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Calculus IChapter 4: Application of DerivativesSection 4-13

Newton’s Method — Question 5

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Question 5

Use Newton’s Method to approximate a root of the equation: cos⁡(x)−x=0\cos(x) - x = 0 Start with an initial guess of x1=0.5x_1 = 0.5, and compute the next three iterations: x2x_2, x3x_3, and x4x_4, rounding to four decimal places.

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Question 5 - Solution

Newton’s iteration is

xn+1=xn−f(xn)f′(xn),f(x)=cos⁡x−x,f′(x)=−sin⁡x−1.x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=\cos x-x,\quad f\prime(x)=-\sin x-1.

Keeping full precision internally gives

x2=0.5000000000−0.3775825619−1.4794255386≈0.7552x3=0.7552224171−−0.0271033119−1.6854506318≈0.7391x4=0.7391416661−−0.0000946154−1.6736538108≈0.7391\begin{aligned}x_{2}&=0.5000000000-\frac{0.3775825619}{-1.4794255386}\approx\boxed{0.7552}\\[6pt]x_{3}&=0.7552224171-\frac{-0.0271033119}{-1.6854506318}\approx\boxed{0.7391}\\[6pt]x_{4}&=0.7391416661-\frac{-0.0000946154}{-1.6736538108}\approx\boxed{0.7391}\end{aligned}

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