Question Statement

Find the real root of the equation $x^3-3x-5=0$ correct to three decimal places using the Newton-Raphson method, with initial approximation $x_0=2$.

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Background and Explanation

The Newton-Raphson method is an iterative numerical technique for approximating roots of real-valued functions. It refines an initial guess by using the function's derivative to find where the tangent line crosses the x-axis, typically converging rapidly when the initial guess is close to the true root.

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Solution

Deriving the Iteration Formula

Given the function: $f(x)=x^3-3x-5$

Differentiate with respect to $x$ to find $f^{\prime }(x)$: $f^{\prime }(x)=3x^2-3$

The general Newton-Raphson formula is: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

where $f^{\prime }(x_n)\ne 0$ and $n=0,1,2,\dots$

Substituting our specific function and its derivative: $f(x_n)=x_n^3-3x_n-5$ $f^{\prime }(x_n)=3x_n^2-3$

Therefore: $x_{n+1}=x_n-\frac{x_n^3-3x_n-5}{3x_n^2-3}$

Simplifying by combining terms over a common denominator: $x_{n+1}=\frac{x_n(3x_n^2-3)-(x_n^3-3x_n-5)}{3x_n^2-3}$

Expanding the numerator: $x_{n+1}=\frac{3x_n^3-3x_n-x_n^3+3x_n+5}{3x_n^2-3}$

$\Rightarrow x_{n+1}=\frac{2x_n^3+5}{3x_n^2-3}$

First Iteration ($n=0$)

Substitute $n=0$ into equation (i) using $x_0=2$:

$x_1=\frac{2x_0^3+5}{3x_0^2-3}=\frac{2(2{)}^3+5}{3(2{)}^2-3}=\frac{2(8)+5}{3(4)-3}=\frac{16+5}{12-3}=\frac{21}{9}=2.3333$

Second Iteration ($n=1$)

Substitute $n=1$ using $x_1=2.3333$:

$x_2=\frac{2x_1^3+5}{3x_1^2-3}=\frac{2(2.3333{)}^3+5}{3(2.3333{)}^2-3}=2.2806$

Third Iteration ($n=2$)

Substitute $n=2$ using $x_2=2.2806$:

$x_3=\frac{2x_2^3+5}{3x_2^2-3}=\frac{2(2.2806{)}^3+5}{3(2.2806{)}^2-3}=2.2790$

Fourth Iteration ($n=3$)

Substitute $n=3$ using $x_3=2.2790$:

$x_4=\frac{2x_3^3+5}{3x_3^2-3}=\frac{2(2.2790{)}^3+5}{3(2.2790{)}^2-3}=2.2790$

Convergence and Final Answer

Comparing the results up to three decimal places:

• $x_3=2.2790$
• $x_4=2.2790$

Since $x_3$ and $x_4$ agree to three decimal places, the iteration has converged.

Therefore, the root of the equation $x^3-3x-5=0$ correct to three decimal places is $\boxed{2.279}$.

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Key Formulas or Methods Used

• Newton-Raphson Iteration Formula: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$
• Polynomial Differentiation: $\frac{d}{dx}(x^3-3x-5)=3x^2-3$
• Simplified Iteration for This Problem: $x_{n+1}=\frac{2x_n^3+5}{3x_n^2-3}$
• Convergence Criterion: Successive approximations agree to the required number of decimal places (three decimal places in this case)

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Summary of Steps

1. Define the function and its derivative: $f(x)=x^3-3x-5$ and $f^{\prime }(x)=3x^2-3$.
2. Derive the specific iteration formula: Substitute into Newton-Raphson to get $x_{n+1}=\frac{2x_n^3+5}{3x_n^2-3}$.
3. Calculate $x_1$: Using $x_0=2$, obtain $x_1=2.3333$.
4. Calculate $x_2$: Using $x_1=2.3333$, obtain $x_2=2.2806$.
5. Calculate $x_3$: Using $x_2=2.2806$, obtain $x_3=2.2790$.
6. Calculate $x_4$: Using $x_3=2.2790$, obtain $x_4=2.2790$.
7. Verify convergence: Confirm $x_3$ and $x_4$ match to three decimal places, giving the final answer $2.279$.