Question 12.9: A root of 3 sin x = x is near to x = 2.5. Use two iterations......
A root of 3 sin x = x is near to x = 2.5. Use two iterations of the Newton-Raphson
technique to find a more accurate approximation.
Step-by-Step
The equation must first be written in the form f(x) = 0, that is
f(x)=3 \sin x-x=0Then
\begin{array}{ll} x_1=2.5 & \\ f(x)=3 \sin x-x & f\left(x_1\right)=-0.705 \\ f^{\prime}(x)=3 \cos x-1 & f^{\prime}\left(x_1\right)=-3.403 \end{array}Then
x_2=2.5-\frac{(-0.705)}{(-3.403)}=2.293The process is repeated with x_1 = 2.293 as the initial approximation:
x_1=2.293 \quad f\left(x_1\right)=-0.042 \quad f^{\prime}\left(x_1\right)=-2.983Then
x_2=2.293-\frac{(-0.042)}{(-2.983)}=2.279Using two iterations of the Newton-Raphson technique, we obtain x = 2.28 as an improved estimate of the root.
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