Find function $f:[0,1]to mathbb R$, whose set of discontinuities is such that $left{ frac{k}{2^n}; quad k,n...












1












$begingroup$


Find function $fcolon[0,1]to mathbb R$, whose set of discontinuities is such that



$$left{frac{k}{2^n}; quad k,n in mathbb N, k le 2^nright }$$



I guess that it is Riemann function or something like that but I have no clue how it would look like






real-analysis continuity







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  • $begingroup$
    The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
    $endgroup$
    – toric_actions
    Jan 3 at 19:13
















1












$begingroup$


Find function $fcolon[0,1]to mathbb R$, whose set of discontinuities is such that



$$left{frac{k}{2^n}; quad k,n in mathbb N, k le 2^nright }$$



I guess that it is Riemann function or something like that but I have no clue how it would look like






real-analysis continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
    $endgroup$
    – toric_actions
    Jan 3 at 19:13














1












1








1





$begingroup$


Find function $fcolon[0,1]to mathbb R$, whose set of discontinuities is such that



$$left{frac{k}{2^n}; quad k,n in mathbb N, k le 2^nright }$$



I guess that it is Riemann function or something like that but I have no clue how it would look like






real-analysis continuity







share|cite|improve this question











$endgroup$




Find function $fcolon[0,1]to mathbb R$, whose set of discontinuities is such that



$$left{frac{k}{2^n}; quad k,n in mathbb N, k le 2^nright }$$



I guess that it is Riemann function or something like that but I have no clue how it would look like






real-analysis continuity




real-analysis continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 19:48









Shaun

8,898113681




8,898113681










asked Jan 3 at 19:07









math.troublemath.trouble

566




566












  • $begingroup$
    The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
    $endgroup$
    – toric_actions
    Jan 3 at 19:13


















  • $begingroup$
    The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
    $endgroup$
    – toric_actions
    Jan 3 at 19:13
















$begingroup$
The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
$endgroup$
– toric_actions
Jan 3 at 19:13




$begingroup$
The set of discontinuities is dense in $[0,1]$. So I think that it might be like Weierstrass function.
$endgroup$
– toric_actions
Jan 3 at 19:13










1 Answer
1






active

oldest

votes


















5












$begingroup$

Let $f(x) = 0$ whenever $x$ is not a dyadic fraction, and $fleft(frac{k}{2^n}right) = 2^{-n}$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Might help to recognize that this is a modified Thomae's function.
    $endgroup$
    – Robert Wolfe
    Jan 3 at 19:23











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Let $f(x) = 0$ whenever $x$ is not a dyadic fraction, and $fleft(frac{k}{2^n}right) = 2^{-n}$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Might help to recognize that this is a modified Thomae's function.
    $endgroup$
    – Robert Wolfe
    Jan 3 at 19:23
















5












$begingroup$

Let $f(x) = 0$ whenever $x$ is not a dyadic fraction, and $fleft(frac{k}{2^n}right) = 2^{-n}$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Might help to recognize that this is a modified Thomae's function.
    $endgroup$
    – Robert Wolfe
    Jan 3 at 19:23














5












5








5





$begingroup$

Let $f(x) = 0$ whenever $x$ is not a dyadic fraction, and $fleft(frac{k}{2^n}right) = 2^{-n}$.






share|cite|improve this answer









$endgroup$



Let $f(x) = 0$ whenever $x$ is not a dyadic fraction, and $fleft(frac{k}{2^n}right) = 2^{-n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 19:17









MathematicsStudent1122MathematicsStudent1122

8,57322466




8,57322466








  • 2




    $begingroup$
    Might help to recognize that this is a modified Thomae's function.
    $endgroup$
    – Robert Wolfe
    Jan 3 at 19:23














  • 2




    $begingroup$
    Might help to recognize that this is a modified Thomae's function.
    $endgroup$
    – Robert Wolfe
    Jan 3 at 19:23








2




2




$begingroup$
Might help to recognize that this is a modified Thomae's function.
$endgroup$
– Robert Wolfe
Jan 3 at 19:23




$begingroup$
Might help to recognize that this is a modified Thomae's function.
$endgroup$
– Robert Wolfe
Jan 3 at 19:23


















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