Solution:

1. f (x) = (1−x)

\[f(x)=\frac {1}{1-x}\]

Apply the Taylor Series Equality formula to calculate the expansions of the given functions.

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(c)}{n!}(n-c)^n\]

Find the first derivatives for the given function f(x)

\[f^0(c)=\frac{1}{1-c}\]

\[f^1(c)=\frac{1}{(1-c)^2}\]

\[f^2(c)=\frac{1}{(1-c)^3}\]

\[f^3(c)=\frac{1}{(1-c)^4}\]

\[f^4(c)=\frac{1}{(1-c)^5}\]

Thus, the Taylor series expansion for f(x) converges over the interval (−1,1) and can be defined as:

\[ f(x) \approx P(x) = \frac{1}{(1-c)^{0}!}(x-c)^0 + \frac{1}{(1-c)^{1}1!}(x-c)^1 + \frac{2}{(1-c)^{2}2!}(x-c)^2+ \frac{2}{(1-c)^{3}3!}(x-c)^3 +m\frac{2}{(1-c)^{4}4!}(x-c)^4+...\]

The sum notation for the expansion can be denoted as:

\[ f(x) \approx P(x) = \sum_{n=0}^{\infty} \frac{1}{(1-c)^{n+1}}(x-c)^n\]

Setting c=0 gives the Maclaurin Series of f(x)

\[ f(x) \approx P(x) = \sum_{n=0}^{\infty}x^n= 1+x+x^2+x^3+x^4\]

Now applying the R function for:

# 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 ) = 4

2. f (x) = e^x

The Taylor Series expansion of the exponential function f(x)=e^x is valid for all real numbers x. It is given 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.

Now finding several derivatives.

\[f(x)^0(c) = e^c\]

\[f(x)^1(c) = e^c\]

\[f(x)^2(c) = e^c\]

\[f(x)^3(c) = e^c\]

\[f(x)^4(c) = e^c\]

\[f(x)^4(c) = e^c\]

\[f(x) = \frac {e^c}{0!}(x-c)^0 + \frac {e^c}{1!}(x-c)^1 + \frac {e^c}{2!}(x-c)^2 + \frac {e^c}{3!}(x-c)^3 +...\]

\[ = \sum_{n=0}^{\infty}\frac {(x-c)^n}{n!}\]

The Maclaurin Series, where c=0

\[ = \sum_{n=0}^{\infty}\frac {x^n}{n!}= 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frac{x^4}{4!}+...\]

Now applying the R function for:

# 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: 7

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

Find first derivatives for the given function f(x)

\[f^0(c)=ln(1+c)\]

\[f^1(c)=\frac{1}{c+1}\]

\[f^2(c)=\frac{1}{(c+1)^2}\]

\[f^3(c)=\frac{2}{(c+1)^3}\]

\[f^4(c)=\frac{6}{(c+1)^4}\]

At x=0, these derivatives simplify:

\[f(0)=ln(1)=0\]

\[f^1(0)=\frac{1}{1} =1\]

\[f^2(0)=\frac{-1}{1^2} =-1\]

\[f^3(0)=\frac{2}{1^3} =2\]

\[f^4(0)=\frac{6}{1^4} =-6\]

Taylor series expansion for f(x)=ln(1+x) converges |x|<1 and can be defined as:

\[ f(x) \approx P(x) = P(x)\]

\[ f(x) \approx P(x) = \frac{ln(1+c)}{0!}(x-c)^0 + \frac{1}{(c+1)1!}(x-c)^1 - \frac{1}{(c+1)^22!}(x-c)^2 + \frac{2}{(c+1)^33!}(x-c)^3\] - (x-c)^4 +… ]

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 c=0 gives the Maclaurin Series of f(x):

\[ f(x) \approx P(x) = ln(1+c)+\sum_{n=1}^{\infty}(-1)^{n+1}\frac {(x)^n}{n}= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +...\]

Now applying the R function for:

# 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: 0.4010417
cat("Actual value of ln(1 +", x_value, ") is:", actual_value)
## Actual value of ln(1 + 0.5 ) is: 0.4054651

4. f(x)=x(1/2)

Taylor 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):

\[ \sqrt[4]{4ac} = \sqrt{4ac}\sqrt{4ac} \]

Now starting with

\[f(x)=\sqrt{x}\]

\[f^1(x)=\frac{1}{2\sqrt{x}}= \frac{1}{2x^{1/2}}\]

\[f^2(x)=\frac{1}{4x^{3/2}}\]

\[f^3(x)=\frac{3}{8x^{5/2}}\]

\[f^4(x)=\frac{15}{16x^{7/2}}\]

Calculate these derivatives at a specific point, say x=1(which simplifies the calculations):

\[f(1) = 1\]

\[f^1(1) = \frac{1}{2}\]

\[f^2(1) = \frac{1}{4}\]

\[f^3(1) = \frac{3}{8}\]

\[f^4(1) = \frac{15}{16}\]

The Taylor series expansion for f(x)=sqrt(x) around x=1 (for instance) is:

\[\sqrt{x}= 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 - \frac{1}{16}(x-1)^3-...=\sum_{n=0}^{\infty}\frac{f^n(1)}{n!}(x-1)^n\]

Taylor series expansion around x=1 for sqrt(x) would be valid for x close to 1 (within the convergence radius of the series).

Now applying the R function for:

# 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: 1.132812
cat("Actual value of sqrt(", x_value, ") is:", actual_value)
## Actual value of sqrt( 1.5 ) is: 1.224745