LAMC Advanced Circle May 21, Jason O Neill and Brad Dirks

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1 LAMC Advanced Circle May 21, 2017 Jason O Neill and Brad Dirks jasonmoneill9@g.ucla.edu, bdirks@ucla.edu Definition 1. Let σ S n be a permutation. We say i is a fixed point of σ if σ(i) = i. In other words, the two line notation of the permutation σ is a matrix which has a column with two i s in it. Definition 2. A permutation σ S n is called a derangement if σ has no fixed points. Example. σ 1 = [ ] σ 2 = [ ] We have that σ 1 S 4 is NOT a derangement, but σ 2 S 4 is a derangement. Problem 3 (Warm-Up). How many permuatations σ S n are derangements? Hint: Recall the Inclusion-Exclusion principle. We will answer this question through the next few problems. Problem 4. Calculate a formula for A B in terms the sets A, B, A B. Hint: Draw a Venn Diagram Problem 5. Calculate a formula for A B C in terms the sets A, B, C, A B, C B, B C, A B C. Hint: Draw a Venn Diagram 1

2 Problem 6 (The Inclusion-Exclusion Principle). Calculate a formula for A 1 A 2 A n in terms the sets A 1, A 2,..., A n, A i A j for some i, j,..., A 1 A 2 A n. Problem 7. How many permutations σ S n have 1 as a fixed point? What about 2 as a fixed point? k as a fixed point? Problem 8. How many permutations σ S n have 1 and 2 as fixed points? What about arbitrary distinct i, j as fixed points? Problem 9. How many permutations σ S n have 1, 2,..., k as fixed points? What about arbitrary distinct i 1, i 2..., i k as fixed points? 2

3 Definition 10. Let F i = {σ S n σ(i) = i} be the set of permutations with i as a fixed point. Problem 11. Using the above definition, describe what the set F 1 F 2 F n represents. Therefore, what does n! F 1 F 2 F n count? Problem 12. Let k be fixed. Calculate the following sum: i 1,i 2,...i k F i1 F i2 F ik (1) Hint: Consider Problem 9 and think about how many ways can we choose k distinct numbers i 1,..., i k out of the set {1, 2,..., n} Using your result from Problem 6, we have that n F 1 F 2 F n = ( 1) k+1 F i1 F i2 F ik (2) 1 i 1,i 2,...i k Problem 13. Use the formula for Inclusion-Exclusion above in (2) and the result from Problem 12 to calculate F 1 F 2 F n 3

4 Problem 14. Now, calculate the number of derangements by subtracting F 1 F 2 F n from n! and bring n! into the sum formula. Problem 15. Use the above result to calculate the number of derangements in S 6 as we first tried to do in the past handout. We now proceed to general properties of functions and an important operation on functions, known as composition. Definition 16. Let f : C D and g : D E be two functions. We define g f (read g of f ), the composition of g and f as the function g f : C E defined by the rule g f(c) = g(f(c)). Note that this composition makes sense as f(c) lies inside of D, by definition of f. Example. If f : R R is defined as f(x) = 2x and g : R R is defined as g(x) = x 5, then the composition f g(x) = f(g(x)) = f(x 5) = 2(x 5) = 2x 10. 4

5 Example. Now, let f : R R be defined as f(x) = x 2. Let g : R 0 R be defined as g(x) = x (this is defined for all x R 0, since if x 0, then x 0, so we can take its square root). Then f g(x) = f(g(x)) = f( x) = ( x) 2 = x, but the composition g f isn t defined. Why isn t it defined? Well, let x R. Then g f(x) = g(x 2 ) = x 2, but since x is a real number, x 2 0, so x 2 0, and thus the function is not defined (in the real numbers) for any x. So you must be careful when taking compositions as sometimes they won t be defined on the entire domain (in fact, g f is only defined at x = 0). Remark. The way we defined the composition g f is such that f is applied first and g is applied second. This is the opposite of how English is read, so be careful! Problem 17. Let f : {1, 2, 3, 4, 5} {1, 2, 3, 4, 5} be the function defined by the following rule: f(n) = n + 1 for n = 1, 2, 3, 4 and f(5) = 1 (Note: this is a permutation as defined last week). Let g : {1, 2, 3, 4, 5} R be defined as g(n) = n 2. First of all, does the composition g f make sense? If so, write it explicitly (as a rule or using a table). Does the composition f g make sense? If so, write it explicitly. 5

