THE BINOMIAL SERIES 63

A

Problem 14. Simplify

√ (^3) (1− 3 x)√(1+x)
(
1 +
x
2
) 3
given that powers ofxabove the first may be
neglected.
√ (^3) (1− 3 x)√(1+x)
(
1 +
x
2
) 3
=(1− 3 x)
1
(^3) (1+x)
1
2
(
1 +
x
2
)− 3
≈
[
1 +
(
1
3
)
(− 3 x)
][
1 +
(
1
2
)
(x)
][
1 +(−3)
(x
2
)]
when expanded by the binomial theorem as far as
thexterm only,
=(1−x)
(
1 +
x
2
)(
1 −
3 x
2
)

(
1 −x+
x
2
−
3 x
2
)
when powers ofxhigher
than unity are neglected
=( 1 − 2 x)
Problem 15. Express
√
(1+ 2 x)
√ (^3) (1− 3 x) as a power
series as far as the term inx^2. State the range of
values ofxfor which the series is convergent.
√
(1+ 2 x)
√ (^3) (1− 3 x)=(1+ 2 x)
1
(^2) (1− 3 x)−
1
3
(1+ 2 x)
1
(^2) = 1 +
(
1
2
)
(2x)

• (1/2)(− 1 /2)
  2!
  (2x)^2 +···
  = 1 +x−
  x^2
  2
  +···which is valid for
  | 2 x|<1, i.e.|x|<
  1
  2
  (1− 3 x)−
  1
  (^3) = 1 +(− 1 /3)(− 3 x)


• 
  

  (− 1 /3)(− 4 /3)
  2!
  (− 3 x)^2 +···
  = 1 +x+ 2 x^2 +···which is valid for
  | 3 x|<1, i.e.|x|<
  1
  3
  Hence
  √
  (1+ 2 x)
  √ (^3) (1− 3 x)=(1+^2 x)
  1
  (^2) (1− 3 x)−
  1
  3

  
  

  (
  1 +x−
  x^2
  2
  +···
  )
  (1+x+ 2 x^2 +···)
  = 1 +x+ 2 x^2 +x+x^2 −
  x^2
  2
  ,
  neglecting terms of higher power than 2,
  = 1 + 2 x+
  5
  2
  x^2
  The series is convergent if−
  1
  3
  <x<
  1
  3
  Now try the following exercise.
  Exercise 34 Further problems on the bino-
  mial series
  In problems 1 to 5 expand in ascending powers
  ofxas far as the term inx^3 , using the binomial
  theorem. State in each case the limits ofxfor
  which the series is valid.

  
  

1. 1
   (1−x)
   [1+x+x^2 +x^3 +···,|x|<1]


2. 
   

   1
   (1+x)^2
   [1− 2 x+ 3 x^2 − 4 x^3 +···,|x|<1]

   
   


3. 
   

   1
   (2+x)^3
   ⎡
   ⎣
   1
   8
   (
   1 −
   3 x
   2

   
   

• 3 x^2
  2
  −
  5 x^3
  4
  +···
  )
  |x|< 2
  ⎤
  ⎦

4. √
   2 +x
   ⎡
   ⎣
   √
   2
   (
   1 +
   x
   4
   −
   x^2
   32

• x^3
  128
  −···
  )
  |x|< 2
  ⎤
  ⎦