For which numbers a,b,c,d will the function f(x) satisfy...

**Mollier** · December 23rd 2009, 04:34 AM

Problem:
For which numbers $a,b,c,d$ will the function
$f(x)=\frac{ax+b}{cx+d}$
satisfy $f(f(x))=x$ for all $x$ ?

------------------------------------------------------------------------
Attempt:
$ f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\ = \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\ =\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} $

By using polynomial division I get the following result:

$ \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)} $

By doing some hopefully correct algebra I get:

$ (a^2+bc)x+(ab+bd)=x $
1. $(ab+bd)=0$
So, $a=-d$

2. $(a^2+bc)=1$
So, $a=\sqrt{1-bc}$

If $b$ and $c$ have opposite signs, their product is negative, and they can be any number.

If on the other hand, their signs are the same, their product is positive and:
$-1 \leq b \leq 1$
$-1 \leq c \leq 1$
-----------------------------------------------------------------------

Am I making this more complicated than it is?
How can I better express my results regarding $b$ , and $c$ ?

Thanks!

**skeeter** · December 23rd 2009, 05:17 AM

Originally Posted by Mollier

Problem:
For which numbers $a,b,c,d$ will the function
$f(x)=\frac{ax+b}{cx+d}$
satisfy $f(f(x))=x$ for all $x$ ?

if $f[f(x)] = x$ , then $f(x)$ is its own inverse, and as such, its graph is symmetrical to the line $y = x$

$\frac{a \cdot f(x) + b}{c \cdot f(x) + d} = x$

$a \cdot f(x) + b = cx \cdot f(x) + dx$

$a \cdot f(x) - cx \cdot f(x) = dx - b$

$f(x)[a - cx] = dx - b$

$f(x) = \frac{dx-b}{a-cx} = \frac{ax+b}{cx+d}$

$\frac{-dx+b}{cx-a} = \frac{ax+b}{cx+d}$

the above equation only requires that $a = -d$ .

**Opalg** · December 23rd 2009, 05:21 AM

Originally Posted by Mollier

Problem:
For which numbers $a,b,c,d$ will the function
$f(x)=\frac{ax+b}{cx+d}$
satisfy $f(f(x))=x$ for all $x$ ?

------------------------------------------------------------------------
Attempt:
$ f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\ = \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\ =\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} $ Correct up to here.

By using polynomial division I get the following result:

$ \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)} $ The trouble with this is that you are about to find that a = –d, so the denominator of the fraction in parentheses is zero.

By doing some hopefully correct algebra I get:

$ (a^2+bc)x+(ab+bd)=x $
1. $(ab+bd)=0$
So, $a=-d$

2. $(a^2+bc)=1$
So, $a=\sqrt{1-bc}$

If $b$ and $c$ have opposite signs, their product is negative, and they can be any number.

If on the other hand, their signs are the same, their product is positive and:
$-1 \leq b \leq 1$
$-1 \leq c \leq 1$
-----------------------------------------------------------------------

Am I making this more complicated than it is?
How can I better express my results regarding $b$ , and $c$ ?

The condition f(f(x)) tells you that $\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} = x$ . Cross-multiply to get $(a^2+bc)x+(ab+bd) = x\bigl((ac+cd)x+(bc+d^2)\bigr)$ . If that quadratic equation holds for all x then you can equate each coefficient to zero. You should find that there are two sets of solutions:

1. $a+d=0$ (b and c can be anything, so long as you don't have all four of a, b, c, d equal to 0);

2. $a = d \ne0,\;b=c=0$ .

**Mollier** · December 23rd 2009, 07:47 PM

Originally Posted by skeeter

if [tex]
the above equation only requires that $a = -d$ .

The way you come to that conclusion is very understandable, thank you.

Originally Posted by Opalg

$\left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)} $

The trouble with this is that you are about to find that a = –d, so the denominator of the fraction in parentheses is zero.

Ah, I did not see that! Thank you!

Merry Christmas

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