Solving Hard Equations with Iteration

What to do when the Quadratic Formula isn't enough!

1. Why do we need Iteration?

Imagine you are asked to solve $x^3 + 2x - 5 = 0$.

When an equation is "too hard" to solve directly, we use a numerical method called Iteration. It is a loop: we guess an answer, put it into a formula to get a better answer, and repeat until the answer stops changing.

2. Creating the Iterative Formula

To start the loop, we must rearrange our impossible equation to make one of the $x$'s the subject.

Example 1: Rearranging an Equation
Show that $x^3 + 2x - 5 = 0$ can be rearranged into $x = \frac{5 - x^3}{2}$.
Step 1: Look at your target The target has division by 2. This means we need to leave the $2x$ where it is and move everything else.
Step 2: Move terms across the equals sign $$2x = 5 - x^3$$ Step 3: Divide by 2 to isolate x $$x = \frac{5 - x^3}{2}$$ Step 4: Add the iteration notation The isolated $x$ becomes the "next" value ($x_{n+1}$). The $x$ inside the formula becomes the "current" value ($x_n$).
$$x_{n+1} = \frac{5 - x_n^3}{2}$$

3. The Calculator Loop

To find the solution, you will usually be given a starting number, called $x_0$. Let's say $x_0 = 1$.

The ANS Button Trick (Crucial for Exams):
1. Type your starting number 1 and press =.
2. Type the fraction exactly as it looks, but replace $x_n$ with the ANS button: (5 - ANS³) ÷ 2.
3. Press = to get $x_1$.
4. Press = again to get $x_2$, again for $x_3$, and so on!
Example 2: Finding the Root
Using $x_{n+1} = \frac{5 - x_n^3}{2}$ and starting with $x_0 = 1$, find the solution to 2 decimal places.
Step 1: Write down the first few iterations $x_1 = 2$
$x_2 = -1.5$
$x_3 = 4.1875$
Wait... these numbers are jumping all over the place! In this specific case, rearranging for $2x$ didn't work well (this is called diverging). Let's try rearranging for the $x^3$ instead!
Example 3: A Better Rearrangement
Let's rearrange $x^3 + 2x - 5 = 0$ into $x = \sqrt[3]{5 - 2x}$ instead.
So, $x_{n+1} = \sqrt[3]{5 - 2x_n}$. Let's start with $x_0 = 1$ again.
Step 1: Use the ANS trick Type 1 =
Type ³√(5 - 2ANS)
Step 2: Press equals repeatedly $x_1 = 1.4422...$
$x_2 = 1.2837...$
$x_3 = 1.3448...$
$x_4 = 1.3220...$
Keep pressing until the first two decimal places stop changing...
Eventually, it settles on $1.328...$
Solution: $x = 1.33$ (to 2 d.p.)

10 Practice Questions & Solutions

Try these on your calculator! Click the question to check your steps.

Q1: Show that $x^3 + 7x - 3 = 0$ can be rearranged into $x_{n+1} = \frac{3 - x_n^3}{7}$
Keep the $7x$ term on the left: $7x = 3 - x^3$
Divide by 7: $x = \frac{3 - x^3}{7}$
Add subscripts: $x_{n+1} = \frac{3 - x_n^3}{7}$
Proof complete.
Q2: Show that $x^3 - 4x + 2 = 0$ can be rearranged into $x_{n+1} = \sqrt[3]{4x_n - 2}$
Keep the $x^3$ term on the left: $x^3 = 4x - 2$
Cube root both sides: $x = \sqrt[3]{4x - 2}$
Add subscripts: $x_{n+1} = \sqrt[3]{4x_n - 2}$
Proof complete.
Q3: Using $x_{n+1} = \frac{10}{x_n + 3}$ and $x_0 = 2$, find the values of $x_1, x_2$, and $x_3$.
$x_1 = \frac{10}{2 + 3} = \frac{10}{5} = 2$
$x_2 = \frac{10}{2 + 3} = 2$
$x_3 = 2$
Answer: $x_1 = 2$, $x_2 = 2$, $x_3 = 2$. (The exact root is 2!)
Q4: Using $x_{n+1} = \sqrt{2x_n + 8}$ and $x_0 = 3$, find the root to 1 decimal place.
$x_1 = 3.7416...$
$x_2 = 3.9348...$
$x_3 = 3.9836...$
$x_4 = 3.9959...$
The value is settling at exactly 4.
Answer: 4.0 (to 1 d.p.)
Q5: An equation $x^2 - 5x + 2 = 0$ is rearranged into $x_{n+1} = \frac{x_n^2 + 2}{5}$. With $x_0 = 0$, find $x_3$.
$x_1 = \frac{0^2 + 2}{5} = 0.4$
$x_2 = \frac{0.4^2 + 2}{5} = 0.432$
$x_3 = \frac{0.432^2 + 2}{5} = 0.43732...$
Answer: $x_3 = 0.437$ (to 3 d.p.)
Q6: Change of Sign: Show that $x^3 + 2x - 1 = 0$ has a root between $x = 0$ and $x = 1$.
Substitute $x = 0$: $0^3 + 2(0) - 1 = -1$
Substitute $x = 1$: $1^3 + 2(1) - 1 = 2$
Because the answer changes from negative (-1) to positive (2), the graph must cross zero.
Proof complete: Change of sign indicates a root.
Q7: Using $x_{n+1} = \frac{4}{x_n^2} + 1$, start with $x_0 = 1$ and find $x_2$ as a fraction.
$x_1 = \frac{4}{1^2} + 1 = 4 + 1 = 5$
$x_2 = \frac{4}{5^2} + 1 = \frac{4}{25} + 1 = \frac{29}{25}$
Answer: $\frac{29}{25}$ (or 1.16)
Q8: Show that $x^3 - x^2 - 4 = 0$ can be rearranged to $x_{n+1} = \sqrt{x_n^3 - 4}$
This is a trick! If you isolate the $x^2$ term:
Move the $x^2$ to the right side: $x^3 - 4 = x^2$
Square root both sides: $\sqrt{x^3 - 4} = x$
Swap sides and add subscripts: $x_{n+1} = \sqrt{x_n^3 - 4}$
Proof complete.
Q9: Find the limit of the iteration $x_{n+1} = \frac{3x_n + 8}{5}$
At the limit, $x_{n+1}$ and $x_n$ are the same. Let's call them $x$.
$x = \frac{3x + 8}{5}$
Multiply by 5: $5x = 3x + 8$
Subtract $3x$: $2x = 8$
Divide by 2: $x = 4$
Answer: The limit is 4.
Q10: Full GCSE Question (3 parts)
a) Show $x^3 - 5x + 3 = 0$ has a root between $x=0$ and $x=1$.
b) Rearrange to $x_{n+1} = \frac{x_n^3 + 3}{5}$.
c) Use $x_0 = 1$ to find the root to 2 d.p.
Part a: Let $f(x) = x^3 - 5x + 3$
$f(0) = 3$
$f(1) = 1 - 5 + 3 = -1$
Change of sign (3 to -1) means a root exists.

Part b: Isolate the $5x$ term.
$x^3 + 3 = 5x$
$x = \frac{x^3 + 3}{5} \Rightarrow x_{n+1} = \frac{x_n^3 + 3}{5}$

Part c: Use the calculator loop.
$x_0 = 1$
$x_1 = 0.8$
$x_2 = 0.7024$
$x_3 = 0.6693...$
$x_4 = 0.6599...$
$x_5 = 0.6574...$
$x_6 = 0.6568...$
The value settles at approximately 0.6566...
Answer: $x = 0.66$ (to 2 d.p.)