How can I show that this infinite product is nonzero?












0












$begingroup$


How would you show that $prod_{k=1}^infty cos( 2 pi/3^k)$ is nonzero?
Wolfram approximates it as about $-0.37$, and I have a guess that
$$
Big vert prod_{k=1}^infty cos( 2 pi/3^k) Big vertgeq c prod_{k=1}^infty 3^{-1/k^2},
$$
although I cannot show it.



This product arises as the modulus of the characteristic function of the Cantor distribution.






real-analysis probability infinite-product







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$endgroup$








  • 1




    $begingroup$
    When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:25






  • 3




    $begingroup$
    @Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
    $endgroup$
    – Clement C.
    Dec 10 '18 at 19:26










  • $begingroup$
    @ClementC. That's true. Thanks.
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:28
















0












$begingroup$


How would you show that $prod_{k=1}^infty cos( 2 pi/3^k)$ is nonzero?
Wolfram approximates it as about $-0.37$, and I have a guess that
$$
Big vert prod_{k=1}^infty cos( 2 pi/3^k) Big vertgeq c prod_{k=1}^infty 3^{-1/k^2},
$$
although I cannot show it.



This product arises as the modulus of the characteristic function of the Cantor distribution.






real-analysis probability infinite-product







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:25






  • 3




    $begingroup$
    @Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
    $endgroup$
    – Clement C.
    Dec 10 '18 at 19:26










  • $begingroup$
    @ClementC. That's true. Thanks.
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:28














0












0








0





$begingroup$


How would you show that $prod_{k=1}^infty cos( 2 pi/3^k)$ is nonzero?
Wolfram approximates it as about $-0.37$, and I have a guess that
$$
Big vert prod_{k=1}^infty cos( 2 pi/3^k) Big vertgeq c prod_{k=1}^infty 3^{-1/k^2},
$$
although I cannot show it.



This product arises as the modulus of the characteristic function of the Cantor distribution.






real-analysis probability infinite-product







share|cite|improve this question









$endgroup$




How would you show that $prod_{k=1}^infty cos( 2 pi/3^k)$ is nonzero?
Wolfram approximates it as about $-0.37$, and I have a guess that
$$
Big vert prod_{k=1}^infty cos( 2 pi/3^k) Big vertgeq c prod_{k=1}^infty 3^{-1/k^2},
$$
although I cannot show it.



This product arises as the modulus of the characteristic function of the Cantor distribution.






real-analysis probability infinite-product




real-analysis probability infinite-product






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 19:22









JMill.JMill.

8617




8617








  • 1




    $begingroup$
    When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:25






  • 3




    $begingroup$
    @Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
    $endgroup$
    – Clement C.
    Dec 10 '18 at 19:26










  • $begingroup$
    @ClementC. That's true. Thanks.
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:28














  • 1




    $begingroup$
    When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:25






  • 3




    $begingroup$
    @Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
    $endgroup$
    – Clement C.
    Dec 10 '18 at 19:26










  • $begingroup$
    @ClementC. That's true. Thanks.
    $endgroup$
    – Dog_69
    Dec 10 '18 at 19:28








1




1




$begingroup$
When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
$endgroup$
– Dog_69
Dec 10 '18 at 19:25




$begingroup$
When does $cos(x)$ vanish? Does $2pi/3^k$ achieve some of those values?
$endgroup$
– Dog_69
Dec 10 '18 at 19:25




3




3




$begingroup$
@Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
$endgroup$
– Clement C.
Dec 10 '18 at 19:26




$begingroup$
@Dog_69 That is not sufficient. Think of $prod_{k=1}^infty (1-1/(2k))$
$endgroup$
– Clement C.
Dec 10 '18 at 19:26












$begingroup$
@ClementC. That's true. Thanks.
$endgroup$
– Dog_69
Dec 10 '18 at 19:28




$begingroup$
@ClementC. That's true. Thanks.
$endgroup$
– Dog_69
Dec 10 '18 at 19:28










2 Answers
2






active

oldest

votes


















3












$begingroup$

For $xapprox 0$, we have $cos xapprox 1-frac 12x^2$ and $lncos xapprox-frac 12x^2 $. This allows you to compare $sumlncos(2pi/3^k) $ with a nicely convergent series






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    It is well-known that an infinite product of positive terms
    $$ prod_{n=1}^infty (1-a_n) text{where} 0 le a_n < 1$$
    converges (to a nonzero limit) if and only if $sum_{n=1}^infty a_n < infty$.
    In this case
    $$a_k = 1 - cos(2pi/3^k) sim frac{(2pi/ 3^{k})^2}{2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
      $endgroup$
      – LutzL
      Dec 10 '18 at 20:16










