Answer: In this problem we establish that the local truncation error for the improved

Chapter 8, Problem 14

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In this problem we establish that the local truncation error for the improved Euler formula is proportional to h3. If we assume that the solution of the initial value problem y = f(t, y), y(t0) = y0 has derivatives that are continuous through the third order (f has continuous second partial derivatives), then it follows that (tn + h) = (tn) + (tn)h + (tn) 2! h2 + (tn) 3! h3 , where tn < tn < tn + h. Assume that yn = (tn). (a) Show that, for yn+1 as given by Eq. (5), en+1 = (tn+1) yn+1 = (tn)h {f[tn + h, yn + hf(tn, yn)] f(tn, yn)} 2! h + (tn)h3 3! . (i) (b) Making use of the facts that (t) = ft[t, (t)] + fy[t, (t)] (t) and that the Taylor approximation with a remainder for a function F(t, y) of two variables is F(a + h, b + k) = F(a, b) + Ft(a, b)h + Fy(a, b)k + 1 2! (h2 Ftt + 2hkFty + k2 Fyy) x= ,y= where lies between a and a + h and lies between b and b + k, show that the first term on the right side of Eq. (i) is proportional to h3 plus higher order terms. This is the desired result. (c) Show that if f(t, y) is linear in t and y, then en+1 = (tn)h3/6, where tn < tn < tn+1.Hint: What are ftt, fty, and fyy?