solve this with newton for x0=0.1, 0.7 and 0.9
x.e^(x^2)-1=0

We are tasked with solving the equation \( x \cdot e^{x^2} - 1 = 0 \) using Newton's method with different initial guesses for \( x_0 = 0.1, 0.7, \) and \( 0.9 \).

### Newton's Method Formula:
Newton's method is an iterative method used to approximate roots of a real-valued function. The formula for the next approximation \( x_{n+1} \) is:

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

Where:
- \( f(x) \) is the function we want to solve.
- \( f'(x) \) is the derivative of \( f(x) \).
- \( x_n \) is the current approximation, and \( x_{n+1} \) is the next approximation.

### Step 1: Define the Function and its Derivative
The function we want to solve is:

\[
f(x) = x \cdot e^{x^2} - 1
\]

To apply Newton's method, we need to compute its derivative \( f'(x) \).

#### Function:
\[
f(x) = x \cdot e^{x^2} - 1
\]

#### Derivative of the Function:
Using the product rule and chain rule for differentiation:

\[
f'(x) = \frac{d}{dx} \left( x \cdot e^{x^2} \right) = e^{x^2} + 2x^2 \cdot e^{x^2}
\]
Thus,
\[
f'(x) = e^{x^2} (1 + 2x^2)
\]

### Step 2: Apply Newton's Method

We will use the following iterative formula:
\[
x_{n+1} = x_n - \frac{x_n \cdot e^{x_n^2} - 1}{e^{x_n^2} (1 + 2x_n^2)}
\]

Now, I will calculate the approximations for \( x_0 = 0.1 \), \( x_0 = 0.7 \), and \( x_0 = 0.9 \) using this formula.

Let me perform the calculations.The solutions obtained using Newton's method for the equation \( x \cdot e^{x^2} - 1 = 0 \) with the different initial guesses are as follows:

- For \( x_0 = 0.1 \), the root is approximately \( 0.65292 \) after 5 iterations.
- For \( x_0 = 0.7 \), the root is approximately \( 0.65292 \) after 3 iterations.
- For \( x_0 = 0.9 \), the root is approximately \( 0.65292 \) after 4 iterations.

In each case, the method converged to the same root \( x \approx 0.65292 \).

Do you need more details or have any questions?

Here are 5 related questions for further practice:
1. What happens if the initial guess is far from the root in Newton's method?
2. How does the choice of tolerance affect the accuracy and speed of convergence in Newton's method?
3. What is the significance of the derivative \( f'(x) \) in determining the convergence of Newton's method?
4. Can Newton's method fail to converge, and under what circumstances?
5. How does Newton's method compare to other root-finding methods like the bisection method?

**Tip:** When using Newton's method, choosing a good initial guess can significantly reduce the number of iterations required for convergence.

Learn how to solve the equation x * e^(x^2) - 1 = 0 using Newton's method with initial guesses 0.1, 0.7, and 0.9. This detailed step-by-step solution involves key numerical methods, exponential functions, and differentiation rules. Suitable for undergraduate students studying numerical methods or calculus.

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