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CIE A-Level Maths Study Notes

2.6.2 Convergence of Approximations

The concept of convergence is a cornerstone, particularly in the context of sequences and their application in approximating the roots of equations. This section explores the intricate process of using sequences to approximate roots, an essential aspect of advanced mathematical studies.

Introduction to Sequences and Convergence

Sequences, as ordered lists of numbers, play a pivotal role in various mathematical concepts, particularly in approximating the roots of equations. This section delves into how sequences can be utilized to estimate these roots, a critical aspect of advanced mathematical analysis.

Approximations

Image courtesy of ResearchGate

1. Defining Sequences:

  • Sequences are ordered lists of numbers, either finite or infinite.

2. Understanding Convergence:

  • A sequence converges if its terms approach a specific value, the limit.

3. Root Approximation with Sequences:

  • Root approximation finds values close to the actual root, often through iterative methods in numerical analysis.

4. Iterative Sequences:

  • Iterative formulas, xn+1=f(xn)x_{n+1} = f(x_n), use previous terms to generate the next, crucial for sequence convergence.

5. Example: Approximating a Root:

  • For x32x5=0x^3 - 2x - 5 = 0:
  • Rearrange for iteration: x=2x+53x = \sqrt[3]{2x + 5}.
  • Iterative formula: xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5}.
  • Start with x0=2x_0 = 2.
  • Iterate:
    • x1=932.0801x_1 = \sqrt[3]{9} \approx 2.0801.
    • x2=2×2.0801+532.0924x_2 = \sqrt[3]{2 \times 2.0801 + 5} \approx 2.0924.
    • x3=2×2.0924+532.0942x_3 = \sqrt[3]{2 \times 2.0924 + 5} \approx 2.0942.
  • Each term refines the approximation of the root.

Worked Example:

Approximate a root of x24x+3.5=0x^2 - 4x + 3.5 = 0 using an iterative sequence (Newton-Raphson Method).

Solution:

1. Initial Setup:

  • Function: f(x)=x24x+3.5f(x) = x^2 - 4x + 3.5
  • Derivative: f(x)=2x4f'(x) = 2x - 4
  • Initial guess x0=1x_0 = 1

2. Newton-Raphson Iterative Formula: xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

3. Perform Iterations (5 iterations):

  • Iteration 1:
    • x0=1x_0 = 1
    • f(x0)=0.5f(x_0) = 0.5
    • f(x0)=2f'(x_0) = -2
    • x1=1.25x_1 = 1.25
  • Iteration 2:
    • x1=1.25x_1 = 1.25
    • f(x1)0.0625f(x_1) \approx 0.0625
    • f(x1)=1.5f'(x_1) = -1.5
    • x21.292x_2 \approx 1.292
  • Iteration 3:
    • x21.292x_2 \approx 1.292
    • f(x2)0.0017f(x_2) \approx 0.0017
    • f(x2)1.416f'(x_2) \approx -1.416
    • x31.293x_3 \approx 1.293
  • Iteration 4:
    • x31.293x_3 \approx 1.293
    • f(x3)1.5×106f(x_3) \approx 1.5 \times 10^{-6}
    • f(x3)1.414f'(x_3) \approx -1.414
    • x41.293x_4 \approx 1.293
  • Iteration 5:
    • x41.293x_4 \approx 1.293
    • f(x4)1.1×1012f(x_4) \approx 1.1 \times 10^{-12} (very close to zero)
    • f(x4)1.414f'(x_4) \approx -1.414
    • x51.293x_5 \approx 1.293

4. Conclusion:

The value of x x converges to approximately 1.293, indicating a root of the equation. The function value f(x)f(x) approaches zero, confirming the convergence.

5. Analysis of Results:

No divergence observed; the sequence converges to the root 1.293. The function value f(x)f(x) nears zero, validating the root's accuracy.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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