Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (m)/Solution 1

The strength of Newton's method is on root finding. We're asked to approximate

$\displaystyle {}x=2^{1/6}$

which we can write as a root finding problem,

$\displaystyle {}x^{6}-2=0.$

With this in mind, let

$\displaystyle {}f(x)=x^{6}-2.$

Taking the derivative gives

$\displaystyle {}f'(x)=6x^{5}.$

We start with $x_{1}=1$ as our first iteration and so if we apply Newton's Method, we get,

$x_{2}=x_{1}-{\frac {f(x_{1})}{f'(x_{1})}}=1-{\frac {f(1)}{f'(1)}}=1-{\frac {-1}{6}}=1+{\frac {1}{6}}={\frac {7}{6}}.$

If we actually compute $2^{1/6}$ we get

$\displaystyle {}2^{1/6}\approx {}1.122462048$

and so ${\frac {7}{6}}=1.166666667$ is already in good agreement.