Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for iteration in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Iteration topic page.
When asked to 'show that the equation has a solution between and ', substitute both values.
When given and a starting value, use the calculator's ANS key to repeat quickly.
If f(a) and f(b) have opposite signs, and f is a continuous curve, a root lies between a and b.
The mark is for the sentence: a root lies between the two values. The two function values are the evidence.
If f(1) = 0, then 1 is the root. Do not call that a sign change between 1 and 2.
f(1) and f(2) can both be negative. The root may sit between 2 and 3 instead.
To iterate x = √(k - x), put the start value in to get the first output. Then put that output back in as the next x, and repeat.
Do the number of steps asked, then round that last value to 1 decimal place. Do not round in the middle.
Let f(x)=x^2-x-3. Then f(2)=4-2-3=-1 and f(3)=9-3-3=3. The values have opposite signs, and f is a polynomial so it has no breaks. Therefore f(x)=0 has a root strictly between 2 and 3. Writing "sign change" without the two calculations does not earn the mark. Write both substitutions in full.
The root is not halfway unless you have a reason. Here f(2) is closer to zero than f(3), so the root is closer to 2, but that is only a hint. A sign change does not give the root to any number of decimal places. It only traps it in an interval. If both values are negative, you have not shown a root in that interval at all.
Method: 1. Substitute both ends of the interval into f. 2. Show the two numbers, including their signs. 3. Opposite signs, for a continuous function, mean a root lies strictly between them. 4. Do not claim the root is the midpoint.
From x^2-x-3=0, add x+3 to get x^2=x+3. If x≠ 0, divide by x: x=1+3/x. The root between 2 and 3 is not zero, so this is allowed. You can check the rearrangement by multiplying back: x=1+3/x gives x^2=x+3, then x^2-x-3=0. Show that line if the question says "show that" the formula comes from the equation.
A different rearrangement gives a different sequence, sometimes one that moves away from the root. Use the formula you are given. Do not invent a neater one in the middle of a numerical question. And do not divide by x if you have not said why x is not zero. Here the interval (2, 3) does that job.
Method: 1. Start from f(x)=0 and show each algebra step. 2. Finish as x=g(x). 3. Check you have not divided by an expression that is zero at the root. 4. The rearrangement is the iteration rule xn+1=g(x_n).
Start with x_0=2. Then x_1=1+3/2=5/2=2.5. Then x_2=1+3/2.5=1+1.2=2.2=11/5. Then x_3=1+dfrac311/5=1+15/11=26/11. The decimals are 2, 2.5, 2.2, 2.3636ldots The sequence is moving about inside the interval. It has not settled to one decimal place yet.
The mistake is to type 2 into the formula for every step, so that x_1, x_2 and x_3 are all 2.5. Each line must use the previous output. On a calculator, leave the unrounded answer in the ANS memory. If the question wants each iterate to 4 decimal places, write those rounded values on the page but keep calculating with the full ANS. Rounding 2.3636 to 2.36 and then dividing 3 by 2.36 shifts every later value.
Method: 1. x_0 is the start. x_1 is the first value the formula produces. 2. Each new input is the full previous output, not the original start. 3. Keep the calculator ANS if the value is not exact. 4. Write the iterates the question asks for, to the accuracy it names.
If the sequence settles, its limit L satisfies L=1+3/L, which is the original equation. So the limit is the root between 2 and 3, not the other root of x^2-x-3=0. That other root is negative, because the product of the roots of x^2-x-3 is -3. A positive starting value and this rearrangement stay positive. Say which root you have found: "the root between 2 and 3".
x_3 is not "correct to 3 decimal places" just because it is the third term. To claim a decimal accuracy you compare two later iterates that agree to that number of places, or you follow the instruction in the question. If it says "find x_3 to 3 decimal places", calculate 26/11=2.3636ldots and round to 2.364. The exact fraction is safer when the question does not demand a rounding.
Method: 1. The sequence is approaching the root you trapped with the sign change. 2. You may round an iterate for the final answer only when the question says so. 3. Two successive iterates that round to the same value support that rounded root. 4. Say which root, if the equation has more than one.
