Define iteration as a method for solving an equation.

Iteration is a repeated process: you start from an initial value and feed each result back in to produce the next estimate of a solution. With a suitable iterative formula, the estimates get closer to the solution the more iterations you perform.

Define interval bisection as a method for solving an equation.

Interval bisection repeatedly halves an interval known to contain a solution: substitute its midpoint and keep whichever half still contains the change of sign . Each halving narrows down where the solution is, so you keep going until you can give it to the accuracy required.

GCSEMathsOCRHigherFlashcards6. AlgebraIteration

Iteration (OCR GCSE Maths: Higher): Flashcards

Exam code: J560

1/13

0Still learning

Know0

FrontIteration
Define iteration as a method for solving an equation.

Was this flashcard helpful?

Cards in this collection (13)

Define iteration as a method for solving an equation.
Iteration is a repeated process: you start from an initial value and feed each result back in to produce the next estimate of a solution.
With a suitable iterative formula, the estimates get closer to the solution the more iterations you perform.
How do you turn the equation $x^{3} + x = 7$ into an iterative formula?
Rearrange the equation to make one $x$ the subject, for example $x = \sqrt[3]{7 - x}$ in this case.
Then write $x_{n + 1}$ for the $x$ on the left and $x_{n}$ for each $x$ on the right, which gives this formula:
$x_{n + 1} = \sqrt[3]{7 - x_{n}}$
Fill in the two missing words:
In $x_{n + 1} = \sqrt[3]{7 - x_{n}}$ the term $x_{n}$ stands for the $______$ value of $x$ and $x_{n + 1}$ stands for the $______$ value of $x$ in the iteration.
The completed sentence is:
In $x_{n + 1} = \sqrt[3]{7 - x_{n}}$ the term $x_{n}$ stands for the current value of $x$ and $x_{n + 1}$ stands for the next value of $x$ in the iteration.
So substituting the latest estimate into the right-hand side gives the next estimate.
True or False?
The values produced by $x_{n + 1} = \sqrt[3]{7 - x_{n}}$ are estimates of the solution to $x = \sqrt[3]{7 - x}$ but not of the solution to $x^{3} + x = 7$ .
False.
The equation $x = \sqrt[3]{7 - x}$ is just a rearrangement of $x^{3} + x = 7$ , so the two equations have the same solution.
The estimates are therefore estimates of the solution to the original equation as well.
When using an iterative formula on a calculator, how does the Ans button help?
Type the starting value and press = so that it is stored as Ans, then type the right-hand side of the formula with Ans in place of $x_{n}$ .
Each press of = then substitutes the latest estimate and gives the next one, so nothing has to be retyped.
True or False?
If a question gives the starting value of an iteration as $x_{1}$ rather than $x_{0}$ the first new estimate you calculate is $x_{2}$ .
True.
The subscripts only count the iterations, so each one adds 1 to the subscript and a start at $x_{1}$ makes the first new estimate $x_{2}$ .
Questions can start at either $x_{0}$ or $x_{1}$ , so check which one you are given.
Use $x_{n + 1} = 5 - 2 \sqrt[3]{x_{n}}$ with $x_{0} = 3$ to find $x_{1}$ correct to 3 decimal places.
Substitute the starting value 3 into the right-hand side of the formula:
$x_{1} = 5 - 2 \sqrt[3]{3} = 2.1155 \dots = 2.116 (3 d.p.)$
Keep the unrounded value $2.1155 \dots$ for the next iteration and round only the values you write down.
How can you show that $x^{3} + x - 7 = 0$ has a solution between $x = 1$ and $x = 2$ ?
Substitute both values into the left-hand side, which gives $- 5$ when $x = 1$ and $3$ when $x = 2$ .
One value is negative and the other is positive, so there is a change of sign and, since the curve $y = x^{3} + x - 7$ has no breaks, a solution must lie between 1 and 2.
Fill in the missing words to show that $x^{3} + x = 7$ has a solution between $x = 1$ and $x = 2$ without rearranging it:
At $x = 1$ the left-hand side is 2, which is $______$ 7, and at $x = 2$ it is 10, which is $______$ 7.
The completed working is:
At $x = 1$ the left-hand side is 2, which is below 7, and at $x = 2$ it is 10, which is above 7.
The left-hand side passes through 7 somewhere between these two values, so a solution lies between 1 and 2.
A solution of $x^{3} + x - 7 = 0$ lies between 1 and 2. How do you carry out a decimal search to narrow it down?
Substitute $x = 1.1$ , $1.2$ , $1.3$ and so on into the left-hand side until the sign changes.
The solution then lies between the two consecutive values either side of that change.
True or False?
If a decimal search finds a change of sign between $x = 1.7$ and $x = 1.8$ then the solution must be 1.7 to 1 decimal place.
False.
The solution could be anywhere between 1.7 and 1.8, and it rounds to 1.8 if it is 1.75 or more.
Substitute the midpoint 1.75 and see whether the sign change is between 1.7 and 1.75 or between 1.75 and 1.8.
Define interval bisection as a method for solving an equation.
Interval bisection repeatedly halves an interval known to contain a solution: substitute its midpoint and keep whichever half still contains the change of sign.
Each halving narrows down where the solution is, so you keep going until you can give it to the accuracy required.
Interval bisection shows that a solution lies in the interval $1.65625 < x < 1.6875$ . What is the solution to 1 decimal place, and how do you know?
The solution is 1.7 to 1 decimal place.
Every number from 1.65625 to 1.6875 rounds to 1.7, so wherever the solution is in that interval it rounds to 1.7.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Simultaneous EquationsNext:Forming & Solving Equations