⟹B=-\frac{1}{5}$

$$\begin{array}{rl}I& =\frac{1}{5}\int \frac{1}{x}dx-\frac{1}{5}\int \frac{1}{x}dx+\frac{1}{5}\int \frac{4x+1}{x^2+5}\\ & =\frac{1}{5}\int \frac{1}{x}dx-\frac{1}{5}\int \frac{1}{x}dx+\frac{1}{5}\int \frac{4x}{x^2+5}+\int \frac{1}{x^2+5}\end{array}$$ $$x=\sqrt{5}\tan \theta ⟹\int \frac{1}{x^2+5}=\frac{1}{\sqrt{5}}\arctan (\frac{x}{\sqrt{5}})$$

Ex:

$$\int \frac{x^3-2x}{x^2+3x+2}dx$$ $$x^3-2x=(x^2+3x+2)(x-3)+5x+6$$ $$\begin{array}{rl}& =\int x-3+\frac{5x+6}{x^2+3x+2}dx\\ & =\int x-3+\frac{A}{x+1}+\frac{B}{x+2}\end{array}$$

1.10 | Improper Integrals

Definition: Improper Integrals Let $a\in \mathbb{R}$, assume $f(x)$ is integrable on $[a,t]$ for all $t\ge a$. Let $L={\lim }_{t\to \infty }{\int }_a^{t}f(x)dx$.

• If $L$ exists (it converges to a number), we say ${\int }_a^{\infty }f(x)dx$ converges to L.
• If $L$ doesn’t exist, we say it diverges.

Type 1 Improper Integral

One of the bounds is $\pm \infty$ Ex:

$${\int }_2^{\infty }\frac{1}{\sqrt{x}}dx$$

Note that the 2 doesn’t affect whether or not this converges, but what it converges to.

$${\int }_2^{\infty }\frac{1}{\sqrt{x}}dx=\underset{c\to \infty }{\lim }({\int }_2^c{x}^{-1/2}dx)=\underset{c\to \infty }{\lim }2\sqrt{x}{\mid }_2^c=\underset{c\to \infty }{\lim }2\sqrt{c}-2\sqrt{2}=DNE$$

Ex:

$${\int }_{-\infty }^0\frac{1}{x^2+1}dx=\underset{c\to \infty }{\lim }{\int }_c^0\frac{1}{x^2+1}dx=\underset{c\to \infty }{\lim }(\arctan (0)-\arctan (c))=(0-(-\frac{\pi }{2}))=\frac{\pi }{2}$$

Ex:

$${\int }_{-\infty }^{\infty }xdx={\int }_{-\infty }^5+{\int }_5^{\infty }$$ $$=\underset{c\to x}{\lim }{\int }_5^{c}xdx=\underset{c\to \infty }{\lim }(\frac{c^2}{2}-\frac{5^2}{2})$$ $$\underset{c\to \infty }{\lim }{\int }_{-c}^{c}xdx=\underset{c\to \infty }{\lim }0=0$$

VERY IMPORTANT NOTE:

$${\int }_{-\infty }^{\infty }f(x)dx\ne \underset{c\to \infty }{\lim }{\int }_{-c}^{c}f(x)dx$$

Ex: Let $p\in \mathbb{R}$. for which $p$ does the following diverge:

$${\int }_1^{\infty }\frac{1}{x^p}dx$$

Case 1: $p=1$

$$\underset{c\to \infty }{\lim }{\int }_1^c\frac{1}{x}dx=\underset{c\to \infty }{\lim }\ln (x){|}_1^c=\underset{c\to \infty }{\lim }\ln (c)=DNE$$

Case 2: $p\ne 1$

$$\underset{c\to \infty }{\lim }{\int }_1^c{x}^{-p}dx=\underset{c\to \infty }{\lim }\frac{1}{1-p}{x}^{1-p}{|}_1^c=\frac{1}{1-p}\underset{c\to \infty }{\lim }(c^{1-p}-1)$$

Theorem: Type-1 p-integrals:

$${\int }_1^{\infty }\frac{1}{x^p}dx$$

converges if and only if $p>1$. If it converges, it converges to $\frac{1}{p-1}$.

Theorem: Improper Type 2 p-integral

$$J={\int }_0^1\frac{1}{x^p}dx$$

1. J converges if and only if $p<1$
2. If J converges, it converges to $J=\frac{1}{1-p}$

$$J={\int }_0^1\frac{1}{x^p}dx=\underset{c\to 0^{+}}{\lim }{\int }_c^1\frac{1}{x^p}dx$$

Let $u=\frac{1}{x},du=-\frac{1}{x^2}dx$ $x=1⟹u=1$ $x\to 0^{+}⟹u=-\infty$

$$\begin{array}{rl}\underset{c\to 0^{+}}{\lim }{\int }_c^1\frac{1}{x^p}dx& =\underset{d\to \infty }{\lim }{\int }_{u=d}^{u=1}{u}^p(-x^{2}du)\\ & =\underset{d\to \infty }{\lim }{\int }_a^d{u}^{p-2}du\\ & ={\int }_1^{\infty }\frac{1}{u^{2-p}du}\end{array}$$

By Improper Type-1 p-integral Theorem, $2-p>1⟺p<1$ Thus $J$ converges, and it converges to $\frac{1}{(2-p)-1}=\frac{1}{1-p}$

Type 2 Improper Integral

Definition: Let $a,b\in \mathbb{R}$ with $a<b$. Assume $f(x)$ is integrable on $[t,b]$ for all $t\in [a,b]$. Assume $f(x)$ is “bad” at a (or b)

$${\int }_a^{b}f(x)dx=\underset{t\to a^{+}}{\lim }{\int }_t^{b}f(x)dx$$

If $f(x)$ is bad $c,a<c<b$, we define

$${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$$

Ex:

$${\int }_0^4\frac{1}{\sqrt{x}}dx$$

Solution

$$=\underset{c\to 0^{+}}{\lim }{\int }_c^4{x}^{-1/2}dx=\underset{c}{\lim }\to 0^{+}2\sqrt{x}{\mid }_c^4=\underset{c\to 0^{+}}{\lim }(4-2\sqrt{c})=4$$

Ex:

$${\int }_0^5\frac{x}{x-2}dx={\int }_0^2\frac{x}{x-2}dx+{\int }_2^5\frac{x}{x-2}dx$$

Solution

$$\frac{x}{x-2}=1+\frac{2}{x-2}\mid \underset{c\to 2^{-}}{\lim }{\int }_0^{c}1+\frac{2}{x-2}dx=DNE$$

1.11 | Area Between Curves

Bounded area of $f(x),g(x)\in [a,b]$:

$${\int }_a^{b}f(x)-g(x)dx$$

If they intersect at $c\in [a,b]$:

$${\int }_a^{c}f(x)-g(x)dx+{\int }_c^{b}g(x)-f(x)dx$$

Example: On $[0,\pi ]$, find the area bounded by $y=\sin x,y=\cos x$

$$A={\int }_0^{\frac{\pi }{4}}\cos x-\sin x+{\int }_{\frac{\pi }{4}}^4\sin x-\cos xdx=2\sqrt{2}$$

Vertical integrals: $x=g(y),x=f(y),y\in [c,d]$

$$A={\int }_c^d{x}_{right}(y)-x_{left}(y)dy={\int }_c^{d}f(y)-g(y)dy$$

Example: Find the area bounded by $x=-y^2,x=-y-2$ $-y^2=-y-2⟹y^2-y-2=0⟹y=-1,2$ $x_{right})y)=-y^2$ $x_{left}(y)=-y-2$

$$A={\int }_{-1}^2{x}_{right}(y)-x_{left}(y)dy={\int }_{-1}^2-y^2-(y-2)dy=\frac{9}{2}$$

1.12 | Volumes of Solids of Revolution

Volume of Revolution (Disc)

$$V={\int }_a^b\pi f(x{)}^{2}dx$$

Ex: Fix $R>0,y=\sqrt{R^2-x^2}$

$$V=\pi {\int }_{-R}^R{\sqrt{R^2-x^2}}^{2}dx=\pi {\int }_{-R}^R{R}^2-x^{2}dx=\frac{4}{3}\pi R^2$$

Proves the volume of a sphere!

Volume of Revolution (Washer)

$$V=\pi {\int }_a^b(R(x{)}^2-r(x{)}^2)dx$$

Where $R$ is the big radius, $r$ is the small one

Ex: Consider the area between $y=3,y=\sqrt{x}$

$$V={\int }_0^4[3^2-{\sqrt{x}}^2]dx=28\pi$$

Ex: Consider $g(x)>f(x)$ rotated around $y=5$

$$V=\pi {\int }_a^b[(5-f(x){)}^2-(5-g(x){)}^2]dx$$

Ex: Consider $f(x),g(x),f(x)<g(x)\in [a,c],f(x)>g(x)\in [c,b]$

$$V=\pi {\int }_a^c{g}^2-f^{2}dx-{\int }_c^b{f}^2-g^{2}dx$$

Ex: Consider $y=x,y=\sqrt{x}$, revolved around the $y$ axis

$$V={\int }_c^d\pi (R(y{)}^2-r(y{)}^2)dy=\pi {\int }_0^1(y^2-y^2{)}^{2}dy=\pi \left(\frac{1}{3}-\frac{1}{5}\right)$$

Ex: Consider a cone with base $R$ and height $H$ revolved around the y axis

$$y=-\frac{H}{R}(x-R)⟹-\frac{Ry}{H}=x-R⟹x=R(1-\frac{y}{H})$$ $$\begin{array}{rl}V& =\pi {\int }_0^H\left(R\left(1-\frac{y}{H}\right)\right)dy\\ & =\pi R^2{\int }_0^{H}1-\frac{2y}{H}+\frac{y^2}{H^2}dy\\ & =\pi R^2(y-\frac{y^2}{H}+\frac{y^3}{3H}){\mid }_0^H\\ & =\pi R^2(\frac{H}{3})\end{array}$$

2 | Differential Equations

2.1 | Introduction to Differential Equations

Definition: Differential Equations

• Differential Equation (DE) = equation involving an unknown function and it’s derivative
• Ordinary Differential Equation (ODE) = single-variable functions
• Partial Differential Equation (PDE) = multivariable functions (multiple inputs)

Definition: Order of a DE

• The order of a DE is the highest derivative that appears in the equations

Definition: Linearity

• An ODE is called linear if it only contains linear functions on $y,y^{\prime },y^{\prime \prime },\dots$

Definition: General and Particular Solution

1. General Solution = Complete collection of solutions to a DE including arbitrary constants
2. Particular Solution = One where all arbitrary constants have been specified