An iteration formula such as xn+1=sqrt[3]5x_n+11 means: take the current value x_n, put it into the right-hand side, and the result is the next value xn+1. If the question gives x_1=3, then x_2=sqrt[3]5× 3+11=2.9625ldots and x_3 comes from putting x_2 in. Read the subscript on the starting value carefully. Starting from x_1 and asking for x_3 needs two steps, while starting from x_0 and asking for x_3 needs three.
On a calculator, type the starting number and press equals. Then type the formula using the Ans key in place of x_n, and press equals each time to move to the next value. This keeps every digit, so rounding never creeps in. Only round at the very end, to the accuracy the question asks for. Writing each value to a few more decimal places than the final answer, such as 2.9625, 2.9554 and 2.9540, shows the working and lets a marker see that the values are settling down.
Method: 1. Check which subscript the starting value has and how many steps you need. 2. Enter the starting value, then type the formula using Ans, and press equals repeatedly. 3. Write down each value to at least one more decimal place than the question needs. 4. Round only the final answer to the requested accuracy.
Tip: Write the subscripts on each line of working, such as x2 = 2.9625, so the marker can see where each value came from.
A show-that question gives you the equation and the target formula. For x^3+2x-10=0 the target might be x=sqrt[3]10-2x. Start from the equation and make x^3 the subject: x^3=10-2x. Then take the cube root of both sides. Write both lines. Each rearrangement has a different target, so you must read it before you start. The equation x^2-3x-1=0 can become x=3+1/x by moving 3x across to get x^2=3x+1 and dividing every term by x.
Take care with signs and with division. Dividing by x must be done to every term: x=3+1/x, not x=3x+1/x. Do not work backwards from the target as your proof. Start from the given equation, and your last line should be the printed formula. Different rearrangements of the same equation give different iteration formulas. Some settle down quickly, some are slow and some do not settle at all, but the exam will give you a formula that works.
Method: 1. Start from the given equation and isolate the power of x or the term the target needs. 2. Apply the inverse operation to every term on the side that needs it. 3. Compare your final line with the printed formula. 4. Do not start from the target and work backwards.
To show that a root is a particular number to a set accuracy, test the two values half a unit either side. To show a root is 3.30 to 2 decimal places, test 3.295 and 3.305, because everything between these rounds to 3.30. For f(x)=x^2-3x-1 you get f(3.295)=-0.028 and f(3.305)=0.008. The sign changes from negative to positive, so the root is between the two values.
The test points are the half-way values, not the numbers 3.29 and 3.31. For one decimal place the test points are 0.05 either side. State both function values with their signs, say there is a change of sign, and write the conclusion in words. This only works if the function is continuous, which is true for every function in GCSE questions. A sign change at the end of an iteration is a strong check on the whole question.
Method: 1. Write the equation as f(x)=0 with everything on one side. 2. Choose the values half a unit of the last decimal place either side of the answer. 3. Work out both function values and state their signs. 4. Conclude that the root lies between them, so it rounds to the answer.
Some iteration questions describe a real quantity. A rectangle with width x and length x+3 has area x^2+3x=20, and that gives the equation x^2+3x-20=0. Rearranging to x=√(20-3x) gives a formula that homes in on the width. Only the positive root is relevant because a length is positive. In a context question, finish with a sentence in the language of the problem and include units where they are given.
If the values of an iteration settle down to a number L, then L must satisfy the formula with xn+1 and x_n both replaced by L. For un+1=(u_n^2+20)/2u_n this gives L=(L^2+20)/2L, so 2L^2=L^2+20 and L^2=20. Then L=√(20)=2√(5). This lets you find an exact value without more iterations. The decimal values 4.5 and 4.4722 are heading towards 4.4721ldots, which supports the answer.
Method: 1. Form the equation from the context, then rearrange to the iteration formula you are given. 2. Reject any negative root if the quantity is a length or a count. 3. To find the limit, replace xn+1 and x_n by the same letter and solve. 4. Compare the exact limit with the decimal values from the iteration.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.