Uniform convergence of $frac{y/(2N)}{sin(y/(2N))}$ towards 1












2












$begingroup$


I can't come up with a proof, why $f_N(y) := frac{frac{y}{2N}}{sinleft(frac{y}{2N}right)}$ converges uniformly against $1$ for $yin(0,pi), Ntoinfty$.



I would be thankful for any advice.






real-analysis uniform-convergence sequence-of-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
    $endgroup$
    – Ian
    Jan 4 at 15:15










  • $begingroup$
    Ian.Want to post an answer with your idea?
    $endgroup$
    – Peter Szilas
    Jan 4 at 15:32










  • $begingroup$
    What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 15:58










  • $begingroup$
    @mathcounterexamples.net For the result to be as they say it must be the former...
    $endgroup$
    – Ian
    Jan 4 at 16:02






  • 1




    $begingroup$
    @stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
    $endgroup$
    – Ian
    Jan 4 at 16:10


















2












$begingroup$


I can't come up with a proof, why $f_N(y) := frac{frac{y}{2N}}{sinleft(frac{y}{2N}right)}$ converges uniformly against $1$ for $yin(0,pi), Ntoinfty$.



I would be thankful for any advice.






real-analysis uniform-convergence sequence-of-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
    $endgroup$
    – Ian
    Jan 4 at 15:15










  • $begingroup$
    Ian.Want to post an answer with your idea?
    $endgroup$
    – Peter Szilas
    Jan 4 at 15:32










  • $begingroup$
    What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 15:58










  • $begingroup$
    @mathcounterexamples.net For the result to be as they say it must be the former...
    $endgroup$
    – Ian
    Jan 4 at 16:02






  • 1




    $begingroup$
    @stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
    $endgroup$
    – Ian
    Jan 4 at 16:10
















2












2








2





$begingroup$


I can't come up with a proof, why $f_N(y) := frac{frac{y}{2N}}{sinleft(frac{y}{2N}right)}$ converges uniformly against $1$ for $yin(0,pi), Ntoinfty$.



I would be thankful for any advice.






real-analysis uniform-convergence sequence-of-function







share|cite|improve this question











$endgroup$




I can't come up with a proof, why $f_N(y) := frac{frac{y}{2N}}{sinleft(frac{y}{2N}right)}$ converges uniformly against $1$ for $yin(0,pi), Ntoinfty$.



I would be thankful for any advice.






real-analysis uniform-convergence sequence-of-function




real-analysis uniform-convergence sequence-of-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 16:06









robjohn♦

270k27313641




270k27313641










asked Jan 4 at 14:59









SlyderSlyder

316




316








  • 4




    $begingroup$
    One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
    $endgroup$
    – Ian
    Jan 4 at 15:15










  • $begingroup$
    Ian.Want to post an answer with your idea?
    $endgroup$
    – Peter Szilas
    Jan 4 at 15:32










  • $begingroup$
    What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 15:58










  • $begingroup$
    @mathcounterexamples.net For the result to be as they say it must be the former...
    $endgroup$
    – Ian
    Jan 4 at 16:02






  • 1




    $begingroup$
    @stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
    $endgroup$
    – Ian
    Jan 4 at 16:10
















  • 4




    $begingroup$
    One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
    $endgroup$
    – Ian
    Jan 4 at 15:15










  • $begingroup$
    Ian.Want to post an answer with your idea?
    $endgroup$
    – Peter Szilas
    Jan 4 at 15:32










  • $begingroup$
    What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
    $endgroup$
    – mathcounterexamples.net
    Jan 4 at 15:58










  • $begingroup$
    @mathcounterexamples.net For the result to be as they say it must be the former...
    $endgroup$
    – Ian
    Jan 4 at 16:02






  • 1




    $begingroup$
    @stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
    $endgroup$
    – Ian
    Jan 4 at 16:10










4




4




$begingroup$
One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
$endgroup$
– Ian
Jan 4 at 15:15




$begingroup$
One idea is to use $x-x^3/6 leq sin(x) leq x$ which holds for $x geq 0$.
$endgroup$
– Ian
Jan 4 at 15:15












$begingroup$
Ian.Want to post an answer with your idea?
$endgroup$
– Peter Szilas
Jan 4 at 15:32




$begingroup$
Ian.Want to post an answer with your idea?
$endgroup$
– Peter Szilas
Jan 4 at 15:32












$begingroup$
What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
$endgroup$
– mathcounterexamples.net
Jan 4 at 15:58




$begingroup$
What do you mean by $y/2N$? $frac{y}{2N}$ or $frac{y}{2}N$?
$endgroup$
– mathcounterexamples.net
Jan 4 at 15:58












$begingroup$
@mathcounterexamples.net For the result to be as they say it must be the former...
$endgroup$
– Ian
Jan 4 at 16:02




$begingroup$
@mathcounterexamples.net For the result to be as they say it must be the former...
$endgroup$
– Ian
Jan 4 at 16:02




1




1




$begingroup$
@stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
$endgroup$
– Ian
Jan 4 at 16:10






$begingroup$
@stressedout The poles are nowhere to be seen here, even when $N=1$, because of the $2$. $1-x^2/6 leq sin(x)/x leq 1$ simply becomes $1 leq x/sin(x) leq frac{1}{1-x^2/6}$ which is valid for $0<x<sqrt{6}$, and $pi/2<sqrt{6}$.
$endgroup$
– Ian
Jan 4 at 16:10












1 Answer
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$begingroup$

A solution using Dini's theorem



The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,pi]$.



For $n ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x in [0, pi]$ the sequence $(1/f_n(x))$ is increasing.






share|cite|improve this answer









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    $begingroup$

    A solution using Dini's theorem



    The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,pi]$.



    For $n ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x in [0, pi]$ the sequence $(1/f_n(x))$ is increasing.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      A solution using Dini's theorem



      The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,pi]$.



      For $n ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x in [0, pi]$ the sequence $(1/f_n(x))$ is increasing.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        A solution using Dini's theorem



        The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,pi]$.



        For $n ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x in [0, pi]$ the sequence $(1/f_n(x))$ is increasing.






        share|cite|improve this answer









        $endgroup$



        A solution using Dini's theorem



        The $f_n$ can be extended by continuity at $0$ by raising $f_n(0)=1$. Hence, we can consider the continuous extended maps on the compact interval $[0,pi]$.



        For $n ge 2$ the sequence $(f_n)$ is uniformly bounded below by a constant strictly positive. Hence proving the uniform convergence of $(f_n)$ is equivalent to the proof of the uniform convergence of $(1/f_n)$. And this is provided by Dini's theorem as for all $x in [0, pi]$ the sequence $(1/f_n(x))$ is increasing.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 16:17









        mathcounterexamples.netmathcounterexamples.net

        26.9k22158




        26.9k22158






























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