Numerical solution of equations
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| root/ruːt/ | Nghiệm |
| numerical method/njuːˈmerɪkl ˈmeθəd/ | Phương pháp số |
| accuracy/ˈækjʊrəsi/ | Độ chính xác |
| sign change/saɪn tʃeɪndʒ/ | Thay đổi dấu |
| iteration/ˌɪtəˈreɪʃn/ | Lặp lại |
| converge/kənˈvɜːdʒ/ | hội tụ |
| iterative formula/ˈɪtərətɪv ˈfɔːmjʊlə/ | công thức lặp |
| rearrangement/ˌriːəˈreɪndʒmənt/ | sắp xếp lại |
| diverge/daɪˈvɜːdʒ/ | phân kỳ |
The root 根 you can't see
- Some equations have no neat algebraic solution. $x^3 + x - 1 = 0$ can't be factorised.
- But you know a root exists between $0$ and $1$ because the function changes sign. Numerical methods 数值方法 let you find it to any accuracy 精度.
Locating a root
- Many equations can't be solved exactly. A root is a solution.
- Sign change 变号: if $f(a)$ and $f(b)$ have opposite signs (and no break between them), a root lies between $a$ and $b$.
Worked example. $f(x) = x^3 + x - 1$. $f(0) = -1$ (negative), $f(1) = 1$ (positive). Sign change → root between $0$ and $1$.
Sign change doesn't guarantee exactly one root. There could be 3 roots (or any odd number) between $a$ and $b$. The sign change only tells you there's at least one.
Iteration 迭代: step up to the curve and across to y = x; the steps converge 收敛 to a root
Where is the root?
y = ax³ + bx² + cx + d
A root is where the curve crosses zero. A sign change in f(x) traps a root between two x-values.
If f(a) and f(b) have opposite signs (and f is continuous between them), then between a and b there is:
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A sign change of a continuous function guarantees a root between a and b.
f(x) = x³ + x − 1. f(0) = −1 and f(1) = 1. Between which two integers is the root?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Sign change between 0 and 1 (f(0) < 0, f(1) > 0), so root is between 0 and 1.
Iteration
- Rearrange the equation into the form $x = F(x)$.
- Use the iterative formula 迭代公式 $x_{n+1} = F(x_n)$ from a first guess $x_0$.
- If the values settle, they converge to a root. Keep going until steady to the asked accuracy.
Iteration: starting from $x_0$, each step applies $F$ and reflects in $y = x$. The cobweb spirals into the fixed point — the root.
An iterative formula has the form:
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Iteration repeatedly applies x_{n+1} = F(x_n) from a first guess.
If the iteration values settle down to a steady number, they have converged to a root.
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Convergence means the sequence approaches a fixed value — a root of the equation.
Using x_{n+1} = ∛(1 − x_n) with x₀ = 0, what is x₁ (2 dp)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
x₁ = ∛(1 − 0) = ∛1 = 1.
Worked example — iteration
- Solve $x^3 + x - 1 = 0$ using $x_{n+1} = \sqrt[3]{1 - x_n}$, starting from $x_0 = 0.5$.
- $x_1 = \sqrt[3]{0.5} = 0.794$, $x_2 = \sqrt[3]{0.206} = 0.591$, $x_3 = \sqrt[3]{0.409} = 0.742$, ...
- The values converge to $x \approx 0.682$.
An iteration x_{n+1} = F(x_n) always converges regardless of the starting value.
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Convergence requires |F′(x)| < 1 near the root. Poor rearrangements or starting values can diverge.
Choosing a rearrangement 重新排列
- Not all rearrangements converge. The iteration $x_{n+1} = F(x_n)$ converges when $|F'(x)| < 1$ near the root.
- If it diverges 发散 (values move away), try a different rearrangement.
You've got it
- a sign change of $f$ between $a$ and $b$ traps a root in between
- iteration: $x_{n+1} = F(x_n)$ from a starting guess
- if the values converge (settle), they approach a root
- convergence requires $|F'(x)| < 1$ near the root
Write from a root between 1 and 2
- f(x) = x cubed minus 7.
- Use a sign change to locate a root.
For f(x) = x cubed minus 7, show there is a root between 1 and 2.
A continuous function that changes sign has a root in the interval.
Where is a root of x cubed minus 7?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
f(1) is negative and f(2) is positive, so the root is between them. Other options — Between 0 and 1 (the lesson interval); Exactly 2 (because f(2) is small).