Logistic function with a slope but no asymptotes?












8












$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    Mar 20 at 16:17










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    Mar 20 at 16:24










  • $begingroup$
    Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
    $endgroup$
    – Aksakal
    Mar 20 at 16:31






  • 6




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    Mar 20 at 16:42








  • 4




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    Mar 20 at 21:39


















8












$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    Mar 20 at 16:17










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    Mar 20 at 16:24










  • $begingroup$
    Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
    $endgroup$
    – Aksakal
    Mar 20 at 16:31






  • 6




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    Mar 20 at 16:42








  • 4




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    Mar 20 at 21:39
















8












8








8





$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$




The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve




sigmoid-curve






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Neil G

9,81012970




9,81012970










asked Mar 20 at 15:44









AksakalAksakal

39k452120




39k452120








  • 2




    $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    Mar 20 at 16:17










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    Mar 20 at 16:24










  • $begingroup$
    Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
    $endgroup$
    – Aksakal
    Mar 20 at 16:31






  • 6




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    Mar 20 at 16:42








  • 4




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    Mar 20 at 21:39
















  • 2




    $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    Mar 20 at 16:17










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    Mar 20 at 16:24










  • $begingroup$
    Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
    $endgroup$
    – Aksakal
    Mar 20 at 16:31






  • 6




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    Mar 20 at 16:42








  • 4




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    Mar 20 at 21:39










2




2




$begingroup$
The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
$endgroup$
– jld
Mar 20 at 16:17




$begingroup$
The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
$endgroup$
– jld
Mar 20 at 16:17












$begingroup$
Basically I want a function that looks like sigmoid but has a slope
$endgroup$
– Aksakal
Mar 20 at 16:24




$begingroup$
Basically I want a function that looks like sigmoid but has a slope
$endgroup$
– Aksakal
Mar 20 at 16:24












$begingroup$
Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
$endgroup$
– Aksakal
Mar 20 at 16:31




$begingroup$
Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
$endgroup$
– Aksakal
Mar 20 at 16:31




6




6




$begingroup$
$operatorname{sign}(x)log(1 + |x|)$?
$endgroup$
– steveo'america
Mar 20 at 16:42






$begingroup$
$operatorname{sign}(x)log(1 + |x|)$?
$endgroup$
– steveo'america
Mar 20 at 16:42






4




4




$begingroup$
Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
$endgroup$
– usεr11852
Mar 20 at 21:39






$begingroup$
Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
$endgroup$
– usεr11852
Mar 20 at 21:39












3 Answers
3






active

oldest

votes


















10












$begingroup$

You could just add a term to a logistic function:



$$
f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
$$



The asymptotes will have slopes $d$.



Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



Sigmoid






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    2 days ago










  • $begingroup$
    this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
    $endgroup$
    – Aksakal
    yesterday



















11












$begingroup$

Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
$$
text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
$$



This is unbounded but grows like $log$ for large $|x|$ and looks like
asinh



I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




Original answer



$newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
$$
lim_{xtopm infty} f(x) = 0.
$$



Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
$$
exists x_1 : x < x_1 implies |f(x)| < e
$$
and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



This means that any such function can't be continuous. Would something like
$$
f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
$$ work?






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    Mar 20 at 18:52












  • $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    Mar 20 at 23:41










  • $begingroup$
    how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
    $endgroup$
    – Aksakal
    yesterday










  • $begingroup$
    @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
    $endgroup$
    – jld
    yesterday












  • $begingroup$
    @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
    $endgroup$
    – jld
    yesterday





















6












$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
$$
which has slope tending towards zero, but is unbounded.



edit by popular demand, a plot, for $|x|le 30$:
enter image description here






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









    10












    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      2 days ago










    • $begingroup$
      this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
      $endgroup$
      – Aksakal
      yesterday
















    10












    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      2 days ago










    • $begingroup$
      this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
      $endgroup$
      – Aksakal
      yesterday














    10












    10








    10





    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$



    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 20 at 17:02









    COOLSerdashCOOLSerdash

    16.5k75294




    16.5k75294








    • 2




      $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      2 days ago










    • $begingroup$
      this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
      $endgroup$
      – Aksakal
      yesterday














