For these exerices will expand this around \(x=0\).

1 f(x) = 1/1-x

\[ f(x) = \frac{1}{1-x}\\ f(0) = 1 \\ f'(x) = \frac{1}{(1-x)^2} \\ f'(0) = 1 \\ f''(x) = \frac{2}{(1-x)^3} \\ f''(0) = 2 \\ f^{(3)}(x) = \frac{6}{(1-x)^4} \\ f^{(3)}(0) = 6 \\ f^{(4)}(x) = \frac{24}{(1-x)^5} \\ f^{(4)}(0) = 24 \\ f^{(5)}(x) = \frac{120}{(1-x)^6} \\ f^{(5)}(0) = 120 \\ f(x) = f(0) + f'(0)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f^{3}(c)}{3!}(x-c)^3+ \frac{f^{4}(c)}{4!}(x-c)^4+ \frac{f^{5}(c)}{5!}(x-c)^5 +...\\ f(x) = 1 + x + x^2 + x^3+ x^4+ x^5 +... \]

c = 0
f <- function(x) {1/(1-x)}
a <- taylor(f,x0=c,5)
a
## [1] 1.000293 1.000029 1.000003 1.000000 1.000000 1.000000
p <- function(x) {a[6] + a[5]*(x-c) + (a[4]*(x-c)^2) + (a[3]*(x-c)^3) + (a[2]*(x-c)^4 +(a[6]*(x-c)^5))}
plot(f, from =-0.25,to=0.75)
par(new=TRUE)
plot(p, from = -0.25,to=0.75,axes=FALSE ,col = 'red')

2 f(x) = e^x

\[ f(x) = e^x\\ f(0) = 1 \\ f^n(x) = e^x \\ f^n(0) = 1 \\ f(x) = f(0) + f'(0)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f^{3}(c)}{3!}(x-c)^3+ \frac{f^{4}(c)}{4!}(x-c)^4+ \frac{f^{5}(c)}{5!}(x-c)^5 +...\\ f(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3+ \frac{1}{24}x^4+ \frac{1}{120}x^5 +... \]

c = 0
f <- function(x) {exp(x)}
a <- taylor(f,x0=c,5) #this divides by the factorial
a
## [1] 0.008334245 0.041666573 0.166666726 0.499999996 1.000000000 1.000000000
p <- function(x) {a[6] + a[5]*(x-c) + (a[4]*(x-c)^2) + (a[3]*(x-c)^3) + (a[2]*(x-c)^4 +(a[6]*(x-c)^5))}
plot(f, from = -0.5,to=0.75)
par(new=TRUE)
plot(p, from = -.5,to=0.75,axes=FALSE ,col = 'red')

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

\[ f(x) = ln(1+x) \\ f(0) = 0 \\ f'(x) = \frac{1}{1+x}\\ f'(0) = 1 \\ f''(x) = \frac{-1}{(1+x)^2} \\ f''(0) = -1 \\ f^{(3)}(x) = \frac{2}{(1+x)^3} \\ f^{(3)} = 2 \\ f^{(4)}(x) = \frac{-6}{(1+x)^4} \\ f^{(4)}(0) = -6 \\ f^{(5)}(x) = \frac{24}{(1+x)^5} \\ f^{(5)}(0) = 24 \\ f(x) = f(0) + f'(0)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f^{3}(c)}{3!}(x-c)^3+ \frac{f^{4}(c)}{4!}(x-c)^4+ \frac{f^{5}(c)}{5!}(x-c)^5 +...\\ f(x) = x + -\frac{1}{2}x^2 + \frac{1}{3}x^3-\frac{1}{4} x^4+ \frac{1}{5}x^5 +... \]

c = 0
f <- function(x) {log(1+x)} #default base is e
a <- taylor(f, x0=c, 5)
a
## [1]  0.2000413 -0.2500044  0.3333339 -0.5000000  1.0000000  0.0000000
p <- function(x) {a[6] + a[5]*(x-c) + (a[4]*(x-c)^2) + (a[3]*(x-c)^3) + (a[2]*(x-c)^4 +(a[6]*(x-c)^5))}
plot(f, from = -0.25,to=0.75)
par(new=TRUE)
plot(p, from = -0.25,to=0.75,axes=FALSE ,col = 'red')