What function symmetric and has unique solution? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need mathematical function for “adding” 0.5 and 0.5 and getting 0.4Show that $frac(1-y^-1)^2(1-y^-1 -c)(1-y^-1+c )$ is decreasing in $y > 1 $.Understanding summations with PoissonIs 'a' differentiable in f when f is a product of a differentiable and non-differentiable function?Reflection with respect to a parabolaIs there a function $f(x,y)$ such that f is equal to “a” when $x>y$ and “b” when $x<y$ and “c”when $x=y$?Symmetric Direct Product Distributive?Evaluating a sum of factorials and exponentialsHow to choose a set of numbers such that sum of any combination of numbers in that set yields a unique result?Wassertein and symmetry
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What function symmetric and has unique solution?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need mathematical function for “adding” 0.5 and 0.5 and getting 0.4Show that $frac(1-y^-1)^2(1-y^-1 -c)(1-y^-1+c )$ is decreasing in $y > 1 $.Understanding summations with PoissonIs 'a' differentiable in f when f is a product of a differentiable and non-differentiable function?Reflection with respect to a parabolaIs there a function $f(x,y)$ such that f is equal to “a” when $x>y$ and “b” when $x<y$ and “c”when $x=y$?Symmetric Direct Product Distributive?Evaluating a sum of factorials and exponentialsHow to choose a set of numbers such that sum of any combination of numbers in that set yields a unique result?Wassertein and symmetry
$begingroup$
I have multiple operands, say a, b, and c. I want an operator acting on a, b, and c, but the result should be invariant of order of operands (symmetric), and there should be no other pair with same result (unique).
What I tried:
Sum: a+b is symmetric, but 5+3 = 7+1.
Similarly, product.
I am unable to prove or disprove whether (a+b/ab) has the property I am asking for.
Does anybody have any ideas about the function I am looking for?
functions summation products symmetry
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
$endgroup$
add a comment |
$begingroup$
I have multiple operands, say a, b, and c. I want an operator acting on a, b, and c, but the result should be invariant of order of operands (symmetric), and there should be no other pair with same result (unique).
What I tried:
Sum: a+b is symmetric, but 5+3 = 7+1.
Similarly, product.
I am unable to prove or disprove whether (a+b/ab) has the property I am asking for.
Does anybody have any ideas about the function I am looking for?
functions summation products symmetry
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
$endgroup$
$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52
add a comment |
$begingroup$
I have multiple operands, say a, b, and c. I want an operator acting on a, b, and c, but the result should be invariant of order of operands (symmetric), and there should be no other pair with same result (unique).
What I tried:
Sum: a+b is symmetric, but 5+3 = 7+1.
Similarly, product.
I am unable to prove or disprove whether (a+b/ab) has the property I am asking for.
Does anybody have any ideas about the function I am looking for?
functions summation products symmetry
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
$endgroup$
I have multiple operands, say a, b, and c. I want an operator acting on a, b, and c, but the result should be invariant of order of operands (symmetric), and there should be no other pair with same result (unique).
What I tried:
Sum: a+b is symmetric, but 5+3 = 7+1.
Similarly, product.
I am unable to prove or disprove whether (a+b/ab) has the property I am asking for.
Does anybody have any ideas about the function I am looking for?
functions summation products symmetry
functions summation products symmetry
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
edited Mar 25 at 20:09
J. W. Tanner
4,7871420
4,7871420
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
asked Mar 25 at 18:31
Ajinkya GawaliAjinkya Gawali
234
234
$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52
add a comment |
$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52
$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you want to define such a function for $a,b$ in the positive integers and with values in the positive integers, you can create it by filling in a table of values without repeating any values, except that the value for $(a,b)$ is the same as the one $(b,a)$. It might start out like this:
$$
beginarrayc
&1&2&3&4&5\
hline
1&1&2&4&7&11\
2&2&3&5&8&12\
3&4&5&6&9&13\
4&7&8&9&10&14\
5&11&12&13&14&15
endarray
$$
In this example, the values in each row up to the diagonal are $1$; then $2,3$; then $4,5,6$, etc.
That defines a function which is symmetric, but otherwise does not repeat value.
