Numerical solution of equations - Further Mathematics - Pearson A-Level

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Welcome to Numerical Solutions! Your Guide to Finding Hidden Roots

Hello future mathematician! This chapter, Numerical solution of equations, is where we stop relying on simple algebraic rearrangement and start using the true power of calculation. You’ve probably learned how to solve quadratic equations, or maybe cubics, but what about something messy like $x^5 - 3x + 1 = 0$?

Most real-world equations are too complicated to solve exactly. Numerical methods are tools that allow us to find a highly accurate approximation of the root (the value of $x$ where the function equals zero). Think of this chapter as learning advanced searching techniques!

Don't worry if this seems tricky at first. We will break down each method step-by-step. All these methods involve iteration – doing a simple calculation repeatedly to refine the answer.

Section 1: Locating the Root (The Change of Sign Rule)

Before we can approximate a root, we need to know roughly where it is. We do this using the properties of continuous functions.

1.1 The Prerequisite: Continuity and the Change of Sign

In Further Maths, unless otherwise stated, we usually deal with functions $f(x)$ that are continuous. This means the graph can be drawn without lifting your pen—there are no gaps or vertical asymptotes.

The root $\alpha$ is where the graph crosses the x-axis, meaning $f(\alpha) = 0$.

The Change of Sign Rule

If you evaluate a continuous function $f(x)$ at two points, $a$ and $b$:

If $f(a)$ is positive and $f(b)$ is negative (or vice versa),
Then a root must lie somewhere within the interval $[a, b]$.

Analogy: Imagine you are walking across a valley (the function). If you start above sea level (positive $f(x)$) and end up below sea level (negative $f(x)$), you must have crossed sea level (the root, $f(x)=0$) at least once!

1.2 Step-by-Step Location

To show that a root $\alpha$ lies in the interval $[a, b]$:

Calculate $f(a)$ and $f(b)$.
Show that $f(a)$ and $f(b)$ have opposite signs (one positive, one negative).
State the conclusion: "Since there is a change of sign and $f(x)$ is continuous, a root must lie in the interval $[a, b]$."

Common Mistake Alert!

You MUST mention continuity! If the function has an asymptote (like $f(x) = 1/x$), it can change sign without having a root. For example, if you check $f(-1)$ and $f(1)$ for $f(x) = 1/x$, the signs change, but the root is not between -1 and 1—it's discontinuous at $x=0$.

Key Takeaway for Section 1

The Change of Sign Rule is the foundation. It guarantees the existence of a root between two points, provided the function is continuous.

Section 2: The Interval Bisection Method

The Bisection method is the simplest, most reliable, but slowest technique for finding a root to a required degree of accuracy.

2.1 The Concept: Halving the Uncertainty

Once we know a root is between $a$ and $b$, the Bisection method repeatedly halves the interval containing the root.

Analogy: Think of searching for a friend hiding in a field. You know they are somewhere in the field. You chop the field in half (bisect it) and ask which half they are in. You then chop that new, smaller half in two, and so on. Every time you cut, your possible error is halved.

2.2 The Bisection Process (Iterative Steps)

Start with the interval $[a, b]$ where $f(a)$ and $f(b)$ have opposite signs.

Step 1: Find the Midpoint
Calculate the midpoint $c$: $$c = \frac{a + b}{2}$$

Step 2: Evaluate $f(c)$
Calculate the value of the function at the midpoint, $f(c)$.

Step 3: Refine the Interval
Compare the sign of $f(c)$ with $f(a)$ and $f(b)$:

If $f(c)$ has the same sign as $f(a)$, the root must be in the new interval $[c, b]$. So, replace $a$ with $c$.
If $f(c)$ has the same sign as $f(b)$, the root must be in the new interval $[a, c]$. So, replace $b$ with $c$.

Step 4: Repeat
Continue Steps 1–3 until the interval $[a, b]$ is small enough to meet the required accuracy (e.g., finding the root correct to 2 decimal places).

Quick Review: Reliability vs. Speed

Pro: Highly reliable. It is guaranteed to work if you start with an interval containing a root.
Con: Very slow convergence. You need many iterations for high accuracy.

Key Takeaway for Section 2

Bisection works by consistently halving the error, making it robust but inefficient. It relies purely on checking the sign of $f(x)$.

Section 3: Linear Interpolation

Linear Interpolation is a faster, geometric approach that attempts to draw a better estimate than just the midpoint.

3.1 The Concept: Using a Straight Line Approximation

Instead of just blindly finding the midpoint, Linear Interpolation assumes that the actual curve $f(x)$ is roughly a straight line between the points $A(a, f(a))$ and $B(b, f(b))$.

The new estimate for the root, $x_1$, is the point where this straight line (called a chord) crosses the x-axis.

3.2 The Interpolation Formula

The estimate $x_1$ divides the interval $[a, b]$ based on the size of the function values, $f(a)$ and $f(b)$. The root is closer to the point whose function value is smaller (i.e., closer to zero).

