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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.
tusaale · 0.1 + 0.2
tallaabo tallaabo
- 0.3 = \frac{3}{10}
Add: 0.1 + 0.2 = 3/10.
- \frac{3}{10} = 0.3
Simplify.
Jawaabta muuji
Haddaad Dooro dhibaato, ama qoraal ama sawirkaaga. Talaab kasta, sawir, jawaabta qarsoon ilaa aad weydiiso.
A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.
Ku soo biir Ku soo galCalaamadaha la isticmaalay halkan
Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.
Sida loo: Floating point and error
- Add: 0.1 + 0.2 = 3/10.
- Simplify.
Su'aalaha dadku waydiiyaan
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.
In ka badan Numerical Methods
Root finding: bisection and Newton's methodNumerical integration: trapezoid and SimpsonInterpolation and Taylor approximation