TRUE/FALSE:
1. T/F: Given a function f(x), Newton’s Method produces an exact solution to f(x) = 0.
ANSWER: False
Newton’s Method does not necessarily produce an exact solution to (f(x) = 0). Instead, it provides an iterative process for approximating the roots of a function. The method starts with an initial guess and refines it through successive iterations.
2. T/F: In order to get a solution to f(x) = 0 accurate to d places after the decimal, at least d + 1 iterations of Newtons’ Method must be used.
ANSWER: True
Newton’s Method is a powerful tool for finding approximate solutions, especially when explicit formulas are not available. It relies on derivatives and tangent line approximations to iteratively refine estimates of the roots.
To obtain a solution accurate to (d) decimal places after the decimal point, you typically need at least (d + 1) iterations of Newton’s Method. Each iteration improves the approximation, and the accuracy increases as you repeat the process. However, the actual number of iterations required can vary based on the function and the initial guess.
TRUE/FALSE:
1.T/F: Given a differentiable function y = f(x), we are generally free to choose a value for dx, which then determines the value of dy.
ANSWER: True
Given a differentiable function (y = f(x)), we can choose a value for (\Delta x) (a small change in the independent variable (x)), which then determines the corresponding change in the dependent variable (y), denoted as (\Delta y).
2. T/F: The symbols “dx” and “∆x” represent the same concept.
ANSWER: False
The symbols “dx” and “(\Delta x)” represent distinct concepts: (dx) represents an infinitesimal change in the independent variable (x). It is used to denote the differential. (\Delta x) represents a finite change in (x), typically describing a small interval. While both relate to changes in the independent variable, they differ in scale: (dx) is infinitesimally small, capturing the behavior of the function at a specific point. (\Delta x) is a finite interval, representing a broader change.
In summary, while both involve changes in the independent variable, the key distinction lies in the scale: (dx) is infinitesimal, and (\Delta x) is finite. Both concepts play essential roles in calculus, allowing us to understand local behavior and broader trends in functions.
3. T/F: The symbols “dy” and “∆y” represent the same concept.
ANSWER: True
The symbols “dy” and “(\Delta y)” represent the same concept. Both denote the change in the dependent variable (y) corresponding to a change in the independent variable (x). While “dy” represents an infinitesimal change, “(\Delta y)” represents a finite change. Both are essential in calculus for understanding rates of change and approximations.
4. T/F: Differentials are important in the study of integration.
ANSWER: True
Differentials play a crucial role in the study of integration. Differentials provide a bridge between the local behavior of a function (as captured by derivatives) and the overall accumulation (as captured by integrals). They are fundamental tools in calculus.
5. How are differentials and tangent lines related?
ANSWER:
Differentials and tangent lines are closely related concepts in calculus. Tangent lines provide a geometric interpretation of derivatives, while differentials help us quantify the local changes in a function. Together, they enhance our understanding of how functions behave near specific points
6. T/F: In real life, differentials are used to approximate function values when the function itself is not known.
ANSWER: True:
In practical applications,
differentials play a crucial role in approximating function values when the exact function is not known. They allow us to navigate the complexities of real-world problems even when we lack precise function definitions. In this regard, they serve as valuable tools for estimation, modeling, and understanding local behavior.
Example Scenarios:
2. ECONOMICS: Economic models often involve unknown functions. Differentials help estimate changes in quantities like demand, supply, or interest rates.
3. ENGINEERING: Engineers use differentials to approximate system responses, such as temperature variations, stress, or fluid flow.In Exercises 3 – 8, the roots of f(x) are known or are easily found. Use 5 iterations of Newton’s Method with the given initial approximation to approximate the root. Compare it to the known value of the root.
3. f(x) = cos x, x0 = 1:5
4. f(x) = sin x, x0 = 1
5. f(x) = x2 + x 2, x0 = 0
6. f(x) = x2 2, x0 = 1:5
7. f(x) = ln x, x0 = 2
8. f(x) = x3 x2 + x 1, x0 = 1
let’s approximate the derivative using a small step size (1e-8) in a forward difference scheme.
# Define the function f(x) for which you want to find the root
f <- function(x) {
# Replace with the actual function expression
# Example: f(x) = x^2 - 2
return(x^2 - 2)
}
# Initial approximation (x0)
x0 <- 1.5
# Number of iterations
num_iterations <- 5
# Apply Newton's Method iteratively
for (i in 1:num_iterations) {
# Compute the derivative of f(x) with respect to x
df_dx <- (f(x0 + 1e-8) - f(x0)) / 1e-8
# Update the approximation using the tangent line
x1 <- x0 - f(x0) / df_dx
# Print the current approximation
cat("Iteration", i, ": x =", x1, "\n")
# Update x0 for the next iteration
x0 <- x1
}## Iteration 1 : x = 1.416667
## Iteration 2 : x = 1.414216
## Iteration 3 : x = 1.414214
## Iteration 4 : x = 1.414214
## Iteration 5 : x = 1.414214# Compare the final approximation to the known root
known_root <- sqrt(2) # Example: Known root for f(x) = x^2 - 2
cat("Final approximation:", x1, "\n")## Final approximation: 1.414214cat("Known root:", known_root, "\n")## Known root: 1.414214Here, we derivative define and use the derivative within the Newton’s Method function.
