Geheugensteuntje - Wiki
Geheugensteuntje - Wiki
Geheugensteuntje - Wiki
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4) Stabiliteit:<br />
a) f(x) = 1 + 2x - 1/(1+x)<br />
Opmerkingen:<br />
1/(1+ε) = 1-ε<br />
Termen O(εi) (i > 1)<br />
verwaarlozen want zeer klein.<br />
eval1:<br />
y = [(1 + 2x(1 + ε1))(1+ ε2) - ([1/((1+x)(1+ε3))]*(1+ε4))](1+ε5)<br />
= [(1+2x+2xε1)(1+ε2) - 1/(1+x) - ε4/(1+x) + ε3/(1+x)] (1+ε5)<br />
= 1+2x -1/(1+x) + 2xε1 + ε2(1 + 2x) - ε4/(1+x) + ε3/(1+x) + ε5(1+2x-1/(1+x))<br />
= y + 2xε1 + ε2 (1+ 2x) + ε3(1/(1+x)) - ε4/(1+x) + ε5y<br />
|Δsy| = | y- y|<br />
= |-2xε1-ε2(1+2x)-ε3(1/(1+x)) + ε4/(1+x) -ε5y|<br />
≤ |-2xεmach - εmach(1+2x) -εmach(1/(1+x)) + εmach/(1+x) - εmachy|<br />
|δsy| = |(y - y)/y|<br />
= |ε1(2x/y) + ε2 (1+ 2x)/y + ε3(1/(y(1+x))) - ε4/(y(1+x)) + ε5 |<br />
≤ εmach [|2x/y| + |(1+2x)/y| + 2/(|1+x||y|) + 1]<br />
Nu,<br />
y = 1 + 2x - 1/(1+x)<br />
= (1+2x)(1+x)-1 (1+x)<br />
= 1 + 3x + 2x2 -1 (1+x)<br />
= x*(3+2x) (1+x)<br />
Dus<br />
|δsy| ≤ εmach [|(2(1+x))/(3+2x)| + |((1+2x)(1+x))/(x*(3+2x)) | + 2/|x*(3+2x)| + 1]<br />
Besluit:<br />
eval1 is instabiel voor x ≈ 0<br />
|δsy| >> |δky| (ligt aan volgorde)<br />
eval1 is (voorwaarts) stabiel voor x ≈ -3/2<br />
|δsy| ≈ |δky| (groot)<br />
(ligt aan probleem zelf)<br />
eval1 is stabiel voor x ≈ -1<br />
|δsy| ≤ 3εmach