1873_solutions

# Of this exercise using sierpinskis example prove that

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: equence E of this sequence is A n is expanding and since the union Ý E  An, n1 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 a1 b1 . ab3 1 B a  2, b  1  a1  Q2 We have ÞÞ x a y b d x, y  Q2 Þ0 Þ0 1 1x x a y b dydx  1 b1 Þ0 xa 1 1 1 B a  1, b  2  1 b1 b1 a1 b1 b1 1 b1 ab3  x b 1 dydx a1 b2 ab3 a1 b1  ab3 2. Prove that if a, b and c are nonnegative numbers then ÞÞÞ x a y b z c d x, y, z a1 b1 c1 . abc4  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 c1 dzdydx. c1 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 c1 1 u c1 1 x dudx  Q3 1 B a  1, b  c  3 B b  1, c  2 c1 a1 bc3 b1 c2 1 c1 abc4 bc3 a1 b1 c1  abc4  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 k1 are nonnegative numbers. Then 1 Þ x a1 x a2 x ak 1 x 1 x 2  x k ak11 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 k1 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.

Ask a homework question - tutors are online