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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
asked 38 mins ago
Lance
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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
asked 38 mins ago
Lance
344
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.
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
asked 38 mins ago
Lance
344
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
asked 38 mins ago
Lance
344
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.
asked 38 mins ago
Lance
344
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.
edited 34 mins ago
asked 38 mins ago
Lance
344
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.
asked 38 mins ago
Lance
344
asked 38 mins ago
Lance
344
344
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
Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
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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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.
answered 33 mins ago
lhf
158k9161375
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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.
answered 26 mins ago
GSofer
371211
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.
answered 33 mins ago
lhf
158k9161375
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.
answered 33 mins ago
lhf
158k9161375
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.
answered 33 mins ago
lhf
158k9161375
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
answered 33 mins ago
lhf
158k9161375
answered 33 mins ago
lhf
158k9161375
answered 33 mins ago
lhf
158k9161375
158k9161375
add a comment |Â
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.
answered 26 mins ago
GSofer
371211
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.
answered 26 mins ago
GSofer
371211
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.
answered 26 mins ago
GSofer
371211
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
answered 26 mins ago
GSofer
371211
answered 26 mins ago
GSofer
371211
answered 26 mins ago
GSofer
371211
371211
add a comment |Â
add a comment |Â
Lance is a new contributor. Be nice, and check out our Code of Conduct.
Lance is a new contributor. Be nice, and 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