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Root finding: bisection and Newton's method
Finding where f(x) = 0 to any accuracy, and how fast each method gets there.
Bisection halves an interval that brackets a root: slow but unfailing. Newton follows the tangent line to the axis, quadratic convergence when it works. Picture it: each tangent line landing closer to the root; the points x₀, x₁, x₂ are marked on the graph. Think it: Newton is a fixed-point iteration of x − f/f′, and its speed comes from that map having zero derivative at the root.
ឧទាហរណ៍ដែលបានធ្វើ · newton's method on x^2 - 2 from 1
ជំហានដោយជំហាន
- f(x) = x^{2} - 2,\quad x_0 = 1
Newton's method: x_{n+1} = x_n − f(x_n)/f′(x_n). Follow the tangent line down to the axis, repeat.
- f'(x) = 2 x
Differentiate once; the same derivative serves every iteration.
- x_{1} = 1.0 - \frac{-1.0}{2.0} = 1.5
Iteration 1.
- x_{2} = 1.5 - \frac{0.25}{3.0} = 1.41666667
Iteration 2.
- x_{3} = 1.41666667 - \frac{0.00694444}{2.83333349} = 1.41421569
Iteration 3.
- x_{4} = 1.41421569 - \frac{6.01 \cdot 10^{-6}}{2.82843161} = 1.41421356
Iteration 4.
- x_{5} = 1.41421356 - \frac{0.0}{2.82842731} = 1.41421356
Iteration 5.
- x_{6} = 1.41421356 - \frac{0.0}{2.82842731} = 1.41421356
Iteration 6.
- \text{converged}
The iterates have stopped changing to 12 decimal places.
- x \approx 1.41421356
Quadratic convergence: the number of correct digits roughly doubles each step.
បង្ហាញចម្លើយ
ឥឡូវអ្នក ជ្រើសបញ្ហា ឬវាយបញ្ចូល ឬគូររបស់អ្នកផ្ទាល់ ។ ជំហាននីមួយៗ រូបភាព ចម្លើយត្រូវបានលាក់រហូតដល់អ្នកសួរ ។
គណនីឥតគិតថ្លៃបន្ថែមចំណាំលើមេរៀននីមួយៗ កំណត់ហេតុនៃអ្វីដែលអ្នកបានបញ្ចប់ បញ្ហាដែលអ្នកបានដោះស្រាយនៅកន្លែងមួយ និងគ្រូដែលអ្នកអាចសួរអំពីទំព័រនេះ ។ គណិតវិទ្យាដោយខ្លួនវាគឺបើកចំហសម្រាប់អ្នកទាំងអស់គ្នា ចុះឈ្មោះចូល ឬមិនចុះឈ្មោះចូល ។
ចុះឈ្មោះ ចូលនិមិត្តសញ្ញាដែលបានប្រើនៅទីនេះ
ចុចនិមិត្តសញ្ញាណាមួយសម្រាប់ការកំណត់ពេញលេញ រូបភាព និងតួអក្សរនីមួយៗមានន័យអ្វី ។
របៀប: Root finding: bisection and Newton's method
- Bracket the root: find a and b with f(a) and f(b) of opposite sign.
- For bisection, halve the interval keeping the sign change; each step gains one bit.
- For Newton, start inside the bracket and iterate x − f(x)/f′(x).
- Stop when successive iterates agree to the precision you need, and confirm f is tiny there.
សំណួរដែលមនុស្សសួរ
Why not just solve exactly?
Most equations have no closed-form solution at all, and many that do are unusable in practice. A numerical method delivers as many correct digits as you need, and a good one tells you how many that is.
Why can Newton's method fail?
If it starts where the tangent is nearly flat it shoots far away; near a repeated root it slows to a crawl; and with several roots it may land on the wrong one. A bracketing method like bisection is slower but cannot fail.
បន្ថែមទៀតក្នុង Numerical Methods
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