446 Answers to Odd-Numbered Exercises
Finallyu(x)=(U−^3 /^2 +( 3 / 2 )
√
2 γ^2 / 5 x)−^2 /^3.
- 459.77 rad/s.
29.u(x)=C 0 e−ax.
31.w(x)= P
2 γ^2
[ 1
4
−x^2 +cosh(γx)−cosh(γ/^2 )
γsinh(γ/ 2 )
]
.
- Thesolutionbreaksdown(bucklingoccurs)iftan(λ/ 2 )=γ/2.
Chapter 1
Section 1.1
- a. 2
(
sin(x)−^1
2
sin( 2 x)+^1
3
sin( 3 x)−···
)
;
b.π
2
−^4
π
(
cos(x)+^1
9
cos( 3 x)+^1
25
cos( 5 x)+···
)
;
c.^1
2
+^2
π
(
sin(x)+^1
3
sin( 3 x)+^1
5
sin( 5 x)+···
)
;
d.^2
π
−^4
π
( 1
3
cos( 2 x)+^1
15
cos( 4 x)+^1
35
cos( 6 x)+···
)
.
3.f(x+p)= 1 =f(x)for anypand allx.
- Ifcis a multiple ofp,thegraphoff(x)betweencandc+pis the same
as that between 0 andp.Otherwise,letkbe the integer such thatkplies
betweencandc+p:
∫c+p
c
f(x)dx=
∫kp
c
f(x)dx+
∫c+p
kp
f(x)dx=
∫p
c∗
f(x)dx+
∫c∗
0
f(x)dx,
wherec∗=c−(k− 1 )p.
- a. cos^2 (x)=
1
2 +
1
2 cos(^2 x);
b. sin
(
x−π
6
)
=cos
(π
6
)
sin(x)−sin
(π
6
)
cos(x);
c. sin(x)cos( 2 x)=−^1
2
sin(x)+^1
2
sin( 3 x).
Section 1.2
- a.
1
2 −
4
π^2
[
cos(πx)+
1
9 cos(^3 πx)+
1
25 cos(^5 πx)+···