Numerical Solution of Equations (Paper 2, Topic 2.6)

Hello Mathematicians! Welcome to one of the most practical chapters in Pure Mathematics 2: Numerical Solution of Equations.

In the real world, many complicated equations (especially those involving mixed functions like \(e^x = \sin x\) or \(x^3 + \ln x = 5\)) cannot be solved exactly using algebra. This chapter teaches us powerful numerical techniques to find highly accurate approximations for the roots of these equations. You will learn how to locate a root and then refine that answer using an iterative process. Let's dive in!

Key Terms Review

  • Root: A solution to the equation \(f(x) = 0\). Graphically, this is the point where the curve crosses the \(x\)-axis.
  • Numerical Method: A mathematical technique used to find an approximate answer to a problem, often relying on repeated calculations.

1. Locating the Root Approximately

Before we can approximate a root, we need to know roughly where it is. The syllabus requires two main methods for root location: graphical consideration and the sign change rule.

1.1 The Graphical Method

This method helps you find a rough estimate for the root by turning the complex equation into two simpler graphs and looking for their intersection point.

Process:

  1. Start with the equation \(f(x) = 0\).
  2. Rearrange it into the form \(g(x) = h(x)\), where both \(g(x)\) and \(h(x)\) are functions you can easily sketch.
  3. Sketch the graphs \(y = g(x)\) and \(y = h(x)\) on the same axes.
  4. The \(x\)-coordinate of the intersection point is the approximate root of the original equation.

Example: To find the root of \(x^3 + 2x - 1 = 0\).
You could rearrange it to \(x^3 = 1 - 2x\).
Sketch \(y = x^3\) and \(y = 1 - 2x\). You will see they intersect near \(x=0.5\). This gives you a starting value.

1.2 The Sign Change Method

This is the formal, accurate way to prove that a root exists within a specific interval.

Concept: If a function \(f(x)\) is continuous (meaning its graph has no breaks, jumps, or asymptotes) over an interval \([a, b]\), and \(f(a)\) and \(f(b)\) have opposite signs, then a root must lie between \(a\) and \(b\).

Analogy: Imagine walking across a hill. If you start below sea level (\(f(a) < 0\)) and end up above sea level (\(f(b) > 0\)), and you never lifted your feet (continuous function), you must have crossed sea level (\(f(x) = 0\)) somewhere in between!

Step-by-Step Procedure:

  1. Define the function \(f(x)\) from your equation \(f(x)=0\).
  2. Identify the interval \([a, b]\) you are testing (e.g., between \(x=1\) and \(x=2\)).
  3. Calculate \(f(a)\) and \(f(b)\).
  4. State your conclusion clearly:
    • If \(f(a)\) and \(f(b)\) have opposite signs (one positive, one negative), state: "Since there is a change of sign and \(f(x)\) is continuous, a root lies between \(a\) and \(b\)."

! Common Mistake to Avoid !
Never forget to mention the word continuous. If the function isn't continuous (like \(y = \frac{1}{x}\) near \(x=0\)), a sign change might occur without a root existing between the two points. For the exams, unless the function is explicitly discontinuous (like having an asymptote), you can usually assume continuity.

Quick Review: Locating Roots

Use the Sign Change method to formally prove a root exists within a narrow interval (like between 1.5 and 1.6). This interval provides your starting value for the next stage.

2. Iteration Using the Formula \(x_{n+1} = F(x_n)\)

Once you know the approximate location of the root, iteration allows you to home in on the exact value using a repetitive calculation process.

2.1 The Iterative Formula

The sequence of approximations is defined by the rule:
$$x_{n+1} = F(x_n)$$
Here, \(x_n\) is the \(n\)-th approximation, and \(x_{n+1}\) is the next, hopefully better, approximation.

The key is understanding where the function \(F(x)\) comes from.

2.2 Rearranging the Equation

To use iteration, the equation \(f(x) = 0\) must be rearranged into the form \(x = F(x)\).

Example: Suppose you want to solve \(x^3 + 2x - 1 = 0\).

