Function with arbitrary small period
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Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that the period of f(x) is $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.
periodic-functions
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up vote
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down vote
favorite
Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that the period of f(x) is $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.
periodic-functions
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Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that the period of f(x) is $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.
periodic-functions
New contributor
Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that the period of f(x) is $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.
periodic-functions
periodic-functions
New contributor
Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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edited 34 mins ago
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asked 38 mins ago


Lance
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344
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New contributor
Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago
add a comment |Â
Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago
Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago
add a comment |Â
2 Answers
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up vote
4
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You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.
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up vote
2
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You're correct.
Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.
We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:
If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.
If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.
So if we choose our period to be $p$, our property is satisfied.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.
add a comment |Â
up vote
4
down vote
You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.
You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.
answered 33 mins ago


lhf
158k9161375
158k9161375
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add a comment |Â
up vote
2
down vote
You're correct.
Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.
We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:
If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.
If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.
So if we choose our period to be $p$, our property is satisfied.
add a comment |Â
up vote
2
down vote
You're correct.
Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.
We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:
If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.
If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.
So if we choose our period to be $p$, our property is satisfied.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You're correct.
Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.
We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:
If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.
If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.
So if we choose our period to be $p$, our property is satisfied.
You're correct.
Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.
We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:
If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.
If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.
So if we choose our period to be $p$, our property is satisfied.
answered 26 mins ago
GSofer
371211
371211
add a comment |Â
add a comment |Â
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Sorry I forgot to add that the function is non costant
– Lance
36 mins ago
F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
23 mins ago