Yk+1 = f(a) + f^(a)(uk)", (2.143) and 1 Uk+1 = n! (2.144) This last equation has the solution Ant Uk |A| < 1, (2.145) [f® (a)/n!]1/(n=1)* where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 t = A"", B1 = (2.146) [f(m)(a)/n!]l/(n-1)' and assume that Yk = z(t) = a+Bịt+B2t2 +.... (2.147) Note that t" = An(*+1) (2.148) %3D Therefore, from yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149) On substituting the expansion of equation (2.147) into this expression, we obtain the result 1 a + Bit" + B2ť²n + ...= f(a) +f (a)(Bịt+ B2ť² %3D n! 1 + B3t° + ...)" + f(a+D(a)(B¡t (2.150) (n + 1)!" + B2t? + B3t³ + ...)"+1 + ....
Yk+1 = f(a) + f^(a)(uk)", (2.143) and 1 Uk+1 = n! (2.144) This last equation has the solution Ant Uk |A| < 1, (2.145) [f® (a)/n!]1/(n=1)* where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 t = A"", B1 = (2.146) [f(m)(a)/n!]l/(n-1)' and assume that Yk = z(t) = a+Bịt+B2t2 +.... (2.147) Note that t" = An(*+1) (2.148) %3D Therefore, from yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149) On substituting the expansion of equation (2.147) into this expression, we obtain the result 1 a + Bit" + B2ť²n + ...= f(a) +f (a)(Bịt+ B2ť² %3D n! 1 + B3t° + ...)" + f(a+D(a)(B¡t (2.150) (n + 1)!" + B2t? + B3t³ + ...)"+1 + ....
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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Explain this and the theorem of part 2 it is from equation2.150
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