88 lines
2.2 KiB
TeX
88 lines
2.2 KiB
TeX
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% Created 2023-09-29 Fri 10:00
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% Intended LaTeX compiler: pdflatex
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\documentclass[11pt]{article}
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\usepackage[utf8]{inputenc}
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\usepackage[T1]{fontenc}
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\usepackage{graphicx}
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\usepackage{longtable}
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\usepackage{wrapfig}
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\usepackage{rotating}
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\usepackage[normalem]{ulem}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{capt-of}
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\usepackage{hyperref}
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\author{Elizabeth Hunt}
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\date{\today}
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\title{}
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\hypersetup{
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pdfauthor={Elizabeth Hunt},
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pdftitle={},
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pdfkeywords={},
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pdfsubject={},
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pdfcreator={Emacs 28.2 (Org mode 9.7-pre)},
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pdflang={English}}
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\begin{document}
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\tableofcontents
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\section{Taylor Series Approx.}
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\label{sec:orgcc72ed1}
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Suppose f has \(\infty\) many derivatives near a point a. Then the taylor series is given by
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\(f(x) = \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)
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For increment notation we can write
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\(f(a + h) = f(a) + f'(a)(a+h - a) + \dots\)
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\(= \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{h!} (h^n)\)
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Consider the approximation
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\(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |f'(a) - \frac{1}{h}(f(a + h) - f(a))|\)
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Substituting\ldots{}
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\(= |f'(a) - \frac{1}{h}((f(a) + f'(a) h + \frac{f''(a)}{2} h^2 + \cdots) - f(a))|\)
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\(f(a) - f(a) = 0\)\ldots{} and \(distribute the h\)
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\(= |-1/2 f''(a) h + \frac{1}{6}f'''(a)h^2 \cdots|\)
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\subsection{With Remainder}
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\label{sec:org7dfd6c7}
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We can determine for some u \(f(a + h) = f(a) + f'(a)h + \frac{1}{2}f''(u)h^2\)
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and so the error is \(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |\frac{h}{2}f''(u)|\)
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\begin{itemize}
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\item\relax [\url{https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series}]
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\begin{itemize}
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\item > Taylor's Theorem w/ Remainder
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\end{itemize}
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\end{itemize}
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\subsection{Of Deriviatives}
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\label{sec:org1ec7c9b}
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Again, \(f'(a) \approx \frac{f(a+h) - f(a)}{h}\),
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\(e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots\)
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\(R_2 = \frac{h}{2} f''(u)\)
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\(|\frac{h}{2} f''(u)| \leq M h^1\)
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\(M = \frac{1}{2}|f'(u)|\)
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\subsubsection{Another approximation}
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\label{sec:org16193b9}
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\(\text{err} = |f'(a) - \frac{f(a) - f(a - h)}{h}|\)
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\(= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))\)
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\(= |f'(a) - \frac{1}{h}(f'(a) + \frac{1}{2}f''(a)h)|\)
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\end{document}
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