Using similar triangles (geometry), we derive the formula for the new estimate $x_1$:

$$x_1 = a + \frac{(b-a)|f(a)|}{|f(a)| + |f(b)|}$$

Where $|f(a)|$ and $|f(b)|$ are the magnitudes (absolute values) of the function evaluations.

Memory Aid / Understanding the Formula:
We start at $a$ and add a fraction of the total width $(b-a)$. The fraction is determined by the ratio of the distances to the x-axis. The point $a$ pulls the root estimate toward itself proportional to the size of $f(b)$.

Since we use absolute values in the denominator, this formula works whether $f(a)$ is positive or negative, as long as they have opposite signs.

3.3 The Interpolation Process

Step 1: Calculate the Estimate
Use the formula above to find $x_1$.

Step 2: Evaluate $f(x_1)$
Calculate the value of the function at the new estimate.

Step 3: Refine the Interval
Just like Bisection, look at the sign of $f(x_1)$ and replace the boundary (either $a$ or $b$) that shares the same sign.

Step 4: Repeat
Continue the process until the required accuracy is achieved.

Quick Review: Speed vs. Reliability

Pro: Usually converges faster than Bisection because it uses information about the magnitude of $f(x)$.
Con: More complex calculation per step.

Key Takeaway for Section 3

Linear Interpolation uses geometry (similar triangles) to draw a straight line between the endpoints, yielding a usually faster estimate than Bisection. It requires using the magnitude of $f(a)$ and $f(b)$.

Section 4: The Newton-Raphson Method

The Newton-Raphson method is the fastest and most powerful numerical technique covered in FP1. However, it requires a knowledge of differentiation.

4.1 The Concept: Using the Tangent

Instead of using a chord (a line between two known points, as in Linear Interpolation), the Newton-Raphson method uses the tangent to the curve at a single starting point, $x_0$.

The slope of the tangent at $x_n$ is given by the derivative, $f'(x_n)$.

The next estimate, $x_{n+1}$, is where this tangent line crosses the x-axis.

4.2 The Newton-Raphson Formula

Using coordinate geometry (the point-slope form of a line), the formula for the next approximation is:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

This formula is essential and must be memorized.

4.3 The Newton-Raphson Process

Step 1: Define the Function and its Derivative
You need both $f(x)$ and $f'(x)$.

Step 2: Choose a Starting Value $x_0$
This is often given in the question, or you must find a sensible starting point near the root (using the Change of Sign rule or a sketch).

Step 3: Calculate the Next Estimate
Substitute $x_0$ into the formula to find $x_1$: $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$$

Step 4: Iterate
Use $x_1$ to find $x_2$, $x_2$ to find $x_3$, and so on, until the required accuracy is reached (usually when the approximations agree to the required number of decimal places).

4.4 Potential Problems and Failures

While incredibly fast (it has quadratic convergence, meaning the number of accurate decimal places roughly doubles each iteration!), Newton-Raphson is not foolproof.

Warning: When the Method Fails

Horizontal Tangent: If $f'(x_n) = 0$, the method fails because you are dividing by zero. This happens if your starting point is too close to a turning point (local maximum or minimum).
Divergence: If the starting value $x_0$ is far away from the root, the tangent might point to a completely different area, causing the estimates to jump further away from the root instead of closer (divergence).
Cycling: The iterations might jump back and forth between two values without converging to the root.

Did you know? The Newton-Raphson method is the core algorithm used in most calculators and computer programs when asked to "solve" a non-linear equation. Its speed makes it invaluable in practical engineering and physics.

Key Takeaway for Section 4

Newton-Raphson is the fastest method, relying on calculus ($f'(x)$). Its formula uses the tangent line. Always check that the derivative is not zero near your estimate!

Chapter Review: Comparing the Methods

You must understand the strengths and weaknesses of each technique.

5.1 Quick Comparison Table

Method	Principle Used	Reliability	Speed (Convergence)	Key Input
Interval Bisection	Change of Sign	Highest (Always works)	Slow (Linear)	$f(a), f(b)$ signs
Linear Interpolation	Straight Line (Chord)	High (Requires change of sign)	Medium	$f(a), f(b)$ magnitudes
Newton-Raphson	Tangent Line (Calculus)	Can fail (if $f'(x)=0$)	Very Fast (Quadratic)	$f(x), f'(x)$

5.2 Proving Accuracy to a Required Degree

When asked to show that a root $\alpha$ is correct to $k$ decimal places (d.p.), you must use the Change of Sign Rule on the boundaries of the possible range.

Example: Show a root is 3.45 (2 d.p.).

The true value must lie between 3.445 and 3.455.
Calculate $f(3.445)$.
Calculate $f(3.455)$.
If there is a change of sign between these two numbers, you have successfully "trapped" the root, proving the accuracy.

This final proof step is crucial regardless of which numerical method (Bisection, Interpolation, or Newton-Raphson) you used to find the estimate!

Keep practicing those iterations! Numerical methods are all about technique and careful calculation. You've got this!