Question Statement

Find the root of the equation $x^2+4\sin (x)=0$ correct to three decimal places using the Newton-Raphson method, starting with the initial approximation $x_0=2$.

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Background and Explanation

The Newton-Raphson method is an iterative technique for finding successively better approximations to the roots of a real-valued function. It requires the function $f(x)$ and its derivative $f^{\prime }(x)$, converging rapidly when the initial guess is sufficiently close to the true root.

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Solution

Deriving the Iteration Formula

The Newton-Raphson formula is given by:

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

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

Given the function: $f(x)=x^2+4\sin x$

Differentiate with respect to $x$: $f^{\prime }(x)=2x+4\cos x$

Therefore: $f\left(x_n\right)=x_n^2+4\sin x_n$ $f^{\prime }\left(x_n\right)=2x_n+4\cos x_n$

Substituting into the Newton-Raphson formula:

$\begin{array}{rl}{x}_{n+1}& =x_n-\frac{x_n^2+4\sin x_n}{2x_n+4\cos x_n}\\ & =\frac{x_n\left(2x_n+4\cos x_n\right)-\left(x_n^2+4\sin x_n\right)}{2x_n+4\cos x_n}\\ & =\frac{2x_n^2+4x_n\cos x_n-x_n^2-4\sin x_n}{2x_n+4\cos x_n}\\ x_{n+1}& =\frac{x_n^2+4\left(x_n\cos x_n-\sin x_n\right)}{2x_n+4\cos x_n}\end{array}$

Iterative Calculations

Iteration 1: Calculating $x_1$ from $x_0=2$

Put $n=0$ in equation (i):

$\begin{array}{rl}{x}_1& =\frac{x_0^2+4\left(x_0\cos x_0-\sin x_0\right)}{2x_0+4\cos x_0}\\ & =\frac{2^2+4(2\cos 2-\sin 2)}{2(2)+4\cos 2}\end{array}$

$\Rightarrow x_1=-1.2702$

Iteration 2: Calculating $x_2$ from $x_1$

Put $n=1$ in equation (i):

$\begin{array}{rl}{x}_2& =\frac{x_1+4\left(x_1\cos x_1-\sin x_1\right)}{2x_1+4\cos x_1}\\ & =\frac{(-1.2702{)}^2+4((-1.2702)\cos (-1.2702)-\sin (-1.2702))}{2(-1.2702)+4\cos (-1.2702)}\end{array}$

$\Rightarrow x_2=-2.8981$

Iteration 3: Calculating $x_3$ from $x_2$

Put $n=2$ in equation (i):

$\begin{array}{rl}{x}_3& =\frac{x_2^2+4\left(x_2\cos x_2-\sin x_2\right)}{2x_2+4\cos x_2}\\ & =\frac{(-2.8981{)}^2+4((-2.8981)\cos (-2.8981)-\sin (-2.8981))}{2(-2.8981)+4\cos (-2.8981)}\end{array}$

$\Rightarrow x_3=-2.1299$

Iteration 4: Calculating $x_4$ from $x_3$

Put $n=3$ in equation (i):

$x_4=\frac{x_3^2+4\left(x_3\cos x_3-\sin x_3\right)}{2x_3+4\cos x_3}$

$=\frac{(-2.1299{)}^2+4((-2.1299)\cos (-2.1299)-\sin (-2.1299))}{2(-2.1299)+4\cos (-2.1299)}$

$\Rightarrow x_4=-1.9504$

Iteration 5: Calculating $x_5$ from $x_4$

Put $n=4$ in equation (i):

$\begin{array}{rl}{x}_5& =\frac{x_4^2+4\left(x_4\cos x_4-\sin x_4\right)}{2x_4+4\cos x_4}\\ & =\frac{(-1.9504{)}^2+4((-1.9504)\cos (-1.9504)-\sin (-1.9504))}{2(-1.9504)+4\cos (-1.9504)}\end{array}$

$\Rightarrow x_5=-1.9339$

Iteration 6: Calculating $x_6$ from $x_5$

Put $n=5$ in equation (i):

$x_6=\frac{x_5^2+4\left(x_5\cos x_5-\sin x_5\right)}{2x_5+4\cos x_5}$

$=\frac{(-1.9339{)}^2+4((-1.9339)\cos (-1.9339)-\sin (-1.9339))}{2(-1.9339)+4\cos (-1.9339)}$

$\Rightarrow x_6=-1.9337$

Conclusion

Upto three decimal places, the values of $x_5$ and $x_6$ are the same ($-1.933$).

Hence, correct upto three decimal places, the root of the equation is $\boxed{-1.933}$.

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Key Formulas or Methods Used

• Newton-Raphson Iteration Formula: $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$
• Function: $f(x)=x^2+4\sin x$
• Derivative: $f^{\prime }(x)=2x+4\cos x$
• Simplified Iteration Formula: $x_{n+1}=\frac{x_n^2+4\left(x_n\cos x_n-\sin x_n\right)}{2x_n+4\cos x_n}$

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Summary of Steps

1. Define the function $f(x)=x^2+4\sin x$ and find its derivative $f^{\prime }(x)=2x+4\cos x$.
2. Set up the iteration formula by substituting $f(x_n)$ and $f^{\prime }(x_n)$ into the Newton-Raphson equation.
3. Simplify the formula to obtain $x_{n+1}=\frac{x_n^2+4(x_n\cos x_n-\sin x_n)}{2x_n+4\cos x_n}$.
4. Start with initial guess $x_0=2$ and calculate successive approximations:
   • $x_1=-1.2702$
   • $x_2=-2.8981$
   • $x_3=-2.1299$
   • $x_4=-1.9504$
   • $x_5=-1.9339$
   • $x_6=-1.9337$
5. Check for convergence: When $x_5$ and $x_6$ agree to three decimal places ($-1.933$), stop the iteration.
6. State the final answer: The root is $-1.933$ (to three decimal places).