This week, we’ll work out some Taylor Series expansions of popular functions. - \(f(x)=\frac{1}{(1-x)}\) - \(f(x)=e^x\) - \(f(x)={\ln (1+x)}\) - \(f(x)=x^{(1 / 2)}\)
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 RMarkdown document.
To calculate the expansions of the given functions, Taylor Series Equality formula is: \[ f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n !}(x-c)^n \]
Find first derivatives for the given function \(\mathrm{f}(\mathrm{x})\) : \[ \begin{aligned} f^{(0)}(c) & =\frac{1}{(1-c)} \\ f^{(1)}(c) & =\frac{1}{(1-c)^2} \\ f^{(2)}(c) & =\frac{2}{(1-c)^3} \\ f^{(3)}(c) & =\frac{6}{(1-c)^4} \\ f^{(4)}(c) & =\frac{24}{(1-c)^5} \end{aligned} \]
Thus, the Taylor series expansion for \(f(x)\) converges over the interval \((-1,1)\) and can be defined as: \[ \begin{aligned} f(x) \approx P(x)= & \frac{1}{(1-c) 0 !}(x-c)^0+\frac{1}{(1-c)^2 1 !}(x-c)^1+\frac{2}{(1-c)^3 2 !}(x-c)^2 \\ & +\frac{6}{(1-c)^3 3 !}(x-c)^3+\frac{24}{(1-c)^4 4 !}(x-c)^4+\cdots \end{aligned} \]
The sum notation for the expansion can be written as: \[ f(x) \approx P(x)=\sum_{n=0}^{\infty} \frac{1}{(1-c)^{n+1}}(x-c)^n \]
Setting \(\mathrm{c}=\mathbf{0}\) gives the Maclaurin Series of \(\mathrm{f}(\mathrm{x})\) : \[ f(x) \approx P(x)=\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+x^4+\cdots \]
# Function of f(x) = 1 / (1 - x)
taylor_series_expansion <- function(x, terms) {
result <- 0
for (n in 0:(terms - 1)) {
term <- x^n
result <- result + term
}
return(result)
}
# Number of terms in the series
num_terms <- 4
# Calculate f(x) = 1 / (1 - x) at different x values
x_values <- seq(-1, 1, by = 0.2) # Range of x values for demonstration
for (x in x_values) {
series_result <- taylor_series_expansion(x, num_terms)
cat("f(", x, ") ≈ P(", x, ") =", series_result, "\n")
}## f( -1 ) ≈ P( -1 ) = 0
## f( -0.8 ) ≈ P( -0.8 ) = 0.328
## f( -0.6 ) ≈ P( -0.6 ) = 0.544
## f( -0.4 ) ≈ P( -0.4 ) = 0.696
## f( -0.2 ) ≈ P( -0.2 ) = 0.832
## f( 0 ) ≈ P( 0 ) = 1
## f( 0.2 ) ≈ P( 0.2 ) = 1.248
## f( 0.4 ) ≈ P( 0.4 ) = 1.624
## f( 0.6 ) ≈ P( 0.6 ) = 2.176
## f( 0.8 ) ≈ P( 0.8 ) = 2.952
## f( 1 ) ≈ P( 1 ) = 4The Taylor Series expansion of the exponential function \(f(x)=e^x\) is valid for all real numbers \(x\). It is giver by: \[ e^x=\sum_{n=0}^{\infty} \frac{x^n}{n !} \]
This expansion represents an infinite sum of powers of \(x\) divided by their factorial. Finding several derivatives. \[ \begin{aligned} & f^{(0)}(c)&=e^c \\ & f^{(1)}(c)&=e^c \\ & f^{(2)}(c)&=e^c \\ & f^{(3)}(c)&=e^c \\ & f^{(4)}(c)&=e^c \end{aligned} \] \[ \begin{aligned} f(x) & =\frac{e^c}{0 !}(x-c)^0+\frac{e^c}{1 !}(x-c)^1+\frac{e^c}{2 !}(x-c)^2+\frac{e^c}{3 !}(x-c)^3+\cdots \\ & =e^c \sum_{n=0}^{\infty} \frac{(x-c)^n}{n !} \end{aligned} \]
The Maclaurin Series, where \(\mathrm{c}=0\) \[ f(x)=\sum_{n=0}^{\infty} \frac{x^n}{n !}=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !}+\cdots . . \]
# Function of e^x
taylor_exp <- function(x, terms = 4) {
result <- 0
for (n in 0:terms) {
result <- result + (x^n) / factorial(n)
}
return(result)
}
# Set specific value of x
x_value <- 2
result <- taylor_exp(x_value)
# Print the result
cat("e^", x_value, "approximated using Taylor series with 4 terms is:", result, "\n")## e^ 2 approximated using Taylor series with 4 terms is: 7Find first derivatives for the given function \(\mathrm{f}(\mathrm{x})\) : \[ \begin{aligned} & f^{(0)}(c)=\ln (1+c) \\ & f^{(1)}(c)=\frac{1}{c+1} \\ & f^{(2)}(c)=-\frac{1}{(c+1)^2} \\ & f^{(3)}(c)=\frac{2}{(c+1)^3} \\ & f^{(4)}(c)=-\frac{6}{(c+1)^4} \end{aligned} \]
