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Chapter 1: Problem 160

In general, composition of functions is not commutative. For example, for the functions \(f\) and \(g\) given by \(f(x)=x+1, g(x)=2 x\), we have \(f(g(x))=2 x+1\) and \(g(f(x))=2 x+2\). Now suppose that we have three functions \(f, g, h\). Then there are six possible compositions of the three, given by \(f(g(h(x))), g(h(f(x))), \ldots\). Give an example of three continuous functions that are defined for all real \(x\) and for which exactly five of the six compositions are the same. (Reprinted with the permission of the Canadian Mathematical Society, this problem was originally published in the Mathematical Mayhem section of Crux Mathematicorum with Mathematical Mayhem, vol. 25,1999, p. 293, problem C87.)

Short Answer

Expert verified
Use functions: \( f(x) = x \), \( g(x) = -x \), and \( h(x) = x \). Five compositions result in \(-x\), one in \(x\).

Step by step solution

01

- Define the simple functions

Start by defining three simple continuous functions for all real numbers. Consider the following functions: 1. \( f(x) = x \) 2. \( g(x) = -x \) 3. \( h(x) = x \)
02

- Compute the compositions

Calculate each of the six possible compositions with the defined functions:1. \( f(g(h(x))) \)2. \( f(h(g(x))) \)3. \( g(f(h(x))) \)4. \( g(h(f(x))) \)5. \( h(f(g(x))) \)6. \( h(g(f(x))) \)
03

- Simplify the compositions

Substitute and simplify each composition.1. \( f(g(h(x))) = f(g(x)) = f(-x) = -x \)2. \( f(h(g(x))) = f(x) = x \)3. \( g(f(h(x))) = g(x) = -x \)4. \( g(h(f(x))) = g(x) = -x \)5. \( h(f(g(x))) = h(-x) = -x \)6. \( h(g(f(x))) = h(-x) = -x \)
04

- Verify the five identical compositions

Notice that five of the six compositions result in \(-x\). These are:1. \( f(g(h(x))) = -x \)2. \( g(f(h(x))) = -x \)3. \( g(h(f(x))) = -x \)4. \( h(f(g(x))) = -x \)5. \( h(g(f(x))) = -x \)
05

- Identify the different composition

The different composition is:\( f(h(g(x))) = x \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composition of Functions
The composition of functions is the process of combining two functions to form a new function. It is denoted as (f o g)(x), which means f(g(x)), where g(x) is computed first, and then f is applied to that result. When composing functions, always start with the innermost function and work outwards. For example, in the given problem, to compute f(g(h(x))), we:
  • Apply h(x), which is just x

  • Then apply g to the result of h(x), giving us g(x) = -x

  • Finally, apply f to g(x), resulting in f(-x) = -x

In general, when combining three or more functions, there are multiple possible ways to compose them, which may yield different results. The original problem highlights this principle: out of six possible compositions of f, g, and h, five compositions turn out to be the same (-x), while one differs (x). This illustrates the non-commutativity of function composition.

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Most popular questions from this chapter

Let \(f_1, f_2, \ldots, f_{2012}\) be functions such that the derivative of each of them is the sum of the others. Let \(F=f_1 f_2 \cdots f_{2012}\) be the product of all these functions. Find all possible values of \(r\), given that \(\lim _{x \rightarrow-\infty} F(x) e^{r x}\) is a finite nonzero number.

For three points \(P, Q\), and \(R\) in \(\mathbb{R}^3\) (or, more generally, in \(\mathbb{R}^n\) ) we say that \(R\) is between \(P\) and \(Q\) if \(R\) is on the line segment connecting \(P\) and \(Q\) ( \(R=P\) and \(R=Q\) are allowed). A subset \(A\) of \(\mathbb{R}^3\) is called convex if for any two points \(P\) and \(Q\) in \(A\), every point \(R\) which is between \(P\) and \(Q\) is also in \(A\). For instance, an ellipsoid is convex, a banana is not. Now for the problem: Suppose \(A\) and \(B\) are convex subsets of \(\mathbb{R}^3\). Let \(C\) be the set of all points \(R\) for which there are points \(P\) in \(A\) and \(Q\) in \(B\) such that \(R\) lies between \(P\) and \(Q\). Does \(C\) have to be convex?

Suppose \(f\) is a continuous, increasing, bounded, real-valued function, defined on \([0, \infty)\), such that \(f(0)=0\) and \(f^{\prime}(0)\) exists. Show that there exists \(b>0\) for which the volume obtained by rotating the area under \(f\) from 0 to \(b\) about the \(x\)-axis is half that of the cylinder obtained by rotating \(y=f(b)\), \(0 \leq x \leq b\), about the \(x\)-axis.

Let $$ c=\sum_{n=1}^{\infty} \frac{1}{n\left(2^n-1\right)}=1+\frac{1}{6}+\frac{1}{21}+\frac{1}{60}+\cdots $$ Show that $$ e^c=\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{8}{7} \cdot \frac{16}{15} \cdots \cdots $$

For \(0 \leq x \leq 1\), let \(T(x)=\left\\{\begin{array}{cl}x & \text { if } x \leq 1 / 2 \\ 1-x & \text { if } x \geq 1 / 2\end{array}\right.\) (You can think of \(T(x)\) as the distance from \(x\) to the nearest integer.) Define \(f(x)=\sum_{n=1}^{\infty} T\left(x^n\right)\). a. Evaluate \(f\left(\frac{1}{\sqrt[3]{2}}\right)\) b. Find all \(x(0 \leq x \leq 1)\) for which \(f(x)=2012\)
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