This week, we’ll work out some Taylor Series expansions of popular functions.

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 an R-Markdown document.

Taylor Series is defined as \(f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\)

library(pracma)

\(f(x) = 1/(1−x)\)

\[ \begin{split} f(x) = \frac{1}{(1-x)} \\ f'(x) = \frac{1}{(1-x)^2} \\ f''(x) = \frac{2}{(1-x)^3} \\ f'''(x) = \frac{6}{(1-x)^4} \\ \end{split} \]

Evaluating at \(x=0\):

\[ \begin{split} f(0) = 1 \\ f'(0) = 1 \\ f''(0) = 2 \\ f'''(0) = 6 \end{split} \]

\[ \begin{split} 1 + \frac{1}{1!}x^1 + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + ... \end{split} \]

\[ \begin{split} 1 + x + x^2 + x^3 + ...x^n \end{split} \]

\[ \begin{split} \sum\limits_{n=0}^{\infty}{x^n} \end{split} \]

# Using Taylor function
taylor_equation1 <- function(x) {1/(1-x)}
problem_1 <- taylor(taylor_equation1, x0 = 0, n = 4)
problem_1
## [1] 1.000029 1.000003 1.000000 1.000000 1.000000

Valid range: \(|x| < 1\)

\(f(x) = e^x\)

\[ \begin{split} f(x) = e^x \\ f'(x) = e^x \\ f''(x) = e^x \\ f'''(x) = e^x \end{split} \]

Evaluating at \(x = 0\):

\[ \begin{split} f(0) = 1 \\ f'(0) = 1 \\ f''(0) = 1 \\ f'''(0) = 1 \end{split} \]

\[ \begin{split} e^x = e^0 + e^0(x-0)^1 + \frac{e^0(x-0)^2}{2!} + \frac{e^0(x-0)^3}{3!}... \end{split} \]

\[ \begin{split} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... \end{split} \]

\[ \begin{split} \sum\limits_{n=0}^{\infty}\frac{x^n}{n!} \end{split} \]

# Using Taylor function
taylor_equation2 <- function(x) {exp(x)}
problem_2 <- taylor(taylor_equation2, x0 = 0, n = 4)
problem_2
## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000

Valid Range: The function is valid for all real values of \(x\), since \(e^x\) is defined for all real numbers.

\(f(x) = ln(1 + x)\)

\[ \begin{split} f(x) = ln(x+1) \\ f'(x) = \frac{1}{(1+x)} \\ f''(x) = -\frac{1}{(1+x)^2} \\ f'''(x) = \frac{2}{(1+x)^3} \\ \end{split} \]

Evaluating at \(x=0\):

\[ \begin{split} f(0) = 0 \\ f'(0) = 1 \\ f''(0) = -1 \\ f'''(0) = 2 \end{split} \]

\[ \begin{split} 0 + \frac{1}{1!}x^1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - ... \end{split} \]

Simplified:

\[ \begin{split} ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \end{split} \]

\[ \begin{split} \sum\limits_{n=0}^{\infty} (-1)^{n+1}\frac{1}{n}x^n \end{split} \]

# Using Taylor function
taylor_equation3 <- function(x) {log(1+x)}
problem_3 <- taylor(taylor_equation3, x0 = 0, n = 4)
problem_3
## [1] -0.2500044  0.3333339 -0.5000000  1.0000000  0.0000000

Valid range: \(|x| < 1\)

\(f(x)=x^{(1/2)}\)

\[ \begin{split} f(x) = x^{(1/2)} \\ f'(x) = \frac{1}{2}x^{-(1/2)} \\ f''(x) = -\frac{1}{4}x^{-(3/2)} \\ f'''(x) = \frac{3}{8}x^{-(5/2)} \\ \end{split} \] Evaluating at \(x=1\):

\[ \begin{split} f(1) = \sqrt1 = 1 \\ f'(1) = \frac{1}{2\sqrt1} = \frac{1}{2}\\ f''(1) = -\frac{1}{4\times 1^\frac{3}{2}} = -\frac{1}{4} \\ f'''(1) = \frac{3}{8\times 1^\frac{5}{2}} = \frac{3}{8} \\ \end{split} \]

\[ \begin{split} f(x) \approx 1 + \frac{1}{2}(x-1) - \frac{1}{4 \times 2!}(x-1)^2 + \frac{3}{8\times3!}(x-1)^3 - ... \end{split} \]

\[ \begin{split} \sum\limits_{n=0}^{\infty} \frac{1}{x \times n!}(x-1)^n \end{split} \]

Valid range: \(x \ge 0\)

# Using Taylor function
a <- 1 
n <- 4  
taylor_equation4 <- function(x) sqrt(x)
problem_4 <- taylor(taylor_equation4, a, n)
problem_4
## [1] -0.03906285  0.21875150 -0.54687738  1.09375167  0.27343706