JChhabria_Assign14_DS605

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 a R-Markdown document.

1. f(x) = 1/(1 - x) Taylor’s formula states that: \[f(x) = \sum_{n=1}^{\infty} f^n(0)x^n/n!\]

where the following represents the nth derivative of the function with respect to x, evaluated at 0 \[f^n(0)\] Evaluating the first few derivatives below:

\[f^0(x) = (1-x)^{-1}\\ f^1(x) = (-1)(1-x)^{-2}(-1) = (1-x)^{-2}\\ f^2(x) = (-2)(1-x)^{-3}(-1) = 2(1-x)^{-3}\\ f^3(x) = (-3)2(1-x)^{-4}(-1) = 6(1-x)^{-4}\\ In\, general\,the\,pattern\,is:\\ f^n(x) = n!(1-x)^{-(n+1)}\\ \] Using the Taylor formula for calculating the co-efficient of the nth term gives:

\[ a_{n} = f^n(0)/n! = (n!(1-0)^{-(n+1)})/n! = 1\\\] This series is bounded by |-1,1| because the radius of convergence is 1.

The derivative term for the first 5 terms is calculated below:

x=0
for(n in 0:4){
  a=factorial(n)/(1-x)^(n+1)
  print(a)
}

## [1] 1
## [1] 1
## [1] 2
## [1] 6
## [1] 24

The derivative term cancels out with the factorial in the denominator. We test the sum of the series upto a given number of terms below, using a R library:

library(pracma)

f<-function(x) {1/(1-x)}
s<-taylor(f,x0=0,n=4)
s

## [1] 1.000029 1.000003 1.000000 1.000000 1.000000

The sum converges to exactly 1 very quickly.

2) f(x) = e^x

\[f^0(x) = e^{x}\,,\, f^0(0) = e^{0} = 1 \\ f^1(x) = e^{x}\,,\, f^1(0) = e^{0} = 1\\ f^2(x) = e^{x}\,,\, f^2(0) = e^{0} = 1\\ f^3(x) = e^{x}\,,\, f^3(0) = e^{0} = 1\\ In\, general\,the\,pattern\,is:\\ f^n(x) = e^{x}\,,\, f^n(0) = e^{0} = 1\\ \] The radius of convergence is in this case is: \[|+\infty,-\infty|\]

Using Taylor’s formula: \[f(x) = \sum_{n=0}^{\infty} f^n(a)(x-a)^n/n!\] Since all the derivatives are = 1, we get: \[f(0) = \sum_{n=0}^{\infty}(x)^n/n!\] Using R to confirm:

f<-function(x) {exp(x)}
s<-taylor(f,x0=0, n=4)
s

## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000

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

\[f^0(x) = ln(1+x)\,,\, f^0(0) = 1/(1+0)= 0 \\ f^1(x) = (1+x)^{-1}\,,\, f^1(0) = 1/(1+0)^{1} = 1\\ f^2(x) = -1(1+x)^{-2}\,,\, f^2(0) = -1/(1+0)^{2} = -1\\ f^3(x) = (-1)(-2)(1+x)^{-3}\,,\, f^3(0) = 2/(1+0)^{3} = 2\\ f^4(x) = (2)(-3)(1+x)^{-4}\,,\, f^3(0) = -6/(1+0)^{4} = -6\\ In\, general\,the\,pattern\,is:\\ f^n(x) = (-1)^{n-1}(n-1)!/(1+x)^n\,,\, f^n(0) = (-1)^{n-1}(n-1)!\\ \] The radius of convergence is |-1,1|

Using Taylor’s formula: \[f(x) = \sum_{n=0}^{\infty} f^n(a)(x-a)^n/n!\] We get: \[f(x) = 0+x-x^2/2!+2x^3/3!-6x^4/4!+....\\ f(x) = x-x^2/2+x^3/3-x^4/4+....\\ \] Using R to confirm:

f<-function(x) {log(1+x)}
s<-taylor(f,x0=0,n=4)
s

## [1] -0.2500044  0.3333339 -0.5000000  1.0000000  0.0000000