The binomial series 63

+

( 1 / 2 )(− 1 / 2 )(− 3 / 2 )

3!

(x

4

) 3

+···

]

= 2

(

1 +

x

8

−

x^2

128

+

x^3

1024

−···

)

= 2 +

x

4

−

x^2

64

+

x^3

512

−···

This is valid when

∣

∣

∣

x

4

∣

∣

∣<1,

i.e. |x|< 4 or− 4 <x< 4

Problem 14. Expand

1

√

( 1 − 2 t)

in ascending

powers oftas far as the term int^3.

State the limits oftfor which the expression

is valid.

1

√

( 1 − 2 t)

=( 1 − 2 t)−

1

2

= 1 +

(

−

1

2

)

(− 2 t)+

(− 1 / 2 )(− 3 / 2 )

2!

(− 2 t)^2

+

(− 1 / 2 )(− 3 / 2 )(− 5 / 2 )

3!

(− 2 t)^3 +···,

using the expansion for( 1 +x)n

= 1 +t+

3

2

t^2 +

5

2

t^3 +···

The expression is valid when| 2 t|<1,

i.e. |t|<

1

2

or−

1

2

<t<

1

2

Problem 15. 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 thex
term 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 16. 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
)
( 2 x)

• ( 1 / 2 )(− 1 / 2 )
  2!
  ( 2 x)^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