NUMERICAL SOLUTIONS OF EQUATIONS
NUMERICAL SOLUTIONS OF EQUATIONS
Concise Theory & Formula Guide
Cambridge AS & A Level Mathematics
1. INTRODUCTION
Numerical methods find approximate solutions to equations that cannot be solved algebraically. Examples include x³ + x - 4 = 0, eˣ = 2x + 1, and sin(x) = x - 1.
Historical Fact: Quintic equations (degree 5 and higher) generally have no algebraic solutions, making numerical methods essential.
2. LOCATING ROOTS (Section 6.1)
Root Definition
Method 1: Graphical Approach
Rearrange equation as g(x) = h(x), sketch both graphs, and find intersection points. Each intersection represents a root.
Method 2: Change of Sign Method
Change of Sign Principle: If f(x) is continuous and f(a) · f(b) < 0, then there exists a root α in (a, b).
Problem: Show f(x) = x⁵ + x - 1 = 0 has a root between 0 and 1
Solution:
f(1) = 1 (positive)
Change of sign → Root exists in (0,1) ✓
⚠️ Conditions: Function must be continuous (no breaks). Method does NOT work for discontinuous functions like tan(x).
3. ITERATIVE METHODS (Section 6.2)
The Iterative Formula
Notation: x₁ = first guess, x₂ = second approximation, xₙ = current value, xₙ₊₁ = next value
Five-Step Process
- Rearrange: Express f(x) = 0 as x = F(x)
- Create formula: xₙ₊₁ = F(xₙ)
- Choose x₁: Select starting value near root
- Iterate: Calculate x₂, x₃, ... until convergence
- Verify: Use change of sign test
Common Rearrangements
| Original Equation | Rearrangement | Iterative Formula |
|---|---|---|
| x³ + x - 4 = 0 | x = ∛(4 - x) | xₙ₊₁ = ∛(4 - xₙ) |
| x² + x - 3 = 0 | x = 3/(x + 1) | xₙ₊₁ = 3/(xₙ + 1) |
| eˣ = 3x + 1 | x = ln(3x + 1) | xₙ₊₁ = ln(3xₙ + 1) |
| x³ - 5x + 2 = 0 | x = (x³ + 2)/5 | xₙ₊₁ = (xₙ³ + 2)/5 |
Problem: Find root of x² + x - 3 = 0 to 2 decimal places
Step 1: Locate root: f(-3) = 3, f(-2) = -1 → root in (-3, -2)
Step 2: Rearrange: x = 3/(x + 1)
Step 3: Formula: xₙ₊₁ = 3/(xₙ + 1), Start: x₁ = -2.5
| n | xₙ | Same to 2 d.p.? |
|---|---|---|
| 1 | -2.5 | - |
| 2 | -2.2 | No |
| 3 | -2.3636 | No |
| 4 | -2.2692 | No |
| 5 | -2.3220 | No |
| 6 | -2.2920 | No |
| 7 | -2.3089 | No |
| 8 | -2.2993 | No |
| 9 | -2.3047 | Yes! |
Answer: α = -2.30 (to 2 d.p.)
Verification: f(-2.305) = 0.00802 (positive), f(-2.295) = -0.0279 (negative) → Change of sign ✓
Calculator Tip
Using ANS Button:
1. Type starting value and press =
2. Type formula using ANS: 3/(ANS + 1)
3. Keep pressing = for automatic iteration
⚠️ Important: Not all rearrangements converge! Some will diverge. Always test or analyze before use.
4. APPLICATIONS (Section 6.3)
Problem: A circular segment and triangle have equal areas. Given angle θ in radians, show θ = 2sin(θ)
Solution:
Area of triangle = ½r²sin(θ)
Area of segment = ½r²θ - ½r²sin(θ)
Given: Area triangle = Area segment
½r²sin(θ) = ½r²θ - ½r²sin(θ)
2sin(θ) = θ → θ = 2sin(θ) ✓
Find θ to 3 s.f.: Use θₙ₊₁ = 2sin(θₙ), Start: θ₁ = 1.85
| n | θₙ | Same to 3 s.f.? |
|---|---|---|
| 1 | 1.85 | - |
| 2 | 1.92255 | No |
| 3 | 1.87753 | No |
| 4 | 1.90664 | No |
| ... | ... | ... |
| 11 | 1.89500 | Yes! |
Answer: θ = 1.90 radians (to 3 s.f.)
5. FORMULA REFERENCE
Core Formulas
If f(a) · f(b) < 0 and f continuous on [a,b]
→ Root exists in (a,b)
xn+1 = F(xn)
Lower bound: a = k - 0.5 × 10-n
Upper bound: b = k + 0.5 × 10-n
Check f(a) and f(b) have opposite signs
Accuracy Requirements
| Required Accuracy | Stop Condition |
|---|---|
| 1 significant figure | xₙ and xₙ₊₁ agree to 1 s.f. |
| 2 decimal places | xₙ and xₙ₊₁ agree to 2 d.p. |
| 3 significant figures | xₙ and xₙ₊₁ agree to 3 s.f. |
| 4 decimal places | xₙ and xₙ₊₁ agree to 4 d.p. |
6. KEY TERMS
Root: Value α where f(α) = 0
Iteration: One step in repetitive process
Convergence: Values approach true root
Divergence: Values move away from root (failure)
Change of Sign: Function changes from positive to negative
Continuous Function: No breaks or jumps in graph
7. ESSENTIAL TIPS
✓ DO:
- Always verify final answer with change of sign test
- Keep full calculator precision during iterations
- Round only at the final answer
- Check function is continuous before using change of sign
- Show all working clearly
✗ DON'T:
- Round intermediate values during iteration
- Stop before consecutive values agree
- Use change of sign on discontinuous functions
- Forget to verify your final answer
- Use inappropriate starting values
8. SUMMARY
Complete Process:
- Locate root using graphical or change of sign method
- Rearrange equation as x = F(x)
- Create iterative formula: xₙ₊₁ = F(xₙ)
- Choose starting value x₁ near root
- Iterate until consecutive values agree
- Verify using change of sign test
- State final answer with accuracy
🎓 END OF CONCISE NOTES 🎓
Master these techniques to solve equations that have challenged mathematicians for centuries!
