In this assignment we will explore 3 typical Taylor Series.
We will first find the Taylor Serie centered at \({ x }_{ 0 }=0\) (McLaurin Serie) for a given function and then write a program to randomly check various value of x in the domain of function and calulate the result using the function definition and the Taylor Serie.
We will first look at function \(f\left( x \right) =\frac { 1 }{ 1-x }\) define over interval (-1,1).
We will expand the Taylor Serie centered at \({ x }_{ 0 }=0\). We will start by finding the first few terms \(f\left( x \right) ,\quad f^{ ' }\left( x \right) ,\quad f^{ (2) }\left( x \right) ,\quad f^{ (3) }\left( x \right) ,\quad and\quad extrapolate\quad f^{ (n) }\left( x \right)\).
We will first find the derivatives and then evaluate them at \({ x }_{ 0 }=0\). Since we have, \(f\left( x \right) =\frac { 1 }{ 1-x }\), let us denote u(x)=1-x, and u’(x)=-1, we can write f(x) as 1/u(x). We will apply chain rules to find derivatives.
\(f^{ ' }\left( x \right) =-1\cdot { u }^{ -2 }\cdot (-1)={ u }^{ -2 }={ (1-x) }^{ -2 }\)
\(f^{ (2) }\left( x \right) =-2\cdot { u }^{ -3 }\cdot (-1)=2\cdot { u }^{ -3 }=2\cdot { (1-x) }^{ -3 }\)
\(f^{ (3) }\left( x \right) =2\cdot (-3)\cdot { u }^{ -4 }\cdot (-1)=2\cdot 3\cdot { u }^{ -4 }=2\cdot 3\cdot { (1-x) }^{ -4 }=3!\cdot { (1-x) }^{ -4 }\)
Extrapolating for n derivatives, we have;
\(f^{ (n) }\left( x \right) =(n-1)!\cdot (-n)\cdot { u }^{ -(n+1) }\cdot (-1)=(n-1)!\cdot n\cdot { u }^{ -(n+1) }=n!\cdot { (1-x) }^{ -(n+1) }\)
Hence since we are evaluating the Serie at \({ x }_{ 0 }=0\), we can write the Tayor Serie for \(f\left( x \right) =\frac { 1 }{ 1-x }\) as follows;
\(\frac { 1 }{ 1-x } =f\left( 0 \right) +f^{ (1) }\left( 0 \right) (x-0)+\frac { 1 }{ 2! } f^{ (2) }\left( 0 \right) { (x-0) }^{ 2 }+\frac { 1 }{ 3! } f^{ (3) }\left( 0 \right) { (x-0) }^{ 3 }+\quad ...\quad +\frac { 1 }{ n! } f^{ (n) }\left( 0 \right) { (x-0) }^{ n }\)
We will now replace the derivative by their value for 0. This will give us;
\(\frac { 1 }{ 1-x } =1+x+\frac { 2! }{ 2! } 1{ x }^{ 2 }+\frac { 3! }{ 3! } 1{ x }^{ 3 }+\quad ...\quad +\frac { n! }{ n! } 1{ x }^{ n }\)
Simplifying the terms and using sigma notations, we obtain the following;
\(\frac { 1 }{ 1-x } =\sum _{ n=0 }^{ \infty }{ { x }^{ n } } ,\quad for\quad x\epsilon (-1,1)\)
We will now look at the funtion \(f\left( x \right) ={ e }^{ x }\) define over Real numbers.
We will expand the Taylor Serie centered at \({ x }_{ 0 }=0\). We will start by finding the first few terms \(f\left( x \right) ,\quad f^{ ' }\left( x \right) ,\quad f^{ (2) }\left( x \right) ,\quad f^{ (3) }\left( x \right) ,\quad and\quad extrapolate\quad f^{ (n) }\left( x \right)\).
