This preview shows page 1. Sign up to view the full content.
Unformatted text preview: equence E
of this sequence is A n is expanding and since the union Ý E An,
n1 the desired result follows at once from Exercise 6. Some Further Exercises on Multiple Integrals
The exercises in this subsection are similar to those that appear above but make use of the gamma function
and beta function that were introduced earlier. They also refer to the standard simplex Q k in R k that is
defined to be the set of all points
Rk
x x 1 , x 2 , , x k
for which x j 0 for each j and
x 1 x 2 x k 1.
1. Prove that if a and b are nonnegative numbers then 410 ÞÞ x a y b d x, y a1 b1
.
ab3 1 B a 2, b 1
a1 Q2 We have ÞÞ x a y b d x, y Q2 Þ0 Þ0
1 1x x a y b dydx 1
b1 Þ0 xa 1
1 1 B a 1, b 2 1
b1
b1
a1 b1 b1
1
b1
ab3 x b 1 dydx a1 b2
ab3
a1 b1
ab3 2. Prove that if a, b and c are nonnegative numbers then ÞÞÞ x a y b z c d x, y, z a1 b1 c1
.
abc4 Q3 We have ÞÞÞ x a y b z c d x, y, z Q3 Þ0 Þ0 Þ0
1 1x 1xy x a y b z c dzdydx
1 Þ 1 Þ 1 x x a y b 1 x y c1 dzdydx.
c1 0 0
In the inside integral we make the substitution u y/ 1 x and we obtain
11
1
ÞÞÞ x a y b z c d x, y, z c 1 Þ 0 Þ 0 x a 1 x b u b 1 x c1 1 u c1 1 x dudx
Q3 1 B a 1, b c 3 B b 1, c 2
c1
a1 bc3 b1 c2
1
c1
abc4 bc3
a1 b1 c1
abc4
3. Given that k is a positive integer and given nonnegative numbers a 1 , a 2 , , a k , guess the value of the integral ÞQ a k a a x 1 1 x 2 2 x k k d x 1 , x 2 , , x k and then prove (by induction, perhaps) that your guess is correct.
We expect the equation
1 a 1
k
ÞQk x a1 x a2 x ak d x 1 , x 2 , , x k a 1a 1 a 2 2 a k ka 1
12
k
1
for every positive integer k. When k 2, the integral ÞQ a k a a x 1 1 x 2 2 x k k d x 1 , x 2 , , x k can be expressed as
1 x1 x2 xk 1 ÞQ Þ 0
k1 1
ak 1
and so the equation
ÞQ a k ÞQ
a a k1 a a a a x 1 1 x 2 2 x k k dx k d x 1 , x 2 , , x k
a x 1 1 x 2 2 x k k 11 1 a x1 x 1 1 x 2 2 x k k d x 1 , x 2 , , x k can be expressed as 411 x2 xk 1 1
a k 1 d x 1 , x 2 , , x k a1 1 a2 1 ak 1
a1 a2 ak k 1 1 a1 1 a2 1 ak 1
a
aa
1Þ
x 1 x 2 x k k 11 1 x 1 x 2 x k 1 a k 1 d x 1 , x 2 , , x k 1
.
a k 1 Qk 1 1 2
a1 a2 ak k 1
We write the assertion that this equation holds for any choice of nonnegative numbers a 1 , , a k as
p k . We have seen that the assertions p 1 and p 2 are true. Now suppose that k is any positive
integer for which the assertion p k is true and suppose that a 1 , , a k1 are nonnegative numbers.
Then
1
Þ x a1 x a2 x ak 1 x 1 x 2 x k ak11 d x 1 , x 2 , , x k
k
a k 1 1 Q k 1 2
1x x x
1
Þ Þ 1 2 k 1 x a1 x a2 x ak 1
12
k
a k 1 1 Q k 1 0
In the inside integral we make the substitution x1 u
and we obtain
1
Þ
a k 1 1 Q k
1 a k 1 1 Þ 0 x a x a x a 1 u a
12
k
1 1 Þ0 ua 1 2 1 k 1 u k1 a k1 1 k 1 x1 xk
x2 1 x1 x2 a a ÞQ du a k1 1
a k 1 1 1
a k 1 1 2 a k a k 1 a k 1 a k 1 2
a...
View
Full
Document
This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details