![Example
Let f : R −→ R be defined as f(x) = sin x.
Let g : R 0 −→ R be defined as g(x) =
√
x.
1. Function composition g ◦ f is not defined
because range(f) = [−1, 1,] contains negative numbers, not in domain(g).
2. Relation composition g ◦ f is defined and results in a partial function on R;
g ◦ f : R −→ R such that
(g ◦ f)(x) =
√
sin x for x ∈ [2kπ, (2k + 1)π] where k ∈ Z.
3. Function composition f ◦ g is defined, and
it is the same as relation composition f ◦ g.
It results in a function f ◦ g : R 0 −→ R such that
(f ◦ g)(x) = sin(
√
x).](https://image.slidesharecdn.com/5-170430015112/85/5-2-Function-composition-7-320.jpg)
5.2. Function composition
- 1. Introduction to set theory and to methodology and philosophy of mathematics and computer programming Function composition An overview by Jan Plaza c 2017 Jan Plaza Use under the Creative Commons Attribution 4.0 International License Version of April 29, 2017
- 2. Definition. Let f and g be functions such that range(f) ⊆ domain(g). The function composition of f and g , denoted g func ◦ f , is the unique function that has the same domain as f, and such that (g ◦ f)(x)=g(f(x)), for every x ∈ domain(f). Note. “g ◦ f” is read “f composed with g” or “function composition of f and g”, as if we were reading the expression “g ◦ f” from the right to the left. While writing “f composed with g” in symbols, write from the right to the left. Note. This definition does not tell what g func ◦ f is when f or g is not a function or when range(f) domain(g), and such cases are considered not iteresting. Proposition. Let f and g be functions such that range(f) ⊆ domain(g), i.e. such that the function composition g ◦ f is defined. Then, (function composition) g func ◦ f = (relation composition) g ◦ f . Convention. Instead of func ◦ we will write just ◦ . The context should tell whether the relation composition or function composition is meant.
- 3. If f : X1 −→ Y1 and g : X2 −→ Y2 and Y1 ⊆ X2 then function composition g ◦ f is defined, and g ◦ f : X1 −→ Y2. f g g ◦ f X1 Y1 X2 Y2
- 4. Equivalently, if f : X −→ Y and g : Y −→ Z then function composition g ◦ f is defined, and g ◦ f : X −→ Z. f g g ◦ f X Y Z
- 5. Example f g g ◦ f 1 2 3 10 11 12 13 14 20 22 24 f = { 1, 11 , 2, 12 , 3, 12 } g = { 10, 20 , 11, 20 , 12, 22 , 13, 24 , 14, 24 } g ◦ f = { 1, 20 , 2, 22 , 3, 22 }
- 6. Example Let f : R −→ R where f(x)=2x. Let g : R −→ R where g(x)=x + 1. We have (f ◦ g)(x)=f(g(x))=2(x + 1) = 2x + 2. We have (g ◦ f)(x)=g(f(x))=2x + 1. Notice that f ◦ g=g ◦ f.
- 7. Example Let f : R −→ R be defined as f(x) = sin x. Let g : R 0 −→ R be defined as g(x) = √ x. 1. Function composition g ◦ f is not defined because range(f) = [−1, 1,] contains negative numbers, not in domain(g). 2. Relation composition g ◦ f is defined and results in a partial function on R; g ◦ f : R −→ R such that (g ◦ f)(x) = √ sin x for x ∈ [2kπ, (2k + 1)π] where k ∈ Z. 3. Function composition f ◦ g is defined, and it is the same as relation composition f ◦ g. It results in a function f ◦ g : R 0 −→ R such that (f ◦ g)(x) = sin( √ x).
- 8. Fact Let f, g be functions such that function composition g ◦ f is defined. Then: 1. domain(g ◦ f) = domain(f) 2. range(g ◦ f) ⊆ range(g) Exercise Let f, g be functions such that function composition g ◦ f is defined. 1. Disprove: domain(g ◦ f) = domain(g) 2. Disprove: range(g ◦ f) = range(g)