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Calculus IChapter 4: Application of DerivativesSection 4-13

Newton’s Method — Question 7

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Question 7

Use Newton’s Method to approximate a solution to the equation: ln⁡(x)+x2=3\ln(x) + x^2 = 3 Start with an initial guess of x1=1.5x_1 = 1.5, and compute x2x_2, x3x_3, and x4x_4. Round each to four decimal places.

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Question 7 - Solution

Newton’s iteration is

xn+1=xn−f(xn)f′(xn),f(x)=ln⁡x+x2−3,f′(x)=1/x+2x.x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=\ln x+x^2-3,\quad f\prime(x)=1/x+2x.

Keeping full precision internally gives

x2=1.5000000000−−0.34453489193.6666666667≈1.5940x3=1.5939640614−0.00694546313.8152948397≈1.5921x4=1.5921436351−0.00000266133.8123713061≈1.5921\begin{aligned}x_{2}&=1.5000000000-\frac{-0.3445348919}{3.6666666667}\approx\boxed{1.5940}\\[6pt]x_{3}&=1.5939640614-\frac{0.0069454631}{3.8152948397}\approx\boxed{1.5921}\\[6pt]x_{4}&=1.5921436351-\frac{0.0000026613}{3.8123713061}\approx\boxed{1.5921}\end{aligned}

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