IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - 2021 This week, we’ll work out some Taylor Series expansions of popular functions.

  1. f (x) = 1/(1−x)
  2. f (x) = e^x
  3. f (x) = ln(1 + x)
  4. f(x)=x^(1/2)

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.

f(x)=∑_n=0^∞ (f^(n)(a))/n! (x−a)^n

f(a)+f(1)(a)(x−a)+(f(2))/2!(a)(x−a)+…

f(x)=1/(1−x)

library(pracma)

equation = function(x) {1/(1-x)}
p = taylor(equation, x0 = 0, n = 4)
p
## [1] 1.000029 1.000003 1.000000 1.000000 1.000000
  1. f(a)=e^a;f(0)=1

f’(a)=e^a;f’(0)=1

f’‘(a)=e^a;f’’(0)=1

f’’‘(a)=e^a;f’’’(0)=1

(f(4))(a)=ea;(f^(4))(0)=1

equation = function(x) {exp(x)}
p = taylor(equation, x0 = 0, n = 4)
p
## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000

f(a)=ln(1+a);=f(0)=0

f’(a)=1/(1+a);=f’(0)=1

f’‘(a)=-1/(1+a)^2;=f’’(0)=−1

f’’‘(a))=2/(1+a)^3;=f’’’(0)=2

f(4)(a)=−6/(1+a)4;=(f(4))(0)=−6

equation = function(x) {log(1+x)}
p = taylor(equation, x0 = 0, n = 4)
p
## [1] -0.2500044  0.3333339 -0.5000000  1.0000000  0.0000000

f(x)=x^(1/2) f(0)=0

f′(x)=x^(1/2)/2 f′(0)=Undefined

The Taylor series expansion for x1/2 is not possible as the first derivative of the function is undefined at f(0).