Composition

3 Key excerpts on "Composition"

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  Discrete Mathematics

  Mathematical Reasoning and Proof with Puzzles, Patterns, and Games

  • Douglas E. Ensley, J. Winston Crawley(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  27. In your own words, explain how you can tell from a list of ordered pairs whether the relation with that list as its rule is a function. 28. Explain how you can tell from a list of ordered pairs whether the inverse of the relation with that list as its rule is a function. 4.2 The Composition Operation The idea of Composition of functions and relations is a very basic one. Mathematics is largely about how complex concepts, structures, and properties can be built logically out of simpler ones. Since functions and relations are fundamental structures in mathematics, it stands to reason that combinations of two or more of these structures could be important. Composition of Functions Before being swept away by a formal definition, let us consider an example of how Composition naturally arises in the English language. One rule that relates pairs of people is the “husband of” relation. Another rule of this type is the “mother of” relation. These relations can be combined to give two distinct meanings: • The relation “mother of the husband of” associates a woman with her mother-in- law. • The relation “husband of the mother of” associates any person with his or her father or stepfather. The preposition “of” naturally ties together English clauses in the same way that Composition ties together mathematical functions, as we see in our formal definition. Definition If f : A → B and g : B → C , then we can build a new function called (g ◦ f ) that has domain A and codomain C , and that follows the rule (g ◦ f )(x) = g( f (x)). We call (g ◦ f ), read “g of f ,” the Composition of g with f . 4.2 The Composition Operation 269 The double of the square root g f Description order for (g ° f ) g ( f ( x )) Second First Evaluation order for (g ° f ) Figure 4-22 How to read (g ◦ f ). Example 1 Given the function f : R ≥0 → R defined by the rule f (x) = √ x, and the function g : R → R defined by the rule g( y) = 2 · y, describe the domain, codomain, and rule for the function (g ◦ f ).

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  A Bridge to Higher Mathematics

  • Valentin Deaconu, Donald C. Pfaff(Authors)
  • 2016(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  62 A bridge to higher mathematics 4.5 Composition and inverse functions Definition 4.41. Suppose f and g are functions. We define a new function f ◦ g called the Composition of f and g with domain {x ∈ dom(g) : g(x) ∈ dom(f )} such that (f ◦ g)(x) = f (g(x)). The domain of f ◦ g may be the empty set, in which case f ◦ g is the empty function (not so interesting). To avoid this situation, many times we assume that ran(g)=dom(f ). If f, g : X → X , then both functions f ◦ g and g ◦ f may be defined. In general, f ◦ g = g ◦ f . Example 4.42. Let f : [0, 4] → [0, 2], f (x) = √ x and let g : R → [−1, ∞), g(x) = x 2 − 1. Then f ◦ g : [− √ 5, −1] ∪ [1, √ 5] → [0, 2], (f ◦ g)(x) =  x 2 − 1 and g ◦ f : [0, 4] → [−1, 3], (g ◦ f )(x) = x − 1. Example 4.43. The function h : (0, π/2) → R, h(x) = ln(tan x) is the compo- sition g ◦ f where g : (0, ∞) → R, g(y) = ln y and f : (0, π/2) → (0, ∞), f (x) = tan x. Notice that (f ◦ g)(z ) = tan(ln z ) makes sense only for z ∈ (1, e π/2 ) and f ◦ g = g ◦ f . Example 4.44. Let f : R → R, f (x) =  x − 1 if x ≤ 1 x 2 if x > 1, g : R → R, g(x) =  x/2 if x ≥ 2 x 3 if x < 2. Let us find f ◦ g and g ◦ f . Solution. We have (f ◦ g)(x) =  g(x) − 1 if g(x) ≤ 1 (g(x)) 2 if g(x) > 1. Notice that g(x) ≤ 1 for x ≤ 1 and for x = 2; otherwise g(x) > 1. It follows that (f ◦ g)(x) =          x 3 − 1 if x ≤ 1 x 6 if 1 < x < 2 0 if x = 2 x 2 4 if x > 2. On the other hand, (g ◦ f )(x) =  f (x) 2 if f (x) ≥ 2 (f (x)) 3 if f (x) < 2. Functions 63 We have f (x) ≥ 2 for x ≥ √ 2 and f (x) < 2 for x < √ 2, hence (g ◦ f )(x) =      x 2 2 if x ≥ √ 2 x 6 if 1 < x < √ 2 (x − 1) 3 if x ≤ 1. The operation of Composition of functions preserves injectivity and surjec- tivity: Theorem 4.45. Let f : Y → Z and let g : X → Y . If f and g are one-to-one functions, then so is f ◦ g : X → Z . If f and g are onto, then f ◦ g : X → Z is also onto. Proof. Assume (f ◦ g)(x) = (f ◦ g)(x ′ ), so f (g(x)) = f (g(x ′ )) for x, x ′ ∈ X .

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  Linear Algebra: Gateway to Mathematics

  Second Edition

  • Robert Messer(Author)
  • 2021(Publication Date)
  • American Mathematical Society

    (Publisher)

  e. Speculate as to problems in extending the homogeneous property to all real numbers. f. 6.2 Compositions and Inverses In your study of algebra and calculus you undoubtedly encountered Compositions and inverses of functions. In this section, we will review these two operations for obtaining new functions from old ones and see how they work for linear functions between vector spaces. The concept of the inverse of a function is motivated by the problem of solving an equation. This can range from something extremely simple, such as 2? + 1 = 7, to an equation that requires a bit of work, 2? 3 − 5? 2 + 5? = 3, or even ? ? − ? 2 = 5, where an approximate solution is the best we can hope to obtain. In each case the value of a function is set equal to a real number, and we want to know if there are any values of the independent variable that satisfy the equation. That is, we want to know if the real number on the right side of the equation is in the image of the function. In the first example, the image of the function defined by ?(?) = 2? + 1 is the set of all real numbers. From this we know that the equation ?(?) = 2? + 1 = ? has a solution for any value of the constant ? . Furthermore, a little algebra or a glance at the graph of this function shows that the only way two real numbers ? 1 and ? 2 will give the same value for ?(? 1 ) and ?(? 2 ) is when ? 1 is equal to ? 2 . Thus, once a constant ? is specified, the equation ?(?) = 2? + 1 = ? has a unique solution. Exercise 3 at the end of this section asks you to show that the functions involved in the other two examples above behave similarly. The two concepts hinted at in these examples are made explicit in the following definition. 6.2. Compositions and Inverses 241 6.4 Definition: Suppose ? ∶ ? → ? is a function from the set ? to the set ? . The function ? is one-to-one if and only if for any ? 1 , ? 2 ∈ ? , ?(? 1 ) = ?(? 2 ) implies ? 1 = ? 2 .

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