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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.
Kugwira ntchito chitsanzo · 0.1 + 0.2
Gawo ndi Gawo
- 0.3 = \frac{3}{10}
Add: 0.1 + 0.2 = 3/10.
- \frac{3}{10} = 0.3
Simplify.
Kusonyeza yankho
Tsopano inu Sankhani vuto, kapena kulemba kapena kujambula wanu. Pakalipano, yankho limeneli lidzapezeka patsamba lino.
M’malo mwake, maakaunti aulere amawonjezera malemba pazophunzira zonse, mnda wa zomwe mwamaliza, mavuto anu omaliza m’malo limodzi, ndi m’bale amene mungafunse za nkhaniyo. Maphunziro a matekinoloje ndi otsegulira kwa aliyense, olembetsa kapena osalembetsa.
Kulembetsa KulowaZithunzi
Pitani pa dzina lililonse lachifaniziro kuti mudziwe tanthauzo lake, chithunzi chake, ndi zimene limatanthauza.
Momwe mungagwiritsire ntchito: Floating point and error
- Add: 0.1 + 0.2 = 3/10.
- Simplify.
Mafunso omwe anthu amafunsa
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.
Zambiri pa Numerical Methods
Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonInterpolation and Taylor approximation