121The last two terms can be combined and represent an effective local error. Therefore,the global error is magnified if|(1−ha/2)/(1 +ha/2)|>1. Conversely, the methodis stable when1−ha/21 +ha/2<1,which holds for allh >0.CHALLENGE 20.7.Recall that Euler’s method isyn+1=yn+hf(tn, yn),and backward Euler isyn+1=yn+hf(tn+1, yn+1).P:y= 1 +.1(12) = 1.1,E:f= (1.1)2−.5 = 0.71,C:y= 1 +.1∗.71 = 1.071,E:f= (1.071)2−0.5.The predicted value is quite close to the corrected value; this is an indicationthat the stepsize is small enough to obtain some accuracy in the computed solution.CHALLENGE 20.8.f(t, y) = 10y2−20.P:yP=y(0) +.1f(0, y(0)) = 1 +.1(−10) = 0.E:fP=f(.1, yP) = 10
This is the end of the preview.
access the rest of the document.
Accuracy and precision, Method acting, eﬀective local error, ha/2