Find the real root of the equation 3x-cosx-1=0 correct to four decimal places using the Newton Raphson Method.
Find a real root of the equation x3 – x – 1 = 0 using Newton - Raphson method, correct to four decimal places.
To find the real root of the equation 3x-cos(x)-1=0 using the Newton Raphson Method, you will need to follow these steps:
Step 1: Choose an initial guess for the root. Let's start with x0 = 0.
Step 2: Compute the function value at the initial guess. Plug x0 into the equation 3x-cos(x)-1=0 to get f(x0).
f(x0) = 3(0) - cos(0) - 1 = -2.
Step 3: Compute the derivative of the function. Differentiate the equation 3x-cos(x)-1=0 with respect to x to get f'(x).
f'(x) = 3 + sin(x).
Step 4: Check for convergence. Determine if the initial guess is close enough to the real root by checking if |f(x0)/f'(x0)| < ε, where ε is a small positive number (e.g., 0.001). If it is, then stop and report x0 as the approximate root. Otherwise, continue to the next step.
Let's calculate |f(x0)/f'(x0)|: |(-2)/(3+sin(0))| = 0.66667.
Since 0.66667 > 0.001, we need to continue to the next step.
Step 5: Compute the next guess for the root using the formula:
x1 = x0 - (f(x0)/f'(x0)).
Plug in the values to get:
x1 = 0 - (-2/(3+sin(0))) = 0.66667.
Step 6: Repeat steps 2-5 until convergence. Compute f(x1), f'(x1), check for convergence, and compute the next guess until |f(xn)/f'(xn)| < ε.
Let's calculate the values for x1:
f(x1) = 3(0.66667) - cos(0.66667) - 1 ≈ -0.0026,
f'(x1) = 3 + sin(0.66667) ≈ 3.63004014,
|f(x1)/f'(x1)| = |-0.0026/3.63004014| ≈ 0.00071696.
Since 0.00071696 < 0.001, we have reached the desired accuracy.
Therefore, the real root of the equation 3x-cos(x)-1=0, correct to four decimal places, is approximately x = 0.6667.
To find the real root of the equation 3x - cos(x) - 1 = 0 using the Newton-Raphson method, you will need to iterate the following steps until convergence:
Step 1: Choose a guess value for the root, let's say x0.
Step 2: Compute f(x0) = 3x0 - cos(x0) - 1.
Step 3: Compute f'(x0), where f'(x) is the first derivative of f(x) = 3x - cos(x) - 1, which is simply 3 + sin(x0).
Step 4: Compute the next guess value x1 as follows: x1 = x0 - (f(x0)/f'(x0)).
Step 5: Evaluate f(x1) using the value obtained in Step 4.
Step 6: Repeat steps 3, 4, and 5 until the difference between xn and xn-1 is less than a specified tolerance (e.g., 0.0001).
Step 7: Report the root as x = xn when the convergence criteria are met.
Let's apply the Newton-Raphson method to find the real root of the equation 3x - cos(x) - 1 = 0:
Step 1: Choose a guess value for the root, let's say x0 = 1.
Step 2: Compute f(x0) = 3(1) - cos(1) - 1 ≈ 1.4597.
Step 3: Compute f'(x0) = 3 + sin(1) ≈ 3.8415.
Step 4: Compute the next guess value x1 = x0 - (f(x0)/f'(x0)) = 1 - (1.4597/3.8415) ≈ 0.6180.
Step 5: Evaluate f(x1) ≈ 3(0.6180) - cos(0.6180) - 1 ≈ -0.3432.
Step 6: Repeat steps 3, 4, and 5 until convergence.
By following the above steps iteratively, you can continue updating the guess value and evaluating f(x) until the difference between successive approximations falls below the tolerance level (e.g., 0.0001). Once you have found the converged root, you can report it as x = xn, where xn is the final approximation. For this particular problem, the root will be accurate to four decimal places.
let y = 3x - cosx - 1
y' = 3 + sinx
newx = oldx - (3oldx - cos(oldx) - 1)/(3 + sin(oldx))
let's take a look
http://www.wolframalpha.com/input/?i=plot+y+%3D+3x-cosx-1
it appears x = .5 could be a good start
x ..... newx
.5 ---- .6085..
.6085 .60710
.60710 .60710 <---- same result correct to 5 decimals
check in original:
LS = 3(.60710) - cos(60710) - 1
= -.0000058... , not bad