Use Newton's Method to determine x_{1} and x_{2} for the f

vousetmoiec

vousetmoiec

Answered question

2021-11-14

Use Newton's Method to determine x1 and x2 for the function f(x)=x37x2+8x3 and the value of x0=5

Answer & Explanation

tnie54

tnie54

Beginner2021-11-15Added 18 answers

Step 1
In numerical analysis, Newton's method, also known as the Newton–Raphson method, is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function.
The most basic version starts with a single-variable function f defined for a real variable x, the function's derivative f ′, and an initial guess \(\displaystyle{x}_{{{0}}}\) for a root of f. If the function satisfies sufficient assumptions and the initial guess is close, then
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right)}}}}}{{{f}'{\left({x}_{{{0}}}\right)}}}}\)
is a better approximation of the root than \(\displaystyle{x}_{{{0}}}\). Geometrically, \(\displaystyle{\left({x}_{{{1}}},\ {0}\right)}\) is the intersection of the x-axis and the tangent of
the graph of f at \(\displaystyle{\left({x}_{{{0}}},\ {f{{\left({x}_{{{0}}}\right)}}}\right)}\): that is, the improved guess is the unique root of the linear approximations at the initial point.
The process is repeated as
\(\displaystyle{x}_{{{n}+{1}}}={x}_{{{n}}}-{\frac{{{f{{\left({x}_{{{n}}}\right)}}}}}{{{f}'{\left({x}_{{{n}}}\right)}}}}\)
Step 2
\(\displaystyle{x}_{{{1}}},\ {x}_{{{2}}}\) for the function \(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\), and the value of \(\displaystyle{x}_{{{0}}}={5}\)
Step 3
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\)
\(\displaystyle{f}'{\left({x}\right)}={3}{x}^{{{2}}}-{14}{x}+{8}\)
\(\displaystyle{f{{\left({x}_{{{0}}}\right)}}}\)
\(\displaystyle={f{{\left({5}\right)}}}\)
\(\displaystyle={5}^{{{3}}}-{7}\times{\left({5}\right)}^{{{2}}}+{8}\times{5}-{3}\)
\(\displaystyle=-{13}\)
\(\displaystyle{f}'{\left({5}\right)}={3}\times{5}^{{{2}}}-{14}\times{5}+{8}\)
\(\displaystyle={13}\)
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right\rbrace}}}{\left\lbrace{f}'{\left({x}_{{{0}}}\right)}\right\rbrace}}}{}}\)
\(\displaystyle={5}-{\frac{{-{13}}}{{{13}}}}\)
\(\displaystyle={6}\)
Step 4
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\)
\(\displaystyle{f}'{\left({x}\right)}={3}{x}^{{{2}}}-{14}{x}+{8}\)
\(\displaystyle{f{{\left({x}_{{{1}}}\right)}}}\)
\(\displaystyle={f{{\left({6}\right)}}}\)
\(\displaystyle={6}^{{{3}}}-{7}\times{6}^{{{2}}}+{8}\times{6}-{3}\)
\(\displaystyle={9}\)
\(\displaystyle{f}'{\left({6}\right)}={3}\times{6}^{{{2}}}-{14}\times{6}+{8}\)
\(\displaystyle={32}\)
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right)}}}}}{{{f}'{\left({x}_{{{0}}}\right)}}}}\)
\(\displaystyle={6}-{\frac{{{9}}}{{{32}}}}\)
\(\displaystyle={5.7188}\)
Step 5
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\)
\(\displaystyle{f}'{\left({x}\right)}={3}{x}^{{{2}}}-{14}{x}+{8}\)
\(\displaystyle{f{{\left({x}_{{{0}}}\right)}}}\)
\(\displaystyle={f{{\left({5.7188}\right)}}}\)
\(\displaystyle={5.7188}^{{{3}}}-{7}\times{\left({5.7188}\right)}^{{{2}}}+{8}\times{5.7188}-{3}\)
\(\displaystyle={0.8479}\)
\(\displaystyle{f}'{\left({5}\right)}={3}\times{5.7188}^{{{2}}}-{14}\times{5.7188}+{8}\)
\(\displaystyle={26.0498}\)
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right)}}}}}{{{f}'{\left({x}_{{{0}}}\right)}}}}\)
\(\displaystyle={5.7188}-{\frac{{{0.8479}}}{{{26.0498}}}}\)
\(\displaystyle={5.6862}\)
Step 6
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\)
\(\displaystyle{f}'{\left({x}\right)}={3}{x}^{{{2}}}-{14}{x}+{8}\)
\(\displaystyle{f{{\left({x}_{{{0}}}\right)}}}\)
\(\displaystyle={f{{\left({5.6862}\right)}}}\)
\(\displaystyle={5.6862}^{{{3}}}-{7}\times{\left({5.6862}\right)}^{{{2}}}+{8}\times{5.6862}-{3}\)
\(\displaystyle={0.0107}\)
\(\displaystyle{f}'{\left({5}\right)}={3}\times{5.6862}^{{{2}}}-{14}\times{5.6862}+{8}\)
\(\displaystyle={25.3919}\)
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right)}}}}}{{{f}'{\left({x}_{{{0}}}\right)}}}}\)
\(\displaystyle={5.6862}-{\frac{{{0.0107}}}{{{25.3919}}}}\)
\(\displaystyle={5.6858}\)
Step 7
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}-{7}{x}^{{{2}}}+{8}{x}-{3}\)
\(\displaystyle{f}'{\left({x}\right)}={3}{x}^{{{2}}}-{14}{x}+{8}\)
\(\displaystyle{f{{\left({x}_{{{0}}}\right)}}}\)
\(\displaystyle={f{{\left({5.6858}\right)}}}\)
\(\displaystyle={5.6858}^{{{3}}}-{7}\times{\left({5.6858}\right)}^{{{2}}}+{8}\times{5.6858}-{3}\)
\(\displaystyle={0}\)
\(\displaystyle{f}'{\left({5}\right)}={3}\times{5.6858}^{{{2}}}-{14}\times{5.6858}+{8}\)
\(\displaystyle={25.3834}\)
\(\displaystyle{x}_{{{1}}}={x}_{{{0}}}-{\frac{{{f{{\left({x}_{{{0}}}\right)}}}}}{{{f}'{\left({x}_{{{0}}}\right)}}}}\)
\(\displaystyle={5.6858}-{\frac{{{0}}}{{{25.3834}}}}\)
\(\displaystyle={5.6858}\)
The roots are \(5.6858,\ 5.6858\)

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