For each function, only consider its valid ranges as indicated in the notes when you arecomputing the Taylor Series expansion.
Derivatives for \(f\left( x \right) = \frac { 1 }{ (1-x) }\) are:
\(f'(x) = \frac{1}{(1-x)^2}\)
\(f''(x) = \frac{2}{(1-x)^3}\)
\(f'''(x) = \frac{6}{(1-x)^4}\)
\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)
\(= 1 + 1x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 +...\)
\(1 + x + x^2 + x^3 + x^4 + …….\)
library(pracma)
eq <- function(x) {1/(1-x)}
p <- taylor(eq, x0 = 0, n = 4)
p
## [1] 1.000029 1.000003 1.000000 1.000000 1.000000
equation <- function(x) {exp(x)}
q <- taylor(equation, x0 = 0, n = 4)
q
## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000
\(f(a) \quad= \quad ln(1+a)\) \(\quad= \quad f(0) = 0\)
\({ f }^{ \prime }(a) \quad= \quad \frac{1}{1+a}\) \(\quad= \quad f(0) = 1\)
\({ f }^{ \prime \prime }(a) \quad= \quad \frac{-1}{(1+a)^2}\) \(\quad= \quad f(0) = -1\)
\({ f }^{ \prime \prime \prime }(a) \quad= \quad \frac{2}{(1+a)^3}\) \(\quad= \quad f(0) = 2\)
eq1 <- function(x) {log(1+x)}
r <- taylor(eq1, x0 = 0, n = 4)
r
## [1] -0.2500044 0.3333339 -0.5000000 1.0000000 0.0000000