Mixing
Algebra - find the formula for the inverse
$begingroup$
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
edited Jan 24 at 17:57
Peter Mortensen
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
$endgroup$
add a comment |
$begingroup$
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
edited Jan 24 at 17:57
Peter Mortensen
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 24 at 17:57
Peter Mortensen
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
$endgroup$
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
edited Jan 24 at 17:57
Peter Mortensen
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
edited Jan 24 at 17:57
Peter Mortensen
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
edited Jan 24 at 17:57
Peter Mortensen
561310
edited Jan 24 at 17:57
Peter Mortensen
561310
edited Jan 24 at 17:57
Peter Mortensen
561310
561310
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
asked Jan 24 at 11:51
Emilia BorkoEmilia Borko
121
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
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 $$x=frac{e^{2y}-1}{e^{2y}+1}$$
edited Jan 24 at 16:05
lioness99a
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
$$dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
$$dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$$dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=dfrac{e^{2y}-1}{e^{2y}+1}$$
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 24 at 11:53
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
add a comment |
add a comment |
$begingroup$
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 $$x=frac{e^{2y}-1}{e^{2y}+1}$$
edited Jan 24 at 16:05
lioness99a
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
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 $$x=frac{e^{2y}-1}{e^{2y}+1}$$
edited Jan 24 at 16:05
lioness99a
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
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 $$x=frac{e^{2y}-1}{e^{2y}+1}$$
edited Jan 24 at 16:05
lioness99a
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
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 $$x=frac{e^{2y}-1}{e^{2y}+1}$$
edited Jan 24 at 16:05
lioness99a
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
edited Jan 24 at 16:05
lioness99a
3,8362727
edited Jan 24 at 16:05
lioness99a
3,8362727
edited Jan 24 at 16:05
lioness99a
3,8362727
3,8362727
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Jan 24 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
add a comment |
add a comment |
$begingroup$
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.
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
$endgroup$
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.
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
edited Jan 24 at 19:03
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
answered Jan 24 at 18:09
AcccumulationAcccumulation
7,1252619
7,1252619
add a comment |
add a comment |
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$begingroup$
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