082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

216 Chapter 6 Sequences and series
Wehaveassumedintheforegoingdiscussionthatnisapositiveintegerinwhichcase
the expansion given by Equation (6.3) will eventually terminate. In Example 6.17 this
wouldoccurwhenwereachedtheterminx 10 .Itcanbeshown,however, thatwhennis
not a positive integer the function (1 +x) n and the expansion given by
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +··· (6.4)
3!
havethesamevalueprovided −1 <x<1.However,whennisnotapositiveintegerthe
seriesdoesnotterminateandwemustdealwithaninfiniteseries.Thisseriesconverges
when −1 < x < 1 and the expansion is then said to be valid. Whenxlies outside this
intervaltheinfiniteseriesdivergesandsobearsnorelationtothevalueof (1 +x) n .The
expansion isthen saidtobe invalid.
Thebinomialtheorem:
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +···
3!
−1<x<1
Example6.18 Usethebinomialtheoremtoexpand
Solution
the terminx 3 .
1
1 +x inascendingpowersofxuptoandincluding
1
1 +x canbewrittenas (1 +x)−1 .UsingthebinomialtheoremgivenbyEquation(6.4)
withn = −1, wefind
(1 +x) −1 = 1 + (−1)x + (−1)(−2) x 2 + (−1)(−2)(−3) x 3 +···
2! 3!
=1−x+x 2 −x 3 +···
provided −1 <x<1.Consequently,ifinfutureapplicationswecomeacrosstheseries
1 −x+x 2 −x 3 + ..., we shall be able to write it in the form (1 +x) −1 . This closed
formavoidstheuseofaninfiniteseriesandsoitiseasiertohandle.Weshallmakeuse
of this technique inChapter 22 when wemeet theztransform.
Example6.19 Obtain a quadraticapproximation to (1 −2x) 1/2 usingthe binomialtheorem.
Solution Using Equation (6.4)withxreplaced by −2x andn = 1 2 wehave
( 1
(1 −2x) 1/2 = 1 + (−2x) +
2)
(1/2)(−1/2) (−2x) 2 + ···
2!
=1−x− 1 2 x2 +···
provided that −1 < −2x < 1, that is − 1 2 < x < 1 . A quadratic approximation is
2
therefore 1 −x− 1 2 x2 .