We will first find the derivatives and then evaluate them at \({ x }_{ 0 }=0\). Since \(f^{ ' }\left( x \right) ={ e }^{ x }\) as well, we will have:
\(f^{ ' }\left( x \right) ={ e }^{ x }\)
\(f^{ (2) }\left( x \right) ={ e }^{ x }\)
\(f^{ (3) }\left( x \right) ={ e }^{ x }\)
Extrapolating for n derivatives, we have;
\(f^{ (n) }\left( x \right) ={ e }^{ x }\)
Hence since we are evaluating the Serie at \({ x }_{ 0 }=0\), we can write the Tayor Serie for \(f\left( x \right) ={ e }^{ x }\) as follows:
\(\frac { 1 }{ 1-x } =f\left( 0 \right) +f^{ (1) }\left( 0 \right) (x-0)+\frac { 1 }{ 2! } f^{ (2) }\left( 0 \right) { (x-0) }^{ 2 }+\frac { 1 }{ 3! } f^{ (3) }\left( 0 \right) { (x-0) }^{ 3 }+\quad ...\quad +\frac { 1 }{ n! } f^{ (n) }\left( 0 \right) { (x-0) }^{ n }\)
We will now replace the derivative by their value for 0. This will give us;
\({ e }^{ x }={ e }^{ 0 }+{ e }^{ 0 }x+\frac { { e }^{ 0 } }{ 2! } { x }^{ 2 }+\frac { { e }^{ 0 } }{ 3! } { x }^{ 3 }+....+\frac { { e }^{ 0 } }{ n! } { x }^{ n }\)
since \({ e }^{ 0 }=1\), this becomes:
\({ e }^{ x }=1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +\quad ...\quad +\frac { { x }^{ n } }{ n! }\)
Simplifying the terms and using sigma notations, we obtain the following;
\({ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }\)
Finally, we will now consider the function \(f\left( x \right) =\ln { (1+x) } \quad with\quad x\epsilon (-1,1)\).
We will expand the Taylor Serie centered at \({ x }_{ 0 }=0\). We will start by finding the first few terms \(f\left( x \right) ,\quad f^{ ' }\left( x \right) ,\quad f^{ (2) }\left( x \right) ,\quad f^{ (3) }\left( x \right) ,\quad and\quad extrapolate\quad f^{ (n) }\left( x \right)\).
We will first find the derivatives and then evaluate them at \({ x }_{ 0 }=0\). Since we have, \(f\left( x \right) =\ln { (1+x) }\), let us denote u(x)=1+x, and u’(x)=1, we can write f(x) as ln(u). We will apply chain rules to find derivatives.
\(f^{ (1) }\left( x \right) =\frac { 1 }{ u } \cdot 1=\frac { 1 }{ 1+x }\)
\(f^{ (2) }\left( x \right) =(-1){ u }^{ -2 }(1)=-\frac { 1 }{ { (1+x) }^{ 2 } }\)
\(f^{ (3) }\left( x \right) =(-1)(-2){ u }^{ -3 }(1)=\frac { 2 }{ { (1+x) }^{ 3 } }\)
\(f^{ (4) }\left( x \right) =(2)(-3){ u }^{ -4 }(1)=-\frac { 3! }{ { (1+x) }^{ 4 } }\)
Extrapolating for n derivatives, we have;
\(f^{ (n) }\left( x \right) ={ (-1) }^{ n-1 }(n-2)!(n-1){ u }^{ -n) }(1)={ (-1) }^{ n-1 }\frac { (n-1)! }{ { (1+x) }^{ n } }\)
Hence since we are evaluating the Serie at \({ x }_{ 0 }=0\), we can write the Tayor Serie for \(f\left( x \right) =\ln { (1+x) }\) as follows:
\(\frac { 1 }{ 1-x } =f\left( 0 \right) +f^{ (1) }\left( 0 \right) (x-0)+\frac { 1 }{ 2! } f^{ (2) }\left( 0 \right) { (x-0) }^{ 2 }+\frac { 1 }{ 3! } f^{ (3) }\left( 0 \right) { (x-0) }^{ 3 }+\quad ...\quad +\frac { 1 }{ n! } f^{ (n) }\left( 0 \right) { (x-0) }^{ n }\)
We will now replace the derivative by their value for 0. This will give us;
\(\ln { (1+x)= } 0+1x-1\frac { { x }^{ 2 } }{ 2! } +2!\frac { { x }^{ 3 } }{ 3! } -3!\frac { { x }^{ 4 } }{ 4! } +\quad ...\quad +{ (-1) }^{ n-1 }(n-1)!\frac { { x }^{ n } }{ n! }\)
Simplifying the terms and using sigma notations, we obtain the following;
\(\ln { (1+x)= } \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n-1 }\frac { { x }^{ n } }{ n } }\)
We will now compare the results for calculation based on functions and taylor series for every functions.
First we will write programs to represent each function and each Taylor Series.
