26. Iterative Methods - Cardiff Tutor Company

Question 1

LEVEL 6

Starting with the value $a_0 = 3$ , use iterative methods to find a solution to $a_{n+1}= \sqrt[3]{16 - (a_{n})^2}$ correct to $1$ decimal place.

Select the correct answer from the list below:

A:

B: $2.3$

C: $1.9$

D: $3.0$

CORRECT ANSWER: A: $2.2$

WORKED SOLUTION:

At each step, we calculate the next value in the iterative process, and then we can put that value back into the formula to find the next one, and so on. The “Ans” button on your calculator is very handy in these types of questions. With $a_0 = 3$ , we get

$a_1= \sqrt[3]{16 - (3)^2} = 1.912931183$

$a_2 = 2.310893147$

$a_3 = 2.200810505$

$a_4 = 2.234473056$

We stop at this point because the latter two terms, $a_3, a_4$ , both round to the same value when rounded to $1$ dp, therefore our solution to $1$ dp is $2.2$ .

Question 2

LEVEL 6

The following formula can be used to find an approximate solution to an equation using iteration. Starting with $x_0 = 1$ , find the values of $x_4$ and $x_5$ to $5$ decimal places using the formula below.

$x_{n+1} = \dfrac{9 - x_{n}^2}{7}$

Select the correct answer from the list below:

A: $x_4 = 1.10870$ and $x_5 = 1.11011$

B: $x_4 = 1.10871$ and $x_5 = 1.11011$

C: $x_4 = 1.113131919$ and $x_5 = 1.11011$

D: $x_4 = 1.299322$ and $x_5 = 1.10001$ .

CORRECT ANSWER:B: $x_4 = 1.10871$ and $x_5 = 1.11011$ .

WORKED SOLUTION:

At each step, we calculate the next value in the iterative process, and then we can put that value back into the formula to find the next one, and so on. The “Ans” button on your calculator is very handy in these types of questions. With $x_0 = 1$ , we get

$x_1 = \dfrac{9 - 1^2}{7} = \frac{8}{7}$ ,
$x_2 = \dfrac{9 - (\frac{8}{7})^2}{7} = 1.099125364$ ,
$x_3 = 1.113131919$ ,
$x_4 = 1.108705333$ ,
$x_5 = 1.110110355$ ,

So, to $5$ decimal places $x_4 = 1.10871$ and $x_5 = 1.11011$ .

Question 3

LEVEL 6

The equation $y^2 - 4y - 19 = 0$ can be rearranged to $y = \sqrt{19 + 4y}$ .

As a result, we can use the iterative formula $y_{n+1} = \sqrt{19 + 4y_n}$ to find an approximate solution to this equation.

Starting with $y_0 = 3$ , use the iterative formula to find a solution to this equation correct to $3$ decimal places.

Select the correct answer from the list below:

A: $6.790$

C: $6.699$

D: $5.796$

CORRECT ANSWER: B:

WORKED SOLUTION:

At each step, we calculate the next value in the iterative process, and then we can put that value back into the formula to find the next one, and so on. The “Ans” button on your calculator is very handy in these types of questions. With $y_0 = 3$ , we get

$y_1 = \sqrt{19 + 4(3)} = 5.567764363$ ,

$y_2 = \sqrt{19 + 4(5.567764363)} = 6.424255401$ ,

$y_3 = 6.685583116$ ,

$y_4 = 6.763307805$ ,

$y_5 = 6.786253106$ ,

$y_6 = 6.793012029$ ,

$y_7 = 6.795001701$ ,

$y_8 = 6.795587304$ ,

$y_9 = 6.79575965$ .

We stop at this point because the latter two terms, $y_8, y_9$ , both round to the same value when rounded to $3$ dp, therefore our solution to $3$ dp is $6.796$ .

Question 4

LEVEL 6

Rearranging the following equation to give a suitable iterative formula:

$x^3 + 15x - 13 = 0$

Select the correct answer from the list below:

A:

B: $x_{n+1} = \sqrt[2]{-15x_{n}+13}$

C: $x_{n+1} = \sqrt[3]{15x_{n}-13}$

D: $x_{n+1} = \sqrt[2]{-x_{n}+13\times15}$

CORRECT ANSWER: A: $x_{n+1} = \sqrt[3]{-15x_{n}+13}$

WORKED SOLUTION:

Rearrange the formula to make $x$ the subject.

$x^3 = -15x + 13$

$x = \sqrt[3]{-15x + 13}$

Add the required notation.

x_{n+1} = \sqrt[3]{-15x_{n} + 13}

Question 5

LEVEL 6

Rearranging the following equation to give a suitable iterative formula:

$x^3 - 15x^2 +1 = 0$

Select the correct answer from the list below:

A: $x_{n+1} = \sqrt[3]{(15x_{n})^2-13}$

B: $x_{n+1} = \sqrt[3]{(15x_{n})^2-1}$

C: $x_{n+1} = \sqrt[3]{15x_{n}-1}$

D: $x_{n+1} = \sqrt[2]{(15x_{n})^3-1}$

CORRECT ANSWER: B: $x_{n+1} = \sqrt[3]{(15x_{n})^2-1}$

WORKED SOLUTION:

Rearrange the formula to make $x$ the subject.

$x^3 = 15x^2 - 1$

$x = \sqrt[3]{15x^2 - 1}$

Add the required notation.

x_{n+1} = \sqrt[3]{15x_{n}^2 - 1}