Functions that Tend To Non-Smooth Functions as Some Parameter Tends to Infinity
$begingroup$
I recently saw a post in which the query was about a function that tends to the Dirac delta function as a parameter in it tends to infinity. The function chosen was $${(1+cos x)^nover C}$$ as $ntoinfty$, with C being the integral of the given function from $-pi$ to $+pi$. The specific query of the post was infact to do with this normalisation constant. Another way of doing this would be with a parametrised Gaussian $$sqrtfrac{alpha}{pi}exp(-alpha x^2)$$ as $alphatoinfty$.
Various other non smooth functions can be represented as limits of parametrised smooth functions as the parameter tends to infinity: |x| can be represented as $${1overalpha}lncosh(alpha x)$$ (or as $${1overalpha}ln(2cosh(alpha x))$$ ... it doesn't really matter whether the 2 is there or not; the sharp-cornered-sigmoid function ( $$xleq -1 ⇒y=-1 ,$$$$ -1<x<+1⇒y=x, &$$$$xgeq 1⇒y=+1$$ by $$frac{1}{alpha}left(lnleft(1+exp(alpha(1+x))right)-lnleft(1+exp(alpha(1-x))right)right)-x .$$ or, equivalently
$$frac{1}{alpha}lnfrac{1+exp(alpha(1+x))}{1+exp(alpha(1-x))}-x .$$
Does anyone know any other limits of parametrised smooth functions for representing either these or other non-smooth functions in this kind of way?
I have found these kinds of function handy in computer graphics; and also in the numerical solution of differential equation: if, say, a forcing function or boundary condition is a non-smooth function of which the being encoded raw into the solution algorithm would cause the propagation of artifacts, then a function such as one of these could be coded-in instead ... and it could even be fine-tuned such that the sharpnesses of it be as much as the algorithm can tolerate.
And plugging the one for $|x|$ into the one for the sharp cornered sigmoid, and turning the result upside-down, you could get a unit tent function with $$1-{1overalpha}lnfrac{1+e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} ,$$ which you might as well reduce to $$1-{1overalpha}lnfrac{e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} $$ or $$1+{1overalpha}ln(e^{-alpha}+operatorname{sech}(alpha x)) .$$
approximation discontinuous-functions piecewise-continuity
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
|
show 6 more comments
$begingroup$
I recently saw a post in which the query was about a function that tends to the Dirac delta function as a parameter in it tends to infinity. The function chosen was $${(1+cos x)^nover C}$$ as $ntoinfty$, with C being the integral of the given function from $-pi$ to $+pi$. The specific query of the post was infact to do with this normalisation constant. Another way of doing this would be with a parametrised Gaussian $$sqrtfrac{alpha}{pi}exp(-alpha x^2)$$ as $alphatoinfty$.
Various other non smooth functions can be represented as limits of parametrised smooth functions as the parameter tends to infinity: |x| can be represented as $${1overalpha}lncosh(alpha x)$$ (or as $${1overalpha}ln(2cosh(alpha x))$$ ... it doesn't really matter whether the 2 is there or not; the sharp-cornered-sigmoid function ( $$xleq -1 ⇒y=-1 ,$$$$ -1<x<+1⇒y=x, &$$$$xgeq 1⇒y=+1$$ by $$frac{1}{alpha}left(lnleft(1+exp(alpha(1+x))right)-lnleft(1+exp(alpha(1-x))right)right)-x .$$ or, equivalently
$$frac{1}{alpha}lnfrac{1+exp(alpha(1+x))}{1+exp(alpha(1-x))}-x .$$
Does anyone know any other limits of parametrised smooth functions for representing either these or other non-smooth functions in this kind of way?
I have found these kinds of function handy in computer graphics; and also in the numerical solution of differential equation: if, say, a forcing function or boundary condition is a non-smooth function of which the being encoded raw into the solution algorithm would cause the propagation of artifacts, then a function such as one of these could be coded-in instead ... and it could even be fine-tuned such that the sharpnesses of it be as much as the algorithm can tolerate.
