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

18.6 Taylor and Maclaurin series 531
The total electricfield isthus
E T
=
Q
+4α 3 +5α 4 +···)
4πε 0
x 2[(1+2α+3α2
−(1−2α+3α 2 −4α 3 +5α 4 +···)]
E T
=
Q +···]
4πε 0
x 2[4α+8α3
Providing x is much larger than d, then d/(2x), that is α is small, and the higher
order terms in the series become increasingly small. Thus we can approximate the
field using only the first term,
E T
∼ =
Q
4πε 0
x 2[4α] =
Q
4πε 0
x 24 d 2x =
Qd
2πε 0
x 3
It can now be seen that the total electric field of the dipole decays asx −3 rather than
x −2 , as would be the case for a singlepoint charge.
EXERCISES18.6
1 Usethe Maclaurin series forsinx to write down the
Maclaurinseries forsin5x.
2 Usethe Maclaurin series forcosx to write down the
Maclaurinseries forcos3x.
3 Usethe Maclaurin series fore x to write down the
Maclaurinseries for 1 e x.
4 Find the Taylor seriesfory(x) = √ x aboutx = 1.
5 (a) Findthe Maclaurin seriesfor
y(x) =x 2 +sinx.
(b) Deduce the Maclaurin series for
y(x) =x n +sinx forany positiveintegern.
6 (a) Obtainthe Maclaurinseries fory(x) =xe x .
(b) Statethe range ofvalues ofxforwhichy(x)
equals itsMaclaurin series.
7 Find the Taylor series fory(x) =x +e x aboutx = 1.
8 Find the Maclaurin seriesfory(x) = ln(1 +x).
Solutions
1 5x− (5x)3
3!
2 1 − (3x)2
2!
+ (5x)5
5!
+ (3x)4
4!
3 1−x+ x2
2! − x3
3! + x4
4! −···
(x −1)2
8
4 1 + x −1 −
2
5(x −1)4
− +···
128
5 (a) x+x 2 − x3
3! + x5
5! − x7
−···
(b) x n +x− x3
3! + x5
5! − x7
7! +···
− (3x)6 +··· 6 (a) x+x 2 + x3
6!
2! + x4
3! + x5
4! +···
(b) Forallvalues ofx
7 p(x) = (1 +e)x
(x −1)3
{
+ (x −1)
16
2 (x −1)3
+e + +
2! 3!
7! +··· 8 x − x2
2 + x3
3 − x4
4 + x5
5 −···
(x −1)4
4!
+···
}