Ex: What constant functions ($y=c$) satisfy:

$$y^{\prime }=y^3+2y^2-80y$$

Solution: We know $f^{\prime }(x)=c^{\prime }=0$

$$0=c(c+10)(c-8)$$ $$c=0,-10,8$$

Ex: Is $y=\sqrt{5-x^3}$ a solution to the DE

$$y^2-2y{y}^{\prime }+x^3=3x^2+5$$

Solution: Yes

$$\begin{array}{rl}{y}^2-2y{y}^{\prime }& =(5-x^3)-2(\sqrt{5-x^3})(\frac{-3x^2}{2\sqrt{5-x^3}})\\ & =5-x^3+3x^2\\ & =-x^3+3x^2+5\end{array}$$

2.2 | Separable Differential Equations

Definition: A first-order differential equation is said to be separable if it can be written in the form

$$\frac{dy}{dx}=g(x)h(y)$$

Method: Solving a separable DE

1. Determine any solutions $y$ with $h(y)=0$
2. Find the solutions $y,h(y)\ne 0$ by evaluating the following. If possible, isolate $y$ is a function of $x$ in the resulting equation.

$$\frac{1}{h(x)}dy=g(x)dx⟹\int \frac{1}{h(y)}dx=\int g(x)dx$$

Ex:

$$\frac{dy}{dx}=\frac{x}{y}$$

Solution: $g(x)=x,h(y)=\frac{1}{y}$ Step 1: $h(y)=0$, no constant solution Step 2: Rearranging and integrating

$$\int ydy=\int xdx$$ $$\frac{y^2}{2}=\frac{x^2}{2}+C$$ $$y=\pm \sqrt{x^2+2C}$$

Checking:

$$\frac{dy}{dx}=\pm \frac{2x}{2\sqrt{x^2+2C}}=\frac{x}{y}$$

Ex:

$$\frac{dy}{dx}=\frac{3x^2+4x+2}{2y-2}$$

Solution:

$g(x)=3x^2+4x+2,h(x)=\frac{1}{2y-2}$ Step 1: $h(y)=0$, no constant solution Step 2: Rearranging and integrating

$$\int [2y-2]dy=\int [3x^2+4x+2]dx$$ $$y^2-2y=x^3+2x^2+2x+C$$

Imagine we know that $y(0)=1$

$$\begin{array}{rl}(-1{)}^2-2(-1)& =0+C\\ C& =3\\ y^2-2y& =x^3+2x^2+3x+3\\ y^2-2y+1& =x^3+2x^2+3x+4\\ (y-1{)}^2& =x^3+2x^2+3x+4\\ y& =1\pm \sqrt{x^2+2x^2+3x+4}\end{array}$$

Ex

$$\frac{dy}{dx}=\frac{y\cos x}{1+2y^2}=\frac{y}{1+2y^2}\cos x$$

Solution $h(y)=0⟹y=0$

$$\int \frac{1+2y^2}{y}dy=\int \cos xdx,y\ne 0$$ $$\ln |x|+y^2=\sin x+C$$ $$\text{ OR }y=0$$

Ex:

$$y^{\prime }=x{\cos }^{2}y$$

Solution

$${\cos }^{2}y=0⟹\cos y=0⟹y=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$$ $$\int \frac{1}{{\cos }^{2}y}dy=\int xdx$$ $$\tan y=\frac{x^2}{2}+C$$ $$y=\arctan \left(\frac{x^2}{2}+C\right),\text{ or }y=\frac{\pi }{2}+k\pi$$

Ex:

$$\frac{dy}{dx}=(x+y{)}^2-1$$

Solution: Let $u=x+y$. $\frac{du}{dx}=1+y^{\prime }$

$$\frac{du}{dx}-1=u^2-1⟹\frac{du}{dx}=u^2$$ $$\begin{array}{rl}\int \frac{1}{u^2}du& =\int 1dx\\ -\frac{1}{u}& =x+C\\ u& =-\frac{1}{x+C}\\ x+y& =-\frac{1}{x+C}\\ y& =-\frac{1}{x+C}-x\end{array}$$

NOTE: we divided both sides by $u$, so $u=0$ is also a solution. $u=0⟹x=-y$

2.3 | Linear First-Order Differential Equations

Definition: A linear differential equation of order $n$ has the form

$$A_n(x)y^n+A_{n-1}(x)y^{n-1}+\cdots +A_1(x)y^{\prime }+A_0(x)y=B(x)$$

where $A_n(x)\ne 0$

Definition: A first-order linear DE of the form

$$y^{\prime }+P(x)y=Q(x)$$

is said to be in standard form.

Example: Consider the first order linear DE:

$$y^{\prime }+\frac{3}{x}y=1$$

Multiply by $x^3$ on both sides, so that the left hand side becomes the derivative of the right side

$$y^{\prime }{x}^3+3x^{2}y=1$$ $$(x^{3}y{)}^{\prime }=x^3$$ $$\int (x^{3}y{)}^{\prime }dx=\int x^{3}dx$$ $$x^{3}y=\frac{x^4}{4}+C$$ $$y=\frac{x}{4}+\frac{C}{x^3},C\in \mathbb{R}$$

NOTE: do not take the $\frac{C}{x^3}$ term and bundle it into a new constant!

Definition: Integrating Factor Given a linear DE of the form $y^{\prime }+P(x)y=Q(x)$, the integrating factor for would be

$$\mu (x)=e^{\int P(x)dx}$$

In the example $y^{\prime }+\frac{3}{x}y=1$, $P(x)=\frac{3}{x},Q(x)=1$

$$\int \frac{3}{x}dx=3\ln |x|⟹\mu (x)=e^{3\ln |x|dx}=e^{\ln |x^3|}=x^3$$

Method: Solving a first order DE

1. Divide by $A_1(x)$ to write the DE in standard form: $y^{\prime }+P(x)y=Q(x)$.
2. Multiply both sides of the equation by the integrating factor $\mu (x)=e^{\int P(x)dx}$.
3. Rewrite the left-hand side of the resulting equation as $(\mu (x)y{)}^{\prime }$.
4. Integrate both sides of the equation with respect to x.
5. Isolate for y.

2.4 | Applications of Differential Equations

Example: Tank has 1000L salt water, initially 0.1kg/L. Salt water of concentration 0.3 kg/L flows in at 10 L/min. Assume ceteris paribus.

1000L = constant volume 0.1 kg/L = initial concentration 1000L x 0.1kg = 100kg of salt

Let $x(t)$ = amount of salt (kg) in the contain at time $t$ minutes

$$x(0)=100$$ $$0.3\frac{kg}{L}\cdot 10\frac{L}{min}=3\frac{kg}{min}$$

Concentration is

$$\frac{x}{100}\frac{kg}{L}$$ $$\begin{array}{rl}\frac{dx}{dt}& =\text{(Rate of salt in) - (Rate of salt out)}\\ & =3-\frac{x}{100}\\ & =300-\frac{x}{100}\\ & =\frac{300-x}{100}=(300-x)\cdot \frac{1}{100}\end{array}$$

Solving the separable DE:

$$\begin{array}{rl}\int \frac{1}{300-x}dx& =\int \frac{1}{100}dt\\ -\ln |300-x|& =\frac{t}{100}+C\end{array}$$

Sub $x(0)=100$

$$\begin{array}{rl}-\ln 300-100& =0+c⟹c=-\ln 200\\ -\ln |300-x|& =\frac{t}{100}-\ln 200\\ \ln |300-x|& =\ln 200-\frac{t}{100}\\ |300-x|& =e^{\ln 200-\frac{t}{100}}\\ |300-x|& =200e^{\frac{-t}{100}}\\ 300-x& =200e^{\frac{-t}{100}}\text{(just kill the absolute value)}\\ x(t)& =300-200e^{-t/100}\end{array}$$

Science: Newton’s Law of Cooling: An object’s temperature changes at a rate proportional to $T_{object}-T_{room}$

$$\frac{dT}{dt}=-k(T-T_{room}),k>0$$

Example: Find the general solution to Netwon’s Law of Cooling

$$\begin{array}{rl}\int \frac{1}{T-T_{room}}dT& =\int -kdt\\ \ln |T-T_{room}|& =-kt+C\\ |T-T_{room}|& =e^{-kt+C}\\ T(t)& =e^c{e}^{-kt+C}\\ T(t)& =A{e}^{-kt+C},A=e^c\end{array}$$

Example: If the object went from 0C to 5C in 10 minutes, solve the ODE

$$\frac{dT}{dt}=-k(T-25)$$

$T_{room}=25$ $T(0)=0$ $T(10)=5$

$$\begin{array}{rl}T(t)& =A{e}^{-kt}+25\\ 0& =A{e}^0+25\\ A& =-25\\ T(t)& =-25e^{-kt}+25\end{array}$$

Solving the general formula:

$$\begin{array}{rl}5& =-25e^{-10k}+25\\ \frac{20}{25}& =e^{-10k}\\ \ln (\frac{4}{5})& =-10k\\ k& =-\frac{\ln (4/5)}{10}\end{array}$$

Solution:

$$T(t)=-25e^{\frac{t\cdot \ln (4/5)}{10}}$$

3 | Numerical Series

3.1 | Introduction to Series

Definition: Series Let $\{a_n{\}}_{n=1}^{\infty }$ be a sequence of real numbers. An infinite series be a symbolic expression of the form

$$a_1+a_2+\cdots ={\int }_{n=1}^{\infty }{a}_n$$

Definition: Partial Sums Given an infinite series $a_n$, we define its sequence of partial sums, $\{s_m{\}}_{m=1}^{\infty }$ as

$$S_m=a_1+a_2+\cdots +a+m=\sum _{n=1}^m{a}_n$$

$S_m$ is called the $m$-th partial sum of the infinite series.

Definition: Geometric Series A geometric series is a series of the form

$$\sum _{n=0}^{\infty }a{x}^n=a+ax+a{x}^2+\cdots$$

The $m$-th partial sum of the geometric series is

$$S_m=\sum _{n=0}^{m}a{x}^n=a+ax+a{x}^2+$$

Theorem: Geometric Series Test Let ${\sum }_{n=0}^{\infty }a{x}_n$ be a geometric series, where $a\ne 0$

1. If $|x|<1$, then ${\int }_{n=0}^{\infty }a{x}^n$ converges to $\frac{a}{1-x}$
2. If $|x|\ge 1$, then ${\int }_{n=0}^{\infty }a{x}^n$ diverges

Example: The series

$$\sum _{n=1}^{\infty }\frac{1}{2^n}$$

Converges to

$$\frac{a}{1-x}=\frac{\frac{1}{2}}{1-\frac{1}{2}=1}$$

Definition: Harmonic Series

$$\sum _{n=1}^{\infty }\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\dots$$

Telescoping Example: The series

$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

Converges to

$$1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\cdots -\frac{1}{\infty }=1$$

3.2 | Arithmetic Properties of Series and the Divergence Test

Definition: Tail Given an infinite series ${\sum }_{n=1}^{\infty }{a}_n$, let $S_m={\int }_{n=1}^m{a}_n=a_1+a_2+\cdots +a_m$ denote its $m$-th partial sum. The difference

$$(\sum _{n=1}^{\infty }{a}_n)-S_m=a_{m+1}+a_{m+2}+\cdots =\sum _{n=m+1}^{\infty }{a}_n$$

Is called a tail of the infinite series $a_n$

Theorem: Tail Convergence / Divergence Theorem Let $\{a_n{\}}_{n=1}^{\infty }$ be a sequence and let $j$ be a positive integer.

