Q12: $x^4-x-10=0$

Question Statement

Find the root of the equation $x^4-x-10=0$ correct to three decimal places using the Newton-Raphson method. Begin with the initial approximation $x_0=2$.

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Background and Explanation

The Newton-Raphson method is an iterative numerical technique for finding successively better approximations to the roots of a real-valued function. Starting from an initial guess $x_0$, each iteration uses the tangent line to the curve at the current point to find the next approximation.

Newton-Raphson Formula: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)},f^{\prime }(x_n)\ne 0,n=0,1,2,\dots$

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Solution

Let $f(x)=x^4-x-10$.

Differentiate: $f^{\prime }(x)=4x^3-1$

Substituting into the Newton-Raphson formula: $x_{n+1}=x_n-\frac{x_n^4-x_n-10}{4x_n^3-1}$

Simplifying over the common denominator: $\begin{array}{rl}{x}_{n+1}& =\frac{x_n(4x_n^3-1)-(x_n^4-x_n-10)}{4x_n^3-1}\\ & =\frac{4x_n^4-x_n-x_n^4+x_n+10}{4x_n^3-1}\\ & =\frac{3x_n^4+10}{4x_n^3-1}\end{array}$

Working iteration formula: $\boxed{x_{n+1}=\frac{3x_n^4+10}{4x_n^3-1}}(i)$

First Iteration ($n=0$): Using $x_0=2$: $x_1=\frac{3(2{)}^4+10}{4(2{)}^3-1}=\frac{48+10}{32-1}=\frac{58}{31}\approx 1.8710$

Second Iteration ($n=1$): Using $x_1=1.8710$: $x_2=\frac{3(1.8710{)}^4+10}{4(1.8710{)}^3-1}=\frac{3(12.2574)+10}{4(6.5496)-1}=\frac{46.7722}{25.1984}\approx 1.8558$

Third Iteration ($n=2$): Using $x_2=1.8558$: $x_3=\frac{3(1.8558{)}^4+10}{4(1.8558{)}^3-1}=\frac{3(11.896)+10}{4(6.389)-1}=\frac{45.688}{24.556}\approx 1.8556$

Since $x_2\approx 1.856$ and $x_3\approx 1.856$ agree to three decimal places:

$\boxed{\text{Root}=1.855}$

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Key Formulas Used

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Q13: $e^x=1+2x$

Question Statement

Find the root of the equation $e^x=1+2x$ correct to three decimal places using the Newton-Raphson method, starting with $x_0=2$.

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Solution

Rearrange: $e^x-2x-1=0$

Let $f(x)=e^x-2x-1$.

Differentiate: $f^{\prime }(x)=e^x-2$

Substituting into Newton-Raphson and simplifying: $\begin{array}{rl}{x}_{n+1}& =x_n-\frac{e^{x_n}-2x_n-1}{e^{x_n}-2}\\ & =\frac{x_n(e^{x_n}-2)-(e^{x_n}-2x_n-1)}{e^{x_n}-2}\\ & =\frac{x_n{e}^{x_n}-e^{x_n}+1}{e^{x_n}-2}\end{array}$

Working iteration formula: $\boxed{x_{n+1}=\frac{(x_n-1)e^{x_n}+1}{e^{x_n}-2}}(ii)$

Iterations starting from $x_0=2$:

Since $x_4=1.2565$ and $x_5=1.2564$ agree to three decimal places:

$\boxed{\text{Root}=1.256}$

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Key Formulas Used

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Q14: $3x-1=\cos x$

Question Statement

Find the root of the equation $3x-1=\cos x$ correct to three decimal places using the Newton-Raphson method with initial approximation $x_0=2$.

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Solution

Rearrange: $3x-\cos x-1=0$

Let $f(x)=3x-\cos x-1$.

Differentiate (recall $\frac{d}{dx}(\cos x)=-\sin x$): $f^{\prime }(x)=3+\sin x$

Substituting into Newton-Raphson and simplifying: $\begin{array}{rl}{x}_{n+1}& =x_n-\frac{3x_n-\cos x_n-1}{3+\sin x_n}\\ & =\frac{x_n(3+\sin x_n)-(3x_n-\cos x_n-1)}{3+\sin x_n}\\ & =\frac{3x_n+x_n\sin x_n-3x_n+\cos x_n+1}{3+\sin x_n}\end{array}$

Working iteration formula: $\boxed{x_{n+1}=\frac{x_n\sin x_n+\cos x_n+1}{3+\sin x_n}}(iii)$

First Iteration ($n=0$): Using $x_0=2$ rad: $x_1=\frac{2\sin (2)+\cos (2)+1}{3+\sin (2)}=\frac{2(0.9093)+(-0.4161)+1}{3+0.9093}=\frac{2.4025}{3.9093}\approx 0.6146$

Second Iteration ($n=1$): Using $x_1=0.6146$: $x_2=\frac{0.6146\sin (0.6146)+\cos (0.6146)+1}{3+\sin (0.6146)}=\frac{0.6146(0.5750)+0.8181+1}{3+0.5750}=\frac{2.1725}{3.5750}\approx 0.6077$

Third Iteration ($n=2$): Using $x_2=0.6077$: $x_3=\frac{0.6077\sin (0.6077)+\cos (0.6077)+1}{3+\sin (0.6077)}\approx \frac{0.6077(0.5693)+0.8221+1}{3+0.5693}=\frac{2.1681}{3.5693}\approx 0.6074$

Since $x_2\approx 0.607$ and $x_3\approx 0.607$ agree to three decimal places:

$\boxed{\text{Root}=0.607}$

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Key Formulas Used