At \(x=0\), these derivatives simplify: \[ \begin{aligned} & \mathrm{f}(0)=\ln (1)=0 \\ & f^{(1)}(0)=\frac{1}{1}=1 \\ & f^{(2)}(0)=-\frac{1}{(1)^2}=-1 \\ & f^{(3)}(0)=\frac{2}{(1)^3}=2 \\ & f^{(4)}(0)=-\frac{6}{(1)^4}=-6 \end{aligned} \]
Thus, the Taylor series expansion for \(\mathrm{f}(\mathrm{x})=\ln (1+x)\) converges \(|x|<1\) and can be defined as: \[ \begin{aligned} f(x) \approx P(x)= & P(x) \\ & =\frac{\ln (1+c)}{0 !}(x-c)^0+\frac{1}{(c+1) 1 !}(x-c)^1-\frac{1}{(c+1)^2 2 !}(x-c)^2 \\ & +\frac{2}{(c+1)^3 3 !}(x-c)^3-\frac{6}{(c+1)^4 4 !}(x-c)^{43}+\cdots \end{aligned} \]
The sum notation for the expansion can be written as: \[ f(x) \approx P(x)=\ln (1+c)+\sum_{n=1}^{\infty}(-1)^{n+1} \frac{(x-c)^n}{n(c+1)^n} \]
Setting \(\mathrm{c}=\mathbf{0}\) gives the Maclaurin Series of \(\mathrm{f}(\mathrm{x})\) : \[ f(x) \approx P(x)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{(x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \]
# Function of ln(1 + x)
maclaurin_ln <- function(x, terms = 4) {
result <- 0
for (n in 1:terms) {
result <- result + ((-1)^(n+1) * (x^n)) / n
}
return(result)
}
# Value of x
x_value <- 0.5
# Calculate the approximation using the Maclaurin series with 4 terms
approximation <- maclaurin_ln(x_value, terms = 4)
# Calculate the actual value of ln(1 + x)
actual_value <- log(1 + x_value)
# Print the results
cat("ln(1 +", x_value, ") approximated using Maclaurin series with 4 terms is:", approximation, "\n")## ln(1 + 0.5 ) approximated using Maclaurin series with 4 terms is: 0.4010417cat("Actual value of ln(1 +", x_value, ") is:", actual_value)## Actual value of ln(1 + 0.5 ) is: 0.4054651Taylor series expansion around a specific point, typically around \(x=c\) (where \(c\) is the center).
The Taylor series expansion of \(f(x)=\sqrt{x}\) around \(x=c\) involves calculating the derivatives of the function at \(x=a\) to determine the coefficients of the series.
Starting with \(f(x)=\sqrt{x}\) : \[ \begin{aligned} f^{(1)}(x) & =\frac{1}{2 \sqrt{x}}=\frac{1}{2 x^{1 / 2}} \\ f^{(2)}(x) & =-\frac{1}{4 x^{3 / 2}} \\ f^{(3)}(x) & =\frac{3}{8 x^{5 / 2}} \\ f^{(4)}(x) & =-\frac{15}{16 x^{7 / 2}} \end{aligned} \]
Calculate these derivatives at a specific point, say \(x=1\) (which simplifies the calculations): \[ \begin{aligned} & f(1)=1 \\ & f^{(1)}(1)=\frac{1}{2} \\ & f^{(2)}(1)=-\frac{1}{4} \\ & f^{(3)}(1)=\frac{3}{8} \\ & f^{(4)}(1)=-\frac{15}{16} \end{aligned} \]
The Taylor series expansion for \(f(x)=\sqrt{x}\) around \(x=1\) (for instance) is: \[ \sqrt{x}=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2+\frac{1}{16}(x-1)^3-\cdots=\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n !}(x-1)^n \]
Taylor series expansion around \(x=1\) for \(\sqrt{x}\) would be valid for \(x\) close to 1 (within the convergetce radius of the series).
# Function of sqrt(x) around x = 1
taylor_sqrt <- function(x, terms = 4) {
result <- 1 # The term for n = 0
for (n in 1:terms) {
result <- result + ((-1)^(n-1) * factorial(2*n-2) * (x-1)^n) / (2^n * factorial(n))
}
return(result)
}
# Value of x
x_value <- 1.5
# Calculate the approximation
approximation <- taylor_sqrt(x_value, terms = 4)
# Calculate the actual value
actual_value <- sqrt(x_value)
# Print the results
cat("sqrt(", x_value, ") approximation around x = 1 with 4 terms is:", approximation, "\n")## sqrt( 1.5 ) approximation around x = 1 with 4 terms is: 1.132812cat("Actual value of sqrt(", x_value, ") is:", actual_value)## Actual value of sqrt( 1.5 ) is: 1.224745