Chapter 1Plugging these values into the recursive equations above yieldst2=0.25 + 0.25 = 0.5andx2.25 + 0.25·[1 + 0.25·sin(0.0625)] = 0.503904.Continuing in this way givest3.75andx3.503904 + 0.25·[1 + 0.5·sin(0.251952)] = 0.785066,t4=1.00andφ(1)≈x4.13920.ForN=8,wehaveh/8=0.125 . Thus, the recursive formulas becometn+1=tn+0.125andxn+1=xn.125·[1 +tnsin(tnxn)].Using these formulas and starting witht0=0andx0= 0, we can ±ll in Table 1-A. From thiswe see thatφ(1)≈x8.19157, which is rounded to ±ve decimal places.Table 1–A: Euler’s method approximations for the solution ofx0=1+tsin(tx),x(0) = 0,att= 1 with 8 steps (
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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 University of California, Berkeley.