1873_solutions

# U n is an upper bound of the set var h h is

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Unformatted text preview: er 2 or 0. Thus Þ 0 s n d  2n 1 n j1 k1 p a kj 3j 1 2n The latter expression can be expressed in the form 2n 1 2n n n  k1 j 1 1 j 2 1 n  1n 2 n j 1 1 j p 1 a kj a kj 1 a kj 2  j pp j1 3 j2 3 3 n 1 3 j1 j 2 1 n 1 1 3 j2 3 jp j p 1 2n a kj 1 a kj 2 a kj p k1 and, in the special cases that follow, we shall write the number 2n 1 2n a kj 1 a kj 2 a kj p k1 as g j 1 , j 2 , j 3 , , j p . As will be explained below in the special cases, the value of the function g at any member j 1 , j 2 , j 3 , , j p of its domain depends upon the number of (distinct) members in the set j 1 , j 2 , j 3 , , j p . 299 1 j 1 , j 2 , j 3 , , j p has p members 2 if j 1 , j 2 , j 3 , , j p has p 1 members 4 if j 1 , j 2 , j 3 , , j p has p 2 members   2p g j 1 , j 2 , j 3 , , j p  if if  j 1 , j 2 , j 3 , , j p has only 1 member The Integral Þ xd x 1 0 When p  1 we have 2n Þ 0 s n d  1 n a kj 1 3j 2n k1 j1 2n n  j1 k1 Now for each j, exactly 2 n give us a kj  0. Therefore 1 Þ0 as n j1 1 3j a kj k1 1, 2, , 2 n give us a kj  2 and the other 2 n of the integers k 1 2n n a kj 1  1n 2 3j 2n 1 values of k n 13 2 1 3j s n d  j1 n 1 2 1 2 Ý, and we conclude that Þ 0 xd x  1. 2 2 2n 1 Some Special Cases The Integral Þ x 2 d x 1 0 When p  2 we have Þ 0 s n d  1 2n n j1 k1 n  1n 2 a kj 3j 1 2n k1 n j1 i1 a kj a ki 3j 3i 2n n j1 i1 n 1 2n 11 3i 3j a kj a ki k1 Now for each i, there are exactly 2 n 1 values of k for which a ki  2 and, whenever j 2 n 1 values of k give us a jk  2. Therefore n 1 2n 2n n j1 i1 11 3i 3j n n a kj a ki  j1 i1 k1 where g i, j  2 if i  j 1 if i and we conclude that 300 j 1 1 g i, j 3i 3j i, exactly 2 n 2 of these Þ 0 s n d  1 n n 1 1 g i, j 3i 3j j1 i1 n n n 11 3i 3j  j1 i1 13 2  n 1 3j j1 1 2 2 n 1 2 13 2 19 8 n 1 8 and we conclude that Þ 0 x 2 d x 1 Þ 0 s n d  1 nÝ lim 3. 8 The Integral Þ x 3 d x 1 0 When p  3 we have Þ 0 s n d  1 2n 3 n a kj 3j j1 k1 2n  1n 2 n n n n k1 r1 s1 t1 n  1n 2 1 2n n a kr a ks a kt 3r 3s 3t 2n n r1 s1 t1 n n  r1 s1 t1 1 3r3s3t a kr a ks a kt k1 1 g r, s, t 3r3s3t where 4 if r  s  t g r, s, t  2 if the set r, s, t has two members 1 if the set r, s, t has three members We observe that n n n n r1 s1 t1 1 g r, s, t  3r3s3t n  r1  13 2 3 n s1 n n 1 3 3r3s3t n 1 2 19 8 n r1 s1 t1 n 1 3r n 1 3s 13 2 1 8 n 1 3t t1 n r1 1 2 13 2 n 1 3r 3 13 2 n 1 2 1 2 and so Þ 0 x 3 d x 1 nÝ lim The Integral Þ x 4 d x 1 0 When p  4 we have 301 Þ 0 s n d  1 5. 16 n 1 3r 23t r1 t1 2 n t1 1 3t 4 2n k1 Þ 0 s n d  n j1 1 2n  1n 2 a kj 3j n 1 2n n n n a kr a ks a kt a ku 3r 3s 3t 3u k1 r1 s1 t1 u1 n  1n 2 n n 2n n 1 3r3s3t3u r1 s1 t1 u1 n n n n a kr a ks a kt a ku k1 1 g r, s, t, u 3r3s3t3u  r1 s1 t1 u1 where 8 if 4 if r, s, t, u has two members 2 if r, s, t, u has three members 1 if g r, s, t, u  r, s, t, u has one member r, s, t, u has four members We observe that n n n n 1 g r,...
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## 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.

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