Problem 59 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that the operation of composition of functions is associative; that is, \(f_{1} \circ\left(f_{2} \circ f_{3}\right)=\left(f_{1} \circ f_{2}\right) \circ f_{3}\).

Step-by-Step Solution

Answer

Function composition is associative as both ways provide the same result.

Step 1: Understand Function Composition

Function composition is the operation where two functions are combined into a single function. Given two functions, say, \(f(x)\) and \(g(x)\), their composition, denoted \(f \circ g\), is defined as \( (f \circ g)(x) = f(g(x))\). In this exercise, we explore the composition of three functions, \(f_1, f_2, \) and \(f_3\).

Step 2: Define the Composition (Inner to Outer)

Consider the composition \(f_1 \circ (f_2 \circ f_3)\). This means that we first apply \(f_3\), then \(f_2\) to the result, and finally \(f_1\) to the outcome. So, for an input \(x\), the entire expression becomes \(f_1(f_2(f_3(x)))\).

Step 3: Define the Composition (Outer to Inner)

Now consider \((f_1 \circ f_2) \circ f_3\). Here, we first apply \(f_3\) to \(x\), then apply \(f_2\) to the result, and finally \(f_1\) to this outcome. Mathematically, it becomes \(f_1(f_2(f_3(x)))\).

Step 4: Compare the Two Compositions

Notice that both compositions, \(f_1 \circ (f_2 \circ f_3)\) and \((f_1 \circ f_2) \circ f_3\), yield the same result, \(f_1(f_2(f_3(x)))\). The operations performed and their order do not change, hence demonstrating that the composition is associative.

Key Concepts

Function Composition

Function composition is the process of taking functions and building a chain of operations, forming a new function. Imagine two functions, like pieces of a puzzle. When you put them together, they create something new. So, if you have two functions, say \( f(x) \) and \( g(x) \), you can "compose" them to create another function, \( (f \circ g)(x) \). This means you first apply \( g \) on \( x \), and then \( f \) on the result: \( f(g(x)) \).

• Think of it like a conveyor belt: the output of \( g(x) \) becomes the input of \( f(x) \).
• This composition can be applied to many functions, not just two.

In the exercise, we look at three functions \( f_1, f_2, \) and \( f_3 \), and see how they work together following the rules of function composition.

Associative Property

The associative property is a fundamental characteristic in mathematics that describes how you group operations. With addition, for example, you know it's okay to add in any order because \( (a + b) + c = a + (b + c) \). The same principle applies when composing functions.
For function composition, the associative property states:

• If you compose three functions \( f_1, f_2, \) and \( f_3 \), it doesn't matter how you group them. You will end up with the same final function.
• This can be expressed mathematically as \( f_1 \circ (f_2 \circ f_3) = (f_1 \circ f_2) \circ f_3 \).

This means whether you first compose \( f_2 \) with \( f_3 \) and then \( f_1 \), or \( f_1 \) with \( f_2 \) and then \( f_3 \), the result will always be the same.

Mathematical Proof

The proof of the associative property for function composition involves ensuring both arrangements give the same result.
Here's how you prove it:

• Start by considering \( f_1(f_2(f_3(x))) \), which is derived by first applying \( f_3 \), then \( f_2 \), and finally \( f_1 \).
• Now, look at \( f_1(f_2(f_3(x))) \) again through a different grouping: apply \( f_3 \), compose \( f_2 \), and finish with \( f_1 \).
• In both cases, every function gets applied in the same order, yielding identical outcomes.

No matter how functions are nested or grouped, as long as the order of applying them remains unchanged, the end result proves the associative property. By equating \( f_1 \circ (f_2 \circ f_3) \) to \( (f_1 \circ f_2) \circ f_3 \), the seamless transition between both operations gets verified.

Operation of Functions

The operation of functions surpasses simple combination and extends to how functions interrelate. When you operate functions together, you are essentially creating a recipe where every function plays a part in the final output.

• Each function takes the result of the previous function and runs its own operation.
• Function operations, while simple alone, become powerful tools when stacked and composed.

This chaining or sequential application of functions forms the backbone of not just theoretical math, but practical applications like algorithms, computer programs, and even everyday problem-solving.
Appreciating function operation allows you to see the bigger picture of how small processes can lead to complex results through composition, demonstrating the profound influence of mathematical structures in systems.