🔥 Numerical Solution of Equations (FP1) 🔥

Hello future mathematicians! Welcome to one of the most practical and fascinating chapters in Further Pure Mathematics. In this chapter, we step away from finding perfect, neat solutions and embrace the reality that most equations we encounter in the real world—like those governing physics, finance, or engineering—cannot be solved exactly using algebra alone.

Instead, we use Numerical Methods to find extremely accurate approximations. Think of it like a smart calculator that needs to zoom in closer and closer to the target answer. By the end of these notes, you’ll be able to locate roots and use iterative processes to find solutions with high precision!


1. Locating Roots: The Sign Change Method

Before we can find a root (a solution where \(f(x) = 0\)), we first need to know approximately where it is. The Sign Change Method is the foundation of numerical analysis.

What is a Root?

A root of the equation \(f(x) = 0\) is simply the \(x\)-value where the graph of \(y = f(x)\) crosses the \(x\)-axis.

The Principle of Sign Change

If you have a function \(f(x)\) that is continuous (meaning the graph has no breaks or jumps) and you evaluate it at two points, \(x = a\) and \(x = b\):

  • If \(f(a)\) is positive (+)
  • And \(f(b)\) is negative (-)
  • OR vice versa,

Then, the function must have crossed the \(x\)-axis at least once between \(a\) and \(b\). Therefore, a root exists in the interval \([a, b]\).

🧠 Analogy: The Mountain Range

Imagine the \(x\)-axis is sea level. If you are standing at point \(a\) (where \(f(a)\)) and you are above sea level (+), and you walk to point \(b\) and end up below sea level (-), you must have crossed sea level (the root) somewhere in between.

Step-by-Step: Using the Sign Change Method

  1. Define the Function: Ensure the equation is in the form \(f(x) = 0\).
  2. Choose the Interval: Use the given interval, say \([1, 2]\), or find small integer values where the function changes sign.
  3. Evaluate: Calculate \(f(a)\) and \(f(b)\).
  4. Conclude: State clearly whether a root exists.

Example: Show that \(e^x - 5x = 0\) has a root between \(x=0.2\) and \(x=0.3\).

Let \(f(x) = e^x - 5x\).
At \(x = 0.2\): \(f(0.2) = e^{0.2} - 5(0.2) \approx 1.2214 - 1.0 = 0.2214\) (Positive)
At \(x = 0.3\): \(f(0.3) = e^{0.3} - 5(0.3) \approx 1.3499 - 1.5 = -0.1501\) (Negative)

Since there is a change of sign and \(f(x)\) is continuous, a root lies between \(0.2\) and \(0.3\).

⚠️ Common Mistakes to Avoid
  • The Discontinuity Trap: The sign change method fails if the function is not continuous in the interval (e.g., if there is an asymptote). Always check if the function involves terms like \(\frac{1}{x}\) which might cause a break.
  • Missing Multiple Roots: If \(f(a)\) is positive and \(f(b)\) is also positive, it doesn't mean there are no roots; it just means there might be zero, two, four, or any even number of roots between \(a\) and \(b\).

Narrowing the Interval for Accuracy

The sign change method is also used to find a root to a specific number of decimal places (d.p.).

Step-by-Step: Finding a Root to 1 Decimal Place

If you know the root is between \(2.4\) and \(2.5\):

  1. We need to check the halfway point to see which side the root lies on. Check \(x = 2.45\).
  2. If \(f(2.45)\) is positive, the root must be between \(2.4\) and \(2.45\). The root, rounded to 1 d.p., is 2.4.
  3. If \(f(2.45)\) is negative, the root must be between \(2.45\) and \(2.5\). The root, rounded to 1 d.p., is 2.5.
  4. Crucial Tip: The final answer (e.g., 2.4) must be consistent with the sign change across the boundary value (e.g., 2.45).
🔑 Key Takeaway for Section 1

The Sign Change Method is a reliable tool for proving the existence of a root within a specific, tight interval, provided the function is continuous.


2. Fixed Point Iteration: \(x_{n+1} = g(x_n)\)

Locating a root tells us where it is, but Iterative Methods help us zoom in and find the value itself to many decimal places. This is often called Fixed Point Iteration in FP1.

The Iterative Formula

The core concept is to take the equation \(f(x) = 0\) and rearrange it into the form: $x = g(x)$

Once in this form, we create an iterative relationship:

$$x_{n+1} = g(x_n)$$

This means we take our current estimate (\(x_n\)), plug it into the function \(g\), and it gives us an improved, new estimate (\(x_{n+1}\)).

We start with an initial estimate, \(x_0\), and repeat the process until the values converge (stop changing) to the required accuracy.


The Importance of Rearrangement

Solving \(f(x) = 0\) can often be rearranged in many different ways, leading to different \(g(x)\) formulas. Only some of these will actually work (i.e., converge to the root).

