標題:
limits
發問:
This exercise concerns the proof of Sterling's formulalim_(n->inf) [n!/ [ (n(1/2)) (n/e)n ]] = (2pi)(1/2)i) Definer_n = (n(1/2)) ((n/e)n) / n!.Express log[ r_(n+1) / r_n ] as simply as possible.ii) Verify that the following limit exists and calculate it.lim_(x->0) [[(1 + x/2)... 顯示更多 This exercise concerns the proof of Sterling's formula lim_(n->inf) [n!/ [ (n(1/2)) (n/e)n ]] = (2pi)(1/2) i) Define r_n = (n(1/2)) ((n/e)n) / n!. Express log[ r_(n+1) / r_n ] as simply as possible. ii) Verify that the following limit exists and calculate it. lim_(x->0) [[(1 + x/2) log(1+x) -x] / (x3)]. iii) Use your answers to parts (i) and (ii), and the comparison test, to show that Sum of_(n=1 to inf) log[(r_(n+1)) / (r_n)] converges. Deduce that lim_(n->inf) (r_n) exists.
最佳解答:
Stirlings formula i) log( r_(n 1) / r_n ) = -1 (n ?) log(1 1/n) ii) Applying lHopital Rule twice, we have lim_(x0) (((1 x/2) log(1 x) -x ) / (x3) ) = 1/12 iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have a_n = -1 (n ?) log(1 1/n) = n ( -1/n ( 1 (1/n)/2 ) log(1 1/n) ) Hence, a_n / (1/n2) = ( -1/n ( 1 (1/n)/2 ) log(1 1/n) ) / (1/n3) and then lim_(n) a_n / (1/n2) = 1/12 ... by (ii) ....(*) Since Σ(1/n2) converges absolutely, (*) implies that Σa_n converges absolutely by the comparison test. Notice that Σa_n = Σlog( r_(n 1) / r_n ) , n=1 to N = log( r_(N 1) / r_1 ) This just means that the limit lim_(n) log( r_(n 1) / r_1 ) exists and hence lim_(n) exp( log( r_(n 1) / r_1 ) ) = lim_(n) ( r_(n 1) / r_1 ) ) exists since exp( ) is a continuous function. This has just shown that lim_(n) r_n exists. 2008-03-09 12:12:34 補充: Stirlings formula i) log( r_(n 1) / r_n ) = -1 plus (n plus ?) log(1 plus 1/n) ii) Applying lHopital Rule twice, we have lim_(x0) (((1 plus x/2) log(1 plus x) -x ) / (x3) ) = 1/12 2008-03-09 12:14:50 補充: iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have a_n = -1 plus (n plus ?) log(1 plus 1/n) = n times ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) ) Hence, a_n / (1/n2) = ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) ) / (1/n3) and then lim_(n) a_n / (1/n2) = 1/12 ... by (ii) ....(*) 2008-03-09 12:17:17 補充: Notice that Σa_n = Σlog( r_(n plus 1) / r_n ) , n=1 to N = log( r_(N plus 1) / r_1 ) This just means that the limit lim_(n) log( r_(n plus 1) / r_1 ) exists and hence lim_(n) exp( log( r_(n plus 1) / r_1 ) ) = lim_(n) ( r_(n plus 1) / r_1 ) )
其他解答:
limits
發問:
This exercise concerns the proof of Sterling's formulalim_(n->inf) [n!/ [ (n(1/2)) (n/e)n ]] = (2pi)(1/2)i) Definer_n = (n(1/2)) ((n/e)n) / n!.Express log[ r_(n+1) / r_n ] as simply as possible.ii) Verify that the following limit exists and calculate it.lim_(x->0) [[(1 + x/2)... 顯示更多 This exercise concerns the proof of Sterling's formula lim_(n->inf) [n!/ [ (n(1/2)) (n/e)n ]] = (2pi)(1/2) i) Define r_n = (n(1/2)) ((n/e)n) / n!. Express log[ r_(n+1) / r_n ] as simply as possible. ii) Verify that the following limit exists and calculate it. lim_(x->0) [[(1 + x/2) log(1+x) -x] / (x3)]. iii) Use your answers to parts (i) and (ii), and the comparison test, to show that Sum of_(n=1 to inf) log[(r_(n+1)) / (r_n)] converges. Deduce that lim_(n->inf) (r_n) exists.
最佳解答:
Stirlings formula i) log( r_(n 1) / r_n ) = -1 (n ?) log(1 1/n) ii) Applying lHopital Rule twice, we have lim_(x0) (((1 x/2) log(1 x) -x ) / (x3) ) = 1/12 iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have a_n = -1 (n ?) log(1 1/n) = n ( -1/n ( 1 (1/n)/2 ) log(1 1/n) ) Hence, a_n / (1/n2) = ( -1/n ( 1 (1/n)/2 ) log(1 1/n) ) / (1/n3) and then lim_(n) a_n / (1/n2) = 1/12 ... by (ii) ....(*) Since Σ(1/n2) converges absolutely, (*) implies that Σa_n converges absolutely by the comparison test. Notice that Σa_n = Σlog( r_(n 1) / r_n ) , n=1 to N = log( r_(N 1) / r_1 ) This just means that the limit lim_(n) log( r_(n 1) / r_1 ) exists and hence lim_(n) exp( log( r_(n 1) / r_1 ) ) = lim_(n) ( r_(n 1) / r_1 ) ) exists since exp( ) is a continuous function. This has just shown that lim_(n) r_n exists. 2008-03-09 12:12:34 補充: Stirlings formula i) log( r_(n 1) / r_n ) = -1 plus (n plus ?) log(1 plus 1/n) ii) Applying lHopital Rule twice, we have lim_(x0) (((1 plus x/2) log(1 plus x) -x ) / (x3) ) = 1/12 2008-03-09 12:14:50 補充: iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have a_n = -1 plus (n plus ?) log(1 plus 1/n) = n times ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) ) Hence, a_n / (1/n2) = ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) ) / (1/n3) and then lim_(n) a_n / (1/n2) = 1/12 ... by (ii) ....(*) 2008-03-09 12:17:17 補充: Notice that Σa_n = Σlog( r_(n plus 1) / r_n ) , n=1 to N = log( r_(N plus 1) / r_1 ) This just means that the limit lim_(n) log( r_(n plus 1) / r_1 ) exists and hence lim_(n) exp( log( r_(n plus 1) / r_1 ) ) = lim_(n) ( r_(n plus 1) / r_1 ) )
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