Question

Find the cube root of \[15\] by using the iteration method.

Answer 1

Hint: First, we will let the cube root of \[15\] be equal to \[x\]. Then we will guess the initial value of the root. Then we will follow Newton-Raphson’s method of iteration and calculate the cube root of \[15\] up to three decimal places. According to Newton-Raphson’s method we can find approximate value as:
\[{x_{n + 1}} = {x_n} - \dfrac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\].

Complete step by step answer:
Let the cube root of \[15\] be \[x\].
\[\therefore x = \sqrt[3]{{15}}\]
Cubing both sides we get;
\[ \Rightarrow {x^3} = 15\]
Shifting the RHS to LHS we get;
\[ \Rightarrow {x^3} - 15 = 0\]
Now let, \[f\left( x \right) = {x^3} - 15\]
We know that \[{2^3} = 8,{3^3} = 27\]. So, the cube root of \[15\] will lie between \[2{\text{ and 3}}\].
Now we will let the initial value of the cube root as \[{x_0} = 2\].
Now we will do iterations according to Newton-Raphson's method.
Iteration 1:
\[f\left( {{x_0}} \right) = x_0^3 - 15\]
Putting the value of \[{x_0}\] we get;
\[f\left( {{x_0}} \right) = {2^3} - 15\]
\[ \Rightarrow f\left( {{x_0}} \right) = - 7\]
Now by differentiating \[f\left( {{x_0}} \right) = x_0^3 - 15\], we get;
\[ \Rightarrow f'\left( {{x_0}} \right) = 3x_0^2\]
Putting the value of \[{x_0}\] we get;
\[ \Rightarrow f'\left( {{x_0}} \right) = 3 \times {2^2} = 12\]
Now according to the formula, we have;
\[{x_{n + 1}} = {x_n} - \dfrac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\].
Putting \[n = 0\], we get;
\[ \Rightarrow {x_1} = {x_0} - \dfrac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_n}} \right)}}\]
Putting the values, we get;
\[ \Rightarrow {x_1} = 2 - \dfrac{{ - 7}}{{12}}\]
Solving we get;
\[ \Rightarrow {x_1} = 2.583\]
Iteration 2:
\[f\left( {{x_1}} \right) = x_{_1}^3 - 15\]
Putting the values, we get;
\[ \Rightarrow f\left( {{x_1}} \right) = {\left( {2.583} \right)^3} - 15\]
\[ \Rightarrow f\left( {{x_1}} \right) = 2.240\]
On differentiating \[f\left( {{x_1}} \right) = x_{_1}^3 - 15\]; we get;
\[ \Rightarrow f'\left( {{x_1}} \right) = 3x_1^2\]
Putting the value, we get;
\[ \Rightarrow f'\left( {{x_1}} \right) = 3 \times {\left( {2.583} \right)^2}\]
On calculating we get;
\[ \Rightarrow f'\left( {{x_1}} \right) = 20.021\]
Again, using the formula;
\[{x_{n + 1}} = {x_n} - \dfrac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\].
Putting \[n = 1\], we get;
\[ \Rightarrow {x_2} = {x_1} - \dfrac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\]
Putting the values, we get;
\[ \Rightarrow {x_2} = 2.583 - \dfrac{{2.240}}{{20.021}}\]
Calculating we get;
\[ \Rightarrow {x_2} = 2..471\]
Iteration 3:
\[f\left( {{x_2}} \right) = x_2^3 - 15\]
Putting the value of \[{x_2}\], we get;
\[ \Rightarrow f\left( {{x_2}} \right) = {\left( {2.471} \right)^3} - 15\]
\[ \Rightarrow f\left( {{x_2}} \right) = 0.096\]
On differentiating \[f\left( {{x_2}} \right) = x_2^3 - 15\], we get;
\[ \Rightarrow f'\left( {{x_2}} \right) = 3x_2^2\]
Putting the value, we get;
\[ \Rightarrow f'\left( {{x_2}} \right) = 3{\left( {2.471} \right)^2}\]
On calculating we get;
\[ \Rightarrow f'\left( {{x_2}} \right) = 18.324\]
Using; \[{x_{n + 1}} = {x_n} - \dfrac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\].
Putting \[n = 2\];
\[ \Rightarrow {x_3} = {x_2} - \dfrac{{f\left( {{x_2}} \right)}}{{f'\left( {{x_2}} \right)}}\]
Putting the value, we get;
\[ \Rightarrow {x_3} = 2.471 - \dfrac{{0.096}}{{18.324}}\]
Calculating we get;
\[ \Rightarrow {x_3} = 2.466\]
Iteration 4:
\[f\left( {{x_3}} \right) = x_3^3 - 15\]
Putting the value, we get;
\[ \Rightarrow f\left( {{x_3}} \right) = {\left( {2.466} \right)^3} - 15\]
Calculating we get;
\[ \Rightarrow f\left( {{x_3}} \right) = 0\]
Differentiating the above equation, we get;
\[ \Rightarrow f'\left( {{x_3}} \right) = 3x_3^2\]
Putting the value, we get;
\[ \Rightarrow f'\left( {{x_3}} \right) = 3 \times {\left( {2.466} \right)^2}\]
\[ \Rightarrow f'\left( {{x_3}} \right) = 18.247\]
Using;
\[{x_{n + 1}} = {x_n} - \dfrac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\].
Putting \[n = 3\], we get;
\[ \Rightarrow {x_4} = {x_3} - \dfrac{{f\left( {{x_3}} \right)}}{{f'\left( {{x_3}} \right)}}\]
Putting the values, we get;
\[ \Rightarrow {x_4} = 2.466 - \dfrac{0}{{18.247}}\]
\[ \Rightarrow {x_4} = 2.466\]
Now since we have got the same value for two consecutive iterations so we will stop further.
Therefore, \[x = 2.466\].

Note:
One thing to keep in mind is that if the initial value is not given in the question, then we will calculate it on our own according to the question, as we have done in this question. Also, we have to continue the iteration until the value gets repeated for the two consecutive iterations.