Assignment 14 Computation of Talor Series

\(\frac{1}{1-x} = \sum _{ n=0 }^{ \infty }{ { x }^{ n } } = 1+x+x^2+x^3+...\) Interval of convergence \((-1, 1)\)

x <- 0.01
firsterm <- 1/(1-x)
firsterm
## [1] 1.010101
secondterm <- 0
for (n in 0:1000) {
  secondterm <- secondterm + x**n
}
secondterm
## [1] 1.010101

\({e}^{x}=\sum _{ n=0 }^{ \infty }\frac{ { x }^{ n } }{n!}= 1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}+\frac{{x}^{4}}{24}+\frac{{x}^{5}}{120}...\) Interval of convergence is \((-\infty, +\infty)\)

x <- 1
e1 <- exp(x)
e1
## [1] 2.718282
e2 <- 0
for (n in 0:1000) {
  e2 <- e2 + (x**2)/factorial(n)
}
e2
## [1] 2.718282

\(ln(1+x)=\sum_{n=1}^{\infty}{{(-1)^{n+1}}\frac{{x}^{n}}{n}}=\frac{x}{1}-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}=...\) Interval of convergence is \((-1,1]\)

x <- 1
ln1 <- log(1+x)
ln1
## [1] 0.6931472
ln2 <- 0
for (n in 1:1000) {
  ln2 <- ln2 + ((-1)**(n+1))*(x**n)/n
}
ln2
## [1] 0.6926474