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CIE A-Level Maths Study Notes

2.6.3 Iterative Solutions and Accuracy

Iterative methods are indispensable in solving equations where direct algebraic solutions are impractical. They use a function repeatedly to approximate a root, offering a practical approach to problem-solving across various mathematical domains.

Understanding Iterative Formulas

Iterative formulas generate a sequence approximating the root of an equation, with the general form xn+1=F(xn)x_{n+1} = F(x_n), where FF is derived from the original equation.

Key Concepts:

  • Iterative Formula: Recursive relation where each term is based on the previous one.
  • Convergence: Sequence values approach a specific value, the root.
  • Divergence: Sequence does not settle, move away, or oscillate.

Connecting Iterative Formulas to Equations:

Iterative formulas are tailored to the equations they solve, often derived by rearranging the original equation.

Example:

For the equation x22=0x^2 - 2 = 0, an iterative formula could be xn+1=2x_{n+1} = \sqrt{2}, derived from rearranging.

Solution:

  • Rearranging the Equation: Rearranged form: x=2x = \sqrt{2}.
  • Developing the Iterative Formula: Iterative formula: xn+1=2x_{n+1} = \sqrt{2}.
  • Using the Iterative Formula: Start with x0x_0, and use the formula for subsequent approximations.

Iteration Process:

1. Initial Guess: x0=1x_0 = 1.

2. First Iteration: x1=21.414x_1 = \sqrt{2} \approx 1.414.

3. Second Iteration: x2=21.414x_2 = \sqrt{2} \approx 1.414.

4. Third Iteration: x3=21.414x_3 = \sqrt{2} \approx 1.414.

5. Subsequent Iterations: Continue applying the formula, each yielding 21.414\sqrt{2} \approx 1.414, indicating convergence.

Conclusion: The iterative formula consistently approximates the square root of 2, demonstrating convergence to 2\sqrt{2}, validating the equation.

Techniques for Achieving Accuracy:

  • Initial Guess: Choose close to the root.
  • Iteration Count: Determine the needed number to reach the desired accuracy.
  • Error Estimation: Calculate the difference between successive iterations to gauge convergence.

Worked Example

Approximate the root of x22=0x^2 - 2 = 0 using the iterative method.

Solution:

1. Setting Up the Problem:

  • Solve x22=0x^2 - 2 = 0 using iteration. The root is 2\sqrt{2}.

2. Choosing an Initial Guess:

  • Initial guess x0=1x_0 = 1, near the actual root 2\sqrt{2}.

3. Iterative Method:

  • Use the Babylonian method: xn+1=12(xn+2xn)x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right).

4. Applying the Iterations:

  • Iteration 1: x1=1.5x_{1} = 1.5
  • Iteration 2: x21.4167x_{2} \approx 1.4167
  • Iteration 3: x31.4143x_{3} \approx 1.4143

5. Error Estimation:

  • Estimate error as xn+1xn|x_{n+1} - x_n|. For the third iteration, error 0.0024 \approx 0.0024.

6. Deciding When to Stop:

  • Continue until error < 10^{-4}. After three iterations, error is within the acceptable range.

Conclusion:

After three iterations, the root of x22=0x^2 - 2 = 0 is approximated as 1.4143\approx 1.4143, illustrating the effectiveness of iterative methods in approximating solutions to equations with desired 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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