    • 2




      $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      2 days ago










    • $begingroup$
      this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
      $endgroup$
      – Aksakal
      yesterday








    2




    2




    $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    2 days ago




    $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    2 days ago












    $begingroup$
    this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
    $endgroup$
    – Aksakal
    yesterday




    $begingroup$
    this seemed to work for my dataset, and I picked it, but the solution is not ideal since the asymptotic slope doesn't decrease
    $endgroup$
    – Aksakal
    yesterday













    11












    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      Mar 20 at 18:52












    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      Mar 20 at 23:41










    • $begingroup$
      how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
      $endgroup$
      – Aksakal
      yesterday










    • $begingroup$
      @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
      $endgroup$
      – jld
      yesterday












    • $begingroup$
      @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
      $endgroup$
      – jld
      yesterday


















    11












    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      Mar 20 at 18:52












    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      Mar 20 at 23:41










    • $begingroup$
      how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
      $endgroup$
      – Aksakal
      yesterday










    • $begingroup$
      @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
      $endgroup$
      – jld
      yesterday












    • $begingroup$
      @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
      $endgroup$
      – jld
      yesterday
















    11












    11








    11





    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$



    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 20 at 18:19

























    answered Mar 20 at 16:15









    jldjld

    12.3k23353




    12.3k23353








    • 2




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      Mar 20 at 18:52












    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      Mar 20 at 23:41










    • $begingroup$
      how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
      $endgroup$
      – Aksakal
      yesterday










    • $begingroup$
      @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
      $endgroup$
      – jld
      yesterday












    • $begingroup$
      @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
      $endgroup$
      – jld
      yesterday
















    • 2




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      Mar 20 at 18:52












    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      Mar 20 at 23:41










    • $begingroup$
      how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
      $endgroup$
      – Aksakal
      yesterday










    • $begingroup$
      @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
      $endgroup$
      – jld
      yesterday












    • $begingroup$
      @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
      $endgroup$
      – jld
      yesterday










    2




    2




    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    Mar 20 at 18:52






    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    Mar 20 at 18:52














    $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    Mar 20 at 23:41




    $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    Mar 20 at 23:41












    $begingroup$
    how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
    $endgroup$
    – Aksakal
    yesterday




    $begingroup$
    how could you parameterize this function to change it's shape? in particular, to regulate the slope at the inflection point
    $endgroup$
    – Aksakal
    yesterday












    $begingroup$
    @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
    $endgroup$
    – jld
    yesterday






    $begingroup$
    @Aksakal if $a > 0$ then just doing $acdottext{asinh}$ would keep the shape and asymptotics the same and the derivative is $frac{a}{sqrt{1+x^2}}$ so the slope at zero is just $a$
    $endgroup$
    – jld
    yesterday














    $begingroup$
    @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
    $endgroup$
    – jld
    yesterday






    $begingroup$
    @Aksakal more generally we could consider the antiderivative of $frac{a}{sqrt{c^2 + (bx)^2}}$ which is $$frac ab logleft(bleft(bx + sqrt{c^2 + (bx)^2}right)right)$$ and allows more ability to change the shape, or just something like $acdottext{asinh}(bx)$
    $endgroup$
    – jld
    yesterday













    6












    $begingroup$

    I will go ahead and turn the comment into an answer. I suggest
    $$
    f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
    $$
    which has slope tending towards zero, but is unbounded.



    edit by popular demand, a plot, for $|x|le 30$:
    enter image description here






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      I will go ahead and turn the comment into an answer. I suggest
      $$
      f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
      $$
      which has slope tending towards zero, but is unbounded.



      edit by popular demand, a plot, for $|x|le 30$:
      enter image description here






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        I will go ahead and turn the comment into an answer. I suggest
        $$
        f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
        $$
        which has slope tending towards zero, but is unbounded.



        edit by popular demand, a plot, for $|x|le 30$:
        enter image description here






        share|cite|improve this answer











        $endgroup$



        I will go ahead and turn the comment into an answer. I suggest
        $$
        f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
        $$
        which has slope tending towards zero, but is unbounded.



        edit by popular demand, a plot, for $|x|le 30$:
        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Mar 20 at 18:49









        steveo'americasteveo'america

        24319




        24319






























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