series.tex

\documentclass[11pt,letterpaper,twocolumn]{article}
%Gummi|065|=)
\title{\textbf{Calculus 2 Notes}}
\author{Tristan Rice}
\date{}
\usepackage{amsmath}
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\usepackage{multicol}
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\begin{document}


\maketitle

\section{Sequences + Series}
\subsection{Series Unit 1}
Sequence + series of numbers

The Fibonacci Sequence is a famouse sequence.
\be
SUM(fibonacci) = \infty
\ee

Xeno's Paradox
\be
\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\ldots=1
\ee
It's not actually a paradox.

\subsection{Series Unit 2}
Series of numbers
\be
1,x,\frac{x^2}2,\frac{x^3}6,\frac{x^4}{24},\frac{x^5}{120},\frac{x^6}{720}\ldots
\ee
\be
\text{n}^{th} \text{ term}: \frac{x^n}{n!}
\ee
Easy to take the derivative or integral of this series.

\be
\int e^{-x^2}
\ee
This function is impossible to integrate.

\be
1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}24 = e^x
\ee
You can use series to represent hard to integrate functions.


\be
\text{a sequence} \rightarrow {a_n}
\ee
Fibonacci below
\begin{tabular}{llllll}
	$n$ & 0 & 1 & 2 & 3 & 4\\
	$a_n$ & 1 & 1 & 2 & 3 & 5\\
\end{tabular}\\
an explicit formula $a_n=\frac1n$

\be
\text{terms} a_n = \{1,\frac12,\frac13,\frac14,\ldots\}
\ee
\be
text{recursively} a_{n+2} = a_n + a_{n+1}
\ee

\subsection{Ways of writing sequences}
explicit formula
\be
a_n = \frac1{3n^2+2}
\ee
recursive formula
\be
a_{n+1}=\frac3{2a_n+r}
\ee
pattern ( easier to understand and write )
\be
a_n = \{0,1,0,1,0,1,0,1,0,1,\ldots\} = \frac{a+(-1)^n}2
\ee

\subsubsection{Example}
Pattern
\be
a_n = \{1,1*3,1*3*5,1*3*5*7,\ldots\}
\ee
Recursive
\be
a_{n+1}=a_n*(2n-1)
\ee
\section{Limits}

The sequence $\{a_n\}$ converges if
\be
\lim_{n\to\infty} a_n \text{ exists}
\ee

\be
a_n = \frac{3n^2+2}{5n^4-7}
\ee
Converges at 0. Bottom has a bigger power.
\be
\lim_{n\to\infty}a_n=0
\ee
\be
a_n=\frac{4n-7}{8n+9}
\ee
Converges at 1/2 since powers are the same
\be
a_n=\frac{4n^5}{5^n}
\ee
Exponentials are faster at going to infinity than polynomials. Converges 0
\be
a_n = \frac{n}{\ln n}
\ee
n is faster than ln(x). Goes to infinity. Diverges

\subsection{l'hopital's rule}
\be
f(x)=\frac{a(x)}{b(x)}
\ee
If the limit isn't $\frac00$ or $\frac\infty\infty$
then
The limit of f(x)
is
\be
\frac{a'(x)}{b'(x)}
\ee
\subsection{Recusive Formulas}
\subsubsection{Monotonic Sequence Theorem}
Monotonic means always going in the same direction (always increasing or always decreasing).

Idea: if a sequence is both monotonic and bounded then the sequence converges.

Bounded means that there is both a largest and a smallest value of a function. The range of function values is contained.

If for all n (the index) greater or equal to N (some particular value of n)
\be
a_{n+1}>=a_n \text{ and } a_n < M \text{ then }\{a_n\} \text{ converges}
\ee
Increasing and has an upper bound M.
\be
a_{n+1}<=a_n \text{ and } a_n > M \text{ then }\{a_n\} \text{ converges}
\ee
Decreasing and has a lower bound M.

\subsubsection{How can we use this to establis convergence for recursively defined sequences?}
\be
a_1 = 2
\ee
\be
a_{n+1}=\frac12(a_n+6)
\ee
Show monotonicily:
\be
a_1 = 2
\ee
\be
a_2 = 4
\ee
\be
a_3 = 5
\ee
\be
a_4 = 5.5
\ee
It appears to be increasing, but we need to prove it.

Show that if $a_{n+1}>a_n$ then $a_{n+2}>a_{n+1}$

Given:
\be
a_{n+1}>a_n
\ee
Add 6 and divide by 2.
\be
\frac12(a_{n+1}+6)>\frac12(a_n+6)
\ee
\be
a_{(n+1)+1} > a_{n+1}
\ee
\be
a_{n+2} > a_{n+1}
\ee

Show it is bounded:

if $a_n<10$ show $a_{n+1}<10$

given:
\be
a_n < 10
\ee
add 6
\be
\frac12(a_n+6)< 8
\ee
\be
a_{n+1} < 8 < 10
\ee
\be
L = \frac12(L+6)
\ee
\be
2L = L+ 6
\ee
\be
L = 6
\ee
\subsection{Series convergence}
\be
\sum_{n=1}^\infty a_n = a_1 + a_2+a_3+\ldots+a_n+\ldots
\ee
This can have 3 possible outcomes.

- Converges: addition problem has an answer not $\infty$.

- Diverge: Answer to problem is infinity, or answer doesn't settle down to one answer.



let $s_n$ be the partial sum
\be
s_n = \sum^n_{i=1} a_i = a_1+1_2+a_3+\ldots+a_{n-1}+a_n
\ee

Infinite series $\sum_{n=1}^\infty a_n$ is convergent if and only iff $\lim_{n\to\infty} s_n = L$ and L is not infinite.

\subsection{Partial sums}

let $a_n = n+3$
\be
\sum_{n=1}^\infty n+3 = (1+3)+(2+3)+(3+3)+\ldots
\ee
\be
s_1 = a_1=4
\ee
\be
s_2=a_1+a_2=4+5=9
\ee
\be
s_3=a_1+a_2+a_3=4+5+6=15
\ee
\be
\{s_n\}=\{4,9,15,\ldots\}
\ee
\be
\lim_{n\to\infty}s_n = \infty
\ee


\subsection{Test for Divergence}
If the terms added together don't go to zero, it will diverge.
If
\be
\lim_{n\to\infty} a_n != 0
\ee
then $\sum^\infty_{n=1}a_n$ diverges.

The inverse is not true.
\section{Inequalities}
\be
x_1>x_2
\ee
\be
-\frac{x_1}a<-\frac{x_2}a
\ee
\be
x_1 > x_2
\ee
\be
\frac1{x_1}<\frac1{x_2}
\ee
















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