Logistic function with a slope but no asymptotes?Has Arcsinh ever been considered as a neural network...
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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 curveFinding 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?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
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
edited Mar 21 at 6:33
Neil G
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
$endgroup$
|
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
edited Mar 21 at 6:33
Neil G
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
$endgroup$
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
|
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
edited Mar 21 at 6:33
Neil G
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
$endgroup$
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
edited Mar 21 at 6:33
Neil G
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
edited Mar 21 at 6:33
Neil G
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
edited Mar 21 at 6:33
Neil G
9,85013070
edited Mar 21 at 6:33
Neil G
9,85013070
edited Mar 21 at 6:33
Neil G
9,85013070
9,85013070
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
asked Mar 20 at 15:44
AksakalAksakal
39.1k452120
39.1k452120
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
|
show 5 more comments
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
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:
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 |
$begingroup$
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?
answered Mar 20 at 16:15
jldjld
12.4k23353
$endgroup$
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
Mar 22 at 17:29
|
show 1 more 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.
edit by popular demand, a plot, for $|x|le 30$:
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:
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:
$$
f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
$$
The asymptotes will have slopes $d$.
Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:
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:
$$
f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
$$
The asymptotes will have slopes $d$.
Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:
answered Mar 20 at 17:02
COOLSerdashCOOLSerdash
16.6k75294
$endgroup$
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 $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:
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
answered Mar 20 at 17:02
COOLSerdashCOOLSerdash
16.6k75294
16.6k75294
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 |
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
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
Mar 21 at 1:30
$begingroup$
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
$begingroup$
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 |
$begingroup$
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?
answered Mar 20 at 16:15
jldjld
12.4k23353
$endgroup$
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
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?
$$
text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
$$
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?
answered Mar 20 at 16:15
jldjld
12.4k23353
$endgroup$
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
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?
$$
text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
$$
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?
answered Mar 20 at 16:15
jldjld
12.4k23353
$endgroup$
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
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?
answered Mar 20 at 16:15
jldjld
12.4k23353
edited Mar 20 at 18:19
answered Mar 20 at 16:15
jldjld
12.4k23353
answered Mar 20 at 16:15
jldjld
12.4k23353
answered Mar 20 at 16:15
jldjld
12.4k23353
12.4k23353
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
Mar 22 at 17:29
|
show 1 more comment
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
Mar 22 at 17:29
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
Mar 22 at 10:48
$begingroup$
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 $acdottext{asinh}$ 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
Mar 22 at 17:04
$begingroup$
@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
Mar 22 at 17:04
$begingroup$
@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
Mar 22 at 17:29
$begingroup$
@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
Mar 22 at 17:29
|
show 1 more 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.
edit by popular demand, a plot, for $|x|le 30$:
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
$endgroup$
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.
edit by popular demand, a plot, for $|x|le 30$:
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
$endgroup$
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.
edit by popular demand, a plot, for $|x|le 30$:
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
$endgroup$
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 $|x|le 30$:
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
edited Mar 21 at 22:04
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
answered Mar 20 at 18:49
steveo'americasteveo'america
24519
24519
add a comment |
add a comment |
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$begingroup$
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