If you want a formula,
$$
f(a,b) = begincasesfrac12 (a^2 - a) + b,& text for a ge b\ \
frac12 (b^2 - b) + a,& text for a le bendcases
$$
or, without the cases,
$$
f(a,b) = frac14(a^2 + b^2 + a + b + (a + b - 3)|a-b|).
$$
answered Mar 25 at 19:49
FredHFredH
3,7401024
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to define such a function for $a,b$ in the positive integers and with values in the positive integers, you can create it by filling in a table of values without repeating any values, except that the value for $(a,b)$ is the same as the one $(b,a)$. It might start out like this:
$$
beginarrayc
&1&2&3&4&5\
hline
1&1&2&4&7&11\
2&2&3&5&8&12\
3&4&5&6&9&13\
4&7&8&9&10&14\
5&11&12&13&14&15
endarray
$$
In this example, the values in each row up to the diagonal are $1$; then $2,3$; then $4,5,6$, etc.
That defines a function which is symmetric, but otherwise does not repeat value.
If you want a formula,
$$
f(a,b) = begincasesfrac12 (a^2 - a) + b,& text for a ge b\ \
frac12 (b^2 - b) + a,& text for a le bendcases
$$
or, without the cases,
$$
f(a,b) = frac14(a^2 + b^2 + a + b + (a + b - 3)|a-b|).
$$
answered Mar 25 at 19:49
FredHFredH
3,7401024
$endgroup$
add a comment |
$begingroup$
If you want to define such a function for $a,b$ in the positive integers and with values in the positive integers, you can create it by filling in a table of values without repeating any values, except that the value for $(a,b)$ is the same as the one $(b,a)$. It might start out like this:
$$
beginarrayc
&1&2&3&4&5\
hline
1&1&2&4&7&11\
2&2&3&5&8&12\
3&4&5&6&9&13\
4&7&8&9&10&14\
5&11&12&13&14&15
endarray
$$
In this example, the values in each row up to the diagonal are $1$; then $2,3$; then $4,5,6$, etc.
That defines a function which is symmetric, but otherwise does not repeat value.
If you want a formula,
$$
f(a,b) = begincasesfrac12 (a^2 - a) + b,& text for a ge b\ \
frac12 (b^2 - b) + a,& text for a le bendcases
$$
or, without the cases,
$$
f(a,b) = frac14(a^2 + b^2 + a + b + (a + b - 3)|a-b|).
$$
answered Mar 25 at 19:49
FredHFredH
3,7401024
$endgroup$
add a comment |
$begingroup$
If you want to define such a function for $a,b$ in the positive integers and with values in the positive integers, you can create it by filling in a table of values without repeating any values, except that the value for $(a,b)$ is the same as the one $(b,a)$. It might start out like this:
$$
beginarrayc
&1&2&3&4&5\
hline
1&1&2&4&7&11\
2&2&3&5&8&12\
3&4&5&6&9&13\
4&7&8&9&10&14\
5&11&12&13&14&15
endarray
$$
In this example, the values in each row up to the diagonal are $1$; then $2,3$; then $4,5,6$, etc.
That defines a function which is symmetric, but otherwise does not repeat value.
If you want a formula,
$$
f(a,b) = begincasesfrac12 (a^2 - a) + b,& text for a ge b\ \
frac12 (b^2 - b) + a,& text for a le bendcases
$$
or, without the cases,
$$
f(a,b) = frac14(a^2 + b^2 + a + b + (a + b - 3)|a-b|).
$$
answered Mar 25 at 19:49
FredHFredH
3,7401024
$endgroup$
If you want to define such a function for $a,b$ in the positive integers and with values in the positive integers, you can create it by filling in a table of values without repeating any values, except that the value for $(a,b)$ is the same as the one $(b,a)$. It might start out like this:
$$
beginarrayc
&1&2&3&4&5\
hline
1&1&2&4&7&11\
2&2&3&5&8&12\
3&4&5&6&9&13\
4&7&8&9&10&14\
5&11&12&13&14&15
endarray
$$
In this example, the values in each row up to the diagonal are $1$; then $2,3$; then $4,5,6$, etc.
That defines a function which is symmetric, but otherwise does not repeat value.
If you want a formula,
$$
f(a,b) = begincasesfrac12 (a^2 - a) + b,& text for a ge b\ \
frac12 (b^2 - b) + a,& text for a le bendcases
$$
or, without the cases,
$$
f(a,b) = frac14(a^2 + b^2 + a + b + (a + b - 3)|a-b|).
$$
answered Mar 25 at 19:49
FredHFredH
3,7401024
answered Mar 25 at 19:49
FredHFredH
3,7401024
answered Mar 25 at 19:49
FredHFredH
3,7401024
answered Mar 25 at 19:49
FredHFredH
3,7401024
3,7401024
add a comment |
add a comment |
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$begingroup$
What is the domain of your operator.
$endgroup$
– Don Thousand
Mar 25 at 18:34
$begingroup$
Integers basically numbers
$endgroup$
– Ajinkya Gawali
Mar 25 at 18:52