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.
Taylor Series is defined as \(f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\).
Finding several derivatives.
\(f^0(c) = \frac{1}{(1-c)}\)
\(f'(c) = \frac{1}{(1-c)^2}\)
\(f''(c) = \frac{2}{(1-c)^3}\)
\(f'''(c) = \frac{6}{(1-c)^4}\)
\(f''''(c) = \frac{24}{(1-c)^5}\)
\[ \begin{split} f(x) &= \frac{1}{(1-c)0!}(x-c)^0 + \frac{1}{(1-c)^2 1!}(x-c)^1 + \frac{2}{(1-c)^3 2!}(x-c)^2 + \frac{6}{(1-c)^4 3!}(x-c)^3 + \frac{24}{(1-c)^5 4!}(x-c)^4 + ... \\ &= \sum\limits_{n=0}^{\infty} \frac{1}{(1-c)^{n+1}}(x-c)^n \end{split} \]
The Maclaurin Series, where c = 0:
\(f(x) = \sum\limits_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + ...\).
This series will only converge if x in the range of \((-1, 1)\). For other x, the series will become infinity.
Finding several derivatives.
\(f^0(c) = e^c\)
\(f'(c) = e^c\)
\(f''(c) = e^c\)
\(f'''(c) = e^c\)
\(f''''(c) = e^c\)
\[ \begin{split} f(x) &= \frac{e^c}{0!}(x-c)^0 + \frac{e^c}{1!}(x-c)^1 + \frac{e^c}{2!}(x-c)^2 + \frac{e^c}{3!}(x-c)^3 + ...\\ &= e^c \sum\limits_{n=0}^{\infty} \frac{(x-c)^n}{n!} \end{split} \]
The Maclaurin Series, where c = 0::
\(f(x) = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\).
Ratio test
\(\frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(n+1)!}\times\frac{n!}{x^n} = \frac{x}{n+1}\)
\(L = \lim\limits_{n\to\infty}\frac{|x|}{n+1} = 0\)
Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers.
Finding several derivatives.
\(f^0(c) = ln(1+c)\)
\(f'(c) = \frac{1}{c+1}\)
\(f''(c) = -\frac{1}{(c+1)^2}\)
\(f'''(c) = \frac{2}{(c+1)^3}\)
\(f''''(c) = -\frac{6}{(c+1)^4}\)
\[ \begin{split} f(x) &= \frac{ln(1+c)}{0!}(x-c)^0 + \frac{1}{(c+1)1!}(x-c)^1 - \frac{1}{(c+1)^2 2!}(x-c)^2 &= ln(1+c) + \sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{(x-c)^n}{n(c+1)^n} \end{split} \]
The Maclaurin Series, where c = 0:
\(f(x) = \sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...\).
Ratio Test
\(\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1+1} x^{n+1}}{n+1}\times\frac{n}{(-1)^{n+1}x^n} = \frac{-xn}{n+1}\)
\(L = \lim\limits_{n\to\infty}|\frac{-xn}{n+1}| = |x|\)
|x| < 1
This series will only converge if \(|x|<1\), that means the valid range is \((-1, 1)\).