## installed_and_loaded.packages.
## prettydoc TRUE
## Ryacas TRUE
## Deriv TRUEComputing the Taylor Series expansion
This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) = 1/(1-x)\)
\(f(x) = e^x\)
\(f(x) = ln(1 + x)\)
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document
\(f(x) = 1/(1-x)\)
\(f(c) = \frac{1}{(1-c)}\)
\(f'(c) = \frac{1}{(1-c)^2}\)
\(f''(c) = \frac{2}{(1-c)^3}\)
\(f'''(c) = \frac{6}{(1-c)^4}\)
\(f''''(c) = \frac{24}{(1-c)^5}\)
\[\begin{equation} f(x) = \frac{1}{(1-c) 0!} (x-c)^0 + \frac{1}{(1-c)^2 1!} (x-c)^1 + \frac{2}{(1-c)^3 2!} (x-c)^2 + \frac{6}{(1-c)^4 3!} (x-c)^3 + \frac{24}{(1-c)^5 4!} (x-c)^4 + ... \end{equation}\] \[\begin{equation} = \frac{1}{(1-c)} + \frac{1}{(1-c)^2} (x-c) + \frac{1}{(1-c)^3} (x-c)^2 + \frac{1}{(1-c)^4} (x-c)^3 + \frac{1}{(1-c)^5} (x-c)^4 +... \end{equation}\]\(= \sum_{n=0}^{\infty} \frac{(x-c)^n}{(1-c)^{n+1}}\)
Ratio test:
\[\begin{equation} L = \lim_{n\to\infty}\frac{a_{n+1}}{a_n} \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} \frac{ \frac{(x-c)^{n+1}}{(1-c)^{n+2}} }{ \frac{(x-c)^n}{(1-c)^{n+1}} } \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} \frac{(x-c)^{n+1}(1-c)^{n+1}}{(x-c)^n(1-c)^{n+2}} = \frac{x-c}{1-c} = x \end{equation}\]The series will only converge (L < 1) when the range is between -1 and 1.
\(f(x) = e^x\)
\(f(c) = e^c\)
\(f'(c) = e^c\)
\(f''(c) = e^c\)
\(f'''(c) = e^c\)
\(f''''(c) = e^c\)
\[\begin{equation} f(x) = \frac{e^c}{0!} (x-c)^0 + \frac{e^c}{1!} (x-c)^1 + \frac{e^c}{2!} (x-c)^2 + \frac{e^c}{3!} (x-c)^3 + \frac{e^c}{4!} (x-c)^4 + ... \end{equation}\] \[\begin{equation} f(x) = e^c + e^c (x-c) + \frac{e^c}{2!} (x-c)^2 + \frac{e^c}{3!} (x-c)^3 + \frac{e^c}{4!} (x-c)^4 + ... \end{equation}\]\(f(x) = \sum_{n=0}^{\infty} \frac{e^c(x-c)^n}{n!}\)
Ratio test:
\[\begin{equation} L = \lim_{n\to\infty}\frac{a_{n+1}}{a_n} \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} \frac{ \frac{e^c(x-c)^(n+1)}{(n+1)!} }{ \frac{e^c(x-c)^n}{n!} } \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} \frac{e^c(x-c)^{n+1}n!}{e^c(x-c)^n(n+1)!} = \frac{x-c}{n!} = 0 \end{equation}\]Since \(lim_{n\to\infty} = 0\), the series will converge (L < 1) when the range is \((-\infty, \infty)\).
\(f(x) = ln(1 + x)\)
\(f(c) = ln(1 + c)\)
\(f'(c) = 1/(1 + c)\)
\(f''(c) = -1/(1 + c)^2\)
\(f'''(c) = 2/(1 + c)^3\)
\(f''''(c) = -6/(1 + c)^4\)
\[\begin{equation} f(x) = \frac{ln(1 + c)}{0!} (x-c)^0 + \frac{1}{(1 + c)1!} (x-c)^1 + \frac{-1}{(1 + c)^2 2!} (x-c)^2 + \frac{2}{(1 + c)^3 3!} (x-c)^3 + \frac{-6}{(1 + c)^4 4!} (x-c)^4 + ... \end{equation}\] \[\begin{equation} f(x) = ln(1 + c) + \frac{x-c}{(1 + c)} - \frac{(x-c)^2}{2(1 + c)^2} + \frac{(x-c)^3}{3(1 + c)^3} - \frac{(x-c)^4}{4(1 + c)^4} + ... \end{equation}\]\(f(x) = ln(1 + c) + \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-c)^n}{n(1 + c)^n}\)
Ratio test:
\[\begin{equation} L = \lim_{n\to\infty}\frac{a_{n+1}}{a_n} \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} \frac{ ln(1 + c) + \frac{(-1)^{n+2}(x-c)^{n+1}}{(n+1)(1 + c)^{n+1}} }{ ln(1 + c) + \frac{(-1)^{n+1}(x-c)^n}{n(1 + c)^n} } \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} 1 + \frac{n(1 + c)^n(-1)^{n+2}(x-c)^{n+1}}{(n+1)(1 + c)^{n+1}(-1)^{n+1}(x-c)^n} = \end{equation}\] \[\begin{equation} L = lim_{n\to\infty} 1 - \frac{n(x-c)}{(n+1)(1 + c)} = lim_{n\to\infty} 1 - \frac{nx}{n+1} = 1 - x \end{equation}\]The series will only converge (L < 1) when the range is between -1 and 1.