Honors Discrete Mathematics
Honors Discrete Mathematics MAD 2104
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This 4 page Class Notes was uploaded by Dino Corwin on Monday October 12, 2015. The Class Notes belongs to MAD 2104 at Florida Atlantic University taught by Jorge Viola-Prioli in Fall. Since its upload, it has received 12 views. For similar materials see /class/221637/mad-2104-florida-atlantic-university in Mathematics Discrete at Florida Atlantic University.
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Date Created: 10/12/15
25COMPOSITIONS OF FUNCTIONS Let fA a B and ng a C be given We can de ne a function from A to C this way for each a in A we take fa which is in B and apply g to fa landing therefore in C This function is called quotf followed by gquot watch the order or the composition of f and g and is denoted by g o fA a C Therefore g o f a gfa An important case occurs when fA a B is bijective In this case we know that the inverse function exists that is we havesz a B and f39lzB a A It follows from the de nition of f391 that f39lof identity of A and also fo f391 identity of B J Viola Prioli If more than two functions are given say fAe B gB a C and hC a D we de ne a function from A to D by composing take f rst followed by g and nally apply h A question emerges is h o g o f h o g o f It is a simple veri cation left as an exercise to show that they coincide The advantage of this result is that we can simply get rid of parenthesis because associativity applies Thus we simply write h o g o fA a D We emphasize that the M is crucial when taking compositions In many applications we face a function fA a A observe that the domain and the target space coincide In a situation like this we can take the composition of f with itself We denote it by f2 Remark this is nothing more than ffollowed by f and so f2 a ffa which is not fa2 For instance if fN a N is given by fx x 1 we have fx2 x12 whereas f2x ffx fx1 x1 1 x 2 J Viola Prioli By induction since associativity holds we can compose fwith itself n times so obtaining fl EXAMPLE Letsz 9 N be such that fa sum of rst and last digits of a Let us apply f repeatedly starting with a 809 We get 80991798916979149591091929498916 etc Observe that we enter into a loop EXAMPLE Let A n E N n gt 1 and de nesz 9 A by means of fa 1p where p is the largest prime divisor of a For instance f10 1 5 6 f6 13 4 etc We claim that for every a E A there exists n such that f a 3 and afterwards the sequence enters into a loop Before proving our assertion let us illustrate the result with a computation starting at 20 We get 20969493949394939 If we start with 29 we obtain 299309694939 Proof of our claim It is easily veri ed that our assertion holds if we start with 2 3 4 10 Please do it We will proceed by Strong Induction J Viola Prioli Assume the result holds for a 2 3 k 1 with k 1 gt 9 due to our remark above Let us prove it is true for a k In what follows capital letters denote the largest prime divisor of the corresponding number Case 1 k is prime say k p Then by applying fwe get fk fp p 1 an even number 50 p 1 quqz Q Thus fp 1 1QSincerp 12Q 15 p232lt p22lt p 1 since 6 lt p We conclude that Q 1 lt k 1 and so the claim is true for Q 1 Therefore wehaveakpep1eQ1e3 Case 2 k is composite so k plp2 P Thus fk 1 P As before we obtain PskZandsolPsk 1 The sequence turns out to be k9 1 P a 3 since the inductive hypothesis guarantees it NOTE There are several additional illustrations on line See Syllabus section 25 J Viola Prioli
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