IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS
This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) = \frac{1}{1-x}\)
\(f(x) = e^x\)
\(f(x) = ln(1 + x)\)
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.
The Taylor Series of f(x), centered at c is
\[\begin{equation} \begin{split} f(c)&= \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)(0)}}{n!} (x-c)^n \end{split} \end{equation}\]library(pracma)Solution
if centered at c=0, Taylor Series of f(x) is
\[\begin{equation} \begin{split} \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \end{split} \end{equation}\] \[\begin{equation} f'(0) = 1/(1-x)^2 =1 \end{equation}\] \[\begin{equation} f''(0) = 2/(1-x)^3 =2 \end{equation}\] \[\begin{equation} f'''(0) = 6/(1-x)^4 =6 \end{equation}\] \[\begin{equation} \frac{d^4}{d(x)^4}\frac{1}{1-x} = 24/(1-x)^5 =24 \end{equation}\]….
\[\begin{equation} \begin{split} \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n &= \frac{1}{(1-0)}+\frac{\frac{1}{(1-0)^2}}{1!}x^1+\frac{\frac{2}{(1-0)^3}}{2!}x^2+\frac{\frac{6}{(1-0)^4}}{3!}x^3+\frac{\frac{24}{(1-0)^5}}{4!}x^4+...\\ &= 1+x+x^2+x^3+x^4+...\\ &=\displaystyle\sum_{n=0}^{\infty} x^n \end{split} \end{equation}\] \[\begin{equation} \begin{split} \displaystyle\sum_{n=0}^{\infty} x^n \end{split} \end{equation}\]f <- function(x) 1/(1-x)
p <- taylor(f, 0, 4)
p ## [1] 1.000029 1.000003 1.000000 1.000000 1.000000if centered at c=0
\[\begin{equation} f'(0) = e^x =1 \end{equation}\] \[\begin{equation} f''(0) = e^x =1 \end{equation}\] \[\begin{equation} f'''(0) = e^x =1 \end{equation}\]…..
\[\begin{equation} \begin{split} \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n &= e^x+\frac{e^x}{1!}x^1+\frac{e^x}{2!}x^2+\frac{e^x}{3!}x^3+...\\ &= 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\\ &=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}x^n \end{split} \end{equation}\]f <- function(x) exp(x)
p <- taylor(f, 0, 4)
p ## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000if centered at c=0
\[\begin{equation} f'(0) = 1/(1+x) =1 \end{equation}\] \[\begin{equation} f''(0) = -1/(1+x)^2 =-1 \end{equation}\] \[\begin{equation} f'''(0) = 2/(1+x)^3 =2 \end{equation}\] \[\begin{equation} \frac{d^4}{d(x)^4} ln(1+x)= -6/(1+x)^4 =-6 \end{equation}\] \[\begin{equation} \frac{d^5}{d(x)^5} ln(1+x)= 24/(1+x)^5 =24 \end{equation}\]…
\[\begin{equation} \begin{split} \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n &= ln(1 + x)+\frac{1}{1!}x^1+\frac{-1}{2!}x^2+\frac{2}{3!}x^3+...\\ &= 0+\frac{1}{1!}x+\frac{-1}{2!}x^2+\frac{2}{3!}x^3+\frac{-6}{4!}x^4+\frac{24}{5!}x^5...\\ &= x-\frac{1}{2!}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5...\\ &=\displaystyle\sum_{n=0}^{\infty}(-1)^{n+1}\frac{1}{n}x^n \end{split} \end{equation}\]f <- function(x) log((1 + x), base = exp(1))
p <- taylor(f, 0, 4)
p ## [1] -0.2500044 0.3333339 -0.5000000 1.0000000 0.0000000