True or False? Iteration is a method for trying to solve an equation.

True. Iteration is a method for trying to solve an equation.

True or False? Iteration will give you the exact solutions to an equation.

False. Iteration only gives you approximate solutions to an equation (called estimates ).

True or False? To use iteration, you need to know a starting (initial) value near to the true solution.

True. To use iteration, you need to know a starting (initial) value near to the true solution. You are normally given this starting value in the question.

True or False? With iteration, the more times you do it, the more accurate the estimate becomes.

True. In general with iteration, the more times you do it, the more accurate the estimate becomes. The estimate gets closer and closer to the true solution of the equation.

Define interval bisection .

Interval bisection is a method for finding an approximate solution by repeatedly halving an interval that contains the solution. Each time, you substitute the midpoint and keep whichever half still shows a change of sign .

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Iteration (WJEC Eduqas GCSE Maths: Higher): Flashcards

Exam code: C300

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True or False?
Iteration is a method for trying to solve an equation.

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Cards in this collection (18)

True or False?
Iteration is a method for trying to solve an equation.
True.
Iteration is a method for trying to solve an equation.
True or False?
Iteration will give you the exact solutions to an equation.
False.
Iteration only gives you approximate solutions to an equation (called estimates).
Explain why you would not use iteration to find the solution to $4 x - 1 = 2$ .
The equation $4 x - 1 = 2$ can be solved algebraically to give $4 x = 3$ so $x = \frac{3}{4}$ .
Iteration is usually used for harder equations which we do not know how to solve.
True or False?
To use iteration, you need to know a starting (initial) value near to the true solution.
True.
To use iteration, you need to know a starting (initial) value near to the true solution.
You are normally given this starting value in the question.
Write the equation $x = \frac{2}{x + 3}$ as an iterative formula.
The equation $x = \frac{2}{x + 3}$ as an iterative formula is $x_{n + 1} = \frac{2}{x_{n} + 3}$ .
$x_{n}$ is the current estimate of $x$ and $x_{n + 1}$ is the next estimate of $x$ .
An iterative formula is given by $x_{n + 1} = \frac{2 x_{n} + 1}{x_{n}}$ .
If the starting (initial) value is $x_{0} = 2$ , how do you find the next value, $x_{1}$ ?
To find $x_{1}$ , substitute $x_{0} = 2$ into the right-hand side of the iterative formula.
This gives $\frac{2 \times 2 + 1}{2}$ which simplifies to $2.5$ .
This means $x_{1} = 2.5$ .
You can also use the "Ans" button on your calculator, by typing "Ans = 2" then typing $\frac{2 \times Ans + 1}{Ans}$ and pressing "=".
True or False?
With iteration, the more times you do it, the more accurate the estimate becomes.
True.
In general with iteration, the more times you do it, the more accurate the estimate becomes.
The estimate gets closer and closer to the true solution of the equation.
An iterative formula with a starting value of $x_{0} = 5$ gives the following results:
$x_{1} = 5.5561221 . . . x_{2} = 5.5458926 . . . x_{3} = 5.5443563 . . x_{4} = 5.5443498 . . .$
What what be an estimate of the solution to 2 decimal places?
$x_{1} = 5.5561221 . . . x_{2} = 5.5445926 . . . x_{3} = 5.5443563 . . x_{4} = 5.5443498 . . .$
An estimate of the solution to 2 decimal places would be 5.54.
This is because it looks like it is settling to 5.544... which rounds to 5.54.
True or False?
Equations must always be rearranged into the form $x = f (x)$ before converting them into an iterative formula.
True.
Equations must always be rearranged into the form $x = f (x)$ before converting them into an iterative formula.
Write down two different iterative formulas that could be made from the equation $x^{3} + x - 1 = 0$ .
