Exercises 1.4x2=1.2+0.2=1.4,y2.400 + 0.2±1.4002+1.4001.2²≈1.960;x3.4+0..63.960 + 0.2±1.9602.9601.4²≈2.789;x4.6+0..84=2.789 + 0.2±2.7892+2.7891.6²≈4.110.7.For this problem notice that the independent variable istand the dependent variable isx.Hence, the recursive formulas given in equations (2) and (3) on page 25 of the text becometn+1=tn+handφ(tn+1)≈xn+1=xn+hf(tn,xn),n=0,1,2,... .For this problem,f(t, x)=1+tsin(tx),t0,andx0= 0. Thus the second recursive formulaabove becomesxn+1=xn+h[1 +tnsin(tnxn)],1,2For the caseN,wehaveh=(1−0)/1=1whichgivesust1=0+1=1andφ(1)≈x1=0+1·(1 + 0·sin 0) = 1.Nh/2=0.5 . Thus we havet1=0+0.5=0.51.5·(1 + 0·sin 0) = 0
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.