6 Example. We can compute the compositions of permutations quite easily (you will see later that the composition [ ] of two permutuations [ ] is a permutation). For example, let σ = and τ =. Now, we will (slowly) compute σ τ. After doing this with a couple of permutations you will become a natural. Well, first we will look at σ τ(1). From the 2-line notation for τ, we observe that τ(1) = 2, so this is the same thing as σ(2), which is 3. Thus, σ τ(1) = 3. We fill in the first column: [ ] σ τ =. 3 Now, σ τ(2) = σ(τ(2)), and we see τ(2) = 1, so this is the same as σ(1) = 1. We can fill in the second column: [ ] σ τ =. 3 1 Finally, we see σ τ(3) = σ(τ(3)) = σ(3) = 2, so finally we have [ ] σ τ = [ ] [ ] [ Problem 18. Let σ =, τ = and ρ = Calculate the compositions of the following pairs of permutations and write it in two-line notation: (Remember: function composition goes from right to left!) σ τ ]. τ σ 6

7 ρ τ σ ρ Definition 19. Let f : D C and g : D C be two functions. We say f = g if f(d) = g(d) for all elements d of D. Example. Using the notation of the example after Definition 19, notice that both compositions f g and g f make sense. This is because the domain of f and the codomain of g are equal, and the domain of g and the codomain of f are equal. If we calculate g f, we see g f(x) = g(f(x)) = g(2x) = 2x 5. This is not the same as f g(x) = 2x 10, as f g(1) = 2(1) 10 = 2 10 = 8 and g f(1) = 2(1) 5 = 2 5 = 3, which is not equal to 8. Definition 20. We say two functions commute with respect to composition if f g = g f. The previous example shows that not every pair of functions commute, and in fact it is quite rare for two functions to commute. Problem 21. For each of the pairs of functions below (each with domain R and codomain R) determine if they commute. In each part, write out explicit rules for f g and g f, and you suspect that these are not equal, give an example of a real number x such that f g(x) g f(x). f(x) = x 2 and g(x) = x + 1 7

8 h(x) = x + 5 and i(x) = x 3 j(x) = sin(x) and h(x) = x + 2π ϕ(x) = 5x and χ(x) = x 2 (For this part of the problem, we must change the domain of ψ to be R >0, but the codomain will remain R) ψ(x) = log 2 (x) and α(x) = 2 x. Remark. Note that in part e we had ψ α(x) = x and α ψ(x) = x. This was no coincidence. Recall from last week s packet that log 2 (x) was an example of a function which is both one-to-one and onto, also known as bijective. Definition 22. For any set S, there always exists a function id S : S S which we call the identity function on f, defined as id S (s) = s for all s inside of S. It is clear that if f : D C is a function, we have id C f = f id D = f. 8

9 Definition 23. A function f : D C which is bijective is also called invertible. We will justify this definition in the upcoming problems. An inverse for f is a function g : C D such that g f = id D and f g = id C. Example. Let f : R R be defined as 16x + 8. I claim f is bijective, and we will compute a function g such that g f = id R and f g = id R. First, we prove f is one-to-one. Assume x, y are two numbers in R such that f(x) = f(y). Then 16x+8 = 16y+8, meaning 16x = 16y by subtracting 8 from each side. Now, dividing both sides by 16 yields x = y, so indeed f is one-to-one. To see that f is onto, let y be a real number. We wish to find some x such that y = f(x) = 16x + 8. Well, we just solve this for x: x = y Then f(x) = 16( y 8 16 ) + 8 = y, so f is onto as y was arbitrarily chosen. So f is bijective. We wish to find a g : R R such that f g = id R. Well, then for any x, we want f(g(x)) = x. But by definition of f, the left hand side is 16g(x) + 8, so we want 16g(x) + 8 = x. Solving for g(x) yields g(x) = x You should check that f g = id R and g f = id R, proving g is an inverse of f. Problem 24. Calculate the inverse function of the following functions (all with domain and codomain equal to R): f(x) = 2x g(x) = x 2 9

10 h(x) = x 3 j(x) = x Remark. The last part of the previous problem hints at an important property of the identity function on any set S (we actually saw this in the previous handout). Namely, the identity function is always bijective, so it has an inverse. This inverse is always equal to the identity function itself. 10

