Solution: e want to find the value of $\sqrt{13}$ by Newton-Raphson's method. For that, take the function $f(x) = x^2 - 13$. Clearly, one of the roots of the equation $f(x) = 0$ is $\sqrt{13}$. We will approximate this using the Newton-Raphson method. We have \[ f(x) = x^2 - 13 \quad \text{and} \quad f'(x) = 2x. \] Thus, the Newton-Raphson iteration scheme will be \begin{align*} x_{n+1} & = x_n - \frac{f(x_n)}{f'(x_n)} \\ & = x_n - \frac{x_n^2 - 13}{2 x_n}, \quad n \geq 0. \end{align*} We will start with an initial guess of $x_0 = 1$. Since $f'(1) = 2 \neq 0$, this guess is a valid guess.

• First iteration: $x_0 = 1$. \[ x_1 = x_0 - \frac{x_0^2 - 13}{2 x_0} = 7. \]

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• Second iteration: $x_1 = 7$. \[ x_2 = x_1 - \frac{x_1^2 - 13}{2 x_1} = 4.4285714. \]

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• Third iteration: $x_2 = 4.4285714$. \[ x_3 = x_2 - \frac{x_2^2 - 13}{2 x_2} = 3.6820276. \]

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• Fourth iteration: $x_3 = 3.6820276$. \[ x_4 = x_3 - \frac{x_3^2 - 13}{2 x_3} = 3.606345. \]

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• Fifth iteration: $x_4 = 3.606345$. \[ x_5 = x_4 - \frac{x_4^2 - 13}{2 x_4} = 3.605551. \]

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• Sixth iteration: $x_5 = 3.605551$. \[ x_6 = x_5 - \frac{x_5^2 - 13}{2 x_5} = 3.605551. \]

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Thus, the value $\sqrt{13}$ correct to 4 decimal places is $3.6055$.