1. If ${\sum }_{n=1}^{\infty }{a}_n$ converges, then its tail ${\sum }_{n=j}^{\infty }{a}_n$ also converges for each $j\ge 1$
2. If the tail ${\sum }_{n=j}^{\infty }{a}_n$ converges for some $j$, then ${\sum }_{n=1}^{\infty }{a}_n$ also converges (There’s also a Tail Divergence Theorem which is just the contrapositive)

Theorem: Convergence Theorem If ${\sum }_{n_1}^{\infty }{a}_n$ converges, then ${\lim }_{n\to \infty }{a}_n=0$

Theorem: Divergence Test (n-th term divergence test) If ${\lim }_{n\to \infty }{a}_n\ne 0$, then ${\sum }_{n=1}^{\infty }{a}_n$ diverges

Katy’s examples 1: converges

$$\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3$$

2: converges

$$r=2^3{3}^{-2}=\frac{8}{9}$$ $$S=\frac{1}{1-\frac{8}{9}}=9$$

3: diverges 4: diverges

$$1+\frac{3^n}{2^n}$$

5: converges, p series $p>1$

$$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi }{6}$$

6: diverges??

$$\underset{n\to \infty }{\lim }arcsin(n)=\frac{\pi }{2}$$

Example: What is the value of $c$ if

$$\sum _{n=2}^{\infty }(1+c{)}^{-n}=2$$

Solution:

$$a=\frac{1}{(1+c{)}^2}$$ $$\sum _{n=2}^{\infty }\frac{1}{(1+c{)}^{n-2}}=\frac{1}{(1+c{)}^2}\sum _{n=0}^{\infty }\frac{1}{(1+c{)}^n}=\frac{1}{(1+c{)}^2}\cdot \frac{1}{1-\frac{1}{(1+c)}}=\frac{1}{(1+c{)}^2}\cdot \frac{1}{\frac{c}{1+c}}$$ $$=\frac{1}{(1+c{)}^2}\cdot \frac{1+c}{c}=\frac{1}{c(1+c)}=2$$ $$0=2c^2+2c-1$$ $$c=\frac{-2\pm 2\sqrt{3}}{4}=\frac{-1\pm \sqrt{3}}{2}$$

Evaluating the +

$$(1+\frac{-1+\sqrt{3}}{2}{)}^{-n}=(\frac{2}{1+\sqrt{3}}{)}^n$$ $$\underset{n\to \infty }{\lim }(\frac{2}{1+\sqrt{3}}{)}^n=0$$

Evaluating the -

$$(1+\frac{-1-\sqrt{3}}{2}{)}^{-n}=(\frac{2}{1-\sqrt{3}}{)}^n$$ $$\underset{n\to \infty }{\lim }(\frac{2}{1-\sqrt{3}}{)}^n=0$$

So the solution is the +

3.3 | Integral Test

Definition: An infinite series ${\sum }_{n=1}^{\infty }{a}_n$ is said to be:

• Positive / Nonnegative if $a_n\ge 0$ for all $n\ge 1$
• Strictly positive if $a_n>0$
• Eventually positive if there exists a positive integer $j$ such that $a_n\ge 0$ for all $n\ge j$
• Eventually strictly positive for … take a wild guess

Lemma: Let ${\sum }_{n=1}^{\infty }{a}_n$ be an eventually positive series. Let $S_m={\sum }_{n=1}^m$ denote its $m$-th partial sum. Then there are only two possibilities.

1. If the sequence of partial sums $\{s_m{\}}_{m=1}^{\infty }$ is bounded from above, the $a_n$ converges
2. If the sequence of partial sums $\{s_m{\}}_{m=1}^{\infty }$ is not bounded from above, then $a_n$ diverges to $\infty$

Theorem: Integral Test Suppose a function $f$ is continuous, positive, and decreasing on the infinite integral $[k,\infty )$ for some positive integer $k$. Then,

1. If ${\int }_k^{\infty }f(x)dx$ converges, then the infinite series $f(n)$ converges
2. If ${\int }_k^{\infty }f(x)$ diverges, then the infinite series $f(n)$ diverges to $\infty$

Example:

$$\sum _{n=1}^{\infty }\frac{1}{n},\sum _{n=1}^{\infty }\frac{1}{n^2}$$

Using the integral test:

$${\int }_k^{\infty }\frac{1}{x}=\ln (x)⟹\sum _{n=k}^{\infty }\ln (x)⟹\text{ diverges}$$ $${\int }_k^{\infty }\frac{1}{x^2}=\ln (x)⟹\sum _{n=k}^{\infty }-\frac{1}{x}⟹\text{ converges}$$

Example:

$$\sum _{n=1}^{\infty }\frac{\ln (n)}{n}$$

Integral Test:

$$f(x)=\ln (x),f^{\prime }(x)=\frac{1-\ln x}{x^2}=0$$ $$\underset{t\to \infty }{\lim }{\int }_1^t\frac{\ln x}{x}dx=\underset{t\to \infty }{\lim }\frac{(\ln x{)}^2}{2}{\mid }_1^t$$ $$\underset{t\to \infty }{\lim }\frac{(\ln (t){)}^2}{2}\to \infty ⟹\text{diverges}$$

Example:

$$\sum _{n=1}^{\infty }\frac{1}{n^2+1}$$

Using the integral test, since $f(x)$ is continuous and positive for $n\ge 1$

$$f(x)=\frac{1}{(n^2+1)}$$ $$\int f(x)=\arctan (x)+C$$ $$\underset{t\to \infty }{\lim }\arctan (t)\to \infty =\pi ⟹\text{converges}$$

Definition: p-series

$$\sum _{n=1}^{\infty }\frac{1}{n^p}=1+\frac{1}{2^p}+\frac{1}{3^p}+\dots$$

• $p>1$ converge
• $p\le 1$ diverge

Example: The following series converges because $2>1$

$$\sum _{n=1}^{\infty }\frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\dots$$

Theorem: Integral Test Estimation Theorem Suppose that a function $f$ is continuous, positive, and decreasing on the infinite interval $[k,\infty )$ for some positive integer $k$, and that ${\int }_k^{\infty }f(x)dx$ converges so that ${\sum }_{n=1}^{\infty }f(n)=S$ for some real number $S$ by the integral test. Let $S_m={\sum }_{n=1}^{m}f(n)$ denote the $m$-th partial sum. Then for any integer $m\ge k$,

$${\int }_{m+1}^{\infty }f(x)dx\le S-S_m\le {\int }_m^{\infty }f(x)dx$$

We denote $R_n=S-S_n$

Example: How many terms are needed to approximate ${\sum }_{n=1}^{\infty }\frac{1}{n^2}$ with an error of at most 0.001

Solution:

$${\int }_1^{\infty }\frac{1}{x^2}dx=\underset{t\to \infty }{\lim }-\frac{1}{x}{\mid }_n^t=\frac{1}{n}\le 0.001$$ $$n\ge 1000$$

Example: Find an upper bound on the error if we choose the partial sum $S_{10}$ to estimate $S={\sum }_{n=1}^{\infty }\frac{1}{n^2}$

Solution:

$$S-S_{10}\le {\int }_{10}^{\infty }\frac{1}{x^2}dx$$

3.4 | Comparison Test and Limit Comparison Test

Theorem: Comparison Test For any two sequences $a_n\le b_n$

• If $b_n$ converges, then any smaller series $a_n$ converges
• If $a_n$ diverges, $b_n$ also diverges

Examples:

$$\sum \frac{n}{n^2+7}$$

Solution: Since $\sum \frac{1}{n^2}$ converges (p-test, $p=2\ge 1$), then by comparison, so does $\sum \frac{n}{n^3+7}$

$$\sum \frac{n+7}{n^2-1}$$

Solution:

$$\frac{n+7}{n^2-1}\ge \frac{n}{n^2-1}\ge \frac{n}{n^2}=\frac{1}{n}\ge 0$$

Since $\sum \frac{1}{n}$ diverges (Harmonic series, p-series), then by comparison, $\sum \frac{n+7}{n^2-1}$ also diverges

$$\sum \frac{n^3-n}{n^4+7}$$

Solution:

$$\frac{n^3-n}{n^4+7}\le \frac{n^3}{n^4+7}\le \frac{n^3}{n^4}\le \frac{1}{n}$$

Since $\sum \frac{1}{n}$ diverges, BUT $\frac{n^3-n}{n^4+7}$ is less than $\frac{1}{n}$, we get no information about it!

Theorem: Limit Comparison Test For any two sequences $a_n\ge 0$, $b_n>0$, let

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}>0$$

1. If $0<L<\infty$ either both converge or both diverge
2. If $L=0$, there exists a positive integer $j$ such that $a_n<b_n$ for all $n\ge j$. Consequently:
   1. If $b_n$ converges, then $a_n$ does too.
   2. If $a_n$ diverges, then $b_n$ does too.
3. If $L=\infty$, then there is a positive integer such that $a_n>b_n$ for all $n\ge j$. Consequently:
   1. If $a_n$ converges, then $b_n$ does too.
   2. If $b_n$ diverges, then $a_n$ does too.
4. If $L=DNE$, we don’t know anything

When to use LCT?

$$\sum \frac{\text{powers of n}}{\text{powers of n}}$$

Almost geometric series:

$$\sum \frac{2^n+n}{3^n+1}$$

Examples:

$$\sum \frac{n^3-n}{n^4+7}\sim \frac{1}{n}$$

Solution:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{n^3-n}{n^4+7}\cdot \frac{n}{1}=1$$

By LCT, since $\frac{1}{n}$ diverges, so does $\frac{n^3-n}{n^4+7}$

$$\sum \frac{2^n-1}{3^n+n}\sim (\frac{2}{3}{)}^n$$

Solution:

$$\underset{n\to \infty }{\lim }\frac{2^n-1}{3^n+n}\cdot \frac{3^n}{2^n}=\underset{n\to \infty }{\lim }\frac{2^n{3}^n}{3^n{2}^n}(\frac{1-{\frac{1}{2}}^n}{1-{\frac{2}{3}}^n})=1$$