    • $begingroup$
      @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
      $endgroup$
      – Clement C.
      Dec 10 '18 at 21:10











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    For $xapprox 0$, we have $cos xapprox 1-frac 12x^2$ and $lncos xapprox-frac 12x^2 $. This allows you to compare $sumlncos(2pi/3^k) $ with a nicely convergent series






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      For $xapprox 0$, we have $cos xapprox 1-frac 12x^2$ and $lncos xapprox-frac 12x^2 $. This allows you to compare $sumlncos(2pi/3^k) $ with a nicely convergent series






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        For $xapprox 0$, we have $cos xapprox 1-frac 12x^2$ and $lncos xapprox-frac 12x^2 $. This allows you to compare $sumlncos(2pi/3^k) $ with a nicely convergent series






        share|cite|improve this answer









        $endgroup$



        For $xapprox 0$, we have $cos xapprox 1-frac 12x^2$ and $lncos xapprox-frac 12x^2 $. This allows you to compare $sumlncos(2pi/3^k) $ with a nicely convergent series







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 19:31









        Hagen von EitzenHagen von Eitzen

        280k23272504




        280k23272504























            2












            $begingroup$

            It is well-known that an infinite product of positive terms
            $$ prod_{n=1}^infty (1-a_n) text{where} 0 le a_n < 1$$
            converges (to a nonzero limit) if and only if $sum_{n=1}^infty a_n < infty$.
            In this case
            $$a_k = 1 - cos(2pi/3^k) sim frac{(2pi/ 3^{k})^2}{2}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
              $endgroup$
              – LutzL
              Dec 10 '18 at 20:16










            • $begingroup$
              @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
              $endgroup$
              – Clement C.
              Dec 10 '18 at 21:10
















            2












            $begingroup$

            It is well-known that an infinite product of positive terms
            $$ prod_{n=1}^infty (1-a_n) text{where} 0 le a_n < 1$$
            converges (to a nonzero limit) if and only if $sum_{n=1}^infty a_n < infty$.
            In this case
            $$a_k = 1 - cos(2pi/3^k) sim frac{(2pi/ 3^{k})^2}{2}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
              $endgroup$
              – LutzL
              Dec 10 '18 at 20:16










            • $begingroup$
              @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
              $endgroup$
              – Clement C.
              Dec 10 '18 at 21:10














            2












            2








            2





            $begingroup$

            It is well-known that an infinite product of positive terms
            $$ prod_{n=1}^infty (1-a_n) text{where} 0 le a_n < 1$$
            converges (to a nonzero limit) if and only if $sum_{n=1}^infty a_n < infty$.
            In this case
            $$a_k = 1 - cos(2pi/3^k) sim frac{(2pi/ 3^{k})^2}{2}$$






            share|cite|improve this answer









            $endgroup$



            It is well-known that an infinite product of positive terms
            $$ prod_{n=1}^infty (1-a_n) text{where} 0 le a_n < 1$$
            converges (to a nonzero limit) if and only if $sum_{n=1}^infty a_n < infty$.
            In this case
            $$a_k = 1 - cos(2pi/3^k) sim frac{(2pi/ 3^{k})^2}{2}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 19:31









            Robert IsraelRobert Israel

            324k23214468




            324k23214468












            • $begingroup$
              Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
              $endgroup$
              – LutzL
              Dec 10 '18 at 20:16










            • $begingroup$
              @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
              $endgroup$
              – Clement C.
              Dec 10 '18 at 21:10


















            • $begingroup$
              Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
              $endgroup$
              – LutzL
              Dec 10 '18 at 20:16










            • $begingroup$
              @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
              $endgroup$
              – Clement C.
              Dec 10 '18 at 21:10
















            $begingroup$
            Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
            $endgroup$
            – LutzL
            Dec 10 '18 at 20:16




            $begingroup$
            Or more precisely, $1-cos(2π/3^k)=2sin^2(π/3^k)le 2min(1,π/3^k)^2$.
            $endgroup$
            – LutzL
            Dec 10 '18 at 20:16












            $begingroup$
            @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
            $endgroup$
            – Clement C.
            Dec 10 '18 at 21:10




            $begingroup$
            @LutzL That's not more precise, it's a different argument. Robert Israel's asymptotic argument is precise.
            $endgroup$
            – Clement C.
            Dec 10 '18 at 21:10


















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