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

Arithmetic · Arithmetic

Evaluate 0.1 + 0.2

0.1 + 0.2

Korak po korak

  1. 0.3 = \frac{3}{10}

    Add: 0.1 + 0.2 = 3/10.

  2. \frac{3}{10} = 0.3

    Simplify.

Otkrij odgovor
0.3

Sada ti Izaberite problem, ili upišite ili nacrtajte svoj, svaki korak, slika, odgovor skriven dok ga ne pitate.

Neka tvoja radi.

Besplatni račun dodaje bilješke na svaku lekciju, zapis onoga što ste završili, vaše riješene probleme na jednom mjestu, i tutora kojeg možete pitati o ovoj stranici. Sama matematika je otvorena za svakoga, prijavljenog ili ne.

Prijavite se Prijava

Simboli korišteni ovdje

Dodirnite bilo koji simbol za potpunu definiciju, sliku i što svako slovo u njemu znači.

Kako: Floating point and error

  1. Add: 0.1 + 0.2 = 3/10.
  2. 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