Since $(\frac{2}{3}{)}^n$ converges (geometric series), by LCT, so does $\frac{2^n+n}{3^n+1}$

$$\sum \frac{\sqrt{n^2+5n}+3}{n^{7/4}+3n-1}\sim \frac{1}{n^{3/4}}⟹\text{diverges}$$

3.5 | Alternating Series Test

Definition: Alternating Series Let $\{a_n{\}}_{n=1}^{\infty }$ be a strictly positive series. An alternating series is an infinite series that is in one of the following forms:

$$(i)\sum _{n=1}^{\infty }(-1{)}^{n-1}{a}_n=a_1-a_2+a_3-a_4+\cdots$$ $$(ii)\sum _{n=1}^{\infty }(-1{)}^n{a}_n=-a_1+a_2-a_3+a_4-\cdots =-\sum _{n=1}^{\infty }(-1{)}^{n-1}{a}_n$$

Theorem: Alternating Series Test $\{a_n{\}}_{n=1}^{\infty }$ converges if

1. ${\lim }_{n\to \infty }{a}_n=0$
2. $a_n$ is eventually decreasing; there exists a positive integer $k$ such that $a_n\ge a_{n+1}>0$ for all $n\ge k$

Example:

$$\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n}=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n}$$

1. ${\lim }_{n\to \infty }\frac{1}{n}=0$
2. $\frac{1}{n}\ge \frac{1}{n+1}$ BY AST this converges

Example

$$\sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{n^2}{n^3+1}$$

1. ${\lim }_{n\to \infty }\frac{1}{n}=0$
2. Use derivative to find decreasing

$$f(x)=\frac{x^3}{x^3+1}$$ $$f^{\prime }(x)=\frac{2x(x^3+1)-x^2(3x^2)}{(x^3+1{)}^2}=\frac{2x-x^4}{(x^3+1{)}^2}=\frac{x(2-x^3)}{(x^3+1{)}^2}$$ $$f^{\prime }(x)=0⟹x=0,x=\sqrt[3]{2}$$ $$f^{\prime }(x)<0⟹x>\sqrt[3]{2}$$

Thus, $\frac{n^2}{n^3+1}$ is decreasing on $n\ge 2$

Example

$$\sum _{n=1}^{\infty }\frac{(-1{)}^{n}3n}{4n-1}$$

1. ${\lim }_{n\to \infty }=\frac{3}{4}\ne 0$ - criteria fails!

Use Divergence Test:

$$\underset{n\to \infty }{\lim }\frac{(-1{)}^{2}3n}{4n-1}\cdot \frac{\frac{1}{n}}{\frac{1}{n}}=\underset{n\to \infty }{\lim }\frac{(-1{)}^{n}3}{4-\frac{1}{n}}=\underset{n\to \infty }{\lim }\frac{(-1{)}^{n}3}{4}=\frac{3}{4}\underset{n\to \infty }{\lim }(-1{)}^n=DNE$$

Since $\lim \ne 0$, this diverges!

Theorem: Alternating Series Estimation Theorem Let $\{a_n{\}}_{n=1}^{\infty }$ be a sequence that satisfies the Alternating Series Test, so that

$$\sum _{n=1}^{\infty }(-1{)}^{n-1}{a}_n\to S,\sum _{n=1}^{\infty }(-1{)}^n{a}_n\to T$$

Let $S_m$ denote the $m$-th partial sum of $S$, and $T_m$ denote the $m$-th partial sum of $T$. Suppose $m\ge k$, where $k$ is the positive integer in part ii) of AST hypothesis. Then,

1. $|S-S_m|\le a_{m+1}$ and $|T-T_m|\le a_{m+1}$
2. If $m$ is even, then $S_m$ is an underestimate of $S$ and $T_m$ is an overestimate of $T$
3. If $m$ is odd, then $S_m$ is an overestimate of $S$ and $T_m$ is an underestimate of $T$

Example: Find an upper bound on the remainder if we use $S_{10}$ to approximate

$$|R_{10}|\le b_{11}=\frac{(-1{)}^{12}}{11^2}=\frac{1}{121}$$

How many terms are needed to approximate ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{n{5}^n}$ with an error of most $10^{-4}$

$$|R_n|\le b_{n+1}=\frac{1}{(n+1)5^{n+1}}\le \frac{1}{10000}⟹$$ $$n=4$$

Is the 121st partial sum of $\frac{(-1{)}^{n+1}{3}^n}{n{4}^n}$ an over or underestimate? $S_1$ is over, $S_2$ is under, so $S_{121}$ is an overestimate

3.6 | Absolute and Conditional Convergence

Definition: Absolutization

$$\sum _{n=1}^{\infty }{a}_n\text{ converges absolutely if }\sum _{n=1}^{\infty }|a_n|\text{ converges}$$

Theorem: Absolute Convergence Test

$$\text{If }\sum _{n=1}^{\infty }|a_n|\text{ converges, then }\sum _{n=1}^{\infty }{a}_n\text{ converges}$$

Example:

$$\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n}\text{ converges}$$ $$\sum _{n=1}^{\infty }\frac{1}{n}\text{ diverges}$$

So this series is conditionally convergent!

Example:

$$\sum _{n=1}^{\infty }\frac{\cos (n)}{n^2}$$

Consider

$$\sum _{n=1}^{\infty }\frac{|\cos (n)|}{n^2}$$

Comparison test:

$$-\frac{1}{n^2}<\frac{\cos (n)}{n}<\frac{1}{n^2}$$

Since $\frac{1}{n^2}$ converges (p-series), we have that $\frac{\cos (n)}{n^2}$ neither diverges to $\infty$ or $-\infty$. Thus, it converges

Order to apply tests

1. Divergence test
2. Check for absolute convergence
   1. p-series
   2. geometric
   3. comparison
   4. integral
3. Check for conditional convergence
   1. alternating series

Examples:

$$\sum _{n=1}^{\infty }\frac{\sqrt{n^3+1}}{3n^3+4n^2+2}$$

1. Divergence test fails (limit = 0)
2. Comparison test with $\frac{1}{3n^{3/2}}$

$$\underset{n\to \infty }{\lim }\frac{\sqrt{n^3+1}}{3n^3+4n^2+2}/\frac{1}{3n^3}=\underset{n\to \infty }{\lim }\frac{3\sqrt{n^6+n^3}}{3n^3+4n^2+2}=1$$

By LCT since $\sum \frac{1}{3n^{3/2}}$ converges (p-series), this also CONVERGES ABSOLUTELY

$$\sum _{n=1}^{\infty }n{e}^{-n^2}$$

3. Divergence test fails
4. Integral test with $u=x^2$

$${\int }_1^{\infty }x{e}^u\frac{du}{2x}=-\frac{1}{2}{\int }_1^{\infty }{e}^{-}u=-\frac{1}{2}[e^{-\infty }-e^{-1}]⟹\text{converrges absolutely}$$ $$\sum _{n=1}^{\infty }\frac{(-1{)}^n{4}^n}{3^n}=(-1{)}^n(\frac{4}{3}{)}^n⟹\text{diverges}$$ $$\sum _{n=1}^{\infty }(-1{)}^n\frac{n^3}{n^4+1}$$

1. Divergence test fails
2. LCT with $\frac{1}{n}$

$$\underset{n\to \infty }{\lim }\frac{n^3}{n^4+1}/\frac{1}{n}=\underset{n\to \infty }{\lim }\frac{n^4}{n^4+1}=1⟹\text{diverges}$$

3. AST

$$\underset{n\to \infty }{\lim }\frac{n^3}{n^4+1}=0$$ $$\frac{(n+1{)}^3}{(n+1{)}^4+1}\le \frac{n^3}{n^4+1}$$

Thus, this series CONVERGES CONDITIONALLY

3.7 | Ratio Test and Root Test

Theorem: Ratio Test Let $k$ be a positive integer. Suppose that $\{a_n{\}}_{n=1}^{\infty }$ is a sequence satisfying $a_n\ne 0$ for every $n\ge k$. Let $L$ be the following, and suppose that either $L\in \mathbb{R}$ or $L=\infty$

$$L=\underset{n\to \infty }{\lim }|\frac{a_{n+1}}{a_n}|$$

1. If $L<1$, then $a_n$ converges absolutely
2. If $L>1$, then $a_n$ diverges
3. If $L=1$, then the test is inconclusive

Example

$$\sum _{n=1}^{\infty }\frac{n^2+3n}{5n}$$ $$\begin{array}{rl}& \underset{n\to \infty }{\lim }|\frac{(n+1{)}^2+3(n+1)}{5^{n+1}}/\frac{n^2+3n}{5^n}\\ =& \underset{n\to \infty }{\lim }|\frac{(n+1{)}^2+3(n+1)}{5^{n+1}}\cdot \frac{5^n}{n^2+3n}|\\ =& \underset{n\to \infty }{\lim }\frac{1}{5}\cdot \frac{(n+1{)}^2+3(n+1)}{n^2+3n}=\frac{1}{5}<1\end{array}$$

Example:

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }|\frac{(-1{)}^{n+1}{9}^{n+1}}{(n+1)2^{n+1}}\cdot \frac{n{2}^n}{(-1{)}^n{9}^n}|\\ =& \underset{n\to \infty }{\lim }\frac{9n}{2(n+1)}\\ =& \frac{9}{2}⟹\text{diverges}\end{array}$$

Example

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }|\frac{(n+1{)}^2+2(n+1)+1}{3(n+1{)}^4}\cdot \frac{3n^4+4}{n^2+2n+1}|\\ =& \underset{n\to \infty }{\lim }\frac{3n^6\cdots }{3n^6\cdots }\\ =& 1⟹\text{inconclusive}\end{array}$$

Example:

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }|\frac{(n+1{)}^{n+1}}{(n+1)!}\frac{n!}{n^n}|\\ =& \underset{n\to \infty }{\lim }|\frac{(n+1{)}^{n+1}}{(n+1)n^n}|\\ =& \underset{n\to \infty }{\lim }|(\frac{n+1}{n}{)}^n|\\ =& \underset{n\to \infty }{\lim }(1+\frac{1}{n}{)}^n\\ =& e>1⟹\text{diverges}\end{array}$$

Example:

$$\underset{n\to \infty }{\lim }|\frac{x^{n+1}}{(n+1)!}\frac{n!}{x^n}=0⟹\text{converges}$$

Theorem: Root Test Let $L$ be the following, and suppose that $L\in \mathbb{R}$ or $L=\infty$

$$L=\underset{n\to \infty }{\lim }\sqrt[n]{|a_n|}$$

1. If $L<1$, then $a_n$ converges absolutely
2. If $L>1$, then $a_n$ diverges
3. If $L=1$, then the test is inconclusive

Example:

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }\sqrt[n]{|\frac{n+1}{3n+7}{|}^n}=\frac{1}{3}<1⟹\text{converges}\end{array}$$

