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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.

Nhazi ahụ ejirila · 0.1 + 0.2

Arithmetic · Arithmetic

Evaluate 0.1 + 0.2

0.1 + 0.2

Nzọụkwụ site n'ụdị

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

    Add: 0.1 + 0.2 = 3/10.

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

    Simplify.

Gosi nzaghachi
0.3

Ugbu a ị Họrọ nsogbu, mọọbụ tinye mọọbụ depụta gị. Nzọụkwụ ọbụla, inyogo, azịza ahụ echekwara ruo mgbe ịjụla.

Kwụsị ọrụ gị

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Akaụntụ Nbanye

Aịkọn ndị ahụ ejirila ebe a

Tinye ọbụla simbụl maka nkọwa zuru ezu, inyogo, na ihe ọbụla ọbụla na-egosi.

Olee otú: Floating point and error

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

Ajụjụ ndị mmadụ jụrụ

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.

Oge Numerical Methods

Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonInterpolation and Taylor approximation