Algebra - find the formula for the inverse












1












$begingroup$


I have the function



$$y=frac12lnleft(frac{1+x}{1-x}right)$$



I have to find the inverse function. I know that



$$e^{2y}=frac{1+x}{1-x}$$



How do I solve this equation for $x$?






functional-analysis inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 11:54










  • $begingroup$
    I have edited the question body for proper formatting. Was this what you meant to ask?
    $endgroup$
    – Shubham Johri
    Jan 24 at 11:55










  • $begingroup$
    Note: the result provided in the answers corresponds to $x = tanh y$
    $endgroup$
    – Damien
    Jan 24 at 13:40
















1












$begingroup$


I have the function



$$y=frac12lnleft(frac{1+x}{1-x}right)$$



I have to find the inverse function. I know that



$$e^{2y}=frac{1+x}{1-x}$$



How do I solve this equation for $x$?






functional-analysis inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 11:54










  • $begingroup$
    I have edited the question body for proper formatting. Was this what you meant to ask?
    $endgroup$
    – Shubham Johri
    Jan 24 at 11:55










  • $begingroup$
    Note: the result provided in the answers corresponds to $x = tanh y$
    $endgroup$
    – Damien
    Jan 24 at 13:40














1












1








1





$begingroup$


I have the function



$$y=frac12lnleft(frac{1+x}{1-x}right)$$



I have to find the inverse function. I know that



$$e^{2y}=frac{1+x}{1-x}$$



How do I solve this equation for $x$?






functional-analysis inverse-function







share|cite|improve this question











$endgroup$




I have the function



$$y=frac12lnleft(frac{1+x}{1-x}right)$$



I have to find the inverse function. I know that



$$e^{2y}=frac{1+x}{1-x}$$



How do I solve this equation for $x$?






functional-analysis inverse-function




functional-analysis inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 17:57









Peter Mortensen

561310




561310










asked Jan 24 at 11:51









Emilia BorkoEmilia Borko

121




121












  • $begingroup$
    Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 11:54










  • $begingroup$
    I have edited the question body for proper formatting. Was this what you meant to ask?
    $endgroup$
    – Shubham Johri
    Jan 24 at 11:55










  • $begingroup$
    Note: the result provided in the answers corresponds to $x = tanh y$
    $endgroup$
    – Damien
    Jan 24 at 13:40


















  • $begingroup$
    Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 11:54










  • $begingroup$
    I have edited the question body for proper formatting. Was this what you meant to ask?
    $endgroup$
    – Shubham Johri
    Jan 24 at 11:55










  • $begingroup$
    Note: the result provided in the answers corresponds to $x = tanh y$
    $endgroup$
    – Damien
    Jan 24 at 13:40
















$begingroup$
Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 11:54




$begingroup$
Is it $$y=frac{1}{2}lnleft(frac{1+x}{1-x}right)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 11:54












$begingroup$
I have edited the question body for proper formatting. Was this what you meant to ask?
$endgroup$
– Shubham Johri
Jan 24 at 11:55




$begingroup$
I have edited the question body for proper formatting. Was this what you meant to ask?
$endgroup$
– Shubham Johri
Jan 24 at 11:55












$begingroup$
Note: the result provided in the answers corresponds to $x = tanh y$
$endgroup$
– Damien
Jan 24 at 13:40




$begingroup$
Note: the result provided in the answers corresponds to $x = tanh y$
$endgroup$
– Damien
Jan 24 at 13:40










3 Answers
3






active

oldest

votes


















8












$begingroup$

$$dfrac{1+x}{1-x}=e^{2y}$$



Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Multiplying $$y=frac12left(frac{1+x}{1-x}right)$$by $2$ we obtain $$2y=lnleft(frac{1+x}{1-x}right)$$



    We then can see that $$e^{2y}=frac{1+x}{1-x}$$



    and so we get $$e^{2y}(1-x)=1+x.$$



    Isolating the variable $x$, we get $$x=frac{e^{2y}-1}{e^{2y}+1}$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      If you have $f(x)=frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $begin{bmatrix}a & b\c&dend{bmatrix}$ and finding its inverse. We have that $e^{2y}=frac{x+1}{-x+1}$, so that gives us $begin{bmatrix}1 & 1\-1&1end{bmatrix}$, and the inverse of that is $frac12begin{bmatrix}1 & -1\1&1end{bmatrix}$. We can ignore the $frac12$ factor, giving the inverse as $x=frac {e^{2y}-1}{e^{2y}+1}$.



      Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=frac{ax+b}{cx+d}$, then $f^{-1}(y)=frac{dy-b}{-cy+a}$.



      Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $frac{x+1}{x-1}<0$, then the log is not real.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        $$dfrac{1+x}{1-x}=e^{2y}$$



        Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$






        share|cite|improve this answer









        $endgroup$


















          8












          $begingroup$

          $$dfrac{1+x}{1-x}=e^{2y}$$



          Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$






          share|cite|improve this answer









          $endgroup$
















            8












            8








            8





            $begingroup$

            $$dfrac{1+x}{1-x}=e^{2y}$$



            Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$






            share|cite|improve this answer









            $endgroup$



            $$dfrac{1+x}{1-x}=e^{2y}$$



            Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 11:53









            lab bhattacharjeelab bhattacharjee

            227k15158275




            227k15158275























                3












                $begingroup$

                Multiplying $$y=frac12left(frac{1+x}{1-x}right)$$by $2$ we obtain $$2y=lnleft(frac{1+x}{1-x}right)$$



                We then can see that $$e^{2y}=frac{1+x}{1-x}$$



                and so we get $$e^{2y}(1-x)=1+x.$$



                Isolating the variable $x$, we get $$x=frac{e^{2y}-1}{e^{2y}+1}$$






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  Multiplying $$y=frac12left(frac{1+x}{1-x}right)$$by $2$ we obtain $$2y=lnleft(frac{1+x}{1-x}right)$$



                  We then can see that $$e^{2y}=frac{1+x}{1-x}$$



                  and so we get $$e^{2y}(1-x)=1+x.$$



                  Isolating the variable $x$, we get $$x=frac{e^{2y}-1}{e^{2y}+1}$$






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Multiplying $$y=frac12left(frac{1+x}{1-x}right)$$by $2$ we obtain $$2y=lnleft(frac{1+x}{1-x}right)$$



                    We then can see that $$e^{2y}=frac{1+x}{1-x}$$



                    and so we get $$e^{2y}(1-x)=1+x.$$



                    Isolating the variable $x$, we get $$x=frac{e^{2y}-1}{e^{2y}+1}$$






                    share|cite|improve this answer











                    $endgroup$



                    Multiplying $$y=frac12left(frac{1+x}{1-x}right)$$by $2$ we obtain $$2y=lnleft(frac{1+x}{1-x}right)$$



                    We then can see that $$e^{2y}=frac{1+x}{1-x}$$



                    and so we get $$e^{2y}(1-x)=1+x.$$



                    Isolating the variable $x$, we get $$x=frac{e^{2y}-1}{e^{2y}+1}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 24 at 16:05









                    lioness99a

                    3,8362727




                    3,8362727










                    answered Jan 24 at 11:59









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    77.8k42866




                    77.8k42866























                        0












                        $begingroup$

                        If you have $f(x)=frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $begin{bmatrix}a & b\c&dend{bmatrix}$ and finding its inverse. We have that $e^{2y}=frac{x+1}{-x+1}$, so that gives us $begin{bmatrix}1 & 1\-1&1end{bmatrix}$, and the inverse of that is $frac12begin{bmatrix}1 & -1\1&1end{bmatrix}$. We can ignore the $frac12$ factor, giving the inverse as $x=frac {e^{2y}-1}{e^{2y}+1}$.



                        Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=frac{ax+b}{cx+d}$, then $f^{-1}(y)=frac{dy-b}{-cy+a}$.



                        Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $frac{x+1}{x-1}<0$, then the log is not real.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          If you have $f(x)=frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $begin{bmatrix}a & b\c&dend{bmatrix}$ and finding its inverse. We have that $e^{2y}=frac{x+1}{-x+1}$, so that gives us $begin{bmatrix}1 & 1\-1&1end{bmatrix}$, and the inverse of that is $frac12begin{bmatrix}1 & -1\1&1end{bmatrix}$. We can ignore the $frac12$ factor, giving the inverse as $x=frac {e^{2y}-1}{e^{2y}+1}$.



                          Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=frac{ax+b}{cx+d}$, then $f^{-1}(y)=frac{dy-b}{-cy+a}$.



                          Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $frac{x+1}{x-1}<0$, then the log is not real.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            If you have $f(x)=frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $begin{bmatrix}a & b\c&dend{bmatrix}$ and finding its inverse. We have that $e^{2y}=frac{x+1}{-x+1}$, so that gives us $begin{bmatrix}1 & 1\-1&1end{bmatrix}$, and the inverse of that is $frac12begin{bmatrix}1 & -1\1&1end{bmatrix}$. We can ignore the $frac12$ factor, giving the inverse as $x=frac {e^{2y}-1}{e^{2y}+1}$.



                            Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=frac{ax+b}{cx+d}$, then $f^{-1}(y)=frac{dy-b}{-cy+a}$.



                            Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $frac{x+1}{x-1}<0$, then the log is not real.






                            share|cite|improve this answer











                            $endgroup$



                            If you have $f(x)=frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $begin{bmatrix}a & b\c&dend{bmatrix}$ and finding its inverse. We have that $e^{2y}=frac{x+1}{-x+1}$, so that gives us $begin{bmatrix}1 & 1\-1&1end{bmatrix}$, and the inverse of that is $frac12begin{bmatrix}1 & -1\1&1end{bmatrix}$. We can ignore the $frac12$ factor, giving the inverse as $x=frac {e^{2y}-1}{e^{2y}+1}$.



                            Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=frac{ax+b}{cx+d}$, then $f^{-1}(y)=frac{dy-b}{-cy+a}$.



                            Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $frac{x+1}{x-1}<0$, then the log is not real.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 24 at 19:03

























                            answered Jan 24 at 18:09









                            AcccumulationAcccumulation

                            7,1252619




                            7,1252619






























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