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Unformatted text preview: simplest way to go about this is to work backwards one year at a time, rst nding p 53 (or q 53 ) and using this to obtain 53 and d 53 , and so on. 2 2. Show that d dx t p x = t p x ( x x + t ). Answer: d dx t p x = d dx S x ( t ) = d dx S ( x + t ) S ( x ) = d dx S ( x + t )[ S ( x )]1 =S ( x + t )[ S ( x )]2 S ( x ) + [ S ( x )]1 S ( x + t ) =S ( x ) S ( x ) S ( x + t ) S ( x ) + S ( x + t ) S ( x ) = x t p x + S ( x + t ) S ( x ) = x t p x + S ( x + t ) S ( x + t ) S ( x + t ) S ( x ) = x t p xS ( x + t ) S ( x + t ) S ( x + t ) S ( x ) = x t p x x + t t p x = t p x ( x x + t ) 3...
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This note was uploaded on 01/27/2011 for the course ACTSC 232 taught by Professor Matthewtill during the Winter '08 term at Waterloo.
 Winter '08
 MATTHEWTILL

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