And plugging the one for $|x|$ into the one for the sharp cornered sigmoid, and turning the result upside-down, you could get a unit tent function with $$1-{1overalpha}lnfrac{1+e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} ,$$ which you might as well reduce to $$1-{1overalpha}lnfrac{e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} $$ or $$1+{1overalpha}ln(e^{-alpha}+operatorname{sech}(alpha x)) .$$
approximation discontinuous-functions piecewise-continuity
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
1
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
1
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
1
$begingroup$
No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:53
$begingroup$
In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:56
|
show 6 more comments
$begingroup$
I recently saw a post in which the query was about a function that tends to the Dirac delta function as a parameter in it tends to infinity. The function chosen was $${(1+cos x)^nover C}$$ as $ntoinfty$, with C being the integral of the given function from $-pi$ to $+pi$. The specific query of the post was infact to do with this normalisation constant. Another way of doing this would be with a parametrised Gaussian $$sqrtfrac{alpha}{pi}exp(-alpha x^2)$$ as $alphatoinfty$.
Various other non smooth functions can be represented as limits of parametrised smooth functions as the parameter tends to infinity: |x| can be represented as $${1overalpha}lncosh(alpha x)$$ (or as $${1overalpha}ln(2cosh(alpha x))$$ ... it doesn't really matter whether the 2 is there or not; the sharp-cornered-sigmoid function ( $$xleq -1 ⇒y=-1 ,$$$$ -1<x<+1⇒y=x, &$$$$xgeq 1⇒y=+1$$ by $$frac{1}{alpha}left(lnleft(1+exp(alpha(1+x))right)-lnleft(1+exp(alpha(1-x))right)right)-x .$$ or, equivalently
$$frac{1}{alpha}lnfrac{1+exp(alpha(1+x))}{1+exp(alpha(1-x))}-x .$$
Does anyone know any other limits of parametrised smooth functions for representing either these or other non-smooth functions in this kind of way?
I have found these kinds of function handy in computer graphics; and also in the numerical solution of differential equation: if, say, a forcing function or boundary condition is a non-smooth function of which the being encoded raw into the solution algorithm would cause the propagation of artifacts, then a function such as one of these could be coded-in instead ... and it could even be fine-tuned such that the sharpnesses of it be as much as the algorithm can tolerate.
And plugging the one for $|x|$ into the one for the sharp cornered sigmoid, and turning the result upside-down, you could get a unit tent function with $$1-{1overalpha}lnfrac{1+e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} ,$$ which you might as well reduce to $$1-{1overalpha}lnfrac{e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} $$ or $$1+{1overalpha}ln(e^{-alpha}+operatorname{sech}(alpha x)) .$$
approximation discontinuous-functions piecewise-continuity
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
I recently saw a post in which the query was about a function that tends to the Dirac delta function as a parameter in it tends to infinity. The function chosen was $${(1+cos x)^nover C}$$ as $ntoinfty$, with C being the integral of the given function from $-pi$ to $+pi$. The specific query of the post was infact to do with this normalisation constant. Another way of doing this would be with a parametrised Gaussian $$sqrtfrac{alpha}{pi}exp(-alpha x^2)$$ as $alphatoinfty$.
Various other non smooth functions can be represented as limits of parametrised smooth functions as the parameter tends to infinity: |x| can be represented as $${1overalpha}lncosh(alpha x)$$ (or as $${1overalpha}ln(2cosh(alpha x))$$ ... it doesn't really matter whether the 2 is there or not; the sharp-cornered-sigmoid function ( $$xleq -1 ⇒y=-1 ,$$$$ -1<x<+1⇒y=x, &$$$$xgeq 1⇒y=+1$$ by $$frac{1}{alpha}left(lnleft(1+exp(alpha(1+x))right)-lnleft(1+exp(alpha(1-x))right)right)-x .$$ or, equivalently
$$frac{1}{alpha}lnfrac{1+exp(alpha(1+x))}{1+exp(alpha(1-x))}-x .$$
Does anyone know any other limits of parametrised smooth functions for representing either these or other non-smooth functions in this kind of way?
