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Sigmoid with a slope but no asymptotes?

Has Arcsinh ever been considered as a neural network activation function?How to do LASSO regression with a dependent variable that is continuous between 0 and 1Approximation of Δoutput in context of Sigmoid functionModification of Sigmoid functionInput 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?Compare two sigmoid shape curvesHow can I even out the output of the sigmoid function?Sigmoid of a sum

4

$begingroup$

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

What is an alternative to a sigmoid 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 3 hours ago

Aksakal

asked 6 hours ago

AksakalAksakal

38.9k452120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically I want a function that looks like sigmoid but has a slope
  $endgroup$
  – Aksakal
  5 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 $lim_xtopminftyf(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
  $endgroup$
  – jld
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
  $endgroup$
  – Aksakal
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $operatornamesign(x)log(1 + |x|)$?
  $endgroup$
  – steveo'america
  5 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

4

$begingroup$

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

What is an alternative to a sigmoid 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 3 hours ago

Aksakal

asked 6 hours ago

AksakalAksakal

38.9k452120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically I want a function that looks like sigmoid but has a slope
  $endgroup$
  – Aksakal
  5 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 $lim_xtopminftyf(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
  $endgroup$
  – jld
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, a sigmoid like shape that doesn’t completely flatten, e.g. log function doesn’t completely flatten
  $endgroup$
  – Aksakal
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $operatornamesign(x)log(1 + |x|)$?
  $endgroup$
  – steveo'america
  5 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

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

What is an alternative to a sigmoid 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 3 hours ago

Aksakal

asked 6 hours ago

AksakalAksakal

38.9k452120

$endgroup$

share|cite|improve this question

edited 3 hours ago

Aksakal

asked 6 hours ago

AksakalAksakal

38.9k452120

share|cite|improve this question

edited 3 hours ago

Aksakal

asked 6 hours ago

AksakalAksakal

38.9k452120

asked 6 hours ago

AksakalAksakal

38.9k452120

asked 6 hours ago

AksakalAksakal

38.9k452120

3 Answers
3

active

oldest

votes

6

$begingroup$

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 3 hours ago

answered 5 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
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

answered 3 hours ago

steveo'americasteveo'america

21318

$endgroup$

add a comment | 

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6

$begingroup$

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 3 hours ago

answered 5 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
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  1 hour ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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 3 hours ago

answered 5 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
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sycorax thanks, i was wondering about that
  $endgroup$
  – jld
  1 hour 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?
$$
textasinh(x) = logleft(x + sqrt1 + x^2right)
$$

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 3 hours ago

answered 5 hours ago

jldjld

12.3k23352

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 5 hours ago

jldjld

12.3k23352

answered 5 hours ago

jldjld

12.3k23352

answered 5 hours ago

jldjld

12.3k23352

4

$begingroup$

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 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

add a comment | 

4

$begingroup$

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 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

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 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

$endgroup$

share|cite|improve this answer

answered 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

answered 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

answered 5 hours ago

COOLSerdashCOOLSerdash

16.4k75293

3

$begingroup$

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

share|cite|improve this answer

answered 3 hours ago

steveo'americasteveo'america

21318

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

answered 3 hours ago

steveo'americasteveo'america

21318

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 3 hours ago

steveo'americasteveo'america

21318

$endgroup$

share|cite|improve this answer

answered 3 hours ago

steveo'americasteveo'america

21318

answered 3 hours ago

steveo'americasteveo'america

21318

answered 3 hours ago

steveo'americasteveo'america

21318