Root finding and numerical integration
Введено| English | Значение |
|---|---|
| iteration/ˌɪtəˈreɪʃn/ | итерация |
How reliable is an approximation?
- A nonlinear equation has no convenient exact solution. An approximation still needs evidence that it is reliable.
- This lesson studies iteration 迭代: A repeated update in which each new approximation is calculated from the previous one.
The method
- A continuous function with opposite signs at two endpoints has a root between them.
- Newton's method uses x next=x-f(x)/f prime(x), with a nonzero derivative.
- The trapezium rule approximates a definite integral using endpoint heights.
f(x)=1/x changes sign between -1 and 1. Why does this NOT prove a root in [-1,1]?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The intermediate value reasoning requires continuity on the whole interval. At the asymptote x=0 the function jumps from negative to positive with no root existing.
A worked case
For f(x)=x²-2 and x₀=1.5, Newton gives x₁=1.5-(2.25-2)/3=1.4166667. With y=x² on [0,2] and two equal strips, h=1 and trapezium area=(1/2)(0+2×1+4)=3; the exact area is 8/3.
Root finding and numerical integration
A continuous function with opposite signs at two endpoints has a root between them
Compare the model with the worked case and explain one change.
After one Newton step for x²-2 from x₀=1.5, find x₁. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
x₁=1.5-(1.5²-2)/(2×1.5)=17/12≈1.416667.
Where it goes wrong
- A sign change across a discontinuity does not prove a root.
- Iteration can diverge or cycle.
- The trapezium rule's overestimate or underestimate depends on curvature, not just whether the function increases.
Not true: Every sign change proves a root, including one across a discontinuity. A sign change shows a root only if the function is continuous there: 1/x changes sign across x=0 but has no root.
Use two strips for the trapezium estimate of integral x² from 0 to 2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
h=1 with heights 0,1,4 gives (1/2)(0+2×1+4)=3.
Every sign change proves a root, including one across a discontinuity.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A sign change shows a root only if the function is continuous there: 1/x changes sign across x=0 but has no root.
Applying it
- Give a stopping criterion and report sensible accuracy.
- Confirm the approximated root by a sign bracket around the stated rounded answer; explain any failure of the chosen iteration.
Find the exact integral of x² from 0 to 2. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The exact integral is [x³/3] from 0 to 2=8/3.
Match each numerical method to its update or rule.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Each method has its own conditions: bisection needs a continuous sign change, Newton needs a nonzero derivative, and the trapezium rule needs equal strip widths.
In your course
- AQA A-level Mathematics (7357), specification section I.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Iteration: A repeated update in which each new approximation is calculated from the previous one.
- A continuous function with opposite signs at two endpoints has a root between them.
- Watch for: A sign change across a discontinuity does not prove a root.