Comparing Values

$$(\ln (n){)}^p<<n^p<<x^n<<n!<<n^n$$

Series Recap

1. Sum of Geometric and Telescoping Series
   1. $\sum a{r}^{n-1},\sum \frac{1}{n}-\frac{1}{n+1}$
2. Divergence Test (any series) 2. ${\lim }_{n\to \infty }{a}_n$ 3. Use this first
3. Integral Test (positive series)
   1. Last resort, must be continuous, positive, decreasing
4. P-series 2. $\sum \frac{1}{n^p}$ 3. Good for comparison and limit comparison
5. Comparison Test
   1. Polynomial / Polynomial
   2. Also last resort, LCT usually better
6. Ratio Test (any series)
   1. Factorials / Exponents
7. Root Test (any series)
   1. Powers of n
8. Alternating Series 2. Only proves conditional convergence

4 | Power Series

4.1 | Introduction to Power Series

Definition: Power Series A power series is a series of the form (centered at 0)

$$\sum _{n=0}^{\infty }{a}_n{x}^n=a_0+a_{1}x+a_2{x}^2+\dots$$

Or (centered at a)

$$\sum _{n=0}^{\infty }{a}_n(x-a{)}^n=a_0+a_1(x-a)+a_2(x-a{)}^2$$

The domain of a power series is the collection of all $x\in \mathbb{R}$ for which the power series converges

Notes:

$$\sum _{n=0}^{\infty }{a}_n(x-a{)}^n$$

1. When $n=0$, the term is $a_0$
2. If the first few coefficients are zero, $a_0=a_1=\cdots =a_k=0$, then

$$\sum _{n=0}^{\infty }{a}_n(x-a{)}^n=\sum _{n=k+1}^{\infty }{a}_n(x-a{)}^n$$

Example: Find the domain of

$$\sum _{n=0}^{\infty }n!x^n$$

Ratio test:

$$\begin{array}{rl}\underset{n\to \infty }{\lim }|\frac{(n+1)!x^{n+1}}{n!x^n}& =\underset{n\to \infty }{\lim }|(n+1)x|\\ & =\underset{n\to \infty }{\lim }(n+1)|x|\\ & =\infty \end{array}$$

Domain: $x=0$

Example: Find the domain of

$$\sum _{n=0}^{\infty }\frac{(x-3{)}^n}{n}$$

Ratio test:

$$\begin{array}{rl}\underset{n\to \infty }{\lim }|\frac{(x-3{)}^{n+1}}{n+1}\frac{n}{(x-3{)}^n}|& =\underset{n\to \infty }{\lim }|\frac{(x-3)n}{n+1}|\\ & =\underset{n\to \infty }{\lim }|(x-3)\frac{n}{n+1}|\\ & =|x-3|\end{array}$$

By the ratio test, $|x-3|<1⟹x\in [2,4)$ If $x=2,\sum \frac{(-1{)}^n}{n}$ converges by AST If $x=4,\sum \frac{1^n}{n}$ diverges by p-series

Theorem: Power Series For any power series ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$, there are only three possibilities

1. The series converges only when $x=a$
2. The series converges for all $x\in \mathbb{R}$
3. There exists $R\in \mathbb{R}$ such that the series converges absolutely for $|x-a|<R$, diverges if $|x-a|>R$, and may converges or diverge if $|x-a|=R$

Definition: Radius and Interval of Convergence The $R$ in the previous theorem is called the radius of convergence, and the domain / interval of convergence is the interval on which the power series converges

1. $R=0$
2. $R=\infty$
3. $R\in (0,\infty )$ - check endpoints in this case

Find the radius and intervals of convergence: Q1

$$\sum _{n=0}^{\infty }\frac{(-3{)}^n{x}^n}{\sqrt{n+1}}$$ $$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{(-3{)}^{n+1}{x}^{n+1}}{\sqrt{n+2}}\frac{\sqrt{n+1}}{(-3{)}^n{x}^n}& =\underset{n\to \infty }{\lim }-3x\frac{\sqrt{n+1}}{\sqrt{n+2}}\\ & =|-3x|<1\\ & ⟹x\in (-\frac{1}{3},\frac{1}{3}]\\ & ⟹R=3\end{array}$$

Q2

$$\sum _{n=0}^{\infty }\frac{n(x+2{)}^n}{3^{n+1}}$$ $$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{(n+1)(x+2{)}^{n+1}}{3^{n+2}}\frac{3^{n+1}}{n(x+2{)}^n}& =\underset{n\to \infty }{\lim }\frac{n+1}{n}\frac{x+2}{3}\\ & =|\frac{x+2}{3}|<1\\ & ⟹|x+2|<3\\ & ⟹x\in (-5,1)\\ & ⟹R=3\end{array}$$

Q3

$$\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{2^{2n}(n!{)}^2}$$ $$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}{x}^{2n+2}}{2^{2n+2}(n+1){!}^2}\frac{2^{2n}(n!{)}^2}{(-1{)}^n{x}^{2n}}& =-x^2\frac{1}{2^2(n+1{)}^2}\\ & =-\frac{x^2}{4(n+1{)}^2}=0<1\\ & ⟹x\in \mathbb{R}\\ & ⟹R=\infty \end{array}$$

4.2 | Representing Functions as Power Series

Theorem: Abel’s Theorem Let $I$ denote the interval of convergence for the power series function

$$f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$$

Then $f$ is continuous on $I$

Properties Let $f(x)=\sum c_n(x-a{)}^n$ and $g(x)=d_n(x-a{)}^n$ Let $R_f,R_g,I_f,I_g$ be the radii and intervals of convergence for $f,g$ respectively. Property 1:

$$h(x)=f(x)\pm g(x)=\sum _{n=0}^{\infty }(c_n\pm d_n)(x-a{)}^n$$

• If $R_f\ne R_g$, then $R_h=\min \{R_f,R_g\}$ and $I_h=I_f\cap I_g$
• If $R_f=R_g$, the $R_h\ge R_f$

Property 2:

$$p(x)=(x-a{)}^{k}f(x)=\sum _{n=0}^{\infty }{x}_n(x-a{)}^{n+k}$$

$R_p=R_f,I_p=I_f$ Property 3:

$$q(x)=f(b{x}^k)=\sum _{n=0}^{\infty }{c}_n{b}^n{x}^{kn}$$

$R_q=(\frac{R_f}{|b|}{)}^{1/k}$, and if $R_f=\infty ,R_q=\infty$ $I_q=\{x\in \mathbb{R}\mid b{x}^k\in I_f\}$

Example: Express $\frac{1}{1+x^2}$ as a power series and find the interval of convergence.

$$\begin{array}{rl}\frac{1}{1+x^2}& =\frac{1}{1-(-x^2)}\\ & =\sum _{n=0}^{\infty }(-x^2{)}^n\\ & =\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}\\ |-x^2|& <1\\ x^2& <1\\ |x|& <1\end{array}$$

Example:

$$\begin{array}{rl}\frac{1}{x+2}& =\frac{1}{2}\frac{1}{1-(-\frac{x}{2})}\\ & =\frac{1}{2}\sum _{n=0}^{\infty }(-\frac{x}{2}{)}^n\\ & =\sum _{n=0}^{\infty }(-1{)}^n\frac{x^n}{2^{n+1}}\end{array}$$

Radius:

$$\begin{array}{rl}|-\frac{x}{2}|& <1\\ |x|& <2\\ R& =2\\ I& =(-2,2)\end{array}$$

Example:

$$\begin{array}{rl}\frac{1}{4-x^2}& =\frac{1}{x}\frac{1}{1-\frac{x^2}{4}}\\ & =\frac{1}{4}\sum _{n=0}^{\infty }{\left(\frac{x^2}{4}\right)}^n\\ & =\sum _{n=0}^{\infty }\frac{x^{2n}}{4^{n+1}}\end{array}$$

Radius

$$\begin{array}{rl}|\frac{x^2}{4}|& <1\\ x^2& <4\\ |x|& <2\\ R& =2\\ I& =(-2,2)\end{array}$$

Example:

$$\begin{array}{rl}\frac{x^3}{x+2}& =x^3\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^n}{2^{n+1}}\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{n+3}}{2^{n+1}}\end{array}$$

Radius $R=2,I=(-2,2)$

Example

$$\begin{array}{rl}\frac{1+x}{1-x}& =1+\frac{2x}{1-x}\\ & =1+2x(\frac{1}{1-x})\\ & =1+2x\sum _{n=0}^{\infty }{x}^n\\ & =1+\sum _{m=0}^{\infty }2x^{n+1}\end{array}$$

Radius

$$|x|<1,I=(-1,1)$$

Example

$$\begin{array}{rl}\frac{3}{x^2-x-2}& =\frac{3}{(x-2)}(x+1)\\ & =\frac{1}{x-2}-\frac{1}{x+1}\\ & =-\frac{1}{2}\left(\frac{1}{1-\frac{x}{2}}\right)-\frac{1}{1-(-x)}\\ & =-\sum _{n=0}^{\infty }\frac{x^n}{2^{n+1}}-\sum _{n=0}^{\infty }(-1{)}^n{x}^n\\ & =\sum _{n=0}^{\infty }\left[(-1{)}^{n+1}-\frac{1}{2^{n+1}}\right]x^m\end{array}$$

4.3 | Differentiating and Integrating Power Series

Theorem: Let $f(x)={\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$ with radius of convergence $R>0$. Then, $f(x)$ is differentiable (and continuous and integrable) on $(a-R,a+R)$. In addition:

$$f^{\prime }(x)=\sum _{n=1}^{\infty }{c}_n(x-a{)}^{n-1},x\in (a-R,a+R)$$ $$\int f(x)dx=C+\sum _{n=0}^{\infty }{c}_n\frac{(x-a{)}^{n+1}}{n+1}$$ $${\int }_b^{d}f(x)dx=\sum _{n=0}^{\infty }{c}_n[\frac{(x-a{)}^{n+1}}{n+1}]$$

Example:

$$\begin{array}{rl}\frac{1}{1-x}& =\sum _{n=1}^{\infty }{x}^n\text{(diffentiate)}\\ -1(1-x{)}^{-2}(-1)& =\sum _{n=1}^{\infty }n{x}^{n-1}\\ \frac{5}{(1-x{)}^2}& =5\sum _{n=1}^{\infty }n{x}^{n-1}\\ & =\sum _{n=1}^{\infty }5n{x}^{n-1},R=1\end{array}$$

$x=-1⟹\sum 5n(-1{)}^{n-1}$ diverges by divergence test $x=1⟹\sum 5n(1)$ diverges by divergence test $I(-1,1)$

Example

$$\begin{array}{rl}\frac{1}{1+x^7}& =\frac{1}{1-(-x^7)}\\ & =\sum _{n=0}^{\infty }(-x^7{)}^n\\ & =\sum _{n=0}^{\infty }(-1{)}^n{x}^{7n}\\ \int \frac{1}{1+x^7}dx& =\int \sum _{n=0}^{\infty }(-1{)}^n{x}^{7n}dx\\ & =C+\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{7n+1}}{7n+1}\end{array}$$