Rearrangement A (Simple):
$$2x = 1 - x^3$$ $$x = \frac{1 - x^3}{2}$$
So, \(F(x) = \frac{1 - x^3}{2}\), and the iteration formula is: $$x_{n+1} = \frac{1 - (x_n)^3}{2}$$

Rearrangement B (Cubic Root):
$$x^3 = 1 - 2x$$ $$x = \sqrt[3]{1 - 2x}$$
So, \(F(x) = \sqrt[3]{1 - 2x}\), and the iteration formula is: $$x_{n+1} = \sqrt[3]{1 - 2x_n}$$

Did you know?
It is possible to rearrange an equation in many different ways, but not all of them will lead to a solution! Some rearrangements will cause the iteration to diverge (get further away from the root). In the exam, you will usually be given a specific iteration formula or a clear instruction on how to rearrange it.

3. The Iteration Process in Practice

Iterative processes are designed for calculator use. This step-by-step guide will ensure you use your calculator efficiently.

Goal: Determine the root to a prescribed accuracy (e.g., 3 decimal places).

Step-by-Step Iteration:

  1. Find a Starting Value (\(x_1\)): This is usually given, or derived from the Sign Change method (e.g., if the root is between 1.5 and 1.6, a good starting value might be \(x_1 = 1.55\)).
  2. Calculate \(x_2\): Substitute \(x_1\) into the formula \(x_{n+1} = F(x_n)\).
    Calculator Trick: Enter \(x_1\), press "=", then type your formula using the "ANS" button for \(x_n\).
  3. Calculate Subsequent Values: Press "=" repeatedly to generate \(x_3, x_4, x_5, \dots\) The calculator automatically substitutes the previous answer (ANS) into the formula.
  4. Check for Convergence and Accuracy: Continue until two successive approximations agree to the required degree of accuracy (D.P. or S.F.).

Example of Convergence Check (Target: 3 d.p.):

  • \(x_5 = 1.23456\)
  • \(x_6 = 1.23481\) (Different at 3 d.p.)
  • \(x_7 = 1.23490\)
  • \(x_8 = 1.23491\)
  • \(x_9 = 1.23491\) (The 4th decimal place is still changing, but the numbers agree to 5 d.p. here, definitely 3 d.p.)

Since \(x_8\) and \(x_9\) both round to 1.235 (3 d.p.), the root is confirmed as 1.235.

Key Takeaway on Accuracy

Always work with at least four or five decimal places during the iterative steps, and only round your final answer to the accuracy requested by the question. If you round too early, your final answer might be incorrect.

4. Understanding Convergence and Divergence

Although you are not required to know the mathematical condition for convergence (the syllabus specifically excludes it!), you must understand the difference between a successful iteration and a failed one.

4.1 Convergent Iteration

A sequence converges if the terms get closer and closer to a fixed number (the root).

When you run a successful iteration on your calculator:

  • The values of \(x_n\) will start settling down.
  • They might oscillate (jump back and forth, getting closer each time—a "cobweb" pattern).
  • They might approach the root monotonically (always increasing or always decreasing—a "staircase" pattern).
4.2 Divergent Iteration (Failure to Converge)

A sequence diverges if the terms get further and further away from the root, or jump wildly without settling.

If your sequence starts diverging, it means the specific rearrangement \(x = F(x)\) you chose or were given is unsuitable for finding that particular root.

How divergence looks on your calculator:

  • The numbers get increasingly large (e.g., \(x_5 = 10, x_6 = 50, x_7 = 200\)).
  • The numbers jump wildly between two or more values without settling (e.g., 2.1, -5.8, 12.5, -30.9...).

Don't worry if this seems tricky at first! The majority of exam questions in P2 will provide you with a convergent formula and a good starting value, allowing you to focus purely on the accuracy process. The critical skills are: confirming the root location via sign change, and carrying out the iterative steps accurately on your calculator.

Chapter Key Takeaway

Numerical methods allow us to find roots of complex equations. We locate the root using a sign change in \(f(x)\), and then use a specific rearrangement, \(x_{n+1} = F(x_n)\), to iterate towards the final, accurate solution.