Use Newton’s method with an initial guess of x0 = 1.1 to complete the table below
and so find an approximate solution (accurate to 3 decimal places) for ln(x
2 + 1) =
0.7.
The numbers in the table should be given to 4 decimal places.Iteration xn xn+1 |xn+1 − xn|
0 1.1
1
2
3
4
5

To solve the equation \(\ln(x^2 + 1) = 0.7\) using Newton's method, we first need to define a function \(f(x)\) whose root we want to find. We start by rearranging the given equation into the form:

\[
f(x) = \ln(x^2 + 1) - 0.7
\]

Newton's method updates guesses for the root using the formula:

\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]

where \(f'(x)\) is the derivative of \(f(x)\).

### Step 1: Define \(f(x)\) and \(f'(x)\)

Given:

\[
f(x) = \ln(x^2 + 1) - 0.7
\]

The derivative \(f'(x)\) is:

\[
f'(x) = \frac{d}{dx} \left( \ln(x^2 + 1) - 0.7 \right) = \frac{2x}{x^2 + 1}
\]

### Step 2: Compute Newton's Iterations

We start with the initial guess \(x_0 = 1.1\).

Let's compute each iteration step-by-step. 

#### Iteration 1:

\[
f(x_0) = \ln(1.1^2 + 1) - 0.7 = \ln(1.21 + 1) - 0.7 = \ln(2.21) - 0.7
\]

\[
f(x_0) \approx 0.7930 - 0.7 = 0.0930
\]

\[
f'(x_0) = \frac{2 \cdot 1.1}{1.1^2 + 1} = \frac{2.2}{2.21} \approx 0.9955
\]

\[
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.1 - \frac{0.0930}{0.9955} \approx 1.0065
\]

\(|x_1 - x_0| = |1.0065 - 1.1| = 0.0935\)

#### Iteration 2:

\[
f(x_1) = \ln(1.0065^2 + 1) - 0.7 = \ln(1.0130 + 1) - 0.7 = \ln(2.0130) - 0.7
\]

\[
f(x_1) \approx 0.7007 - 0.7 = 0.0007
\]

\[
f'(x_1) = \frac{2 \cdot 1.0065}{1.0065^2 + 1} = \frac{2.0130}{2.0130} = 1.0000
\]

\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.0065 - \frac{0.0007}{1.0000} \approx 1.0058
\]

\(|x_2 - x_1| = |1.0058 - 1.0065| = 0.0007\)

#### Iteration 3:

\[
f(x_2) = \ln(1.0058^2 + 1) - 0.7 = \ln(1.0116 + 1) - 0.7 = \ln(2.0116) - 0.7
\]

\[
f(x_2) \approx 0.7001 - 0.7 = 0.0001
\]

\[
f'(x_2) = \frac{2 \cdot 1.0058}{1.0058^2 + 1} = \frac{2.0116}{2.0116} = 1.0000
\]

\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.0058 - \frac{0.0001}{1.0000} \approx 1.0057
\]

\(|x_3 - x_2| = |1.0057 - 1.0058| = 0.0001\)

#### Iteration 4:

\[
f(x_3) = \ln(1.0057^2 + 1) - 0.7 = \ln(1.0114 + 1) - 0.7 = \ln(2.0114) - 0.7
\]

\[
f(x_3) \approx 0.7000 - 0.7 = 0.0000
\]

\[
f'(x_3) = \frac{2 \cdot 1.0057}{1.0057^2 + 1} = \frac{2.0114}{2.0114} = 1.0000
\]

\[
x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = 1.0057 - \frac{0.0000}{1.0000} \approx 1.0057
\]

\(|x_4 - x_3| = |1.0057 - 1.0057| = 0.0000\)

Since the difference \(|x_4 - x_3|\) is now sufficiently small, we can stop.

### Completed Table:

| Iteration \(n\) | \(x_n\)   | \(x_{n+1}\) | \(|x_{n+1} - x_n|\) |
|----------------|---------|-----------|------------------|
| 0              | 1.1000  | 1.0065    | 0.0935           |
| 1              | 1.0065  | 1.0058    | 0.0007           |
| 2              | 1.0058  | 1.0057    | 0.0001           |
| 3              | 1.0057  | 1.0057    | 0.0000           |
| 4              | 1.0057  | 1.0057    | 0.0000           |

The approximate solution, accurate to three decimal places, is \(x \approx 1.006\).

Would you like more details or have any questions?

### Related Questions:

1. What is the convergence criteria in Newton's method?
2. How do we derive the formula for Newton's method?
3. Can Newton's method fail to find a root? If so, under what conditions?
4. How does the choice of the initial guess affect the convergence of Newton's method?
5. What are other numerical methods for finding roots of equations?

### Tip:
Newton's method is highly efficient when the function is well-behaved near the root, but it may fail to converge or give misleading results if the function is not differentiable or if the derivative is zero at some point in the iteration process.

Explore how Newton's method is used to find an approximate solution to ln(x^2 + 1) = 0.7, accurate to three decimal places. Follow detailed iterations and calculations to understand the application of numerical methods in solving equations.

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