To test endpoints, choose $C=0$ (this doesn’t affect convergence / divergence) $x=-1$

$$\sum \frac{(-1{)}^n(-1{)}^{7n+1}}{7n+1}=\sum \frac{(-1{)}^{8n+1}}{7n+1}-\sum \frac{-1}{7n+1}$$

Diverges by LCT $x=1$

$$\sum \frac{(-1{)}^n(1{)}^{7n+1}}{7n+1}=\sum \frac{(-1{)}^n}{7n+1}$$

Converges by AST $I=(-1,1]$

Example:

$$\begin{array}{rl}\frac{1}{1-x}& =\sum x^n\\ \frac{1}{-(-x)}& =\sum (-x{)}^n\\ \int \frac{1}{1-(-x)}& =\int \sum (-1{)}^n{x}^{n}dx\\ \ln |1+x|& =C+\sum \frac{(-1{)}^n{x}^{n+1}}{n+1}\\ R& =1\end{array}$$

Let $x=0,\ln |1+0|=...$ Endpoints: $x=-1$

$$\sum \frac{(-1{)}^n(-1{)}^{n+1}}{n+1}=\sum \frac{(-1{)}^{2n+1}}{n+1}=\sum -\frac{1}{n+1}$$

Diverges by LCT $x=1$

$$\sum \frac{(-1{)}^n}{n+1}$$

Converges by AST $I=(-1,1]$

Theorem:

$$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$

Proof:

$$g(x)=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$ $$g^{\prime }(x)=\sum _{n=1}^{\infty }\frac{n{x}^{n-1}}{n!}=\sum _{n=1}^{\infty }\frac{x^{n-1}}{(n-1)!}=\sum _{n=0}^{\infty }\frac{x^n}{n!}=g(x)$$ $$\begin{array}{rl}\frac{dg(x)}{dx}& =g(x)\\ \int \frac{1}{g(x)}dg(x)& =\int 1dx\\ \ln |g(x)|& =x+C\\ |g(x)& =e^c{e}^x|\\ g(x)& =\pm e^c{e}^x\\ g(x)& =e^x\end{array}$$

4.4 | Taylor Series and Taylor Polynomials

3b1b Explanation

Pendulum approximation

$$\cos (\theta )\approx 1-\frac{{\theta }^2}{2}$$

Taylor Series approximate non-polynomial functions with polynomials.

Ex: Approximate $\cos (x)=0$ with a function $P(x)=c_0+c_{1}x+c_2{x}^2$ We know that $P(0)=1$, so $P(x)=1+c_{1}x+c_2{x}^2⟹c_0=1$

We want the derivative of $\cos (x)$ to be similar to the derivative of $P(x)$ ${\cos }^{\prime }(0)=-\sin (0)=0$ $P^{\prime }(0)=c_1+2c_2(0)=c_1=0⟹c_1=0$ $P(x)=1+0x+c_2{x}^2$

We want the second derivative of $\cos (x)$ to be similar as well ${\cos }^{\prime \prime }(0)=-\cos (0)=-1$ $P^{\prime \prime }(0)=0+2c_2=2c_2=-1⟹c_2=-\frac{1}{2}$ $P(x)=1+0x-\frac{1}{2}{x}^2=1-\frac{1}{2}{x}^2$

Observe:

• We can add a term $\frac{1}{24}{x}^4$ if we wanted to get even closer
• There are many factorial terms ($4!=24$)
• Adding new terms doesn’t change the previous terms
• Observations about higher-order derivatives make the approximation more accurate

Taylor Polynomial

$$P(x)=f(0)+\frac{df}{dx}(0)\frac{x^1}{1!}+\frac{d^{2}f}{d{x}^2}(0)\frac{x^2}{2!}+\frac{d^{3}f}{d{x}^3}(0)\frac{x^3}{3!}+\cdots$$

Taylor Polynomial around a point $a$:

$$P(x)=f(a)+\frac{df}{dx}(a)\frac{(x-a{)}^1}{1!}+\frac{d^{2}f}{d{x}^2}(a)\frac{(x-a{)}^2}{2!}+\cdots$$

Example: Approximating $e^x$ around 0

$$P(x)=1+1\frac{x^1}{1!}+1\frac{x^2}{2!}+\cdots$$

Example: Approximating the integral function of $f(x)$

Base = $(x-a)$ Height = $\frac{d^2{f}_{area}}{d{x}^2}(a)(x-a)$ Triangle = $\frac{1}{2}\frac{d^2{f}_{area}}{d{x}^2}(a)(x-a{)}^2$

$$f(x)\approx f(a)+\frac{df}{dx}(a)(x-a)+\frac{1}{2}\frac{d^{2}f}{d{x}^2}(a)(x-a{)}^2$$

Does it make sum to keep approximating without stopping?

• Taylor Series = Infinitely long Taylor Polynomial
• Certain series converge to a certain value
• For instance, the Taylor Polynomial for $e^x$ converges to $e$ at all inputs $x$
• The Taylor Polynomial for $\ln (x)$ converges to $x$ on $[0,1]$, otherwise it diverges
• This is called the Radius of Convergence
• Lagrange Error Bound, convergence tests, etc.

Textbook

Definition: The $n$-th derivative of $f$ will be denoted by $f^{(n)}(x)$

Theorem: Uniqueness of Power Series Representation Suppose that $f$ has a power series representation centered at $x=a$, meaning

$$f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$$

holds for all $x$ satisfying $|x-a|<R$, where $R>0$. Then,

$$c_n=\frac{f^{(n)}(a)}{n!}$$

Definition: Taylor and Maclaurin Series Let $f$ be differentiable arbitrarily many times at $x=a$, i.e., $f^{(n)}(a)$ is a well-defined real number for every nonnegative integer $n$. The power series

$$\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$

is called the Taylor Series for $f$ centered at $x=a$. The Maclaurin Series is a Taylor Series centered at 0.

$$\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$$

Definition: Taylor Polynomial Assume that $f$ is $m$-times differentiable at $x=a$. The $m$-th degree Taylor Polynomial for $f$ centered at $x=a$ is

$$\begin{array}{rl}{T}_{m,a}(x)& =\sum _{n=0}^m\frac{f^{(n)}(a)}{n!}(x-a{)}^n\\ & =f(a)+f^{\prime }(a)(x-a)+\dots +\frac{f^{(m)}(a)}{m!}(x-a{)}^m\end{array}$$

We note that $T_{m,a}(x)$ is the $m$-th partial sum of the Taylor series for $f$ centered at $x=a$

Theorem: For every integer $j\in \{0,1,2,\dots ,m\}$, we have that

$$T_{m,a}^{(j)}(a)=F^{(j)}(a)$$

Theorem: Lagrange Remainder Formula Let $m$ be a nonnegative integer. Suppose that $f^{(m+1)}$ is continuous on an open interval $I$ that contains $a$. If $x\in I$, then there exists a real number $c\in [a,x]$ such that

$$f(x)-T_{m,a}(x)=\frac{f^{(m+1)}(c)}{(m+1)!}(x-a{)}^{m+1}$$

Theorem: Taylor’s Inequality Let $m$ be a nonnegative integer. Suppose that $f^{(m+1)}$ is continuous on an open interval $I$ that contains $a$. If $|f^{(m+1)}(x)\le K$ for every $x\in I$, then we have

$$|f(x)-T_{m,a}(x)|\le \frac{K|x-a{|}^{m+1}}{(m+1)!}$$

Theorem: Convergence Theorem for Taylor Series Assume that $f$ has derivatives of all orders on an open interval $I$ that contains $a$. Further assume that there is a constant $K$ such that $f^{(n)}(x)\le K$ for every nonnegative integer $n$ and for every $x\in I$. Then for every $x\in I$

$$f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$

4.5 | Some examples of Taylor Series

Examples: Taylor Series

Write a Maclaurin Series for $f(x)=\sin x$ $f(x)=\sin x⟹f(0)=0$ $f^{\prime }(x)=\cos x⟹f^{\prime }(0)=1$ $f^{(2)}(x)=-\sin x⟹f^{\prime (2)}=0$ $f^{(3)}(x)=-\cos x⟹f^{(3)}(0)=-1$

$$\begin{array}{rl}\sin (x)& =x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}\end{array}$$

Write a Maclaurin series for $f(x)=\cos x$ $f(x)=\cos x$ $f^{\prime }(x)=-\sin x$ $f^{\prime \prime }(x)=-\cos x$ $f^{\prime \prime \prime }(x)=\sin x$

$$\begin{array}{rl}\frac{d\sin (x)}{dx}& =\frac{d}{dx}\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)}!\\ \cos (x)& =\sum _{n=0}^{\infty }\frac{(-1{)}^n(2n+1)x^{2n}}{(2n+1)!}\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}\end{array}$$

Known Power Series

$$\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^n,R=1$$ $$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!},R=\infty$$ $$\sin x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!},R=\infty$$ $$\cos x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!},R=\infty$$

Example: Find the Maclaurin Series and radius for $f(x)=x{e}^{-x}$

$$e^{-x}=\sum _{n=0}^{\infty }\frac{(-x{)}^n}{n!}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^n}{n!}$$ $$f(x)=x{e}^{-x}=x\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^n}{n!}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{n+1}}{n!},R=\infty$$

Example: Find the first three nonzero terms of the Maclaurin Series for $f(x)=e^{-x^2}\cos (x)$

$$e^{-x^2}=\sum _{n=0}^{\infty }\frac{(-x^2{)}^n}{n!}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{n!}$$ $$\cos (x)=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n+1)!}$$ $$\begin{array}{rl}f(x)& =(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots )\cdot (1-\frac{x^3}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots )\\ & =1-\frac{3}{2}{x}^3+\frac{25}{24}{x}^4+\cdots \end{array}$$

Example: Write the following as an infinite sum

$$\begin{array}{rl}{\int }_0^1\cos (x^2)dx& ={\int }_0^1\sum _{n=0}^{\infty }\frac{(-1{)}^n(x^2{)}^{2n}}{(2n)!}\\ & ={\int }_0^1\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{4n}}{(2n)!}\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{4n+1}}{(4n+1)(2n)!}{\mid }_0^1\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(4n+1)(2n)!}\end{array}$$