I have found these kinds of function handy in computer graphics; and also in the numerical solution of differential equation: if, say, a forcing function or boundary condition is a non-smooth function of which the being encoded raw into the solution algorithm would cause the propagation of artifacts, then a function such as one of these could be coded-in instead ... and it could even be fine-tuned such that the sharpnesses of it be as much as the algorithm can tolerate.
And plugging the one for $|x|$ into the one for the sharp cornered sigmoid, and turning the result upside-down, you could get a unit tent function with $$1-{1overalpha}lnfrac{1+e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} ,$$ which you might as well reduce to $$1-{1overalpha}lnfrac{e^alpha cosh(alpha x)}{e^alpha+cosh(alpha x)} $$ or $$1+{1overalpha}ln(e^{-alpha}+operatorname{sech}(alpha x)) .$$
approximation discontinuous-functions piecewise-continuity
approximation discontinuous-functions piecewise-continuity
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
edited Dec 2 '18 at 22:45
AmbretteOrrisey
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
asked Dec 2 '18 at 20:39
AmbretteOrriseyAmbretteOrrisey
54210
54210
$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
1
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
1
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
1
$begingroup$
No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:53
$begingroup$
In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:56
|
show 6 more comments
$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
1
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
1
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
1
$begingroup$
No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:53
$begingroup$
In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:56
$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
1
1
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
1
1
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
1
1
$begingroup$
No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:53
$begingroup$
No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:53
$begingroup$
In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:56
$begingroup$
In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:56
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To get the heavyside step function, any unboundedly 'encelerated' sigmoid function can be used, with a little scaling & vertical displacement, such as $$frac{1}{2}+frac{1}{pi}operatorname{atn}(alpha x)$$ or $$frac{1}{2}(1+operatorname{tanh}(alpha x)) .$$ The function $tanh$ has the theoretical advantage of being one that cleaves more closely to the limits $y=-1$ & $y=+1$; but on the other hand, it's worth keeping both of these in mind, as in a use of such as these in practice for the kind of thing I have given a use of them for, the greater 'slackness' of $operatorname{atn}$ could actually be an advantage (in fact I have found that sometimes it is). Also in similar fashion a square pulse can be represented by $$frac{1}{pi}(operatorname{atn}(alpha (1-x))+operatorname{atn}(alpha (1+x)))$$ or by $$frac{1}{2}(operatorname{tanh}(alpha (1-x))+operatorname{tanh}(alpha (1+x))) .$$
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
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$begingroup$
To get the heavyside step function, any unboundedly 'encelerated' sigmoid function can be used, with a little scaling & vertical displacement, such as $$frac{1}{2}+frac{1}{pi}operatorname{atn}(alpha x)$$ or $$frac{1}{2}(1+operatorname{tanh}(alpha x)) .$$ The function $tanh$ has the theoretical advantage of being one that cleaves more closely to the limits $y=-1$ & $y=+1$; but on the other hand, it's worth keeping both of these in mind, as in a use of such as these in practice for the kind of thing I have given a use of them for, the greater 'slackness' of $operatorname{atn}$ could actually be an advantage (in fact I have found that sometimes it is). Also in similar fashion a square pulse can be represented by $$frac{1}{pi}(operatorname{atn}(alpha (1-x))+operatorname{atn}(alpha (1+x)))$$ or by $$frac{1}{2}(operatorname{tanh}(alpha (1-x))+operatorname{tanh}(alpha (1+x))) .$$
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
