Higher Engineering Mathematics, Sixth Edition

64 Higher Engineering Mathematics

Now try the following exercise

Exercise 30 Further problems on the

binomial series

In problems1 to5 expand inascending powers ofx

as far as the term inx^3 , using the binomial theorem.

Statein each casethelimits ofxfor which theseries

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

⎤

⎦

5.

1

√

1 + 3 x

⎡

⎢

⎢

⎣

(

1 −

3

2

x+

27

8

x^2 −

135

16

x^3 +···

)

|x|<

1

3

⎤

⎥

⎥

⎦

6. Expand( 2 + 3 x)−^6 to three terms. For what
   values ofxis the expansion valid?
   ⎡
   ⎢
   ⎢
   ⎣

1

64

(

1 − 9 x+

189

4

x^2

)

|x|<

2

3

⎤

⎥

⎥

⎦

7. Whenxis very small show that:

(a)

1

( 1 −x)^2

√

( 1 −x)

≈ 1 +

5

2

x

(b)

( 1 − 2 x)

( 1 − 3 x)^4

≈ 1 + 10 x

(c)

√

1 + 5 x

√ (^31) − 2 x≈ 1 +
19
6
x

8. Ifxis very small such thatx^2 and higher pow-
   ers may be neglected, determine the power

series for

√

x+ 43

√

8 −x

√ (^5) ( 1 +x) 3
[
4 −
31
15
x
]

9. Express the following as power series in
   ascending powers ofxas far as the term in
   x^2. State in each case the range ofxfor which
   the series is valid.

(a)

√(

1 −x

1 +x

)

(b)

( 1 +x)^3

√

( 1 − 3 x)^2

√

( 1 +x^2 )

⎡

⎢

⎢

⎣

(a) 1−x+

1

2

x^2 ,|x|< 1

(b) 1−x−

7

2

x^2 ,|x|<

1

3

⎤

⎥

⎥

⎦

7.5 Practical problemsinvolving the

binomial theorem

Binomial expansionsmaybeusedfornumerical approx-

imations, for calculations with small variations and in

probability theory (see Chapter 57).

Problem 17. The radius of a cylinder is reduced

by 4% and its height is increased by 2%. Determine

the approximate percentage change in (a) its

volume and (b) its curved surface area, (neglecting

the products of small quantities).

Volume of cylinder=πr^2 h.

Let r and h be the original values of radius and

height.

The new values are 0.96ror( 1 − 0. 04 )rand 1.02hor

( 1 + 0. 02 )h.

(a) New volume=π[( 1 − 0. 04 )r]^2 [( 1 + 0. 02 )h]

=πr^2 h( 1 − 0. 04 )^2 ( 1 + 0. 02 )

Now( 1 − 0. 04 )^2 = 1 − 2 ( 0. 04 )+( 0. 04 )^2

=( 1 − 0. 08 ),

neglecting powers of small terms.