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
    14

    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
    19,240
    Thanks
    2835

    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 x(y)= \frac{me^{ny}}{e^{ny}+ 1}. An obvious first step is to multiply both sides by e^{ny}+ 1 to get x(e^{ny}+ 1)= me^{ny} and then xe^{ny}+ x= me^{ny} which we can then write as xe^{ny}- me^{ny}= e^{ny}(x- m)= -1. Divide both sides by x- 1 to get e^{ny}= \frac{-1}{x- m}= \frac{1}{m- x}.

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

    Finally, 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
    14

    Re: How to get a value from sigmoid function?

    Quote Originally Posted by HallsofIvy View Post
    Finally, 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
    19,240
    Thanks
    2835

    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 x(y)= \frac{me^{ny}}{e^{ny}+ 1}. An obvious first step is to multiply both sides by e^{ny}+ 1 to get x(e^{ny}+ 1)= me^{ny} and then xe^{ny}+ x= me^{ny} which we can then write as xe^{ny}- me^{ny}= e^{ny}(x- m)= -1.
    This is an error. It should be e^{ny}(x- m)= -x, not -1.
    Then e^{ny}= -\frac{x}{x- m}= \frac{x}{m- x}

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

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

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

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

    Finally, 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