maths.free › Numerical Methods › Floating point and error
Floating point and error
Why 0.1 + 0.2 ≠ 0.3 on a computer, and how error propagates.
Computers store numbers in binary with ~16 significant digits; 0.1 is not representable exactly. Round-off accumulates, and subtracting nearly equal numbers destroys digits (cancellation). Picture it: 255 in binary is 11111111, and 0.1 in binary never terminates. Think it: a stable algorithm keeps error growth linear; conditioning is a property of the problem, stability of the method.
Radni primjer · 0.1 + 0.2
Korak po korak
- 0.3 = \frac{3}{10}
Add: 0.1 + 0.2 = 3/10.
- \frac{3}{10} = 0.3
Simplify.
Otkrij odgovor
Now you Izaberite problem, ili tipkanje ili nacrtati vlastiti. Svaki korak, slika, odgovor skriven dok ne pitate.
Besplatan račun dodaje bilješke na svakoj lekciji, zapis onoga što ste završili, rješeni problemi na jednom mjestu, i tutor možete pitati o ovoj stranici. Sama matematika je otvorena za sve, potpisani ili ne.
Prijavi se PrijavaSimboli koji se koriste ovdje
Dodirnite bilo koji simbol za punu definiciju, sliku, i što znači svako slovo u njoj.
Kako: Floating point and error
- Add: 0.1 + 0.2 = 3/10.
- Simplify.
Pitanja koja ljudi postavljaju
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.
Više u Numerical Methods
Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonInterpolation and Taylor approximation