fixpont-iteráció¶
y′=f(t,y) y(a)=b
y(t)=b+∫taf(s,y(s))ds
y→T(y), ahol T(y)(t)=b+∫taf(s,y(s))ds
yn→yn+1=T(yn), ahol T(yn)(t)=b+∫taf(s,yn(s))ds
f(t,y)=sin(t)−y, y0(t)=1
y1(t)=1+∫t0(sin(s)−1)ds=1−cos(t)+1−t=2−t−cos(t)
y2(t)=1+∫t0(sin(s)−2+s+cos(s))ds=2−2t+t22−cos(t)+sin(t)
y3(t)=1+∫t0(sin(s)−2+2s−s22+cos(s)−sin(s))ds=1−2t+t2−t36+sin(t)
y4(t)=1+∫t0(sin(s)−1+2s−s2+s36−sin(s))ds=1−t+t2−t33+t424