Examples: Find the Taylor Series and interval of convergence for:

$$\begin{array}{rl}{2}^x& =(e^{\ln 2}{)}^x\\ & =\sum _{n=0}^{\infty }\frac{(\ln (2)x{)}^n}{n!}\\ & =\sum _{n=0}^{\infty }\frac{(\ln 2{)}^n{x}^n}{n!}\end{array}$$ $$\begin{array}{rl}{\sin }^2(x)& =\frac{1}{2}(1-\cos (2x))\\ & =\frac{1}{2}-\frac{1}{2}\cos (2x)\\ & =\frac{1}{2}-\frac{1}{2}\sum _{n=0}^{\infty }\frac{(-1{)}^n(2x{)}^{2n}}{(2n)!}\\ & =\frac{1}{2}+\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}{2}^{2n}{x}^{2n}}{2(2n)!}\\ & =\frac{1}{2}=\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}{2}^{2n-1}{x}^{2}n}{(2n)!}\end{array}$$ $$\begin{array}{rl}\ln \left(\frac{1+x}{1-x}\right)& =\ln (1+x)-\ln (1-x)\\ \int \frac{1}{1-(-x)}-\frac{1}{1-x}& =\int \sum _{n=0}^{\infty }(-x{)}^n-\sum _{n=0}^{\infty }{x}^n\\ \ln (1+x)-\ln (1-x)& =\sum _{n=0}^{\infty }\frac{(-x{)}^{n+1}}{n+1}-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}+C\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}{x}^{n+1}}{n+1}+\frac{x^{n+1}}{n+1}+C\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}-1}{n+1}{x}^{n+1}\end{array}$$

Note: all odd $n$ have 0 as a coefficient, all even $n$ have 2

$$\ln \left(\frac{1+x}{1-x}\right)=\sum _{n=0}^{\infty }-\frac{2x^{2n+1}}{2n+1}$$

Ex: Determine an explicit formula of the power series

$$\begin{array}{rl}1-(\ln 2)x+\frac{(\ln 2{)}^2}{2}{x}^2-\frac{(\ln 2{)}^3}{6}{x}^3+\dots & =\sum _{n=0}^{\infty }\frac{(-\ln (2)x{)}^n}{n!}\\ & =e^{(-\ln (2)x)}\\ & =e^{\ln 2^{-x}}\\ & =\frac{1}{2^x}\end{array}$$ $$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{3^n}{5^{n}n!}& =\sum _{n=0}^{\infty }\frac{(3/5{)}^n}{n!}\\ & =e^{3/5}\end{array}$$

Examples: Taylor Polynomials

Find the 2, 4, 6th degree Taylor Polynomial centered at $x=0$ for

$$\begin{array}{rlrl}f(x)& =x\sin x& & \\ & =x\sum \frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}& & \\ & =\sum \frac{(-1{)}^n{x}^{2n+2}}{(2n+1)!}& & \\ T_{2,0}(x)& =x^2& & (n=0)\\ T_{4,0}(x)& =x^2-\frac{x^4}{4}& & (n=0,1)\\ T_{6,0}(x)& =x^2-\frac{x^4}{6}+\frac{x^6}{5!}& & (n=0,1,2)\end{array}$$

Find the 4th degree Taylor Polynomial centered at $x=\frac{\pi }{2}$ for $f(x)=x\sin x$ $f(x)=x\sin (x)⟹f\left(\frac{\pi }{2}\right)=\frac{\pi }{2}$ $f^{\prime }(x)=\sin (x)+x\cos (x)⟹f^{\prime }(\frac{\pi }{2})=1$ $f^{\prime \prime }(x)=2\cos (x)+x\sin (x)⟹f^{\prime \prime }(\frac{\pi }{2})=\frac{\pi }{2}$ $f^{\prime \prime \prime }(x)=-3\sin (x)-x\cos (x)⟹f^{\prime \prime \prime }(x)=-3$ $f^{\prime \prime \prime \prime }(x)=-4\cos (x)+x\sin (x)⟹f^{\prime \prime \prime \prime }(x)=\frac{\pi }{2}$

$$T_{4,\frac{\pi }{2}}(x)=\frac{\pi }{2}+1\left(x-\frac{\pi }{2}\right)-\frac{\pi /2}{2!}{\left(x-\frac{\pi }{2}\right)}^2-\frac{3}{3!}(x-\frac{\pi }{2}{)}^3+\frac{\pi /2}{4!}(x-\frac{\pi }{2}{)}^4$$

Approximate the function $f(x)=\sqrt[3]{x}$ by a Taylor polynomial of degree 2 at $a=8$ How accurate is this when $7\le x\le 9$? $f(x)=x^{\frac{1}{3}}⟹f(8)=2$ $f^{\prime }(x)=\frac{1}{3}{x}^{-\frac{2}{3}}⟹f^{\prime }(8)=\frac{1}{12}$ $f^{\prime \prime }(x)=-\frac{2}{9}{x}^{-\frac{5}{3}}⟹f^{\prime \prime }(8)=-\frac{1}{144}$ $f^{\prime \prime \prime }(x)=\frac{10}{27}{x}^{-\frac{8}{5}}$

$$T_{2,8}(x)=2+\frac{1}{12}(x-8)-\frac{1}{144\cdot 2}(x-8{)}^2$$

Finding Error

$$\begin{array}{rl}|f^{(3)}(x)|& <k\\ \frac{|10}{27}{x}^{-8/3}|& <k\\ |\frac{10}{27}{x}^{-8/3}|& <\frac{10}{27}(7{)}^{-8/3}<0.0021\end{array}$$ $$7\le x\le 9$$ $$-1\le x-8\le 1$$ $$|x-8|\le 1$$ $$\begin{array}{rl}|R_2(x)|& \le \frac{k}{3!}|x-8{|}^3\\ & \le \frac{0.0021}{3!}(1{)}^3\\ & \le \frac{0.0021}{6}\\ & \le 0.0004\end{array}$$

4.6 | Binomial Series

Example: Binomial Coefficient Let $k$ be a real number and let $n$ be a strictly positive integer. We define

$$\left(\genfrac{}{}{0ex}{}{k}{n}\right)=\frac{(k-0)(k-1)\cdots (k-(n-1))}{n!}=\frac{k(k-1)\cdots (k-n+1)}{n!}$$

We note that the numerator of $\left(\genfrac{}{}{0ex}{}{k}{n}\right)$ has $n$ factors. We also define $\left(\genfrac{}{}{0ex}{}{k}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{k}{-n}\right)=0$

Remark: Choose Notation

$$\left(\genfrac{}{}{0ex}{}{k}{n}\right)=\frac{k!}{n!(k-n)!}$$

The Binomial Theorem states that

$$(1+x{)}^k=\sum _{n=0}^k(\genfrac{}{}{0ex}{}{k}{n})x^n$$

Theorem: Let $k$ be a real number and $n$ be a strictly positive integer. Then

$$\left(\genfrac{}{}{0ex}{}{k}{n+1}\right)(n+1)+\left(\genfrac{}{}{0ex}{}{k}{n}\right)n=\left(\genfrac{}{}{0ex}{}{k}{n}\right)k$$

Theorem: Generalized Binomial Theorem Let $k$ be a real number. If $|x|<1$, then

$$(1+x{)}^k=\sum _{n=0}^{\infty }(\genfrac{}{}{0ex}{}{k}{n})x^n=1+kx+\frac{k(k-1)}{2}{x}^2+\frac{k(k-1)(k-2)}{6}{x}^3+\cdots$$

If $|x|>1$ and $a$ is not a nonnegative integer, then this sum diverges.

Prove: If $f(x)={\sum }_{n=0}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)x^n$, then ${}^{\prime }(x)+x{f}^{\prime }(x)=kf(x),x\in (-1,1)$

$$\begin{array}{rl}{f}^{\prime }(x)+x{f}^{\prime }(x)& =\sum _{n=0}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)n{x}^{n-1}+x\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)n{x}^{n-1}\\ & =\left(\genfrac{}{}{0ex}{}{k}{1}\right)+\sum _{n=2}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)n{x}^{n-1}+\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)n{x}^n\\ & =k+\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n+1}\right)(n+1)x^n+\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)n{x}^n\\ & =k+\sum _{n=1}^{\infty }[(\genfrac{}{}{0ex}{}{k}{n+1})(n+1)+(\genfrac{}{}{0ex}{}{k}{n})n]x^n\\ & =k+\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)k{x}^n\\ & =k(1+\sum _{n=1}^{\infty }(\genfrac{}{}{0ex}{}{k}{n})x^n)\\ & =k\sum _{n=0}^{\infty }\left(\genfrac{}{}{0ex}{}{k}{n}\right)x^n\\ & =kf(x)\end{array}$$

Prove: Let $g(x)=\frac{f(x)}{(1+x{)}^k}$ then $g^{\prime }(x)=0,x\in (-1,1)$

$$\begin{array}{rl}{g}^{\prime }(x)& =\frac{f^{\prime }(x)(1+x{)}^k-f(x)k(1+x{)}^{k-1}}{(1+x{)}^{2k}}\\ & =\frac{f^{\prime }(x)(1+x{)}^k-f^{\prime }(x)(1+x)(1+x{)}^{k-1}}{(1+x{)}^{2k}}\\ & =\frac{f^{\prime }(x)(1+x{)}^k-f^{\prime }(x)(1+x{)}^k}{(1+x{)}^{2k}}\\ & =0\end{array}$$

Ex: Write $\frac{1}{(1+x{)}^2}$ as a power series.

$$(1+x{)}^{-2}=\sum _{n=0}^{\infty }(\genfrac{}{}{0ex}{}{-2}{n})x^n$$ $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{-2}{n}\right)& =\frac{(-2)(-3)(-4)\cdots (-2-n+1)}{n!}\\ & =\frac{(-1{)}^n(2)(3)(4)\cdots (n+1)}{n!}\\ & =(-1{)}^n(n+1)\end{array}$$ $$\frac{1}{(1+x{)}^2}=\sum _{n=0}^{\infty }(-1{)}^n(n+1)x^n$$

Ex: Find the Maclaurin series for the function $f(x)=\frac{1}{\sqrt{4-x}}$

$$\begin{array}{rl}\frac{1}{\sqrt{4-x}}& =\frac{1}{\sqrt{4\left(1+\left(-\frac{x}{4}\right)\right)}}\\ & =\frac{1}{2}{\left(1+\left(-\frac{x}{4}\right)\right)}^{-1/2}\\ & =\frac{1}{2}\sum _{n=0}^{\infty }\left(\genfrac{}{}{0ex}{}{-\frac{1}{2}}{n}\right){\left(-\frac{x}{4}\right)}^n\end{array}$$

4.7 | Applications of Taylor Series

Ex: Determine 2 different series that can be used to evaluate the following functions $f(x)=3^x$

1. $e^x⟹e^{x\ln 3}$
2. $(1+2{)}^x$ $f(x)=\frac{1}{(3+x{)}^4}$
3. $(3+x{)}^{-4}$
4. geometric $f(x)=-\frac{1}{2}\sin (2x)$
5. $\sin (x)$
6. $\cos (x)$ with derivative

