If \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\) and f(f(x)) = x then
d = - a
To determine the condition under which \(f(f(x)) = x\) for the given function \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\), we need to substitute \(f(x)\) into itself and then equate the result to \(x\).
Function Composition: Evaluating \(f(f(x))\)
Let's first find the expression for \(f(f(x))\):
\[f(f(x)) = f\left( \frac{{ax + b}}{{cx + d}} \right)\]
Substitute \(\frac{{ax + b}}{{cx + d}}\) into the function \(f(x)\) wherever \(x\) appears:
\[f(f(x)) = \frac{{a\left( \frac{{ax + b}}{{cx + d}} \right) + b}}{{c\left( \frac{{ax + b}}{{cx + d}} \right) + d}}\]
To simplify this complex fraction, we multiply the numerator and the denominator by \((cx+d)\):
\[f(f(x)) = \frac{{a(ax + b) + b(cx + d)}}{{c(ax + b) + d(cx + d)}}\]
Now, expand the terms in the numerator and the denominator:
\[f(f(x)) = \frac{{a^2x + ab + bcx + bd}}{{acx + bc + cdx + d^2}}\]
Group the terms with \(x\) and constant terms in both the numerator and the denominator:
\[f(f(x)) = \frac{{(a^2 + bc)x + (ab + bd)}}{{(ac + cd)x + (bc + d^2)}}\]
Equating \(f(f(x))\) to \(x\)
Given the condition \(f(f(x)) = x\), we set our derived expression equal to \(x\):
\[\frac{{(a^2 + bc)x + (ab + bd)}}{{(ac + cd)x + (bc + d^2)}} = x\]
Multiply both sides by the denominator \(((ac + cd)x + (bc + d^2))\) to clear the fraction:
\[(a^2 + bc)x + (ab + bd) = x \left( (ac + cd)x + (bc + d^2) \right)\]
Expand the right side of the equation:
\[(a^2 + bc)x + (ab + bd) = (ac + cd)x^2 + (bc + d^2)x\]
Parameter Analysis: Comparing Coefficients
For this equation to hold true for all values of \(x\) (i.e., to be an identity), the coefficients of corresponding powers of \(x\) on both sides of the equation must be equal.
Let's compare the coefficients:
- Coefficient of \(x^2\):
On the left side, the coefficient of \(x^2\) is \(0\). On the right side, it is \((ac + cd)\).
Therefore, \(ac + cd = 0\)
Factor out \(c\): \(c(a + d) = 0\) - Coefficient of \(x\):
On the left side, the coefficient of \(x\) is \((a^2 + bc)\). On the right side, it is \((bc + d^2)\).
Therefore, \(a^2 + bc = bc + d^2\)
Subtract \(bc\) from both sides: \(a^2 = d^2\) - Constant term:
On the left side, the constant term is \((ab + bd)\). On the right side, the constant term is \(0\).
Therefore, \(ab + bd = 0\)
Factor out \(b\): \(b(a + d) = 0\)
| Power of \(x\) | Left Side Coefficient | Right Side Coefficient | Equation |
|---|---|---|---|
| \(x^2\) | \(0\) | \(ac + cd\) | \(ac + cd = 0 \implies c(a+d)=0\) |
| \(x\) | \(a^2 + bc\) | \(bc + d^2\) | \(a^2 + bc = bc + d^2 \implies a^2 = d^2\) |
| Constant | \(ab + bd\) | \(0\) | \(ab + bd = 0 \implies b(a+d)=0\) |
Deriving the Condition
From the equation \(a^2 = d^2\), we can conclude that \(a = d\) or \(a = -d\).
Let's analyze these two possibilities with the other two equations:
-
If \(a = d\):
- From \(c(a+d)=0\), substituting \(d=a\), we get \(c(a+a) = 0 \implies 2ac = 0\). This implies either \(a=0\) or \(c=0\).
- From \(b(a+d)=0\), substituting \(d=a\), we get \(b(a+a) = 0 \implies 2ab = 0\). This implies either \(a=0\) or \(b=0\).
- If \(a=0\), then \(d=0\). In this specific scenario, \(f(x) = \frac{b}{cx}\). Then \(f(f(x)) = f\left(\frac{b}{cx}\right) = \frac{b}{c(\frac{b}{cx})} = \frac{b}{\frac{bc}{cx}} = \frac{bcx}{bc} = x\), provided \(b \neq 0\) and \(c \neq 0\). Here, \(d=a=0\), which satisfies both \(d=a\) and \(d=-a\).
- If \(a \neq 0\), then from \(2ac=0\), we must have \(c=0\). From \(2ab=0\), we must have \(b=0\). So, if \(a=d\) and \(a \neq 0\), then \(b=0\) and \(c=0\). In this case, \(f(x) = \frac{ax}{d}\). Since \(a=d\), \(f(x) = x\), which naturally satisfies \(f(f(x))=x\). So, \(d=a\) *can* be a solution but requires additional constraints (\(b=0, c=0\) unless \(a=0\)).
-
If \(a = -d\) (or equivalently \(d = -a\)):
- From \(c(a+d)=0\), substituting \(d=-a\), we get \(c(a-a) = 0 \implies c(0) = 0\). This equation is always true, regardless of the value of \(c\).
- From \(b(a+d)=0\), substituting \(d=-a\), we get \(b(a-a) = 0 \implies b(0) = 0\). This equation is always true, regardless of the value of \(b\).
The condition \(d = -a\) satisfies all three derived coefficient equations unconditionally (as long as \(a^2+bc \neq 0\) to ensure the expression does not become \(0/0\) in a way that breaks the identity for all \(x\)). In contrast, \(d=a\) requires additional conditions on \(b\) and \(c\) (like \(b=0\) and \(c=0\) unless \(a=0\)). Therefore, \(d = -a\) is the most general condition for \(f(f(x))=x\).
The final answer is \(\text{d = - a}\).
Important Questions from Differential Equations
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The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
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