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

530 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Thetotalelectricfield,E T
,canbeobtainedbyaddingthetwocontributionstogether:
E T
=E LEFT
+E RIGHT
=
⎡
= Q ⎣
4πε 0
1
(
x − d 2
−Q Q
(
4πε 0
x + d ) 2
+ (
4πε
2 0
x − d ) 2
2
) 2
−
1
⎤
(
x + d ) 2 ⎦
2
It is possible to gain further insight into the properties of the electrostatic dipole by
carrying out a power series expansion. To prepare the equation we take a factor of
1/x 2 outside of the brackets
E T
=
⎡
Q ⎣
4πε 0
x 2
1
(
1 − d 2x
) 2
−
1
(
1 + d 2x
) 2
⎤
⎦
1
Now consider the Maclaurin series of
(1−α) 2.Following
the process explained in
the previous examples, the first five termsare calculated:
1
y(α) =
(1−α) 2, y(0)=1
y ′ 2!
(α) =
(1−α) 3, y′ (0)=2!
y ′′ 3!
(α) =
(1−α) 4, y′′ (0) =3!
y (3) 4!
(α) =
(1−α) 5, y(3) (0) =4!
y (4) 5!
(α) =
(1−α) 6, y(4) (0) =5!
1
So, = 1+2!α+3! α2
(1−α) 2 2! +4!α3 3! +5!α4 4! +··· = 1+2α+3α2 +4α 3 +
5α 4 +···.
Following a similarprocess for the Maclaurin series of 1/(1 + α) 2 ,
1
y(α) =
(1+α) 2, y(0)=1
y ′ (α) = −2!
(1+α) 3, y′ (0) = −2!
y ′′ 3!
(α) =
(1+α) 4, y′′ (0) =3!
y (3) (α) = −4!
(1+α) 5, y(3) (0) = −4!
y (4) 5!
(α) =
(1+α) 6, y(4) (0) =5!
1
weobtain =1+ (−2!)α +3!α2
(1+α)
2
2! + (−4!)α3 3! +5!α4 4! +···
=1−2α+3α 2 −4α 3 +5α 4 +···.
These two expansions may be used to approximate the total electric field by substituting
α =d/(2x).