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MIT6_042JS10_lec19_sol

MIT6_042JS10_lec19_sol - Massachusetts Institute of...

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Massachusetts Institute of Technology 6.042J/18.062J, Spring ’10 : Mathematics for Computer Science March 17 Prof. Albert R. Meyer revised March 15, 2010, 675 minutes Solutions to In-Class Problems Week 7, Wed. Problem 1. The Elementary 18.01 Functions ( F18 ’s) are the set of functions of one real variable defined recur- sively as follows: Base cases: • The identity function, id( x ) ::= x is an F18 , • any constant function is an F18 , • the sine function is an F18 , Constructor cases: If f, g are F18 ’s, then so are 1. f + g , fg , e g (the constant e ), 2. the inverse function f ( 1) , 3. the composition f g . (a) Prove that the function 1 /x is an F18 . Warning: Don’t confuse 1 /x = x 1 with the inverse, id ( 1) of the identity function id( x ) . The inverse id ( 1) is equal to id . Solution. log x is the inverse of e x so log x F18 . Therefore so is c log x for any constant c , and · hence e c log x = x c F18 . Now let c = 1 to get x 1 = 1 /x F18 . 1 (b) Prove by Structural Induction on this definition that the Elementary 18.01 Functions are closed under taking derivatives . That is, show that if f ( x ) is an F18 , then so is f ::= df/dx . (Just work out 2 or 3 of the most interesting constructor cases; you may skip the less interesting ones.) Solution. Proof. By Structural Induction on def of f F18 . The induction hypothesis is the above statement to be shown. Creative Commons 2010, Prof. Albert R. Meyer . 1 There’s a little problem here: since log x is not real-valued for x 0 , the function f ( x ) ::= 1 /x constructed in this way is only defined for x > 0 . To get an F18 equal to 1 /x defined for all x = 0 , use ( x/ | x | ) · f ( | x | ) , where | x | = x 2 .

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2 Solutions to In-Class Problems Week 7, Wed. Base Cases: We want to show that the derivatives of all the base case functions are in F18 . This is easy: for example, d id( x ) /dx = 1 is a constant function, and so is in F18 . Similarly, d sin( x ) /dx = cos( x ) which is also in F18 since cos( x ) = sin( x + π/ 2) F18 by rules for constant functions, the identity function, sum, and composition with sine. This proves that the induction hypothesis holds in the Base cases. Constructor Cases: ( f ( 1) ). Assume f, df/dx F18 to prove d f ( 1) ( x ) /dx F18 . Letting y = f ( x ) , so x = f ( 1) ( y ) , we know from Leibniz’s rule in calculus that 1 df ( 1) ( y ) /dy = dx/dy = . (1) dy/dx For example, d sin ( 1) ( y ) /dy = 1 / ( d sin( x ) /dx ) = 1 / cos( x ) = 1 / cos(sin ( 1) ( y )) . Stated as in ( 1 ), this rule is easy to remember, but can easily be misleading because of the variable switching between x and y . It’s more clearly stated using variable-free notation: ( f ( 1) ) = (1 /f ) f ( 1) . (2) Now, since f F18 (by assumption), so is 1 /f (by part ( a )) and f ( 1) (by constructor rule 2 .), and therefore so is their composition (by rule 3 ). Hence the righthand side of equation (
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MIT6_042JS10_lec19_sol - Massachusetts Institute of...

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