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

528 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
1
Itcan beshown thatthisseries converges to for |x| < 1.Hence,
1 +x
1
∞∑
1 +x =1−x+x2 −x 3 +···= (−1) n x n for |x| < 1
0
1
Forvaluesofxoutside (−1,1)thevaluesof
1 +x and ∑ (−1) n x n aresimplynotequal;
try evaluating the l.h.s. and r.h.s.with,say,x = −2.
Example18.15 Findthe Taylor series fory(x) = e −x aboutx = 1.
Solution y=e −x , y(1)=e −1
y ′ = −e −x ,
y ′′ =e −x ,
and soon. Hence,
y ′ (1) = −e −1
y ′′ (1) =e −1
e −x = e −1 − (e −1 −1
(x −1)2
)(x −1) +e
2!
{
= e −1 (x −1)2
1 −(x−1)+ −
2!
∑ ∞
e −x = e −1 n
(x −1)n
(−1)
n!
0
−1
(x −1)3
−e
3!
}
(x −1)3
3!
+···
+···
Example18.16 Findthe Maclaurinseries fory(x) =xcosx.
Solution TheMaclaurinseries, p(x), forcosx is
cosx=p(x)=1− x2
2! + x4
4! − x6
6! +···
So the Maclaurin series forxcosx isxp(x), that is
xcosx=xp(x)=x− x3
2! + x5
4! − x7
6! +···
An alternative form of Taylor series is often used in numerical analysis. We know
that the Taylor series fory(x) generatedaboutx =aisgiven by
y(x) =y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a) +y (3) (x −a)3
(a) +···
2! 3!
Replacingabyx 0
we obtain
y(x) =y(x 0
) +y ′ (x 0
)(x −x 0
) +y ′′ (x 0
) (x−x 0 )2
2!
If we now letx−x 0
=h, wesee that
+y (3) (x 0
) (x−x 0 )3
3!
y(x 0
+h) =y(x 0
) +y ′ (x 0
)h +y ′′ (x 0
) h2
2! +y(3) (x 0
) h3
3! +···
To interpret thisformof Taylor series werefer toFigure 18.6.
+···