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Logistic function with a slope but no asymptotes?

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Logistic function with a slope but no asymptotes?

Has Arcsinh ever been considered as a neural network activation function?Effect of e when using the Sigmoid Function as an activation functionApproximation of Δoutput in context of Sigmoid functionModification of Sigmoid functionFinding the center of a logistic curveInput and Output range of the composition of Gaussian and Sigmoidal functions and it's entropyFinding the slope at different points in a sigmoid curveQuestion about Sigmoid Function in Logistic RegressionHas Arcsinh ever been considered as a neural network activation function?The link between logistic regression and logistic sigmoidHow can I even out the output of the sigmoid function?

6

$begingroup$

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 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
  $endgroup$
  – jld
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically I want a function that looks like sigmoid but has a slope
  $endgroup$
  – Aksakal
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
  $endgroup$
  – jld
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $operatorname{sign}(x)log(1 + |x|)$?
  $endgroup$
  – steveo'america
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  
  
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

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 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
  $endgroup$
  – jld
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically I want a function that looks like sigmoid but has a slope
  $endgroup$
  – Aksakal
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
  $endgroup$
  – jld
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $operatorname{sign}(x)log(1 + |x|)$?
  $endgroup$
  – steveo'america
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  
  
  

  
  
  
  

 | 
show 3 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 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

$endgroup$

share|cite|improve this question

edited 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

share|cite|improve this question

edited 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

edited 21 mins ago

Neil G

9,79012970

edited 21 mins ago

Neil G

9,79012970

asked 15 hours ago

AksakalAksakal

39k452120

asked 15 hours ago

AksakalAksakal

39k452120

3 Answers
3

active

oldest

votes

8

$begingroup$

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

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

edited 12 hours ago

answered 14 hours ago

jldjld

12.3k23352

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:

share|cite|improve this answer

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

answered 12 hours ago

steveo'americasteveo'america

23318

$endgroup$

add a comment | 

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8

$begingroup$

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

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

edited 12 hours ago

answered 14 hours ago

jldjld

12.3k23352

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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

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

edited 12 hours ago

answered 14 hours ago

jldjld

12.3k23352

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a 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?
$$
text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
$$

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 $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ 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

edited 12 hours ago

answered 14 hours ago

jldjld

12.3k23352

$endgroup$

share|cite|improve this answer

edited 12 hours ago

answered 14 hours ago

jldjld

12.3k23352

answered 14 hours ago

jldjld

12.3k23352

answered 14 hours ago

jldjld

12.3k23352

6

$begingroup$

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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:

share|cite|improve this answer

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:

share|cite|improve this answer

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:

share|cite|improve this answer

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

share|cite|improve this answer

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

answered 13 hours ago

COOLSerdashCOOLSerdash

16.4k75293

5

$begingroup$

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.

share|cite|improve this answer

answered 12 hours ago

steveo'americasteveo'america

23318

$endgroup$

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

answered 12 hours ago

steveo'americasteveo'america

23318

$endgroup$

add a comment | 

$begingroup$

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.

share|cite|improve this answer

answered 12 hours ago

steveo'americasteveo'america

23318

$endgroup$

share|cite|improve this answer

answered 12 hours ago

steveo'americasteveo'america

23318

answered 12 hours ago

steveo'americasteveo'america

23318

answered 12 hours ago

steveo'americasteveo'america

23318