SOLUTION 2: We are given the equation $ x^3=x^2+2 $, or $ x^3-x^2-2=0 $, so let's define function $ f(x) = x^3-x^2-2 $, whose graph is given below.
The derivative of $f$ is $ f'(x) = 3x^2-2x $. Now use Newton's Method:
$$ x_{n+1} = x_{n} - { f(x_{n}) \over f'(x_{n}) } \ \ \ \ \longrightarrow $$
$$ x_{n+1} = x_{n} - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow $$
(Let's simplify the right-hand side of this equation. First get a common denominator.)
$$ x_{n+1} = x_{n} \ { 3x_{n}^2-2x_{n} \over 3x_{n}^2-2x_{n} } - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow $$
$$ x_{n+1} = { 3x_{n}^3-2x_{n}^2 \over 3x_{n}^2-2x_{n} } - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow $$
$$ x_{n+1} = { 3x_{n}^3-2x_{n}^2 - ( x_{n}^3-x_{n}^2-2 ) \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow $$
$$ x_{n+1} = { 2x_{n}^3-x_{n}^2+2 \over 3x_{n}^2-2x_{n} } $$
a.) Let $ x_{0}=1 $. Then using Newton's Method formula we get that
$$ x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(1)^3-(1)^2+2 \over 3(1)^2-2(1) } = 3 $$
and
$$ x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(3)^3-(3)^2+2 \over 3(3)^2-2(3) } = { 47 \over 21 } \approx 2.238095238 $$
Using Newton's Method formula for 7 iterations in a spreadsheet results in :
Thus the solution $r$ to the original equation to five decimal places is $ r \approx 1.69562 $.
b.) Let $ x_{0}=2 $. Then using Newton's Method formula we get that
$$ x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(2)^3-(2)^2+2 \over 3(2)^2-2(2) } = { 14 \over 8 } = 1.75 $$
and
$$ x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(1.75)^3-(1.75)^2+2 \over 3(1.75)^2-2(1.75) } = { 618 \over 364 } \approx 1.697802198 $$
Using Newton's Method formula for 5 iterations in a spreadsheet results in :
Thus the solution $r$ to the original equation to five decimal places is $ r \approx 1.69562 $.
c.) Let $ x_{0}=0.5 $. Then using Newton's Method formula we get that
$$ x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(0.5)^3-(0.5)^2+2 \over 3(0.5)^2-2(0.5) } = -8 $$
and
$$ x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(-8)^3-(-8)^2+2 \over 3(-8)^2-2(-8) } = { -1086 \over 208 } \approx -5.221153846 $$
Using Newton's Method formula for 13 iterations in a spreadsheet results in :
Thus the solution $r$ to the original equation to five decimal places is $ r \approx 1.69562 $.
d.) Let $ x_{0}=1000 $. Then using Newton's Method formula we get that (Use technology.)
$$ x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(1000)^3-(1000)^2+2 \over 3(1000)^2-2(1000) } \approx 666.7778526 $$
and
$$ x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(666.7778526 )^3-(666.7778526 )^2+2 \over 3(666.7778526 )^2-2(666.7778526 ) } = \approx 444.6297922 $$
Using Newton's Method formula for 21 iterations in a spreadsheet results in :
Thus the solution $r$ to the original equation to five decimal places is $ r \approx 1.69562 $.
e.) Let $ x_{0}=-500 $. Then using Newton's Method formula we get that (Use technology.)
$$ x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(-500)^3-(-500)^2+2 \over 3(-500)^2-2(-500) } \approx -333.2223675 $$
and
$$ x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(-333.2223675)^3-(-333.2223675)^2+2 \over 3(-333.2223675)^2-2(-333.2223675) } = \approx -222.0373498 $$
Using Newton's Method formula for 25 iterations in a spreadsheet results in :
Thus the solution $r$ to the original equation to five decimal places is $ r \approx 1.69562 $.
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