Problem 76
Question
Let \(\mathcal{T}\) be the collection of functions of the form $$ x \mapsto \frac{a x+b}{c x+d} $$ with \(a d-b c=1 .\) (These functions are called linear fractional transformations.) Show that the composition of two functions in \(\mathcal{T}\) is also in \(\mathcal{T}\). By setting \(a=d=1\) and \(b=c=0,\) we see that the identity function \(I(x)=x\) is in \(\mathcal{T}\). Show that every function in \(\mathcal{T}\) has an inverse that is also in \(\mathcal{T}\) and that is obtained by swapping \(a\) and \(d,\) and negating \(b\) and \(c\)
Step-by-Step Solution
Verified Answer
The composition, identity, and inverse of functions in \( \mathcal{T} \) retain the structure of \( \mathcal{T} \).
1Step 1: Understand linear fractional transformations
A linear fractional transformation is of the form \( f(x) = \frac{ax + b}{cx + d} \), where the condition \( ad - bc = 1 \) holds. Functions in the set \( \mathcal{T} \) are defined by this rule.
2Step 2: Show composition is in \( \mathcal{T} \)
Consider two functions \( f_1(x) = \frac{a_1x + b_1}{c_1x + d_1} \) and \( f_2(x) = \frac{a_2x + b_2}{c_2x + d_2} \) from \( \mathcal{T} \). Their composition is \( f_1(f_2(x)) = \frac{a_1(a_2x+b_2) + b_1(c_2x+d_2)}{c_1(a_2x+b_2) + d_1(c_2x+d_2)} \), which simplifies to \( \frac{(a_1a_2 + b_1c_2)x + (a_1b_2 + b_1d_2)}{(c_1a_2 + d_1c_2)x + (c_1b_2 + d_1d_2)} \). For the new set of coefficients, the condition \( ad - bc = 1 \) must still hold. Calculating, we find that \((a_1a_2 + b_1c_2)(d_1d_2 + c_1b_2) - (a_1b_2 + b_1d_2)(c_1a_2 + d_1c_2) = 1 \), confirming that the composition is in \( \mathcal{T} \).
3Step 3: Confirm identity function is in \( \mathcal{T} \)
By setting \( a = d = 1 \) and \( b = c = 0 \), the function becomes \( I(x) = \frac{1 \cdot x + 0}{0 \cdot x + 1} = x \). This is the identity function. Also, \( ad - bc = 1 \cdot 1 - 0 \cdot 0 = 1 \), so the identity function satisfies the condition.
4Step 4: Show inverse is in \( \mathcal{T} \)
For a function \( f(x) = \frac{ax + b}{cx + d} \), its inverse is \( f^{-1}(x) = \frac{dx - b}{-cx + a} \). By swapping \( a \) and \( d \) and negating \( b \) and \( c \), the determinant \( da - (-b)(-c) = ad - bc = 1 \) confirms the inverse function is also in \( \mathcal{T} \).
Key Concepts
Function CompositionInverse FunctionsIdentity FunctionDeterminant Condition
Function Composition
Function composition is a process where you combine two functions to create a new one. Imagine you have two functions, say \( f(x) \) and \( g(x) \). You want to know what happens when you take \( g(x) \) and then apply \( f \) to the result. That's what composition does. It’s like taking a piece of clay (output of \( g \)) and molding it further using another tool (\( f \)).
In mathematical terms, if you have two linear fractional transformations \( f_1(x) = \frac{a_1x + b_1}{c_1x + d_1} \) and \( f_2(x) = \frac{a_2x + b_2}{c_2x + d_2} \), the composition \( f_1(f_2(x)) \) can be simplified to a new transformation:
\[ f_1(f_2(x)) = \frac{(a_1a_2 + b_1c_2)x + (a_1b_2 + b_1d_2)}{(c_1a_2 + d_1c_2)x + (c_1b_2 + d_1d_2)} \]This stays within the family of functions \( \mathcal{T} \) because it satisfies the special condition \( ad - bc = 1 \).
In mathematical terms, if you have two linear fractional transformations \( f_1(x) = \frac{a_1x + b_1}{c_1x + d_1} \) and \( f_2(x) = \frac{a_2x + b_2}{c_2x + d_2} \), the composition \( f_1(f_2(x)) \) can be simplified to a new transformation:
\[ f_1(f_2(x)) = \frac{(a_1a_2 + b_1c_2)x + (a_1b_2 + b_1d_2)}{(c_1a_2 + d_1c_2)x + (c_1b_2 + d_1d_2)} \]This stays within the family of functions \( \mathcal{T} \) because it satisfies the special condition \( ad - bc = 1 \).
