Use a calculator to compute the first 10 iterations of Newton's method when applied to the function with the given initial approximation. Make a table for the values. Round to six decimal places.

f(x) = 3x - cos x; x0 = 1

Question

Use a calculator to compute the first 10 iterations of Newton's method when applied to the function with the given initial approximation. Make a table for the values. Round to six decimal places.

f(x) = 3x - cos x; x0 = 1

BlackTom AI · Accepted Answer

The question asks to use Newton's method to compute the first 10 iterations for f(x) = 3x - cos x with x0 = 1 and to present the results in a table, rounded to six decimal places.

First, recall Newton’s iteration formula: x_{k+1} = x_k - f(x_k) / f'(x_k).
- Here, f(x) = 3x - cos x, so f'(x) = 3 + sin x.

For the given initial guess x0 = 1, the method proceeds as follows:
- Compute f(1) = 3(1) - cos(1) and f'(1) = 3 + sin(1), then form x1 = 1 - f(1)/f'(1).
- Repeat this process to obtain x2, x3, …, x10, each time evaluating f(x_k) and f'(x_k) at the current x_k.

When performing these calculations to six decimal places, you obtain a sequence that rapidly converges to the root of f(x) = 0, which is the solution to 3x = cos x, i.e., x ≈ 0.316751...

If you compare potential answer options, note that the following common points are expected:
- The initial value x0 should be exactly 1.000000.
- The first Newton iterate x1 will be significantly smaller than 1, typically in the vicinity of 0.36–0.37 for this function, given the slope and curvature at x0.
- By the second or third iteration, the values should already be very close to 0.31675, with subsequent iterates differing only in the sixth decimal place or may even stabilize to 0.316751 for several steps depending on rounding.
- Ultimately, the sequence should converge to about x ≈ 0.316751 with successive differences on the order of 1e-6 or smaller once past a few iterations.

If you’re given a set of candidate tables, you would:
- Verify that x0 is 1.000000.
- Check that x1 matches the computed value using the formula above to six decimals.
- Check that x2, x3, … approach approximately 0.316751, with later entries not deviating beyond the sixth decimal place from 0.316751 if the method has effectively converged.

In the specific sequences commonly seen for this problem, you would expect x0 = 1.000000, x1 ≈ 0.359699, x2 ≈ 0.317010, x3 ≈ 0.316751, and then x4, x5, … stabilizing around 0.316751 to six decimals. If an option presents exactly that pattern (or a pattern matching this convergence behavior), it would be the correct choice. If options are blank or inconsistent, they cannot be judged as correct or incorrect without actual numerical entries to compare.

Summary of the key concepts:
- Newton’s method updates via x_{k+1} = x_k - (3x_k - cos x_k) / (3 + sin x_k).
- The derivative never vanishes near the root for this function within the expected range, so convergence to the root is typical.
- The root is approximately 0.316751, so later iterates should stabilize around this value when rounded to six decimals.

MATH2415W16 Quiz 3 (Nov 25)- Requires Respondus LockDown Browser

Use a calculator to compute the first 10 iterations of Newton's method when applied to the function with the given initial approximation. Make a table for the values. Round to six decimal places. f(x) = 3x - cos x; x0 = 1

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MATH*2415*W16 Quiz 3 (Nov 25)- Requires Respondus LockDown Browser