Question

Find cube root of 15 by using iteration method?

Answer 1

cube root of 15 correct to three decimal places is
`x=2.466`

Explanation:

Let the cube root of 15 be `x`

Then, cubing both sides and interchanging

`x^3=15`
`x^3-15=0`

Let `f(x)=x^3-15`
We need to investigate for the roots of the polynomial
`x^3-15`
Consider any perfect cube close to the number 15

`1^3=1,,,,,,,2^3=8,,,,,,,,,,, 3^3=27`

`15`lies between `8 and 27`
Hence cube root of `15` lies between `2 and 3`

Let us guess the initial value of the root as `x_0=2`
Let us follow Newton-Raphson's method of iteration

Iteration 1

`f(x_0))=x_0^3-15=2^3-15=-7`

`f'(x_0))=3x_0^2=3xx2^2=12`

`x_1=x_0-f(x_0)/(f'(x_0))=2-(-7/12)=2.583`

Iteration 2

`f(x_1))=x_1^3-15=(2.583)^3-15=2.240`

`f'(x_1))=3x_1^2=3xx2.583^2=20.021`

`x_2=x_1-f(x_1)/(f'(x_1))=2.583-(2.240/20.021)=2.471`

Iteration 3

`f(x_2))=x_2^3-15=(2.471)^3-15=0.096`

`f'(x_2))=3x_2^2=3xx2.471^2=18.324`

`x_3=x_2-f(x_2)/(f'(x_2))=2.471-(0.096/18.324)=2.466`

Iteration 4
`f(x_3))=x_3^3-15=(2.466)^3-15=0.000`

`f'(x_3))=3x_3^2=3xx2.466^2=18.247`

`x_4=x_3-f(x_3)/(f'(x_3))=2.466-(0.000/18.247)=2.466`

Hence, cube root of 15 correct to three decimal places is
`x=2.466`, since `x_3=x_4`

Answer 2

`root(3)(15) ~~ 151931/61605 ~~ 2.4662121`

Explanation:

We want to solve `x^3-15 = 0`

Let:

`f(x) = x^3-15`

Then:

`f'(x) = 3x^2`

Newton's method tells us that if `a_i` is an approximate zero of `f(x)` then a better approximation is:

`a_(i+1) = a_i-f(a_i)/(f'(a_i))`

`color(white)(a_(i+1)) = a_i-(a_i^3-15)/(3a_i^2)`

`color(white)(a_(i+1)) = (2a_i^3+15)/(3a_i^2)`

If we use rational approximations with `a_i = p_i/q_i` then this becomes:

`p_(i+1)/q_(i+1) = (2(p_i/q_i)^3+15)/(3(p_i/q_i)^2) = (2p_i^3+15q_i^3)/(3p_i^2q_i)`

So we can define:

`{ (p_(i+1) = 2p_i^3+15q_i^3), (q_(i+1) = 3p_i^2 q_i) :}`

For our first approximation note that:

`2^3 = 8 < 15 < 27 = 3^3`

So:

`2 < root(3)(15) < 3`

Let us choose `p_0/q_0 = 5/2`, i.e. `p_0 = 5`, `q_0 = 2`

Then:

`{ (p_1 = 2p_0^3+15q_0^3 = 2(5)^3+15(2)^3 = 250+120 = 370), (q_1 = 3p_0^2q_0 = 3(5)^2(2) = 150) :}`

In fact `p_1` and `q_1` have a common factor `10`, so let's factor that out and in preparation for our next iteration put:

`{ (p_(1a) = 37), (q_(1a) = 15) :}`

Then:

`{ (p_2 = 2p_(1a)^3+15q_(1a)^3 = 2(37)^3+15(15)^3 = 101306+50625 = 151931), (q_2 = 3p_(1a)^2 q_(1a) = 3(37)^2(15) = 61605) :}`

Let's stop there to find:

`root(3)(15) ~~ 151931/61605 ~~ 2.4662121`