### First Function
func_1 <- function(x){
result <- 1/(1-x)
return(result)
}
func_taylor_1 <- function(x, d=50){
i <- 0
result <- 0
while (i <= d){
result <- result + x^i
i<-i+1
}
return(result)
}
### 2nd Function
func_2 <- function(x){
result <- exp(x)
return(result)
}
func_taylor_2 <- function(x, d=50){
i <- 0
result <- 0
while (i <= d){
result <- result + (x^i/factorial(i))
i<-i+1
}
return(result)
}
### 3rd Function
func_3 <- function(x){
result <- log(1+x)
return(result)
}
func_taylor_3 <- function(x, d=50){
i <- 1
result <- 0
while (i <= d){
result <- result + ((-1)^(i+1))*(x^i/i)
i<-i+1
}
return(result)
}We will now write a function that will allow us to compare the calcualtions between each function and their corresponding Taylor Series.
test_taylor <- function(func_indicator=1, degree=50, n=100){
# This function will test the taylor series for various functions denoted by func_indicator
# For function 1 or 3, the range for x values will be (-1,1) exclusive
remove <-c(-1,1)
if (func_indicator == 1 || func_indicator==3){
x_universe <- seq(-1,1,length.out = n+2)
x_universe <-setdiff(x_universe, remove)
}else{
x_universe <- seq(-100.0,100.0,length.out = n)
}
func_results <- vector(mode='numeric', length=n)
taylor_results <-vector(mode='numeric', length=n)
for (i in 1:n){
x<-x_universe[i]
if(func_indicator == 1){
func_results[i] <- func_1(x)
taylor_results[i] <- func_taylor_1(x, degree)
}else if (func_indicator == 2){
func_results[i] <-func_2(x)
taylor_results[i] <- func_taylor_2(x, degree)
}else{
func_results[i] <-func_3(x)
taylor_results[i] <- func_taylor_3(x, degree)
}
} #End of For loop
r_list<-list(x_universe, func_results, taylor_results)
return(r_list)
}We will now test by calling this test function…
# Set allowed Error
error_allowed <- 0.0001
# Testing function1
res1_1 <- test_taylor(1, 10, 50)
res1_2 <- test_taylor(1, 50, 50)
x_values1_1 <- res1_1[[1]]
func_values1_1 <- res1_1[[2]]
taylor_values1_1 <- res1_1[[3]]
x_values1_2 <- res1_2[[1]]
func_values1_2 <- res1_2[[2]]
taylor_values1_2 <- res1_2[[3]]
compare_values1_1 <- func_values1_1 - taylor_values1_1
compare_values1_2 <- func_values1_2 - taylor_values1_2
compare_values1_1 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [34] TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [45] FALSE FALSE FALSE FALSE FALSE FALSEcompare_values1_2 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [34] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [45] TRUE TRUE FALSE FALSE FALSE FALSEAs expected, we get better results when we performed Taylor Series with more terms (10 vs 50). We will repeat the test for the other 2 fucntions.
# Testing function2
res2_1 <- test_taylor(2, 10, 50)
res2_2 <- test_taylor(2, 50, 50)
x_values2_1 <- res2_1[[1]]
func_values2_1 <- res2_1[[2]]
taylor_values2_1 <- res2_1[[3]]
x_values2_2 <- res2_2[[1]]
func_values2_2 <- res2_2[[2]]
taylor_values2_2 <- res2_2[[3]]
compare_values2_1 <- func_values2_1 - taylor_values2_1
compare_values2_2 <- func_values2_2 - taylor_values2_2
compare_values2_1 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [34] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [45] FALSE FALSE FALSE FALSE FALSE FALSEcompare_values2_2 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE
## [34] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [45] FALSE FALSE FALSE FALSE FALSE FALSE# Testing function3
res3_1 <- test_taylor(3, 10, 50)
res3_2 <- test_taylor(3, 50, 50)
x_values3_1 <- res3_1[[1]]
func_values3_1 <- res3_1[[2]]
taylor_values3_1 <- res3_1[[3]]
x_values3_2 <- res3_2[[1]]
func_values3_2 <- res3_2[[2]]
taylor_values3_2 <- res3_2[[3]]
compare_values3_1 <- func_values3_1 - taylor_values3_1
compare_values3_2 <- func_values3_2 - taylor_values3_2
compare_values3_1 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [34] TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
## [45] FALSE FALSE FALSE FALSE FALSE FALSEcompare_values3_2 <= error_allowed## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [34] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [45] TRUE TRUE TRUE TRUE FALSE FALSE