# Define the function and its derivative
f <- function(x) {
cos(x) # Change this function according to the problem
}
f_prime <- function(x) {
-sin(x) # Change this derivative according to the problem
}
# Newton's Method function
newton_method <- function(f, f_prime, x0, n_iterations) {
x <- x0
for (i in 1:n_iterations) {
x <- x - f(x) / f_prime(x)
}
return(x)
}
# Initial approximations and number of iterations
x0_list <- c(1.5, 1, 0, 1.5, 2, 1)
n_iterations <- 5
# Applying Newton's Method and comparing with known roots
for (i in 1:length(x0_list)) {
root_approx <- newton_method(f, f_prime, x0_list[i], n_iterations)
cat("Root approximation for problem", i + 2, ":", root_approx, "\n")
}## Root approximation for problem 3 : 1.570796
## Root approximation for problem 4 : 1.570796## Warning in cos(x): NaNs produced## Warning in sin(x): NaNs produced## Root approximation for problem 5 : NaN
## Root approximation for problem 6 : 1.570796
## Root approximation for problem 7 : 1.570796
## Root approximation for problem 8 : 1.570796In Exercises 13 – 16, use Newton’s Method to approximatewhen the given functions are equal, accurate to 3 places afterthe decimal. Use technology to obtain good initial approximations.13. f(x) = x2, g(x) = cos x14. f(x) = x2 1, g(x) = sin x15. f(x) = ex2 , g(x) = cos x16. f(x) = x, g(x) = tan x on [6; 6]17. Why does Newton’s Method fail in
# Define the functions f(x) and g(x)
f <- function(x, problem) {
if (problem == 13) {
return(x^2)
} else if (problem == 14) {
return(x^2 - 1)
} else if (problem == 15) {
return(exp(x^2))
} else if (problem == 16) {
return(x)
}
}
g <- function(x, problem) {
if (problem == 13) {
return(cos(x))
} else if (problem == 14) {
return(sin(x))
} else if (problem == 15) {
return(cos(x))
} else if (problem == 16) {
return(tan(x))
}
}
# Define the derivatives of f(x) and g(x)
f_prime <- function(x, problem) {
if (problem == 13) {
return(2 * x)
} else if (problem == 14) {
return(2 * x)
} else if (problem == 15) {
return(2 * x * exp(x^2))
} else if (problem == 16) {
return(1)
}
}
g_prime <- function(x, problem) {
if (problem == 13) {
return(-sin(x))
} else if (problem == 14) {
return(cos(x))
} else if (problem == 15) {
return(-sin(x))
} else if (problem == 16) {
return(1 + tan(x)^2)
}
}
# Newton's Method function
newton_method <- function(f, f_prime, g, g_prime, x0, n_iterations, tolerance) {
x <- x0
for (i in 1:n_iterations) {
fx <- f(x, problem)
gx <- g(x, problem)
if (abs(fx - gx) < tolerance) {
break # If the difference between f(x) and g(x) is close to zero, break the loop
}
x <- x - (fx - gx) / (f_prime(x, problem) - g_prime(x, problem))
}
return(x)
}
# Set the problem number
problem <- 13
# Set the initial approximation and number of iterations
x0 <- 1 # You can change this initial approximation
n_iterations <- 1000 # Maximum number of iterations
tolerance <- 0.0001 # Tolerance for stopping criterion
# Apply Newton's Method
root_approx <- newton_method(f, f_prime, g, g_prime, x0, n_iterations, tolerance)
# Print the root approximation
cat("Root approximation for problem", problem, ":", round(root_approx, 3), "\n")## Root approximation for problem 13 : 0.824# Set the problem number
problem <- 14
# Set the initial approximation and number of iterations
x0 <- 1 # You can change this initial approximation
n_iterations <- 1000 # Maximum number of iterations
tolerance <- 0.0001 # Tolerance for stopping criterion
# Apply Newton's Method
root_approx <- newton_method(f, f_prime, g, g_prime, x0, n_iterations, tolerance)
# Print the root approximation
cat("Root approximation for problem", problem, ":", round(root_approx, 3), "\n")## Root approximation for problem 14 : 1.41# Set the problem number
problem <- 15
# Set the initial approximation and number of iterations
x0 <- 1 # You can change this initial approximation
n_iterations <- 1000 # Maximum number of iterations
tolerance <- 0.0001 # Tolerance for stopping criterion
# Apply Newton's Method
root_approx <- newton_method(f, f_prime, g, g_prime, x0, n_iterations, tolerance)
# Print the root approximation
cat("Root approximation for problem", problem, ":", round(root_approx, 3), "\n")## Root approximation for problem 15 : 0.006# Set the problem number
problem <- 16
# Set the initial approximation and number of iterations
x0 <- 1 # You can change this initial approximation
n_iterations <- 1000 # Maximum number of iterations
tolerance <- 0.0001 # Tolerance for stopping criterion
# Apply Newton's Method
root_approx <- newton_method(f, f_prime, g, g_prime, x0, n_iterations, tolerance)
# Print the root approximation
cat("Root approximation for problem", problem, ":", round(root_approx, 3), "\n")## Root approximation for problem 16 : 0.053