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Logistic function with a slope but no asymptotes?

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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

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  
  
  

  
  $begingroup$
  $operatornamesign(x)log(1 + |x|)$?
  $endgroup$
  – 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
  
  

  
  
  
  

 | 
show 5 more comments

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

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  
  
  

  
  $begingroup$
  $operatornamesign(x)log(1 + |x|)$?
  $endgroup$
  – 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
  
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

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

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

$endgroup$

share|cite|improve this question

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

share|cite|improve this question

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

edited Mar 21 at 6:33

Neil G

9,85013070

edited Mar 21 at 6:33

Neil G

9,85013070

asked Mar 20 at 15:44

AksakalAksakal

39k452120

asked Mar 20 at 15:44

AksakalAksakal

39k452120

3 Answers
3

active

oldest

votes

10

$begingroup$

You could just add a term to a logistic function:

$$
f(x; a, b, c, d, e)=fraca1+bexp(-cx) + dx + e
$$

The asymptotes will have slopes $d$.

Here is an example with $a=10, b = 1, c = 2, d = frac120, e = -5$:

share|cite|improve this answer

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

$endgroup$

• 
  
  
  

  
  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
  Mar 21 at 1:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 22 at 10:46
  

  
  
  
  

add a comment | 

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?
$$
textasinh(x) = logleft(x + sqrt1 + x^2right)
$$

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

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 $textasinh'(x) = frac1sqrt1+x^2$ so it has a nice simple derivative.

---

Original answer

$newcommandevarepsilon$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
$$
f(x) = begincases x^-1 & xneq 0 \ 0 & x = 0endcases
$$ work?

share|cite|improve this answer

edited Mar 20 at 18:19

answered Mar 20 at 16:15

jldjld

12.3k23353

$endgroup$

• 
  
  
  

  
  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
  Mar 22 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal if $a > 0$ then just doing $acdottextasinh$ would keep the shape and asymptotics the same and the derivative is $fracasqrt1+x^2$ so the slope at zero is just $a$
  $endgroup$
  – jld
  Mar 22 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal more generally we could consider the antiderivative of $fracasqrtc^2 + (bx)^2$ which is $$frac ab logleft(bleft(bx + sqrtc^2 + (bx)^2right)right)$$ and allows more ability to change the shape, or just something like $acdottextasinh(bx)$
  $endgroup$
  – jld
  Mar 22 at 17:29
  
  

  
  
  
  

 | 
show 1 more comment

6

$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatornamesign(x)logleft(1 + ,
$$
which has slope tending towards zero, but is unbounded.

edit by popular demand, a plot, for $|x|le 30$:

share|cite|improve this answer

edited Mar 21 at 22:04

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

$endgroup$

add a comment | 

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10

$begingroup$

You could just add a term to a logistic function:

$$
f(x; a, b, c, d, e)=fraca1+bexp(-cx) + dx + e
$$

The asymptotes will have slopes $d$.

Here is an example with $a=10, b = 1, c = 2, d = frac120, e = -5$:

share|cite|improve this answer

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

$endgroup$

• 
  
  
  

  
  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
  Mar 21 at 1:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 22 at 10:46
  

  
  
  
  

add a comment | 

10

$begingroup$

You could just add a term to a logistic function:

$$
f(x; a, b, c, d, e)=fraca1+bexp(-cx) + dx + e
$$

The asymptotes will have slopes $d$.

Here is an example with $a=10, b = 1, c = 2, d = frac120, e = -5$:

share|cite|improve this answer

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

$endgroup$

• 
  
  
  

  
  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
  Mar 21 at 1:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 22 at 10:46
  

  
  
  
  

add a comment | 

$begingroup$

You could just add a term to a logistic function:

$$
f(x; a, b, c, d, e)=fraca1+bexp(-cx) + dx + e
$$

The asymptotes will have slopes $d$.

Here is an example with $a=10, b = 1, c = 2, d = frac120, e = -5$:

share|cite|improve this answer

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

$endgroup$

share|cite|improve this answer

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

answered Mar 20 at 17:02

COOLSerdashCOOLSerdash

16.6k75294

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?
$$
textasinh(x) = logleft(x + sqrt1 + x^2right)
$$

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

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 $textasinh'(x) = frac1sqrt1+x^2$ so it has a nice simple derivative.

