1. Let L be Laplace transform operator. Find
L1{
15
}.
s  4 s  13
2
Solution.
We have
15
15
15
5i / 2
5i / 2
.




2
s  4 s  13 ( s  2)  9 ( s  2  3i )( s  2  3i ) s  2  3i s  2  3i
2
Let's use the formula

e
 st t
e dt  L(e t ) 
0
1
.
s 
Then
L1{
15
5i
}  {e ( 23i ) t  e ( 23i ) t }  5e 2 t sin( 3t ) .
2
s  4 s  13
2
2. The Bessel differential equation is given by
x 2 y' ' xy'( x 2  p 2 ) y  0 .
For p  2 verify that
sin( x )
.
y1 ( x ) 
x
Find a second linearly independent solution.
Solution
We have
y1 ' 
2 x cos( x )  sin( x )
 4 x cos( x )  4 x 2 sin( x )  3sin( x )
,
.
y
'
'

1
2x 3/ 2
4x5/ 2
So the left hand side of Bessel equation is
1
( 4 x cos( x )  4 x 2 sin( x )  3sin( x )  4 x cos( x ) 
1/ 2
4x
1
 2 sin( x )  4( x 2  p 2 ) sin( x ))  1 / 2 (1  4 p 2 ) sin( x ) .
4x
1
So this function ( y1 ) satisfy Bessel equation with p   (!).
2
x 2 y1 ' ' xy1 '( x 2  p 2 ) y1 
Let now find a second solution in the form
y( x)  y1 ( x) z( x) .
We have
y'  y1 ' z  y1 z' , y' '  y1 ' ' z  2 y1 ' z' y1 z' ' . So the left hand side of Bessel equation is
1
x 2 ( y1 ' ' z  2 y1 ' z ' y1 z ' ' )  x ( y1 ' z  y1 z ' )  ( x 2  ) y1 z
4
1
 z ( x 2 y1 ' ' xy1 '( x 2  ) y1 )  x 2 y1 z ' '( 2 x 2 y1 ' xy1 ) z '
4
2
2
 x y1 z' '(2 x y1 ' xy1 ) z'  x 3 / 2 (sin( x) z' '2 cos( x ) z' ) .
It is sufficient to find a partial solution of an equation
sin( x ) z ' '2 cos( x ) z '  0 .
We can take for instance
z
cos( x )
. Then we have a second solution of Bessel equation
sin( x )
cos( x )
.
x1/ 2
y2 ( x) 
3. Ten engineering schools in the US were surveyed. A sample contained 250 electrical engineers, 80
of which were women; 175 chemical engineers, 40 of which were women. Compute the 90%
confidence interval for the difference between the proportion of women in these two fields of
engineering. Is the difference significant?
Solution.
80
40
 0.32 , p 2^ 
 0.22857 . Further (this is an approximation)
250
175
0.32  0.68
0.22857  0.77143
p1^  N ( p1 ,
) , p2^  N ( p2 ,
) and
250
175
We have p1 
^
p1^  p 2^  N ( p1  p 2 , 0.00087  0.00101) . So we receive the 90%-confidence interval (the
table value is 1.645)
0.09143  0.0713 for p1  p2 and this difference is significant.
4. Two components of a microcomputer have the following pdf for their useful lifetimes X and Y:
f ( x, y )  xe x (1 y ) for x  0 , y  0 . Otherwise zero.
What is the probability that the lifetime X of the first component exceeds 3?
What are the marginal pdf's of X and Y? Are the two lifetimes independent?
What is the probability that the lifetime of at least one component exceeds 3?
Solution.
Let's find pdf for X (x>0):

f ( x )   xe x (1 y ) dy  xe x  (  x 1e  xy ) |0  e  x .
0
Analogously pdf for Y (y>0):


z  z dz
1
.
e

2
1

y
1

y
(
1

y
)
0
g ( y )   xe x (1 y ) dx  ( z  x(1  y ))  
0
Now

P( X  3)   e  x dx  e 3  0.049787 .
3
Obviously X and Y are dependent: joint pdf is not a product of marginal pdf's.
Let's find
3
3
P(max( X , Y )  3)  1  P(max( X , Y )  3)  1   xe (  e  xy dy )dx
0
x
0
3
 1   e  x (1  e 3 x )dx  e 3  (1  e 12 ) / 4  0.049787  0.249998  0.299785 .
0
5. Verify that
y  x  n J n (x )
is a particular solution of
xy' '(1  2n ) y ' xy  0 , x  0 .
Solution.
Bessel function
J n (x ) is (by definition) a solution of
1
n2
u' ' u'(1  2 )u  0 .
x
x
n
We can take u  x y and substitute this u into the previous equation. We receive
1
n2
LeftHandSide  n(n  1) x n 2 y  2nx n 1 y ' x n y ' ' (nx n 1 y  x n y ' )  (1  2 ) x n y
x
x
n
n 1
n
n 1
= x y  (2n  1) x y ' x y ' '  x ( xy  (2n  1) y ' xy' ' ) . So
xy  ( 2n  1) y ' xy' '  0 qed.
6. Show that
E[( X  a ) 2 ] is minimized at a  E[ X ] .
Solution.
E[( X  a ) 2 ]  E[(( X  E[ X ])  ( E[ X ]  a )) 2 ]  E[( X  E[ X ]) 2 ]  ( E[ X ]  a ) 2
 2( E[ X ]  a )  E[ X  E[ X ]]  E[( X  E[ X ]) 2 ]  ( E[ X ]  a ) 2  E[( X  E[ X ]) 2 ] .
7. Use the results from Chapter 8 to prove that
U (148)  U (190) . Chapter 8 was over external direct products.
Solution.
U (n ) is the multiplicative group of all invertible elements of the ring
Let n have a decomposition on prime factors:
n  p1k1  ...  prkr .
Z n  Z / nZ .
The general result is
U (n)  U ( p1k1 )  ...  U ( prkr ) .
Consequently
U (148)  U (4)  U (37), U (190)  U ( 2)  U (5)  U (19)  U (5)  U (19) .
Further
U ( 4)  {1, 3}  Z 2 , U ( p)  {1,..., p  1}  Z p1 ( p  5,19,37 ).
So
U (148)  Z 2  Z 36  Z 2  Z 4  Z 9
(another general result:
Z 36  Z 4  Z 9 ).
Analogously
U (190)  Z 4  Z18  Z 4  Z 2  Z 9
and
U (148)  U (190) .
8. The mean of a Poisson random variable X is
  9 . Compute the probability
P(   3  X    3 ) .
Solution.
We have for Poisson variable X
   2  9 , so
P(   3  X    3 )  P(0  X  18)  0.994557
(from the table).