Ex: Find the values of the following series

$$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{e^n}{2^{n+1}}& =\frac{1}{2}\frac{1}{1-\left(\frac{e}{2}\right)}\\ & =\frac{1}{2-e}\end{array}$$

Ex 2

$$\sum \frac{k(k-1)\cdots (k-n)1}{n!}{x}^n$$ $$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{\pi (\pi -1)(\pi -2)\cdots (\pi -n+1)}{7^{n}n!}& =\sum _{n=0}^{\infty }\frac{\pi (\pi -1)(\pi -2)\cdots (\pi -n+1)}{n!}{\left(\frac{1}{7}\right)}^n\\ & ={\left(1+\frac{1}{7}\right)}^{\pi }\\ & ={\left(\frac{8}{7}\right)}^{\pi }\end{array}$$

Ex 3

$$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{(-1{)}^n(1)(5)(9)\cdots (4n-3)}{4^n{3}^{4n}n!}\cdot 19^n& =\sum _{n=0}^{\infty }\frac{\left(-\frac{1}{4}\right)\left(-\frac{5}{4}\right)\cdots \left(-n+\frac{3}{4}\right)}{n!}\cdot \frac{19^n}{3^{4n}}\\ & =\sum _{n=0}^{\infty }\frac{\left(-\frac{1}{4}\right)\left(-\frac{5}{4}\right)\cdots \left(-n+\frac{3}{4}\right)}{n!}\cdot {\left(\frac{19}{81}\right)}^n\\ & ={\left(1+\frac{19}{81}\right)}^{-1/4}\\ & ={\left(\frac{100}{81}\right)}^{-1/4}\\ & =\frac{3}{\sqrt{10}}\end{array}$$

Ex: Find the sum of

$$\begin{array}{rl}f(x)& =\sum _{n=0}^{\infty }\left(\frac{(-1{)}^n{x}^{2n+1}}{2n+1}\right)+4\\ f^{\prime }(x)& =\sum _{n=0}^{\infty }\frac{(-1{)}^n(2n+1)x^{2n}}{2n+1}\\ & =\sum _{n=0}^{\infty }(-x^2{)}^n\\ & =\frac{1}{1-(-x^2)}\\ & =\frac{1}{1+x^2}\\ \int f^{\prime }(x)dx& =\int \frac{1}{1+x^2}⟹4+\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}\\ f(x)& =\arctan (x)+C\\ f(0)& =4⟹f(x)=\arctan (x)+4\end{array}$$

Ex: Express as a power series

$$e^{(}-x^2)=\sum _{n=0}^{\infty }\frac{(-x^2{)}^n}{n!}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{n!}$$ $$\begin{array}{rl}{\int }_0^1{e}^{-x^2}dx& ={\int }_0^1\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{n!}dx\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{n!(2n+1)}{\mid }_0^1\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n}{n!(2n+1)}\end{array}$$

Ex: How many terms to get this approximation within an error of $\frac{1}{1000}?$

$$\begin{array}{rl}& =1-\frac{1}{3\cdot 1!}+\frac{1}{5\cdot 2!}+\frac{1}{7\cdot 3!}+\frac{1}{9\cdot 4!}-\frac{1}{11\cdot 5!}\\ & =1-\frac{1}{3}+\frac{1}{10}-\frac{1}{43}+\frac{1}{216}\le \frac{1}{1320}\end{array}$$

Ex: Evaluate as power series

$$x\sin (x^3)dx=x\sum _{n=0}^{\infty }\frac{(-1{)}^n(x^3{)}^{2n+1}}{(2n+1)!}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{6n+4}}{(2n+1)!}$$ $$\begin{array}{rl}\int x\sin (x^3)dx& =\int \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{6n+4}}{(2n+1)!}\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{6n+5}}{(2n+1)!(6n+5)}+C\end{array}$$

Ex: Evaluate

$$\begin{array}{rl}\cos (x)& =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}\\ \frac{\cos (x)-1}{x}& =\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n-1}}{(2n)!}\end{array}$$ $$\begin{array}{rl}\int \frac{\cos (x)-1}{x}dx& =\int \sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n-1}}{(2n!)}dx\\ & =\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)(2n)!}+C\end{array}$$

Ex: Evaluate

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{e^x-1-x}{x^2}& =\underset{x\to 0}{\lim }\frac{\left(1+x+\frac{x^2}{2}+\cdots \right)-1-x}{x^2}\\ & =\underset{x\to 0}{\lim }\frac{\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots }{x^2}\\ & =\underset{x\to 0}{\lim }\frac{1}{2}+\frac{x}{3!}+\frac{x^2}{4+\cdots }=\frac{1}{2}\end{array}$$

Ex:

$$\arctan (x)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}$$ $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{x-\arctan (x)}{x^3}& =\frac{x-(x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots )}{x^3}\\ & =\frac{\frac{x^3}{3}-\frac{x^5}{5}+\cdots }{x^3}\\ & =\frac{1}{3}-\frac{x^2}{5}+\frac{x^5}{+}\cdots \\ & =\frac{1}{3}\end{array}$$

Ex:

$$\sin (x)=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$$ $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\sin (x)-x+\frac{1}{6}{x}^3}{x^5}& =\frac{1}{x^5}((x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\cdots )-x+\frac{1}{6}{x}^3)\\ & =\frac{1}{x^5}(\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9!}{9})\\ & =\frac{1}{5!}\\ & =\frac{1}{20}\end{array}$$

Ex:

$$f(x)=\frac{1}{\sqrt{1-x}}-1$$

$f(x)=(1-x{)}^{-\frac{1}{2}}-1⟹f(0)=0$ $f^{\prime }(x)-\frac{1}{2}(1-x{)}^{-\frac{3}{2}}⟹f^{\prime }(0)=-\frac{1}{2}$ $f^{\prime \prime }(x)=\frac{3}{4}(1-x{)}^{-5/2}⟹f^{\prime \prime }(0)=\frac{3}{4}$

$$\begin{array}{rl}f(x)& =0+\frac{1}{2}x+\frac{3}{4\cdot 2!}{x}^2+\cdots \\ f(x)& =\frac{1}{2}x+\frac{3}{8}x+\cdots \\ f(\frac{v^2}{c^2})& =\frac{1}{2}(\frac{v^2}{c^2})+\frac{3}{8}(\frac{v^2}{c^2})+\cdots \\ m{c}^{2}f(\frac{v^2}{c^2})& =\frac{m{c}^2}{2}(\frac{v^2}{c^2})+\frac{3m{c}^2}{8}(\frac{v^2}{c^2})+\cdots \\ m{c}^{2}f(\frac{c^2}{c^2})& =\frac{1}{2}m{v}^2+\cdots \end{array}$$

4.8 | Big-O Notation

Definition: Big-O Notation Let $a$ denote a real number or $\infty$. Assume that $f(x),g(x)$ are two functions that are defined for all real $x$ values that are near $a$ but are not necessarily when $x=a$. More precisely, if $a\in \mathbb{R}$, we assume that $f(x),g(x)$ are both defined for all $x$ in the union

$$I=(a-ϵ,a)\cup (a,a+ϵ)=\{x\in \mathbb{R}\mid 0<|x-a|<ϵ\}$$

where $ϵ$ is some strictly positive real number. If $a=\infty$, we assume that $f(x),g(x)$ are both defined for all $x$ in some open interval $I=(c\infty )$, where $c\in \mathbb{R}$. We write

$$f(x)=O(g(x))\text{ as }x\to a$$

if there exists a real constant $M>0$ such that for all $x\in I$,

$$|f(x)|\le M|g(x)|$$

Ex: Express $|7x^5+4x^3+x|\le 14|x^5|$ in Big-O notation

$$7x^5+4x^3+x=O(x^5)$$

Ex: Show that if $f(x)+3x-2$ then $f(x)=O(x^3)$ $x^2\le x^3$ $3x-2\le x^3$ $x^2+3x-2\le x^3+x^3=2x^3$

Ex:

$$\begin{array}{rl}{1}^2+2^2+\cdots +n^2=& \frac{n(n+1)(2n+1)}{6}\\ & =\frac{2n^3+3n+n}{6}\\ & =\frac{1}{3}{n}^3+\frac{1}{2}n+\frac{n}{6}\\ & =O(n^3)\end{array}$$

Ex:

1. F
2. T
3. F
4. F
5. T

Ex: Consider $f(x)=\sin (x)$

$$\begin{array}{rl}|\sin (x)-T_{1,0}(x)& =|\sin (x)-x|\\ & =|\frac{f^{\prime \prime }(c)}{2!}{x}^2|\\ & =|-\frac{\sin (c)}{2}{x}^2|\\ & \le \frac{1}{2}|x^2|\\ \sin (x)-x& =O(x^2)\\ \sin (x)& =x+O(x^2)\end{array}$$

Theorem Let $a$ denote a real number or $\infty$. If $f(x)=O(g(x))$ as $x\to a$ and $g(x)=O(h(x))$ as $x\to a$, then $f(x)=O(h(x))$ as $x\to a$.

Theorem Let m and n be nonnegative integers. As $x\to 0$, the following statements hold. (i) $O(x^m)\cdot O(x^n)=O(x^{m+n})$ (ii) $O(x^m)\pm O(x^n)=O(x^k)$, where $k=\min \{m,n\}$. (iii) $x\cdot O(x^n)=O(x^{n+1})$. (iv) $\frac{1}{x}\cdot O(x^n)=O(x^{n-1})$ when $n\ge 1$. (v) $C\cdot O(x^n)=O(C{x}^n)=O(x^n)$ for any real constant C. (vi) ${\lim }_{x\to 0}O(x^n)=0$ when $n\ge 1$.

Theorem: Taylor’s Inequality, the Big-O Version Let $m$ be a nonnegative integer. Suppose that $f^{(m+1)}$ is continuous on an open interval $I$ that contains $a\in \mathbb{R}$. If there is a real constant $K>0$ such that $f^{(m+1)}(x)\le K$ for every $x\in I$, then

$$f(x)-T_{m,a}(x)=O((x-a{)}^{m+1})$$

as $x\to a$. Moreover, if $P(x)$ is a polynomial of degree $\le m$ such that

$$f(x)-P(x)=O((x-a{)}^{m+1})$$

as $x\to a$, then $P(x)=T_{m,a}(x)$

Ex

1. $O(x^2)+2$
2. $O(x^3)+4x^2+2x-3$
3. $O(x^4)+\frac{3}{2}x$

Ex

$$\sin (x)=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\cdots =x-\frac{x^3}{3!}+O(x^5)$$ $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\sin (x)}{x}& =\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{3!}+O(x^5)}{x}\\ & =\underset{x\to 0}{\lim }1-\frac{x^3}{3!}+O(x^4)\\ & =1\end{array}$$

thanks katy howell escobar!