---

Original answer

$newcommandevarepsilon$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
$$
f(x) = begincases x^-1 & xneq 0 \ 0 & x = 0endcases
$$ work?

share|cite|improve this answer

edited Mar 20 at 18:19

answered Mar 20 at 16:15

jldjld

12.3k23353

$endgroup$

• 
  
  
  

  
  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
  Mar 22 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal if $a > 0$ then just doing $acdottextasinh$ would keep the shape and asymptotics the same and the derivative is $fracasqrt1+x^2$ so the slope at zero is just $a$
  $endgroup$
  – jld
  Mar 22 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal more generally we could consider the antiderivative of $fracasqrtc^2 + (bx)^2$ which is $$frac ab logleft(bleft(bx + sqrtc^2 + (bx)^2right)right)$$ and allows more ability to change the shape, or just something like $acdottextasinh(bx)$
  $endgroup$
  – jld
  Mar 22 at 17:29
  
  

  
  
  
  

 | 
show 1 more comment

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?
$$
textasinh(x) = logleft(x + sqrt1 + x^2right)
$$

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

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 $textasinh'(x) = frac1sqrt1+x^2$ so it has a nice simple derivative.

---

Original answer

$newcommandevarepsilon$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
$$
f(x) = begincases x^-1 & xneq 0 \ 0 & x = 0endcases
$$ work?

share|cite|improve this answer

edited Mar 20 at 18:19

answered Mar 20 at 16:15

jldjld

12.3k23353

$endgroup$

• 
  
  
  

  
  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
  Mar 22 at 10:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal if $a > 0$ then just doing $acdottextasinh$ would keep the shape and asymptotics the same and the derivative is $fracasqrt1+x^2$ so the slope at zero is just $a$
  $endgroup$
  – jld
  Mar 22 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aksakal more generally we could consider the antiderivative of $fracasqrtc^2 + (bx)^2$ which is $$frac ab logleft(bleft(bx + sqrtc^2 + (bx)^2right)right)$$ and allows more ability to change the shape, or just something like $acdottextasinh(bx)$
  $endgroup$
  – jld
  Mar 22 at 17:29
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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?
$$
textasinh(x) = logleft(x + sqrt1 + x^2right)
$$

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

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 $textasinh'(x) = frac1sqrt1+x^2$ so it has a nice simple derivative.

---

Original answer

$newcommandevarepsilon$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
$$
f(x) = begincases x^-1 & xneq 0 \ 0 & x = 0endcases
$$ work?

share|cite|improve this answer

edited Mar 20 at 18:19

answered Mar 20 at 16:15

jldjld

12.3k23353

$endgroup$

share|cite|improve this answer

edited Mar 20 at 18:19

answered Mar 20 at 16:15

jldjld

12.3k23353

answered Mar 20 at 16:15

jldjld

12.3k23353

answered Mar 20 at 16:15

jldjld

12.3k23353

6

$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatornamesign(x)logleft(1 + ,
$$
which has slope tending towards zero, but is unbounded.

edit by popular demand, a plot, for $|x|le 30$:

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edited Mar 21 at 22:04

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

$endgroup$

add a comment | 

6

$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatornamesign(x)logleft(1 + ,
$$
which has slope tending towards zero, but is unbounded.

edit by popular demand, a plot, for $|x|le 30$:

share|cite|improve this answer

edited Mar 21 at 22:04

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

$endgroup$

add a comment | 

$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatornamesign(x)logleft(1 + ,
$$
which has slope tending towards zero, but is unbounded.

edit by popular demand, a plot, for $|x|le 30$:

share|cite|improve this answer

edited Mar 21 at 22:04

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

$endgroup$

share|cite|improve this answer

edited Mar 21 at 22:04

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

answered Mar 20 at 18:49

steveo'americasteveo'america

24319

answered Mar 20 at 18:49

steveo'americasteveo'america

24319