11 Notice we actually haven t proved that if f is bijective then an inverse exists. That is the content of the next theorem: Theorem 25. Let f : D C be an invertible function. Then we can define an inverse of f, called g : C D. Start of Proof of Theorem 25 Let c be an element of C. Then, because f is onto, we can find some d c (the subscript c is just there to help us remember to which c the d maps) in D such that f(d) = c. Define g(c) = d c. Why is this a function? Well, if g(c) = d c and g(c) = d c for some d c which is not equal to d c, then by how we chose d c, we must have f(d c ) = c and f(d c) = c. But by f being one-to-one, we have d c = d c, and so g is indeed a function! Problem 26. Prove that the g defined in the proof above is an inverse of f (that is, prove g f = id D and f g = id C ). Problem 27. Let f : D C be an invertible function. Assume we can find two inverses of f, call them g 1 : C D and g 2 : C D (we know at least one exists by the previous theorem). Prove g 1 = g 2. (Hint: g 1 = g 1 id C and g 2 = id D g 2 or can prove by contradiction: assume there exists an x such that g 1 (x) g 2 (x) and proceed from there.) 11

12 Remark. By what was just proven, it makes sense to say that an inverse of f can be called the inverse of f. From now on, we will denote the inverse of f as f 1. Also, by definition of f 1, we must have that f and f 1 commute under composition. Definition 28. Let f : D C be a function. A function g : C D is called a left inverse of f if g f = id D. A function h : C D is called a right inverse of f if f h = id C. Remark. Remember the definitions of one-to-one and onto from last time. A function f : D C is one-to-one if f(x) = f(y) implies x = y. A function f : D C is onto if for any y C, we can find an x in D such that f(x) = y. Problem 29. Let f : D C be a function. Prove that if f has a left inverse g : C D, then f is one-to-one. Problem 30. Let f : D C be a function. Prove that if f has a right inverse g : C D, then f is onto. Remark. Note that if f is invertible then f 1 is simultaneously a left and a right inverse, and so we see that f being invertible means it is both one-toone and onto. Thus, it makes sense to say a function is bijective if and only if it is invertible. 12

13 Problem 31. Let f : D C and g : C D be two invertible functions. Prove f g is bijective. (Hint: Using the previous remark, you only need to give an example of an inverse function of f g.) Problem 32. (Warning: Harder than previous problems) Let f : D C and g : C B be two functions. Assume g f is one-to-one. Prove that f is one-to-one. Problem 33. (Warning: Don t overthink the problem) Let f : D C and g : E D be two functions. Assume f g is onto. Prove that f is onto. Remark. The permutations as defined in the last handout are bijective, thus invertible. Earlier you proved that the composition of two invertible functions is invertible, so we know that the composition of permutations is a permutation. 13

14 [ Example. Let σ = ]. We compute σ 1. Intuitively, σ just sends each number to itself plus one (and sends 4 to 1), so its inverse should send each number to itself minus 1 (and 1 to 4). We will show that this is indeed the case. Our definition of σ 1 will come from the fact that σ 1 σ = id {1,2,3,4}. Well, in order for σ 1 (σ(1)) = 1, we need σ 1 (4) = 1. Simiilarly, σ 1 (σ(2)) = 2, so σ 1 (3) = 2. In the same way we see that σ 1 (4) = 3 and σ 1 (1) = 4, so σ 1 = [ Problem 34. For the following permutations, give their inverses in two-line notation: [ ] σ = ]. τ = [ ] ρ = [ ] 14

15 Now we move on from invertible functions to an extremely fundamental property of function composition: associativity. Definition 35. Let f : D C, g : C B and h : B A be three functions. We know that g f : D B and h g : C A are functions, so it makes sense to look at (h g) f : D A and h (g f) : D A. Prove that (h g) f = h (g f). (Recall the definition of equality for functions) Definition 36. The above property of function composition is called associativity, you have probably heard this word before. Multiplication and addition of real numbers are associative operations as well. Thanks to the proof above, we can forget about the parentheses when we are writing h g f, as the order of operations don t matter. Problem 37. Can you think of any mathematical or real life operations which are not associative? Example. Here s one: exponentiation. Indeed, we have (2 2 ) 3 = 4 3 = 256 and 2 (23) = 2 8 = 4 4 = There exists a much simpler example, however. Theorem 38. Using induction, one can show that for any number of functions f 1 : S 1 S 2, f 2 : S 2 S 3,..., f n : S n S n+1, the composition f n f n 1... f 1 : S 1 S n+1 doesn t need parentheses (i.e. associativity extends to more than 3 functions). This is called the general associativity law. 15

16 Definition 39. A permutation σ S n is alternating or zig-zag or updown-up-down if σ(1) < σ(2), σ(2) > σ(3), σ(3) < σ(4), etc. Example. For instance, the permutation [ ] σ 1 = (3) Problem 40. How many alternating permutations σ are there in S 4. HINT: You can write them all down. Problem 41. How many alternating permutations σ are there in S 6. HINT: You probably can NOT write them all down. 16