- Start with two known transformations.
- Plug the output of the first into the second.
- The resulting function also fits the pattern and rules.
Inverse Functions
An inverse function essentially reverses the process of the original function. If \( f \) takes you one way, \( f^{-1} \) leads you back. Consider it like retracing your steps.
For a linear fractional transformation \( f(x) = \frac{ax + b}{cx + d} \), finding its inverse is straightforward. You simply swap \( a \) and \( d \), and change the signs of \( b \) and \( c \):
\[ f^{-1}(x) = \frac{dx - b}{-cx + a} \]
To verify this inverse still lies in our collection \( \mathcal{T} \), you check the condition \( ad - bc = 1 \). This assures the reversed function is indeed valid among these special types of transformations.
For a linear fractional transformation \( f(x) = \frac{ax + b}{cx + d} \), finding its inverse is straightforward. You simply swap \( a \) and \( d \), and change the signs of \( b \) and \( c \):
\[ f^{-1}(x) = \frac{dx - b}{-cx + a} \]
To verify this inverse still lies in our collection \( \mathcal{T} \), you check the condition \( ad - bc = 1 \). This assures the reversed function is indeed valid among these special types of transformations.
- Swap coefficients \( a \) and \( d \).
- Negate \( b \) and \( c \).
- Confirm it satisfies the determinant rule.
Identity Function
The identity function acts like a mirror, reflecting each input to itself. With \( I(x) = x \), you get out exactly what you throw in.
In the context of linear fractional transformations \( \mathcal{T} \), setting \( a = d = 1 \) and \( b = c = 0 \) crafts this familiar function. When expressed in the typical transformation form, it looks like:
\[ I(x) = \frac{1x + 0}{0x + 1} = x \]
Here, the determinant condition \( ad - bc = 1 \) holds as well, ensuring that this function belongs to \( \mathcal{T} \). This alignment makes the identity function a part of the team; it’s the unchanged friend among shifty peers.
In the context of linear fractional transformations \( \mathcal{T} \), setting \( a = d = 1 \) and \( b = c = 0 \) crafts this familiar function. When expressed in the typical transformation form, it looks like:
\[ I(x) = \frac{1x + 0}{0x + 1} = x \]
Here, the determinant condition \( ad - bc = 1 \) holds as well, ensuring that this function belongs to \( \mathcal{T} \). This alignment makes the identity function a part of the team; it’s the unchanged friend among shifty peers.
- Set coefficients appropriately.
- Observe input equals output.
- Determine condition holds true.
Determinant Condition
The determinant condition plays a crucial role by imposing a constraint on the coefficients of our transformations. It’s the check-and-balance guardian for linear fractional transformations.
For any function in \( \mathcal{T} \), the calculation \( ad - bc = 1 \) ensures it behaves properly and stays in line with its peers. This condition is essential for guaranteeing that transformations can compose, have inverses, and reflect identity properly.
In practice, this calculation is done as follows:
For any function in \( \mathcal{T} \), the calculation \( ad - bc = 1 \) ensures it behaves properly and stays in line with its peers. This condition is essential for guaranteeing that transformations can compose, have inverses, and reflect identity properly.
In practice, this calculation is done as follows:
- Multiply the diagonal coefficients \( a \) and \( d \).
- Subtract the product of the other diagonal \( b \) and \( c \).
- Confirm the result equals 1 to stay within \( \mathcal{T} \).
Other exercises in this chapter
Problem 75
Plot the locus of points \(P\) with an ordinate that is equal to the distance of the point \(P\) to the point (0,1)
View solution Problem 75
Plot the given curve in a viewing window containing the given point \(P\). Zoom in on the point \(P\) until the graph of the curve appears to be a straight line
View solution Problem 76
Suppose that \(\mathrm{A} \neq 0 .\) Let \(s=-\frac{B}{2 A} .\) Let \(\mathcal{P}\) be the parabola whose equation is \(y=A x^{2}+B x+C\). Suppose that the hori
View solution Problem 76
Plot the given curve in a viewing window containing the given point \(P\). Zoom in on the point \(P\) until the graph of the curve appears to be a straight line
View solution