Assignment 6 Solution

We get 5 1 5 and 13 0 017610

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Unformatted text preview: £  £   £ 2 0 6 6 6 6 6 6 4 4 4 4  £ ¨ 0 4 4 4 4 2 2 2 2 2 2 £ ¡ ¢ 4( 2 2 0 0 0 0 0 0 0 0  £ £ 3( )= 2( )= 1(  !£ ¦ §¥ ¤¥ we get 5( © =1 =  £ £ ¡ £  £ 5 ( )= ) " and 13( 0 0176)+10( 0 1584) 11(0 9504)+9 7(0 1056) 12(0 0144) = #  # # #  #  #   # " which is in perfect agreement with question #1, as expected.  6) = 10 95808 # 4 (3 ENGM3052 Assignment #6: Solutions Fall 1999: pg 2 3) For the natural cubic spline, we have ( )( )( 0 3 =20 =20 =20 =20 1 ( 0 4 ( 0 2 )( ' ' ' ' &% &% &% &% $ $ $ $ = 0, we solve the following for ; B ( ( ' @B BD BF B G ( ( ( ' ' H H H HP H PP H ' ( 6 2(0 9) = 0 975 8 4 95+2(0 9) =09 3= 75 6 72 09 4 = 7 467 = = 1 1 2 = 6 3 45 54 G 1 ' ( 786 '( I I I ' ' 6 00 4 95 6 72 10 + 13) 11 + 10) 9 7 + 11) BF ' 9 ( ( ( ' ' @B ' E C BF BD B G 82 0 0 75 2 0 0 7 467 3 2( 3 2( 3 2( E 4 11 + 10) 9 7 + 11) 12 + 9 7) C = ' ' 1 A9 @B 1C 7 86 3 3 2( 3 2( 3 2( ' 2 E 1 345 ( ( 1 1 345 BD 1 4 (1) 2 7 5 (2) 12...
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This note was uploaded on 01/16/2014 for the course EGM 3052 taught by Professor Fenton during the Fall '09 term at Dalhousie.

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