Iteration
Iteration is a powerful method for finding approximate solutions to equations that are difficult or impossible to solve exactly. It’s a process of making a series of increasingly accurate guesses, a technique used in everything from engineering design to computer graphics.
The Core Concepts of Iteration
1. The Sign Change Rule
This rule helps you confirm that a solution (a root) exists between two values. If a continuous function $f(x)$ is positive at one value and negative at another, it must cross the x-axis somewhere in between.
To prove a root lies between $a$ and $b$, you must show that $f(a)$ and $f(b)$ have opposite signs.
2. The Iterative Formula
To iterate, you first rearrange an equation like $f(x)=0$ into the form $x=g(x)$. This is then written using suffix notation:
This means: “The next guess ($x_{n+1}$) is found by substituting the current guess ($x_n$) into the formula.”
The Iteration Process
You start with an initial guess, $x_0$, and repeat the process until your answer is accurate enough.
Worked Examples
Example 1: Showing a Root Exists
Show that $x^3 – 4x + 1 = 0$ has a root between $x=0$ and $x=1$.
- Let $f(x) = x^3 – 4x + 1$.
- Substitute $x=0$:
$f(0) = (0)^3 – 4(0) + 1 = 1$ (Positive). - Substitute $x=1$:
$f(1) = (1)^3 – 4(1) + 1 = -2$ (Negative).
Conclusion: Since there is a sign change for a continuous function, a root must lie between 0 and 1.
Example 2: Using an Iterative Formula
Starting with $x_0 = 3$, use the formula $x_{n+1} = \sqrt[3]{7x_n – 4}$ to find $x_1$, $x_2$, and $x_3$ to 3 d.p.
Calculator Tip: Type 3 and press =. Then type the formula using the “Ans” button for $x_n$. Pressing = repeatedly will generate the sequence!
- $x_1 = \sqrt[3]{7(3) – 4} = \sqrt[3]{17} \approx 2.571$
- $x_2 = \sqrt[3]{7(2.57128…) – 4} \approx 2.411$
- $x_3 = \sqrt[3]{7(2.41071…) – 4} \approx 2.343$
Tutor Insights
🤔 Common Misunderstandings
- Rounding too early: You must use the full, unrounded, previous answer from your calculator for the next calculation to maintain accuracy.
- The Sign Change conclusion: Forgetting to write the full concluding sentence, including “sign change” and “continuous function”.
📝 Common Exam Mistakes
- Calculation errors: Typing the formula into the calculator incorrectly.
- Not showing sufficient working: You must write down the value of each iteration ($x_1, x_2, x_3…$) before your final answer.
- Stopping too soon: Not iterating enough times to find a solution to the required degree of accuracy.
Practice Questions
- Show that the equation $x^3 + x – 3 = 0$ has a root between $x=1$ and $x=2$.
- The equation $x^3 – 2x – 5 = 0$ can be written as $x = \sqrt[3]{2x + 5}$. Starting with $x_0 = 2$, find the values of $x_1$, $x_2$, and $x_3$ to 3 decimal places.
- An iterative formula is given as $x_{n+1} = \frac{x_n^2+10}{2x_n+5}$. Starting with $x_0=1$, find the solution correct to 2 decimal places.
Show Answers
- Working: Let $f(x) = x^3 + x – 3$.
$f(1) = 1+1-3 = -1$ (Negative).
$f(2) = 8+2-3 = 7$ (Positive).
Since there is a sign change for a continuous function, a root lies between 1 and 2. - Working:
$x_1 = \sqrt[3]{2(2)+5} = \sqrt[3]{9} \approx 2.080$
$x_2 = \sqrt[3]{2(2.080…) + 5} \approx 2.092$
$x_3 = \sqrt[3]{2(2.092…) + 5} \approx 2.095$ - Working:
$x_0 = 1$
$x_1 = 1.5714…$
$x_2 = 1.5238…$
$x_3 = 1.5284…$
$x_4 = 1.5279…$
$x_5 = 1.5280…$
Rounding to 2 d.p., both $x_4$ (1.53) and $x_5$ (1.53) are the same. Answer: 1.53.
FAQs
What if my values don’t get closer together?
If your values are getting further apart (diverging), it means the rearrangement you’re using isn’t suitable for finding that specific root. In GCSE exams, you will almost always be given a formula that works (converges). If it doesn’t seem to be working, double-check that you have typed it into your calculator correctly.
How many decimal places should I use during calculations?
Always keep as many decimal places as your calculator allows during the intermediate steps. The best way to do this is by using the ‘ANS’ button. Only round your *final* answer to the accuracy requested in the question.