To find iterative formulas, you need to rearrange the equation $x^{3} + x - 1 = 0$ into the form $x = f (x)$ first.
One possible rearrangement is $x = 1 - x^{3}$ (by making the middle term the subject).
Another possible rearrangement is $x = \sqrt[3]{1 - x}$ (by first making $x^{3}$ the subject).
So two possible iterative formulas are $x_{n + 1} = 1 - {x_{n}}^{3}$ and $x_{n + 1} = \sqrt[3]{1 - x_{n}}$ .
There are other possible harder rearrangements, for example by factorising the first two terms to get $x (x^{2} + 1) = 1$ which gives $x = \frac{1}{x^{2} + 1}$ or $x = \pm \sqrt{\frac{1}{x} - 1}$ .
True or False?
The iterative formula $x_{n + 1} = \sqrt{x_{n} + 1}$ could have come from the equation $x^{2} + x + 1 = 0$ .
False.
The iterative formula $x_{n + 1} = \sqrt{x_{n} + 1}$ could not have come from the equation $x^{2} + x + 1 = 0$ .
The iterative formula $x_{n + 1} = \sqrt{x_{n} + 1}$ comes from the equation $x = \sqrt{x + 1}$ .
Squaring both sides gives $x^{2} = x + 1$ .
Bringing all the terms to the left gives $x^{2} - x - 1 = 0$ .
The question said $x^{2} + x + 1 = 0$ , which is incorrect.
How would you show that the equation $x^{3} + x = 1$ has a solution between $x = 0$ and $x = 1$ ?
To show that the equation $x^{3} + x = 1$ has a solution between $x = 0$ and $x = 1$ :
Either substitute $x = 0$ and $x = 1$ into the left-hand side and compare this with the right-hand side (one value should be bigger and the other should be smaller).
Or rearrange the equation to $x^{3} + x - 1 = 0$ , then substitute $x = 0$ and $x = 1$ into the left-hand side and compare to the new right-hand side of zero (one should give a positive value and the other should give a negative value). This is called a change of sign.
For example, $x = 0$ gives $0^{3} + 0 - 1 < 0$ and $x = 1$ gives $1^{3} + 1 - 1 > 0$ , so there is a change in sign from negative to positive, meaning a solution is between $x = 0$ and $x = 1$ .
In a decimal search for a solution between 1 and 2, to 1 decimal place, which values of $x$ do you substitute?
Substitute $x = 1.0$ , then $1.1$ , $1.2$ , $1.3$ and so on in turn, stopping at the first change of sign.
The solution then lies between the last two values you substituted.
Define interval bisection.
Interval bisection is a method for finding an approximate solution by repeatedly halving an interval that contains the solution.
Each time, you substitute the midpoint and keep whichever half still shows a change of sign.
A decimal search on $x^{3} + x - 7 = 0$ finds a change of sign between $x = 1.7$ and $x = 1.8$ . Why is this not yet enough to give the solution to 1 decimal place?
The solution could be anywhere between 1.7 and 1.8, so it could round to either of them.
Substituting the midpoint, $x = 1.75$ , shows which half of the interval the solution is in, and so which way it rounds.
True or False?
When $x = 2.3$ , the value of $x^{3} - 2 x - 7$ is $0.567$ , so $x = 2.3$ is a solution of $x^{3} - 2 x - 7 = 0$ .
False.
A solution makes $x^{3} - 2 x - 7$ equal to exactly zero, and $0.567$ is not zero.
To locate a solution, you look for a change of sign between two values of $x$ instead.
When an equation is rearranged to have 0 on the right-hand side, its left-hand side is negative at $x = 2.4$ and $x = 2.45$ but positive at $x = 2.5$ . Fill in the gaps.
There is a change of sign between $x =______$ and $x = 2.5$ so the solution there is $______$ to 1 decimal place.
The completed statement is:
There is a change of sign between $x = 2.45$ and $x = 2.5$ so the solution there is $2.5$ to 1 decimal place.
Every number strictly between 2.45 and 2.5 rounds up to 2.5.
When using interval bisection to find a solution to 1 decimal place, how do you know when to stop?
Stop when every number in the interval rounds to the same value at the required accuracy.
For example, if the solution lies in $1.65625 < x < 1.6875$ , every value in that interval rounds to 1.7, so the solution is 1.7 to 1 decimal place.

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