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