Higher Engineering Mathematics, Sixth Edition

70 Higher Engineering Mathematics

f(x)=sinxf( 0 )=sin0= 0

f′(x)=cosxf′( 0 )=cos0= 1

f′′(x)=−sinxf′′( 0 )=−sin0= 0

f′′′(x)=−cosxf′′′( 0 )=−cos0=− 1

fiv(x)=sinxfiv( 0 )=sin0= 0

fv(x)=cosxfv( 0 )=cos0= 1

fvi(x)=−sinxfvi( 0 )=−sin0= 0

fvii(x)=−cosxfvii( 0 )=−cos0=− 1

Substitutingthe above values into Maclaurin’s series of

equation (5) gives:

sinx= 0 +x( 1 )+

x^2

2!

( 0 )+

x^3

3!

(− 1 )+

x^4

4!

( 0 )

+

x^5

5!

( 1 )+

x^6

6!

( 0 )+

x^7

7!

(− 1 )+······

i.e.sinx=x−

x^3

3!

+

x^5

5!

−

x^7

7!

+ ···

Problem 4. Using Maclaurin’s series, find the

first five terms for the expansion of the function

f(x)=e^3 x.

f(x)=e^3 x f( 0 )=e^0 = 1

f′(x)=3e^3 x f′( 0 )=3e^0 = 3

f′′(x)=9e^3 x f′′( 0 )=9e^0 = 9

f′′′(x)=27e^3 x f′′′( 0 )=27e^0 = 27

fiv(x)=81e^3 x fiv( 0 )=81e^0 = 81

Substitutingthe above values into Maclaurin’s series of

equation (5) gives:

e^3 x= 1 +x( 3 )+

x^2

2!

( 9 )+

x^3

3!

( 27 )

+

x^4

4!

( 81 )+······

e^3 x= 1 + 3 x+

9 x^2

2!

+

27 x^3

3!

+

81 x^4

4!

+···

i.e. e^3 x= 1 + 3 x+

9 x^2

2

+

9 x^3

2

+

27 x^4

8

+···

Problem 5. Determine the power series for tanx

as far as the term inx^3.

f(x)=tanx

f( 0 )=tan0= 0

f′(x)=sec^2 x

f′( 0 )=sec^20 =

1

cos^20

= 1

f′′(x)=(2secx)(secxtanx)

=2sec^2 xtanx

f′′( 0 )=2sec^2 0tan0= 0

f′′′(x)=(2sec^2 x)(sec^2 x)

+(tanx)(4secxsecxtanx),by the

product rule,

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

f′′′( 0 )=2sec^40 +4sec^2 0tan^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 6. 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

fiv(x)=

− 6

( 1 +x)^4

fiv( 0 )=

− 6

( 1 + 0 )^4

=− 6

fv(x)=

24

( 1 +x)^5

fv( 0 )=

24

( 1 + 0 )^5

= 24