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

  1. #1
    Member Mollier's Avatar
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    For which numbers a,b,c,d will the function f(x) satisfy...

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

    ------------------------------------------------------------------------
    Attempt:
    $\displaystyle
    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:

    $\displaystyle
    \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:

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

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

    If $\displaystyle b$ and $\displaystyle c$ 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:
    $\displaystyle -1 \leq b \leq 1 $
    $\displaystyle -1 \leq c \leq 1 $
    -----------------------------------------------------------------------

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

    Thanks!
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by Mollier View Post
    Problem:
    For which numbers $\displaystyle a,b,c,d$ will the function
    $\displaystyle f(x)=\frac{ax+b}{cx+d} $
    satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$?
    if $\displaystyle f[f(x)] = x$ , then $\displaystyle f(x)$ is its own inverse, and as such, its graph is symmetrical to the line $\displaystyle y = x$

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

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

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

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

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

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


    the above equation only requires that $\displaystyle a = -d$.
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by Mollier View Post
    Problem:
    For which numbers $\displaystyle a,b,c,d$ will the function
    $\displaystyle f(x)=\frac{ax+b}{cx+d} $
    satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$?

    ------------------------------------------------------------------------
    Attempt:
    $\displaystyle
    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:

    $\displaystyle
    \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:

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

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

    If $\displaystyle b$ and $\displaystyle c$ 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:
    $\displaystyle -1 \leq b \leq 1 $
    $\displaystyle -1 \leq c \leq 1 $
    -----------------------------------------------------------------------

    Am I making this more complicated than it is?
    How can I better express my results regarding $\displaystyle b$, and $\displaystyle c$?
    The condition f(f(x)) tells you that $\displaystyle \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} = x$. Cross-multiply to get $\displaystyle (a^2+bc)x+(ab+bd) = x\bigl((ac+cd)x+(bc+d^2)\bigr)$. 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. $\displaystyle a+d=0$ (b and c can be anything, so long as you don't have all four of a, b, c, d equal to 0);

    2. $\displaystyle a = d \ne0,\;b=c=0$.
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  4. #4
    Member Mollier's Avatar
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    Quote Originally Posted by skeeter View Post
    if [tex]
    the above equation only requires that $\displaystyle a = -d$.
    The way you come to that conclusion is very understandable, thank you.

    Quote Originally Posted by Opalg View Post


    $\displaystyle \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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