Example: If we have the equation \(x^3 - 4x - 1 = 0\), here are two possible rearrangements:

Rearrangement A (Converges):
\(x^3 = 4x + 1\)
\(x = \sqrt[3]{4x + 1}\)
$$x_{n+1} = \sqrt[3]{4x_n + 1}$$

Rearrangement B (Diverges):
\(4x = x^3 - 1\)
\(x = \frac{x^3 - 1}{4}\)
$$x_{n+1} = \frac{x_n^3 - 1}{4}$$

The key challenge in FP1 is often choosing the rearrangement that leads to convergence.

Step-by-Step: Performing an Iteration

Let's use the formula \(x_{n+1} = \sqrt[3]{4x_n + 1}\) and an initial estimate \(x_0 = 2\).

  1. Start \(n=0\): Set \(x_0 = 2\).
  2. Find \(x_1\): \(x_1 = \sqrt[3]{4(2) + 1} = \sqrt[3]{9} \approx 2.080084\)
  3. Find \(x_2\): \(x_2 = \sqrt[3]{4(2.080084) + 1} \approx 2.100650\)
  4. Find \(x_3\): \(x_3 = \sqrt[3]{4(2.100650) + 1} \approx 2.105828\)
  5. ... Keep going until the required precision is met.

🔥 Calculator Trick (Essential for FP1):
To save time and avoid rounding errors, use your calculator's ANS (Answer) function:

  1. Type the starting value (\(x_0\)) and press =.
  2. Type the iteration formula, replacing \(x_n\) with ANS. (e.g., \(\sqrt[3]{4 \times \text{ANS} + 1}\))
  3. Press = repeatedly to generate \(x_1, x_2, x_3, \ldots\)
🔑 Key Takeaway for Section 2

Iteration turns solving an equation into a process of successive approximation: \(x_{n+1} = g(x_n)\). The crucial step is rearranging \(f(x)=0\) into a convergent \(x=g(x)\) form.


3. Visualizing Iteration and Convergence

Why do some rearrangements work while others fail? The answer lies in the graph of \(y = x\) and \(y = g(x)\).

The Fixed Point

The root \(\alpha\) is the point where \(x = g(x)\). Graphically, this is the intersection point of the line \(y = x\) and the curve \(y = g(x)\).

When we iterate, we are creating a path that moves us closer (or further away) from this intersection point.

Staircase and Cobweb Diagrams

We trace the iterative process on the graphs:

  1. Start at \(x_n\) on the \(x\)-axis.
  2. Move vertically to \(y = g(x)\) to find the value of \(g(x_n)\) (which is \(x_{n+1}\)).
  3. Move horizontally from \(y = g(x)\) to the line \(y = x\). This maps the value of \(x_{n+1}\) onto the \(x\)-axis for the next step.

Repeating this process creates a distinctive pattern:

Type 1: Staircase Convergence

The values approach the root steadily, like climbing a staircase. This happens when the slope of \(g(x)\) is positive and shallow near the root.

Type 2: Cobweb Convergence (Oscillating)

The values oscillate around the root, spiraling inwards like a cobweb. This happens when the slope of \(g(x)\) is negative and shallow near the root.

Type 3: Divergence (Fails)

The values move further and further away from the intersection point. This means the rearrangement chosen for \(g(x)\) will not find the root.


The Condition for Convergence (Advanced Insight)

For the iterative process \(x_{n+1} = g(x_n)\) to converge to a root \(\alpha\), the following condition must be met in the neighborhood of the root:

$$|\ g'(\alpha)\ | < 1$$

Where \(g'(x)\) is the derivative (gradient) of \(g(x)\) with respect to \(x\).

Simplifying the Concept: Gradient is Key!

The rule is: The gradient of \(y=g(x)\) must be shallower than the gradient of \(y=x\).

  • Since the gradient of \(y = x\) is exactly 1, the gradient of \(g(x)\) must be between \(-1\) and \(+1\).
  • If the gradient \(|g'(x)| > 1\), the steps in the iteration become larger, and the sequence diverges (moves away from the root).
  • If the gradient \(|g'(x)| < 1\), the steps become smaller, and the sequence converges (moves toward the root).

In examination questions, you will often be asked to differentiate \(g(x)\) and show that \(|g'(x)| < 1\) for a given interval to prove that the iteration will work.

Did you know? This method is sometimes called the Fixed Point Theorem, and it forms the basis for how computers solve complex non-linear systems!

🔑 Key Takeaway for Section 3

Iteration is visualized by tracing steps between \(y=g(x)\) and \(y=x\). Convergence occurs only if the gradient of \(g(x)\) near the root is between -1 and 1 (\(|g'(\alpha)| < 1\)).


Quick Review Box: Numerical Solution Fundamentals

Root Location (Sign Change):
  • Test \(f(a)\) and \(f(b)\). If signs differ, a root exists in \([a, b]\).
  • Must assume \(f(x)\) is continuous.
Iteration (\(x_{n+1} = g(x_n)\)):
  • Requires rearranging \(f(x)=0\) into a suitable \(x=g(x)\) form.
  • Use the ANS button on your calculator!
Convergence:
  • The iteration must trace towards the intersection of \(y=x\) and \(y=g(x)\).
  • Convergence is guaranteed if the magnitude of the gradient of \(g(x)\) is less than 1 near the root: \(|g'(x)| < 1\).