Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy

Thread: How to get a value from sigmoid function?

  1. #1
    Newbie
    Joined
    Aug 2017
    From
    London
    Posts
    23

    Question How to get a value from sigmoid function?

    $$x(y) = m\sigma(ny)$$

    where $m$ and $n$ are constants and $\sigma(ny)$ the sigmoid function. How can we right the equation for $y(x)$?

    $$y(x) = ?$$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035

    Re: How to get a value from sigmoid function?

    The term 'sigmoid function' can refer to a variety of functions but I suspect you are referring to $\displaystyle x(y)= \frac{me^{ny}}{e^{ny}+ 1}$. An obvious first step is to multiply both sides by $\displaystyle e^{ny}+ 1$ to get $\displaystyle x(e^{ny}+ 1)= me^{ny}$ and then $\displaystyle xe^{ny}+ x= me^{ny}$ which we can then write as $\displaystyle xe^{ny}- me^{ny}= e^{ny}(x- m)= -1$. Divide both sides by x- 1 to get $\displaystyle e^{ny}= \frac{-1}{x- m}= \frac{1}{m- x}$.

    Now take the natural logarithm of both sides: $\displaystyle ny= ln\left(\frac{1}{m- x}\right)= -ln(m- x)$.

    Finally, $\displaystyle y= -\frac{ln(m- x)}{n}$. Of course, that is a real number only for x< m.
    Thanks from brianx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2017
    From
    London
    Posts
    23

    Re: How to get a value from sigmoid function?

    Quote Originally Posted by HallsofIvy View Post
    Finally, $\displaystyle y= -\frac{ln(m- x)}{n}$. Of course, that is a real number only for x< m.
    I probably misunderstand a point.

    Consider $n=1, y=0$, then $x=\frac{1}{2}m$ but the $y(x)$ equation does not give the same.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035

    Re: How to get a value from sigmoid function?

    Quote Originally Posted by HallsofIvy View Post
    The term 'sigmoid function' can refer to a variety of functions but I suspect you are referring to $\displaystyle x(y)= \frac{me^{ny}}{e^{ny}+ 1}$. An obvious first step is to multiply both sides by $\displaystyle e^{ny}+ 1$ to get $\displaystyle x(e^{ny}+ 1)= me^{ny}$ and then $\displaystyle xe^{ny}+ x= me^{ny}$ which we can then write as $\displaystyle xe^{ny}- me^{ny}= e^{ny}(x- m)= -1$.
    This is an error. It should be $\displaystyle e^{ny}(x- m)= -x$, not -1.
    Then $\displaystyle e^{ny}= -\frac{x}{x- m}= \frac{x}{m- x}$

    $\displaystyle ny= ln\left(\frac{x}{m- x}\right)= ln(x)- ln(m- x)$

    $\displaystyle y= \frac{ln(x)- ln(m- x)}{n}$ or $\displaystyle y= \frac{1}{n}ln\left(\frac{x}{m- x}\right)$.

    Divide both sides by x- 1 to get $\displaystyle e^{ny}= \frac{-1}{x- m}= \frac{1}{m- x}$.

    Now take the natural logarithm of both sides: $\displaystyle ny= ln\left(\frac{1}{m- x}\right)= -ln(m- x)$.

    Finally, $\displaystyle y= -\frac{ln(m- x)}{n}$. Of course, that is a real number only for x< m.
    Thanks for catching that.
    Last edited by HallsofIvy; Aug 27th 2017 at 03:53 AM.
    Thanks from brianx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to simply sigmoid function?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Aug 24th 2017, 09:26 AM
  2. Replies: 2
    Last Post: Aug 16th 2012, 07:43 PM
  3. Derivative of sigmoid function query
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 14th 2011, 07:06 AM
  4. Sigmoid function question...
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Apr 28th 2009, 10:22 AM
  5. sigmoid function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Aug 7th 2006, 10:18 PM

/mathhelpforum @mathhelpforum