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Chapter 5: Problem 6

. Let \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) where \(f(x)=a x+b\) and \(g(x)=c x+d\) for all \(x \in \mathbf{R}\), with \(a, b, c, d\) real constants. What relationship(s) must be satisfied by \(a, b, c, d\) if \((f \circ g)(x)=(g \circ f)(x)\) for all \(x \in \mathbf{R} ?\)

Short Answer

Expert verified
The relationships that must be satisfied by a, b, c, and d are \(ad = cb\) and \(b = d\).

Step by step solution

01

Find f(g(x))

First, find the composition of f and g, denoted as \(f(g(x))\). In this case, substitute \(g(x)=cx+d\) into \(f(x)=ax+b\), replacing every instance of x with \(g(x)\). This gives us \(f(g(x)) = a(cx+d) + b = acx + ad + b\).
02

Find g(f(x))

Now, find the composition of g and f, denoted as \(g(f(x))\). Substitute \(f(x)=ax+b\) into \(g(x)=cx+d\), replacing every instance of x with \(f(x)\). This gives us \(g(f(x)) = c(ax+b) + d = cax + cb + d\).
03

Set f(g(x)) equal to g(f(x))

Now, set \(f(g(x))\) equal to \(g(f(x))\) and solve for the unknowns a, b, c, and d. This gives us the equation \(acx + ad + b = cax + cb + d\). Equating the coefficients, we obtain the following relationships: \(ac=ca\) (which is always true because multiplication is commutative), \(ad = cb\), and \(b = d\).

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Key Concepts

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

Commutativity
Commutativity is a fundamental property of certain operations like addition and multiplication. It means you can swap the order of the operands, and the result remains unchanged. For instance, in multiplication, \(a \times b = b \times a\). In the context of this exercise, the commutativity refers to the condition \( ac = ca \), which is inherently true due to the nature of real number multiplication. However, it emphasizes that even when considering complex expressions, the basic arithmetic rules apply. Understanding and recognizing where this property holds can simplify problem solving, as it reduces the need to rearrange terms unnecessarily.
Composition of Functions
Function composition involves combining two functions to form a new function. This entails taking the output of one function and using it as the input for another. If you have two functions, \( f(x) \) and \( g(x) \), their composition is written as \( (f \circ g)(x) = f(g(x)) \).
  • For the function \( f(x) = ax + b \) and \( g(x) = cx + d \), composing them as \( f(g(x)) \) involves substituting \( g(x) \) into \( f \). This results in \( acx + ad + b \).
  • Similarly, composing \( g(f(x)) \) means substituting \( f(x) \) into \( g \), which results in \( cax + cb + d \).

When working with compositions, carefully follow the substitution step by step, ensuring all terms line up correctly. This is crucial for verifying equality between two function compositions or working through function-related proofs and problems.
Equation Solving
Equation solving involves finding values for variables that make an equation true. In this scenario, we set two composed functions equal to each other: \( f(g(x)) = g(f(x)) \).
  • The first step is to expand each side: from \( f(g(x)) \), you get \( acx + ad + b \), and from \( g(f(x)) \), you get \( cax + cb + d \).
  • By aligning these two, you notice the coefficients of each power of \( x \) must match for the equation to hold true for all \( x \).
  • The equation \( acx + ad + b = cax + cb + d \) yields key conditions: \( ac=ca \) (which is already satisfied due to commutativity), \( ad = cb \), and \( b = d \).

Solving equations of this nature requires understanding how terms are distributed and balanced across both sides. Make sure you methodically find relationships between the variables for consistent solutions.

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Most popular questions from this chapter

For \(m, n \in \mathbf{Z}^{+}\), prove that if \(m\) pigeons occupy \(n\) pigeonholes, then at least one pigeonhole has \(\lfloor(m-1) / n\rfloor+1\) or more pigeons roosting in it.

The following is analogous to the "big-Oh" notation introduced in conjunction with Definition \(5.23\). For \(f, g: \mathbf{Z}^{+} \rightarrow \mathbf{R}\) we say that \(f\) is of order atleast \(g\) if there exist constants \(M \in \mathbf{R}^{+}\)and \(k \in \mathbf{Z}^{+}\)such that \(|f(n)| \geq M|g(n)|\) for all \(n \in \mathbf{Z}^{+}\), where \(n \geq k\). In this case we write \(f \in \Omega(g)\) and say that \(f\) is "big Omega of \(g\)." So \(\Omega(g)\) represents the set of all functions with domain \(\mathbf{Z}^{+}\)and codomain \(\mathbf{R}\) that dominate \(g\). Suppose that \(f, g, h: \mathbf{Z}^{+} \rightarrow \mathbf{R}\), where \(f(n)=5 n^{2}+3 n\), \(g(n)=n^{2}, h(n)=n\), for all \(n \in \mathbf{Z}^{+}\). Prove that (a) \(f \in \Omega(g)\); (b) \(g \in \Omega(f)\); (c) \(f \in \Omega(h)\); and (d) \(h \notin \Omega(f)-\) that is, \(h\) is not "big Omega of \(f\) ""

As in the previous exercise, \(s(m, n)\) denotes a Stirling number of the first kind. a) For \(m \geq n>\mathbb{1}\) prove that $$ s(m, n)=(m-1) s(m-1, n)+s(m-1, n-1) $$ b) Verify that for \(m \geq 2\) $$ s(m, 2)=(m-1) ! \sum_{i=1}^{m-1} \frac{1}{i} $$

For \(f, g: \mathbf{Z}^{+} \rightarrow \mathbf{R}\), we say that \(f\) is "big Theta of \(g\)," and write \(f \in \Theta(g)\), when there exist constants \(m_{1}, m_{2} \in \mathbf{R}^{+}\)and \(k \in \mathbf{Z}^{+}\)such that \(m_{1}|g(n)| \leq|f(n)| \leq m_{2}|g(n)|\), for all \(n \in \mathbf{Z}^{+}\), where \(n \geq k\). Prove that \(f \in \Theta(g)\) if and only if \(f \in \Omega(g)\) and \(f \in O(g)\)

a) For \(A=(-2,7] \subseteq \mathbf{R}\) define the functions \(f, g: A \rightarrow \mathbf{R}\) by $$ f(x)=2 x-4 \text { and } g(x)=\frac{2 x^{2}-8}{x+2} $$ Verify that \(f=g\). b) Is the result in part (a) affected if we change \(A\) to \([-7,2) ?\)
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