Max function with absolute values for arbitrary number of arguments
$begingroup$
The maximum between two numbers $x$ and $y$ can be easily written as
$$
max(x,y) = frac12left(x+y +|x-y|right).
$$
We can obviously generalize this to any number of arguments as
$$
max(x_1,dots,x_n) = max(x_1, max(... ,max(x_{n-1},x_n)...)) = text{ a mess with absolute values}
$$
but I do not like this. I would like to have a nicely written formula and the formula should look the same if we shuffle $x_1,dots,x_n$.
Why? You might ask? I'm doing some numerical computations where I need to smooth out the $max$ function, I just trivially smooth out the absolute value:
$$
max_{varepsilon}(x,y) = frac12left(x+y +|x-y|_{varepsilon}right) = frac12left(x+y +sqrt{ (x-y)^2 + varepsilon^2} - varepsilonright).
$$
However, I actually need to smooth out the $max$ function for arbitrary number of arguments. The basic requirement on $max_varepsilon(x_1,dots, x_n)$ is that it is invariant under any permutation of its arguments. Therefore if I have $max(x_1,dots,x_n)$ written out with absolute values and the expression is symmetrical in $x_1,dots,x_n$ then I can just replace $|cdot|$ with smoothed version of absolute value $|cdot |_varepsilon$ and I get the desired result.
Thus, is there an expression with just absolute values which is symmetrical in $x_1,dots,x_n$ and when evaluated it yields the maximum?
absolute-value maxima-minima
asked Jan 18 at 12:56
tomtom
2,92711131
$endgroup$
add a comment |
$begingroup$
The maximum between two numbers $x$ and $y$ can be easily written as
$$
max(x,y) = frac12left(x+y +|x-y|right).
$$
We can obviously generalize this to any number of arguments as
$$
max(x_1,dots,x_n) = max(x_1, max(... ,max(x_{n-1},x_n)...)) = text{ a mess with absolute values}
$$
but I do not like this. I would like to have a nicely written formula and the formula should look the same if we shuffle $x_1,dots,x_n$.
Why? You might ask? I'm doing some numerical computations where I need to smooth out the $max$ function, I just trivially smooth out the absolute value:
$$
max_{varepsilon}(x,y) = frac12left(x+y +|x-y|_{varepsilon}right) = frac12left(x+y +sqrt{ (x-y)^2 + varepsilon^2} - varepsilonright).
$$
However, I actually need to smooth out the $max$ function for arbitrary number of arguments. The basic requirement on $max_varepsilon(x_1,dots, x_n)$ is that it is invariant under any permutation of its arguments. Therefore if I have $max(x_1,dots,x_n)$ written out with absolute values and the expression is symmetrical in $x_1,dots,x_n$ then I can just replace $|cdot|$ with smoothed version of absolute value $|cdot |_varepsilon$ and I get the desired result.
Thus, is there an expression with just absolute values which is symmetrical in $x_1,dots,x_n$ and when evaluated it yields the maximum?
absolute-value maxima-minima
asked Jan 18 at 12:56
tomtom
2,92711131
$endgroup$
1
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09
add a comment |
$begingroup$
The maximum between two numbers $x$ and $y$ can be easily written as
$$
max(x,y) = frac12left(x+y +|x-y|right).
$$
We can obviously generalize this to any number of arguments as
$$
max(x_1,dots,x_n) = max(x_1, max(... ,max(x_{n-1},x_n)...)) = text{ a mess with absolute values}
$$
but I do not like this. I would like to have a nicely written formula and the formula should look the same if we shuffle $x_1,dots,x_n$.
Why? You might ask? I'm doing some numerical computations where I need to smooth out the $max$ function, I just trivially smooth out the absolute value:
$$
max_{varepsilon}(x,y) = frac12left(x+y +|x-y|_{varepsilon}right) = frac12left(x+y +sqrt{ (x-y)^2 + varepsilon^2} - varepsilonright).
$$
However, I actually need to smooth out the $max$ function for arbitrary number of arguments. The basic requirement on $max_varepsilon(x_1,dots, x_n)$ is that it is invariant under any permutation of its arguments. Therefore if I have $max(x_1,dots,x_n)$ written out with absolute values and the expression is symmetrical in $x_1,dots,x_n$ then I can just replace $|cdot|$ with smoothed version of absolute value $|cdot |_varepsilon$ and I get the desired result.
Thus, is there an expression with just absolute values which is symmetrical in $x_1,dots,x_n$ and when evaluated it yields the maximum?
absolute-value maxima-minima
asked Jan 18 at 12:56
tomtom
2,92711131
$endgroup$
The maximum between two numbers $x$ and $y$ can be easily written as
$$
max(x,y) = frac12left(x+y +|x-y|right).
$$
We can obviously generalize this to any number of arguments as
$$
max(x_1,dots,x_n) = max(x_1, max(... ,max(x_{n-1},x_n)...)) = text{ a mess with absolute values}
$$
but I do not like this. I would like to have a nicely written formula and the formula should look the same if we shuffle $x_1,dots,x_n$.
Why? You might ask? I'm doing some numerical computations where I need to smooth out the $max$ function, I just trivially smooth out the absolute value:
$$
max_{varepsilon}(x,y) = frac12left(x+y +|x-y|_{varepsilon}right) = frac12left(x+y +sqrt{ (x-y)^2 + varepsilon^2} - varepsilonright).
$$
However, I actually need to smooth out the $max$ function for arbitrary number of arguments. The basic requirement on $max_varepsilon(x_1,dots, x_n)$ is that it is invariant under any permutation of its arguments. Therefore if I have $max(x_1,dots,x_n)$ written out with absolute values and the expression is symmetrical in $x_1,dots,x_n$ then I can just replace $|cdot|$ with smoothed version of absolute value $|cdot |_varepsilon$ and I get the desired result.
Thus, is there an expression with just absolute values which is symmetrical in $x_1,dots,x_n$ and when evaluated it yields the maximum?
absolute-value maxima-minima
absolute-value maxima-minima
asked Jan 18 at 12:56
tomtom
2,92711131
asked Jan 18 at 12:56
tomtom
2,92711131
asked Jan 18 at 12:56
tomtom
2,92711131
asked Jan 18 at 12:56
tomtom
2,92711131
asked Jan 18 at 12:56
tomtom
2,92711131
2,92711131
1
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09
add a comment |
1
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09
1
1
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09
add a comment |
0
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1
$begingroup$
Would you be happy with en.wikipedia.org/wiki/Smooth_maximum ?
$endgroup$
– Ru Hasha
Jan 18 at 15:02
$begingroup$
Cool! That should solve my problem, but I'm still curious about the question I have posted.
$endgroup$
– tom
Jan 18 at 18:09