To get the heavyside step function, any unboundedly 'encelerated' sigmoid function can be used, with a little scaling & vertical displacement, such as $$frac{1}{2}+frac{1}{pi}operatorname{atn}(alpha x)$$ or $$frac{1}{2}(1+operatorname{tanh}(alpha x)) .$$ The function $tanh$ has the theoretical advantage of being one that cleaves more closely to the limits $y=-1$ & $y=+1$; but on the other hand, it's worth keeping both of these in mind, as in a use of such as these in practice for the kind of thing I have given a use of them for, the greater 'slackness' of $operatorname{atn}$ could actually be an advantage (in fact I have found that sometimes it is). Also in similar fashion a square pulse can be represented by $$frac{1}{pi}(operatorname{atn}(alpha (1-x))+operatorname{atn}(alpha (1+x)))$$ or by $$frac{1}{2}(operatorname{tanh}(alpha (1-x))+operatorname{tanh}(alpha (1+x))) .$$
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
To get the heavyside step function, any unboundedly 'encelerated' sigmoid function can be used, with a little scaling & vertical displacement, such as $$frac{1}{2}+frac{1}{pi}operatorname{atn}(alpha x)$$ or $$frac{1}{2}(1+operatorname{tanh}(alpha x)) .$$ The function $tanh$ has the theoretical advantage of being one that cleaves more closely to the limits $y=-1$ & $y=+1$; but on the other hand, it's worth keeping both of these in mind, as in a use of such as these in practice for the kind of thing I have given a use of them for, the greater 'slackness' of $operatorname{atn}$ could actually be an advantage (in fact I have found that sometimes it is). Also in similar fashion a square pulse can be represented by $$frac{1}{pi}(operatorname{atn}(alpha (1-x))+operatorname{atn}(alpha (1+x)))$$ or by $$frac{1}{2}(operatorname{tanh}(alpha (1-x))+operatorname{tanh}(alpha (1+x))) .$$
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
To get the heavyside step function, any unboundedly 'encelerated' sigmoid function can be used, with a little scaling & vertical displacement, such as $$frac{1}{2}+frac{1}{pi}operatorname{atn}(alpha x)$$ or $$frac{1}{2}(1+operatorname{tanh}(alpha x)) .$$ The function $tanh$ has the theoretical advantage of being one that cleaves more closely to the limits $y=-1$ & $y=+1$; but on the other hand, it's worth keeping both of these in mind, as in a use of such as these in practice for the kind of thing I have given a use of them for, the greater 'slackness' of $operatorname{atn}$ could actually be an advantage (in fact I have found that sometimes it is). Also in similar fashion a square pulse can be represented by $$frac{1}{pi}(operatorname{atn}(alpha (1-x))+operatorname{atn}(alpha (1+x)))$$ or by $$frac{1}{2}(operatorname{tanh}(alpha (1-x))+operatorname{tanh}(alpha (1+x))) .$$
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
edited Dec 3 '18 at 10:56
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 3 '18 at 10:25
AmbretteOrriseyAmbretteOrrisey
54210
54210
add a comment |
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$begingroup$
Sequence $f_n$ with $f_n(x)=frac{2}{pi}mathrm(atan)(nx)$ converges to a variant of Heaviside step function.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:13
1
$begingroup$
Ah yes, thanks - I see that: & you could get the Heavyside function itself with lim{α→∞}(1/2+(1/π)atn(αx))
$endgroup$
– AmbretteOrrisey
Dec 2 '18 at 21:25
1
$begingroup$
Yes. About the first function you give, a simpler version is : $Kcos(x)^n$. See the answer of yves Daoust in math.stackexchange.com/q/2293384 given for convergence towards a gauss function with a different normalization.
$endgroup$
– Jean Marie
Dec 2 '18 at 21:42
1
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No doubt though the "1+" business is an expedient for obviating negative values ... and also it results in a function that meets the x-axis tangentially in the 'wings'. The post is "Proving a sequence of functions is an aporoximation to the identity". I havent got the URL, as I'm using the 'app', & the URL is not displayed in it. But I'll check your link out & look at the counter reasons.
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– AmbretteOrrisey
Dec 2 '18 at 21:53
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In fact, due to formula $1+cos(2a)=2cos(a)^2$, it amount to the same...
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– Jean Marie
Dec 2 '18 at 21:56