Inverse Function - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are: Inverse Function, One-To-One Correspondence, Unique Element, Invertible Function, Codomain of Function, Non-Negative Real Number, Composition of Functions, Commutative Law, Identity Function, Graphs of Functions

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2012/2013

Uploaded on 04/27/2013

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CSE115/ENGR160 Discrete Mathematics
02/21/12
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CSE115/ENGR160 Discrete Mathematics 02/21/

2.3 Inverse function

  • Consider a one-to-one correspondence f from A to B
  • Since f is onto, every element of B is the image of some element in A
  • Since f is also one-to-one, every element of B is the image of a unique element of A
  • Thus, we can define a new function from B to A that reverses the correspondence given by f

One-to-one correspondence and

inverse function

  • If a function f is not one-to-one correspondence, cannot define an inverse function of f
  • A one-to-one correspondence is called invertible (^) 4

Example

  • f is a function from {a, b, c} to {1, 2, 3} with f(a)=2, f(b)=3, f(c)=1. Is it invertible? What is it its inverse?
  • Let f: Z→Z such that f(x)=x+1, Is f invertible? If so, what is its inverse? y=x+1, x=y-1, f-1(y)=y-
  • Let f: R→R with f(x)=x^2 , Is it invertible?
    • Since f(2)=f(-2)=4, f is not one-to-one, and so not invertible

Composition of functions

  • Let g be a function from A to B and f be a function from B to C, the composition of the functions f and g, denoted by f ◦ g, is defined by (f ◦ g)(a)=f(g(a)) - First apply g to a to obtain g(a) - Then apply f to g(a) to obtain (f ◦ g)(a)=f(g(a))

Composition of functions

  • Note f ◦ g cannot be defined unless the range of g is a subset of the domain of f

Example

  • f(x)=2x+3, g(x)=3x+2. What are f ◦ g and g ◦ f?
  • (f ◦ g)(x)=f(g(x))=f(3x+2)=2(3x+2)+3=6x+
  • (g ◦ f)(x)=g(f(x))=g(2x+3)=3(2x+3)+2=6x+
  • Note that f ◦ g and g ◦ f are defined in this example, but they are not equal
  • The commutative law does not hold for composition of functions

f and f -

  • f and f-1^ form an identity function in any order
  • Let f: A →B with f(a)=b
  • Suppose f is one-to-one correspondence from A to B
  • Then f-1^ is one-to-one correspondence from B to A
  • The inverse function reverses the correspondence of f, so f-1(b)=a when f(a)=b, and f(a)=b when f-1(b)=a
  • (f-1^ ◦f)(a)=f-1(f(a))=f-1(b)=a, and
  • (f ◦ f-1^ )(b)=f(f-1^ )(b))=f(a)=b

f f 11

f f A f f B A B

= = − −

− − 1 1

1 1 ( )

 ι ,  ι ,ι , ι areidentityfunctionsfoA andB

Example

13

floor : y =   x ceiling : y =   x

2.4 Sequences

  • Ordered list of elements
    • e.g., 1, 2, 3, 5, 8 is a sequence with 5 elements
    • 1, 3, 9, 27, 81, …, 30, …, is an infinite sequence
  • Sequence {a (^) n }: a function from a subset of the set of integers (usually either the set of {0, 1, 2, …} or the set {1, 2, 3, …}) to a set S
  • Use a (^) n to denote the image of the integer n
  • Call a (^) n a term of the sequence

Geometric progression

  • Geometric progression : a sequence of the form

a, ar, ar^2 , ar^3 ,…, arn where the initial term a and common ratio r are real numbers

  • Can be written as f(x)=a ∙ rx
  • The sequences {bn } with bn =(-1) n^ , {c (^) n } with c (^) n =2∙5n, {dn } with dn=6 ∙(1/3) n^ are geometric progression - bn : 1, - 1, 1, - 1, 1, … - c (^) n: 2, 10, 50, 250, 1250, … - dn: 6, 2, 2/3, 2/9, 2/27, …

Arithmetic progression

  • Arithmetic progression : a sequence of the form a, a+d, a+2d, …, a+nd where the initial term a and the common difference d are real numbers
  • Can be written as f(x)=a+dx
  • {sn } with s (^) n =-1+4n, {tn } with t (^) n=7-3n
    • {sn }: -1, 3, 7, 11, …
    • {tn}: 7, 4, 1, 02, …

Recurrence relation

  • Express a (^) n in terms of one or more of the previous terms of the sequence
  • Example: a (^) n =a (^) n-1+3 for n=1,2,3,… and a 1 =
    • a 2 =a 1 +3=2+3=5, a 3 =a 2 +3=(2+3)+3=2+3x2=8 , a 4 =a 3 +3=(2+3+3)+3=2+3+3+3=2+3x3=
    • a (^) n=2+3(n-1)
    • a (^) n =a (^) n-1+3=(an-2+3)+3=an-2+3x =(a (^) n-3+3)+3x2=an-3+3x =a 2 +3(n-2)=(a 1 +3)+3(n-2)=2+3(n-1)

Fibonacci sequence

  • • f 0 =0, f 1 =1, fn =fn-1+fn-2, for n=2, 3,
    • – f 2 =f 1 +f 0 =1+0=
    • – f 3 =f 2 +f 1 =1+1=
    • – f 4 =f 3 +f 2 =2+1=
    • – f 5 =f 4 +f 3 =3+2=
    • – f 6 =f 5 +f 4 =5+3=