What is the derivation of the limit form of the Dirichlet function?

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The Dirichlet function is defined as the indicator function of rational numbers. I have also seen this function described by:
$$f(x)=lim_ktoinftylim_jtoinftycos^2jk!pi x$$
How does this limit act as the indicator, and how does it yield an answer if cosine is limited to Infinity?






limits irrational-numbers rational-numbers rationality-testing







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  • What do you mean by "cosine is limited to Infinity"?
    – user10354138
    3 hours ago










  • See (math.stackexchange.com/questions/264889/…)
    – Chinnapparaj R
    2 hours ago














up vote
4
down vote

favorite
1












The Dirichlet function is defined as the indicator function of rational numbers. I have also seen this function described by:
$$f(x)=lim_ktoinftylim_jtoinftycos^2jk!pi x$$
How does this limit act as the indicator, and how does it yield an answer if cosine is limited to Infinity?






limits irrational-numbers rational-numbers rationality-testing







share|cite|improve this question









New contributor




mtung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • What do you mean by "cosine is limited to Infinity"?
    – user10354138
    3 hours ago










  • See (math.stackexchange.com/questions/264889/…)
    – Chinnapparaj R
    2 hours ago












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





The Dirichlet function is defined as the indicator function of rational numbers. I have also seen this function described by:
$$f(x)=lim_ktoinftylim_jtoinftycos^2jk!pi x$$
How does this limit act as the indicator, and how does it yield an answer if cosine is limited to Infinity?






limits irrational-numbers rational-numbers rationality-testing







share|cite|improve this question









New contributor




mtung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











The Dirichlet function is defined as the indicator function of rational numbers. I have also seen this function described by:
$$f(x)=lim_ktoinftylim_jtoinftycos^2jk!pi x$$
How does this limit act as the indicator, and how does it yield an answer if cosine is limited to Infinity?






limits irrational-numbers rational-numbers rationality-testing




limits irrational-numbers rational-numbers rationality-testing






share|cite|improve this question









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mtung 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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edited 3 hours ago









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  • What do you mean by "cosine is limited to Infinity"?
    – user10354138
    3 hours ago










  • See (math.stackexchange.com/questions/264889/…)
    – Chinnapparaj R
    2 hours ago
















  • What do you mean by "cosine is limited to Infinity"?
    – user10354138
    3 hours ago










  • See (math.stackexchange.com/questions/264889/…)
    – Chinnapparaj R
    2 hours ago















What do you mean by "cosine is limited to Infinity"?
– user10354138
3 hours ago




What do you mean by "cosine is limited to Infinity"?
– user10354138
3 hours ago












See (math.stackexchange.com/questions/264889/…)
– Chinnapparaj R
2 hours ago




See (math.stackexchange.com/questions/264889/…)
– Chinnapparaj R
2 hours ago










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Suppose $x$ is rational and $x=frac pq$ for $gcd(p,q)=1$. If $kge q$, $k!x$ will be an integer, so $cos k!pi x=pm1$ and $cos^2jk!pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.



Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|cos k!pi x|<1$ for all $k$. As $jtoinfty$, this variable being in the exponent yields $cos^2jk!pi xto0$, so the function evaluates to 0.






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    Suppose $x$ is rational and $x=frac pq$ for $gcd(p,q)=1$. If $kge q$, $k!x$ will be an integer, so $cos k!pi x=pm1$ and $cos^2jk!pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.



    Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|cos k!pi x|<1$ for all $k$. As $jtoinfty$, this variable being in the exponent yields $cos^2jk!pi xto0$, so the function evaluates to 0.






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      Suppose $x$ is rational and $x=frac pq$ for $gcd(p,q)=1$. If $kge q$, $k!x$ will be an integer, so $cos k!pi x=pm1$ and $cos^2jk!pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.



      Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|cos k!pi x|<1$ for all $k$. As $jtoinfty$, this variable being in the exponent yields $cos^2jk!pi xto0$, so the function evaluates to 0.






      share|cite|improve this answer






















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        up vote
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        Suppose $x$ is rational and $x=frac pq$ for $gcd(p,q)=1$. If $kge q$, $k!x$ will be an integer, so $cos k!pi x=pm1$ and $cos^2jk!pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.



        Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|cos k!pi x|<1$ for all $k$. As $jtoinfty$, this variable being in the exponent yields $cos^2jk!pi xto0$, so the function evaluates to 0.






        share|cite|improve this answer












        Suppose $x$ is rational and $x=frac pq$ for $gcd(p,q)=1$. If $kge q$, $k!x$ will be an integer, so $cos k!pi x=pm1$ and $cos^2jk!pi x=1$. Thus the function is 1 for sufficiently large $k,j$, so it evaluates to 1.



        Suppose $x$ is irrational, then $k!x$ will never be an integer, so $|cos k!pi x|<1$ for all $k$. As $jtoinfty$, this variable being in the exponent yields $cos^2jk!pi xto0$, so the function evaluates to 0.







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