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Indicator Function appearing when solving functional equation
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Let $I_A(x)=begincases 0 & x in A\1 & textelse endcases$ where $A subseteq mathbbR.$
A particularly hard functional equation boiled down to determining the sets $A$ satisfying $I_A(x)I_A(x+y)=I_A(y)I_A(x-y).$ Due to analysis of the original functional equation, we know that $I_A(0)=1,$ or $0 notin A.$
I have determined the following properties:
$I_A(x) = I_A(-x),$ or $x in A iff -x in A$
$x in A Rightarrow x/2 in A$
If $a in A,$ then $I_A(x)I_A(a-x)=I_A(x)I_A(a+x)=0.$
I've noticed that if A is non-empty, it will contain arbitrarily large and small values. Thus, I've conjectured that the only possible $A$ that satisfy the aforementioned indicational equation are $emptyset, , mathbbR setminus 0.$
functions functional-equations
asked Jul 14 at 22:44
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623211
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up vote
1
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Let $I_A(x)=begincases 0 & x in A\1 & textelse endcases$ where $A subseteq mathbbR.$
A particularly hard functional equation boiled down to determining the sets $A$ satisfying $I_A(x)I_A(x+y)=I_A(y)I_A(x-y).$ Due to analysis of the original functional equation, we know that $I_A(0)=1,$ or $0 notin A.$
I have determined the following properties:
$I_A(x) = I_A(-x),$ or $x in A iff -x in A$
$x in A Rightarrow x/2 in A$
If $a in A,$ then $I_A(x)I_A(a-x)=I_A(x)I_A(a+x)=0.$
I've noticed that if A is non-empty, it will contain arbitrarily large and small values. Thus, I've conjectured that the only possible $A$ that satisfy the aforementioned indicational equation are $emptyset, , mathbbR setminus 0.$
functions functional-equations
asked Jul 14 at 22:44
Display name
623211
Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
1
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $I_A(x)=begincases 0 & x in A\1 & textelse endcases$ where $A subseteq mathbbR.$
A particularly hard functional equation boiled down to determining the sets $A$ satisfying $I_A(x)I_A(x+y)=I_A(y)I_A(x-y).$ Due to analysis of the original functional equation, we know that $I_A(0)=1,$ or $0 notin A.$
I have determined the following properties:
$I_A(x) = I_A(-x),$ or $x in A iff -x in A$
$x in A Rightarrow x/2 in A$
If $a in A,$ then $I_A(x)I_A(a-x)=I_A(x)I_A(a+x)=0.$
I've noticed that if A is non-empty, it will contain arbitrarily large and small values. Thus, I've conjectured that the only possible $A$ that satisfy the aforementioned indicational equation are $emptyset, , mathbbR setminus 0.$
functions functional-equations
asked Jul 14 at 22:44
Display name
623211
Let $I_A(x)=begincases 0 & x in A\1 & textelse endcases$ where $A subseteq mathbbR.$
A particularly hard functional equation boiled down to determining the sets $A$ satisfying $I_A(x)I_A(x+y)=I_A(y)I_A(x-y).$ Due to analysis of the original functional equation, we know that $I_A(0)=1,$ or $0 notin A.$
I have determined the following properties:
$I_A(x) = I_A(-x),$ or $x in A iff -x in A$
$x in A Rightarrow x/2 in A$
If $a in A,$ then $I_A(x)I_A(a-x)=I_A(x)I_A(a+x)=0.$
I've noticed that if A is non-empty, it will contain arbitrarily large and small values. Thus, I've conjectured that the only possible $A$ that satisfy the aforementioned indicational equation are $emptyset, , mathbbR setminus 0.$
functions functional-equations
asked Jul 14 at 22:44
Display name
623211
asked Jul 14 at 22:44
Display name
623211
asked Jul 14 at 22:44
Display name
623211
asked Jul 14 at 22:44
Display name
623211
623211
Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
1
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08
add a comment |Â
Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
1
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08
Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
1
1
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
add a comment |Â
up vote
2
down vote
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
answered Jul 14 at 23:21
Kavi Rama Murthy
21k2830
21k2830
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
add a comment |Â
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
I was just fixing to answer this with the link: en.wikipedia.org/wiki/Dyadic_rational
– Mason
Jul 14 at 23:22
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
No I think you are reading the "indicator function" backwards tho... $I_mathbbQ(pi)I_mathbbQ(pi+pi)=1times 1 neq I_mathbbQ(pi)I_mathbbQ(pi-pi)=1times 0 $
– Mason
Jul 14 at 23:27
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
The complements of these sets do the trick. This confusion is the result of the OP defining the indicator function in precisely the opposite of the conventional notation.
– Mason
Jul 14 at 23:28
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
Very strange definition of indicator function by the OP. Thanks to Mason. My answer can be corrected by taking complements. I would not correct my answer because the OP has confused us by not sticking to standard convention.
– Kavi Rama Murthy
Jul 14 at 23:30
add a comment |Â
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Your notation looks a little backwards. What you have written for as $I_A$ is a function that takes on $1$ when $x not in A$. This is fine it's just not the typical definition. Is that what you intended? Typically we might write such a function as $I_barA$ where $barA$ refers to the complement of $A$.
– Mason
Jul 14 at 22:56
1
@Mason Yes. The solution to the original functional equation is $I_A(x)gamma(x),$ where $gamma>0$ is a positive function that we already found.
– Display name
Jul 14 at 22:58
Can $A$ be the set of irrational numbers?
– Mason
Jul 14 at 23:08