By Martin Gardner, Silvanus P. Thompson
Calculus Made effortless has lengthy been the preferred calculus primer, and this significant revision of the vintage math textual content makes the topic to hand nonetheless extra understandable to readers of all degrees. With a brand new advent, 3 new chapters, modernized language and techniques all through, and an appendix of hard and relaxing perform difficulties, Calculus Made effortless has been completely up-to-date for the fashionable reader.
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Additional resources for Calculus Made Easy
If x0 (t) = f (t, x(t)), show that x00 (t) = ft (t, x) + f (t, x)fx (t, x), where ft and fx are the partial derivatives of f (t, x). 10.??? Prove that ex 1+x+ 1 2 2! x + ··· + 1 p p! x for all x 0. 11.??? 4) of x(tn +h) may be obtained by repeated di↵erentiation of the ODE: x0 (tn ) = x(tn ) + g(tn ), x00 (tn ) = x0 (tn ) + g 0 (tn ), .. x(p) (tn ) = x(p 1) (tn ) + g(tn ). 42 3. 5) for the same IVP satisfies en+1 = r( h)en + Tn+1 , n = 0, 1, 2, . . , 1 2 1 p where r(s) = 1 + s + 2! s + · · · + p!
E. what is needed for methods to have order at least p = 1. 14) we find (collecting terms appropriately) ⇥ ⇤ Lh z(t) = 1 + ↵1 + ↵0 z(t) + h 2 + ↵1 ( 2 + 1 + 0 ) z 0 (t) + O(h2 ). 15) 0. These conditions can be written more concisely if we introduce two polynomials. 16) respectively. 2 For order p all terms up to those containing hp+1 must be retained but, since the -terms are already multiplied by h, the expansions of the z 0 terms need only include terms up to hp . 52 4. 15) in terms of the characteristic polynomials.
1 for p = 1, 2, 3? 01? ) Comment on your answer. 8.?? For the IVP u0 (t) = v(t), v 0 (t) = u(0) = 1, u(t), t > 0, v(0) = 0, use the chain rule to di↵erentiate u2 (t) + v 2 (t) with respect to t. Hence prove that u2 (t) + v 2 (t) = 1 for all t 0. Use the TS(2) method to derive a means of computing un+1 and vn+1 in terms of un and vn . Prove that this TS(2) approximation satisfies u2n + vn2 = (1 + 14 h4 )n when u0 = 1 and v0 = 0. 9.? If x0 (t) = f (t, x(t)), show that x00 (t) = ft (t, x) + f (t, x)fx (t, x), where ft and fx are the partial derivatives of f (t, x).