72 Higher Engineering Mathematics
Problem 11. Develop a series for sinhxusing
Maclaurin’s series.f(x)=sinhxf( 0 )=sinh0=e^0 −e−^0
2= 0f′(x)=coshxf′( 0 )=cosh0=e^0 +e−^0
2
= 1
f′′(x)=sinhxf′′( 0 )=sinh0= 0
f′′′(x)=coshxf′′′( 0 )=cosh0= 1
fiv(x)=sinhxfiv( 0 )=sinh0= 0
fv(x)=coshxfv( 0 )=cosh 0= 1
Substituting in equation (5) gives:sinhx=f( 0 )+xf′( 0 )+x^2
2!f′′( 0 )+x^3
3!f′′′( 0 )+x^4
4!fiv( 0 )+x^5
5!fv( 0 )+···= 0 +(x)( 1 )+x^2
2!( 0 )+x^3
3!( 1 )+x^4
4!( 0 )+x^5
5!( 1 )+···i.e.sinhx=x+x^3
3!+x^5
5!+···(as obtained in Section 5.5, page 49)Problem 12. Produce a power series for cos^22 x
as far as the term inx^6.From double angle formulae, cos2A=2cos^2 A−1(see
Chapter 17).from which, cos^2 A=1
2( 1 +cos2A)and cos^22 x=1
2( 1 +cos 4x)From Problem 1,cosx= 1 −x^2
2!+x^4
4!−x^6
6!+···hence cos4x= 1 −( 4 x)^2
2!+( 4 x)^4
4!−( 4 x)^6
6!+···= 1 − 8 x^2 +32
3x^4 −256
45x^6 +···Thus cos^22 x=1
2( 1 +cos 4x)=1
2(
1 + 1 − 8 x^2 +32
3x^4 −256
45x^6 +···)i.e.cos^22 x= 1 − 4 x^2 +16
3x^4 −128
45x^6 +···Now try the following exerciseExercise 32 Further problemson
Maclaurin’s series- Determine the first four terms of the power
series for sin2x⎡using Maclaurin’s series.
⎢
⎣sin2x= 2 x−4
3x^3 +4
15x^5−8
315x^7 +···⎤
⎥
⎦- Use Maclaurin’s series to produce a power
series for cosh 3xas far as the term inx^6.
[
1 +
9
2x^2 +27
8x^4 +81
80x^6]- Use Maclaurin’stheorem to determine the first
three terms of the power series for ln[ ( 1 +ex).
ln2+
x
2+x^2
8]- Determine the power series for cos 4tas far as
the term int^6.
[
1 − 8 t^2 +
32
3t^4 −256
45t^6]- Expand e
3
2 xin a power series as far as the terminx^3.[
1 +3
2x+9
8x^2 +9
16x^3]- Develop, as far as the term inx^4 , the power
series for sec2x.
[
1 + 2 x^2 +10
3x^4]- Expand e^2 θcos3θas far as the term inθ^2 using
Maclaurin’s series.
[
1 + 2 θ−5
2θ^2]- Determine thefirst threeterms of the series for
sin^2 xby applying Maclaurin’s theorem.
[
x^2 −
1
3x^4 +2
45x^6 ···]- UseMaclaurin’s series to determinetheexpan-
sion of( 3 + 2 t)^4.
[
81 + 216 t+ 216 t^2 + 96 t^3 + 16 t^4
]