Higher Engineering Mathematics

68 NUMBER AND ALGEBRA

on; thus cosxmeets this condition. However,

iff(x)=lnx,f′(0)=^10 =∞, thus lnxdoes not

meet this condition.

(c)The resultant Maclaurin’s series must be

convergent

In general, this means that the values of the

terms, or groups of terms, must get progres-

sively smaller and the sum of the terms must

reach a limiting value.

For example, the series 1+^12 +^14 +^18 + ···is

convergent since the value of the terms is getting

smaller and the sum of the terms is approaching

a limiting value of 2.

8.4 Worked problems on Maclaurin’s

series

Problem 1. Determine the first four terms of

the power series for cosx.

The values of f(0), f′(0), f′′(0),... in the
Maclaurin’s series are obtained as follows:

f(x)=cosxf(0)=cos 0= 1

f′(x)=−sinxf′(0)=−sin 0= 0

f′′(x)=−cosxf′′(0)=−cos 0=− 1

f′′′(x)=sinxf′′′(0)=sin 0= 0

fiv(x)=cosxfiv(0)=cos 0= 1

fv(x)=−sinxfv(0)=−sin 0= 0

fvi(x)=−cosxfvi(0)=−cos 0=− 1

Substituting these values into equation (5) gives:

f(x)=cosx= 1 +x(0)+

x^2

2!

(−1)+

x^3

3!

(0)

+

x^4

4!

(1)+

x^5

5!

(0)+

x^6

6!

(−1)+···

i.e. cosx= 1 −

x^2

2!

+

x^4

4!

−

x^6

6!

+···

Problem 2. Determine the power series for

cos 2θ.

Replacingxwith 2θin the series obtained in Prob-

lem 1 gives:

cos 2θ= 1 −

(2θ)^2

2!

+

(2θ)^4

4!

−

(2θ)^6

6!

+···

= 1 −

4 θ^2

2

+

16 θ^4

24

−

64 θ^6

720

+···

i.e.cos 2θ= 1 − 2 θ^2 +

2

3

θ^4 −

4

45

θ^6 +···

Problem 3. Determine the power series for

tanxas far as the term inx^3.

f(x)=tanx

f(0)=tan 0= 0

f′(x)=sec^2 x

f′(0)=sec^20 =

1

cos^20

= 1

f′′(x)=(2 secx)( secxtanx)

=2 sec^2 xtanx

f′′(0)=2 sec^2 0 tan 0= 0

f′′′(x)=(2 sec^2 x)( sec^2 x)

+(tanx)(4 secxsecxtanx),

by the product rule,

=2 sec^4 x+4 sec^2 xtan^2 x

f′′′(0)=2 sec^40 +4 sec^2 0 tan^20 = 2

Substituting these values into equation (5) gives:

f(x)=tanx= 0 +(x)(1)+

x^2

2!

(0)+

x^3

3!

(2)

i.e. tanx=x+

1

3

x^3

Problem 4. Expand ln (1+x) to five terms.

f(x)=ln (1+x) f(0)=ln (1+0)= 0

f′(x)=

1

(1+x)

f′(0)=

1

1 + 0

= 1

f′′(x)=

− 1

(1+x)^2

f′′(0)=

− 1

(1+0)^2

=− 1

f′′′(x)=

2

(1+x)^3

f′′′(0)=

2

(1+0)^3

= 2