Composition Of Functions Composition Of Functions The composition of functions is always associative. That is, if f, g, and h are three functions with suitably chosen domains and codomains, then f ∘ (g ∘ h) = (f ∘ g) ∘ h, where the parentheses serve to indicate that composition is to be performed first for the parenthesized functions. Since there is no distinction between the choices of placement of parentheses, they may be safely left off. The functions g and f are said to commute with each other if g ∘ f = f ∘ g. In general, composition of functions will not be commutative. Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, only when . Considering functions as special cases of relations (namely functional relations), one can analogously define composition of relations, which gives the formula for in terms of and . Derivatives of compositions involving differentiable functions can be found using the chain rule. Higher derivatives of such functions are given by Faà di Bruno's formula.
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The structures given by composition are axiomatized and generalized in category theory. Until now, given a function f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1. In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points: Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}. Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1). (i) This type of exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so: f (1) = –1 (ii) The point in the g(x) set of point with x = –1 is the point (–1, –2), so: g(–1) = –2 (iii) What is "(g o f )(1)"? This is read as "g-compose-f of 1", and means "plug 1 into f, evaluate, and then plug the result into g". The computation can feel a lot easier if I use the following, more intuitive, formatting: (g o f )(1) = g( f(1)) Now I'll work in steps, keeping in mind that, while I may be used to doing things from the left to the right (because that's how we read), composition works from the right to the left (or, if you prefer, from the inside out). So I'll start with the x = 1. I am plugging this into f(x), so I look in the set of f(x) points for a point with x = 1. The point is (1, –1).
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(g o f )(1) = g( f(1)) = g(–1) Working from the right back toward the left, I am now plugging x = –1 (from "f(1) = –1") into g(x), so I look in the set of g(x) points for a point with x = –1. That point is (–1, –2). This tells me that g(–1) = –2, so now I have my answer: (g o f )(1) = g( f(1)) = g(–1) = –2 Note that they never told us what were the formulas, if any, for f(x) or g(x); we were only given a list of points. But this list was sufficient for answering the question, as long as we keep track of our x- and y-values. Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}. Find (i) ( f o g)(0), (ii) ( f o g)(–1), and (iii) (g o f )(–1). (i) To find ( f o g)(0), ("f-compose- g of zero"), I'll rewrite the expression as: ( f o g)(0) = f(g(0)) This tells me that I'm going to plug zero into g(x), simplify, and then plug the result into f(x). Looking at the list of g(x) points, I find (0, 2), so g(0) = 2, and I need now to find f(2). Looking at the list of f(x) points, I find (2, –3), so f(2) = –3. Then: ( f o g)(0) = f(g(0)) = f(2) = –3 (ii) The second part works the same way: ( f o g)(–1) = f(g(–1)) = f(–2) = 3
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