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Does a function with periodicity properties opposite to Dirichlet function exist?
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Dirichlet function is known to be periodic and every rational number appears to be its period thus no fundamental period is present.On the other side every irrational number is not its period.
I'm interested whether there exists a function with the opposite properties considering periodicity:
$$
f(x) = f(x + r), ; r in mathbb I \
f(x) ne f(x+q), ; q in mathbb Q
$$
Where $mathbb I$ is a set of irrational numbers and $mathbb Q$ is a set of rational ones.
I'm looking for a way to either prove or disprove such a function may be defined. Proving Dirichlet function properties is pretty straight forward but i'm not even sure where to start from for the opposite case.
My try is below:
I started from trying out different values of $x$. For example let some $rin mathbb I$ and $x = 0$. Put $r = t$, that means $f(0) = f(t)$, on the other hand when $t in mathbb Q$, let $x = t$ and $r = sqrt2 - t$, therfore $f(t) = f(sqrt2)$. Eventually we got that $f(sqrt2) = f(0)$. So for any irrational number the function is constant, and constant function does not have a fundamental period. But i'm not sure the above proves anything.
algebra-precalculus periodic-functions
asked Jul 23 at 13:06
roman
4391412
add a comment |Â
up vote
0
down vote
favorite
Dirichlet function is known to be periodic and every rational number appears to be its period thus no fundamental period is present.On the other side every irrational number is not its period.
I'm interested whether there exists a function with the opposite properties considering periodicity:
$$
f(x) = f(x + r), ; r in mathbb I \
f(x) ne f(x+q), ; q in mathbb Q
$$
Where $mathbb I$ is a set of irrational numbers and $mathbb Q$ is a set of rational ones.
I'm looking for a way to either prove or disprove such a function may be defined. Proving Dirichlet function properties is pretty straight forward but i'm not even sure where to start from for the opposite case.
My try is below:
I started from trying out different values of $x$. For example let some $rin mathbb I$ and $x = 0$. Put $r = t$, that means $f(0) = f(t)$, on the other hand when $t in mathbb Q$, let $x = t$ and $r = sqrt2 - t$, therfore $f(t) = f(sqrt2)$. Eventually we got that $f(sqrt2) = f(0)$. So for any irrational number the function is constant, and constant function does not have a fundamental period. But i'm not sure the above proves anything.
algebra-precalculus periodic-functions
asked Jul 23 at 13:06
roman
4391412
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Dirichlet function is known to be periodic and every rational number appears to be its period thus no fundamental period is present.On the other side every irrational number is not its period.
I'm interested whether there exists a function with the opposite properties considering periodicity:
$$
f(x) = f(x + r), ; r in mathbb I \
f(x) ne f(x+q), ; q in mathbb Q
$$
Where $mathbb I$ is a set of irrational numbers and $mathbb Q$ is a set of rational ones.
I'm looking for a way to either prove or disprove such a function may be defined. Proving Dirichlet function properties is pretty straight forward but i'm not even sure where to start from for the opposite case.
My try is below:
I started from trying out different values of $x$. For example let some $rin mathbb I$ and $x = 0$. Put $r = t$, that means $f(0) = f(t)$, on the other hand when $t in mathbb Q$, let $x = t$ and $r = sqrt2 - t$, therfore $f(t) = f(sqrt2)$. Eventually we got that $f(sqrt2) = f(0)$. So for any irrational number the function is constant, and constant function does not have a fundamental period. But i'm not sure the above proves anything.
algebra-precalculus periodic-functions
asked Jul 23 at 13:06
roman
4391412
Dirichlet function is known to be periodic and every rational number appears to be its period thus no fundamental period is present.On the other side every irrational number is not its period.
I'm interested whether there exists a function with the opposite properties considering periodicity:
$$
f(x) = f(x + r), ; r in mathbb I \
f(x) ne f(x+q), ; q in mathbb Q
$$
Where $mathbb I$ is a set of irrational numbers and $mathbb Q$ is a set of rational ones.
I'm looking for a way to either prove or disprove such a function may be defined. Proving Dirichlet function properties is pretty straight forward but i'm not even sure where to start from for the opposite case.
My try is below:
I started from trying out different values of $x$. For example let some $rin mathbb I$ and $x = 0$. Put $r = t$, that means $f(0) = f(t)$, on the other hand when $t in mathbb Q$, let $x = t$ and $r = sqrt2 - t$, therfore $f(t) = f(sqrt2)$. Eventually we got that $f(sqrt2) = f(0)$. So for any irrational number the function is constant, and constant function does not have a fundamental period. But i'm not sure the above proves anything.
algebra-precalculus periodic-functions
asked Jul 23 at 13:06
roman
4391412
asked Jul 23 at 13:06
roman
4391412
asked Jul 23 at 13:06
roman
4391412
asked Jul 23 at 13:06
roman
4391412
4391412
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Suppose $r$ is irrational, and $q$ is rational, then $q-r$ is irrational and we have $$f(x)neq f(x+q)=fleft((x+r)+(q-r)right)=f(x+r)=f(x)$$so your function can't have the properties you want.
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
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
accepted
Suppose $r$ is irrational, and $q$ is rational, then $q-r$ is irrational and we have $$f(x)neq f(x+q)=fleft((x+r)+(q-r)right)=f(x+r)=f(x)$$so your function can't have the properties you want.
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
add a comment |Â
up vote
2
down vote
accepted
Suppose $r$ is irrational, and $q$ is rational, then $q-r$ is irrational and we have $$f(x)neq f(x+q)=fleft((x+r)+(q-r)right)=f(x+r)=f(x)$$so your function can't have the properties you want.
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose $r$ is irrational, and $q$ is rational, then $q-r$ is irrational and we have $$f(x)neq f(x+q)=fleft((x+r)+(q-r)right)=f(x+r)=f(x)$$so your function can't have the properties you want.
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
Suppose $r$ is irrational, and $q$ is rational, then $q-r$ is irrational and we have $$f(x)neq f(x+q)=fleft((x+r)+(q-r)right)=f(x+r)=f(x)$$so your function can't have the properties you want.
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
answered Jul 23 at 13:17
Mark Bennet
76.4k773170
76.4k773170
add a comment |Â
add a comment |Â
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