Explain the deretmaine l stc ksa2:26 AM83%Q5 = (1+4 · 5)(3)(21)(3) = (21)(20)(19) = 7980,(1.185)we would obtain the indicated wrong value, 7980, rather than the correctvalue, 4641. So, be careful!1.7.2 Example B49Prove the following (Newton's theorem): If fk is a polynomial of the nthdegree, then it can be written in the formfk = fo +1!Afo .(1) +A" fo 1(n).2!+...+n!(1.186)Assume that fk has the representationfk= ao + a1k() + a2k(²) + ·..+ ank(n).(1.187)where ao, a1, ..., an are constants. Differencing fk n times gives= a1 + 2a2k(1)+ 3azk(2)+ па,k(п-1),+...A? fk = 2.1. a2 + 3 · 2 · azk(1)+..+n(n – 1)ank(n-2).(1.188)A" fk = ann(n – 1) ... (1).-THE DIFFERENCE CALCULUS25Setting k= 0 in the original function and its differences allows us to concludethatAm foат0, 1,(1.189)m =...., n.т!To illustrate the use of this theorem, consider the functionfk = k4.(1.190)NowAfk = 4k° + 6k² + 4k + 1,A² fk = 12k? + 24k + 14,(1.191)A³ fk = 24k + 36,A* fk = 24,andfo = 0, Afo = 1,A² fo = 14, A³ fo = 36, A fo = 24.(1.192)Therefore, from equation (1.189),ао — 0,a1 = 1,a2 = 7,аз — 6,a41,and fkk4 has the factorial polynomial representationfk = k4 = k(1) + 7k(2) + 6k(3) + k(4).(1.194)

Name: Explain the deretmaine l stc ksa2:26 AM83%Q5 = (1+4 · 5)(3)(21)(3) = (
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: Explain the deretmaine l stc ksa2:26 AM83%Q5 = (1+4 · 5)(3)(21)(3) = (21)(20)(19) = 7980,(1.185)we would obtain the indicated wrong value, 7980, rather than the correctvalue, 4641. So, be careful!1.7.2 Example B49Prove the following (Newton's theorem