Prove that $ B_{n}(x)=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}$ converges...












1












$begingroup$


Let $ngeq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by



begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]subseteq Bbb{R}.$



MY WORK



Let $[a,b]subseteq Bbb{R}$ be arbitrary, $xin [a,b]$ and $nin Bbb{N}$ be fixed. Then,



begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq left( dfrac{left| xright|} {2}right)^{n+2k}leq left(dfrac{c}{2}right)^{n+2k}.end{align}
where $c=max{|a|,|b|}.$ Now,
begin{align} sum^{infty}_{k=0} left(dfrac{c}{2}right)^{n+2k} =left(dfrac{c}{2}right)^{n}sum^{infty}_{k=0} left(dfrac{c}{2}right)^{2k} =left(dfrac{c}{2}right)^{n} left(dfrac{4}{4-c^2}right)<infty,;;text{where};;c<2.end{align}
Hence,
begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
converges uniformly on $[a,b]subseteq Bbb{R}$ with $c<2.$



QUESTION:



The question asks for a proof for any closed interval of $ Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $cgeq 2$? Or I'm I missing something in the proof?






real-analysis sequences-and-series analysis uniform-convergence







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$endgroup$












  • $begingroup$
    Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
    $endgroup$
    – Jakobian
    Dec 31 '18 at 17:39


















1












$begingroup$


Let $ngeq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by



begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]subseteq Bbb{R}.$



MY WORK



Let $[a,b]subseteq Bbb{R}$ be arbitrary, $xin [a,b]$ and $nin Bbb{N}$ be fixed. Then,



begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq left( dfrac{left| xright|} {2}right)^{n+2k}leq left(dfrac{c}{2}right)^{n+2k}.end{align}
where $c=max{|a|,|b|}.$ Now,
begin{align} sum^{infty}_{k=0} left(dfrac{c}{2}right)^{n+2k} =left(dfrac{c}{2}right)^{n}sum^{infty}_{k=0} left(dfrac{c}{2}right)^{2k} =left(dfrac{c}{2}right)^{n} left(dfrac{4}{4-c^2}right)<infty,;;text{where};;c<2.end{align}
Hence,
begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
converges uniformly on $[a,b]subseteq Bbb{R}$ with $c<2.$



QUESTION:



The question asks for a proof for any closed interval of $ Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $cgeq 2$? Or I'm I missing something in the proof?






real-analysis sequences-and-series analysis uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
    $endgroup$
    – Jakobian
    Dec 31 '18 at 17:39
















1












1








1


1



$begingroup$


Let $ngeq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by



begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]subseteq Bbb{R}.$



MY WORK



Let $[a,b]subseteq Bbb{R}$ be arbitrary, $xin [a,b]$ and $nin Bbb{N}$ be fixed. Then,



begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq left( dfrac{left| xright|} {2}right)^{n+2k}leq left(dfrac{c}{2}right)^{n+2k}.end{align}
where $c=max{|a|,|b|}.$ Now,
begin{align} sum^{infty}_{k=0} left(dfrac{c}{2}right)^{n+2k} =left(dfrac{c}{2}right)^{n}sum^{infty}_{k=0} left(dfrac{c}{2}right)^{2k} =left(dfrac{c}{2}right)^{n} left(dfrac{4}{4-c^2}right)<infty,;;text{where};;c<2.end{align}
Hence,
begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
converges uniformly on $[a,b]subseteq Bbb{R}$ with $c<2.$



QUESTION:



The question asks for a proof for any closed interval of $ Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $cgeq 2$? Or I'm I missing something in the proof?






real-analysis sequences-and-series analysis uniform-convergence







share|cite|improve this question











$endgroup$




Let $ngeq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by



begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]subseteq Bbb{R}.$



MY WORK



Let $[a,b]subseteq Bbb{R}$ be arbitrary, $xin [a,b]$ and $nin Bbb{N}$ be fixed. Then,



begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq left( dfrac{left| xright|} {2}right)^{n+2k}leq left(dfrac{c}{2}right)^{n+2k}.end{align}
where $c=max{|a|,|b|}.$ Now,
begin{align} sum^{infty}_{k=0} left(dfrac{c}{2}right)^{n+2k} =left(dfrac{c}{2}right)^{n}sum^{infty}_{k=0} left(dfrac{c}{2}right)^{2k} =left(dfrac{c}{2}right)^{n} left(dfrac{4}{4-c^2}right)<infty,;;text{where};;c<2.end{align}
Hence,
begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
converges uniformly on $[a,b]subseteq Bbb{R}$ with $c<2.$



QUESTION:



The question asks for a proof for any closed interval of $ Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $cgeq 2$? Or I'm I missing something in the proof?






real-analysis sequences-and-series analysis uniform-convergence




real-analysis sequences-and-series analysis uniform-convergence






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edited Dec 31 '18 at 19:59







Omojola Micheal

















asked Dec 31 '18 at 17:23









Omojola MichealOmojola Micheal

1,749324




1,749324












  • $begingroup$
    Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
    $endgroup$
    – Jakobian
    Dec 31 '18 at 17:39




















  • $begingroup$
    Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
    $endgroup$
    – Jakobian
    Dec 31 '18 at 17:39


















$begingroup$
Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
$endgroup$
– Jakobian
Dec 31 '18 at 17:39






$begingroup$
Your estimation was too big. Try this one $left|frac{(-1)^k}{k!(n+k)!}left(frac{x}{2}right)^{n+2k}right|leq frac{(c/2)^{n+2k}}{k!} $
$endgroup$
– Jakobian
Dec 31 '18 at 17:39












2 Answers
2






active

oldest

votes


















1












$begingroup$

No, your proof surely does not work for all intervals $[a,b]$ of $mathbb R$. It only works for those intervals $[a,b]$ such that $lvert arvert,lvert brvert<2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 19:56





















1












$begingroup$

Happy New Year! Credits to Jakobian



begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq dfrac{1} {k!}left( dfrac{left| xright|} {2}right)^{n+2k}leq dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}.end{align}
Since,
begin{align} limlimits_{ktoinfty}left|dfrac{k!} {(k+1)!}left(dfrac{c}{2}right)^{n+2k+2} left(dfrac{2}{c}right)^{n+2k} right|&=limlimits_{ktoinfty}left|dfrac{1} {(k+1)}left(dfrac{c}{2}right)^{2} right|\&=dfrac{c^2}{4}limlimits_{ktoinfty}dfrac{1} {(k+1)} \&=0<1end{align}
Thus, $sum^{infty}_{k=0} dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
converges uniformly on any $[a,b]subseteq Bbb{R}$






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    active

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    1












    $begingroup$

    No, your proof surely does not work for all intervals $[a,b]$ of $mathbb R$. It only works for those intervals $[a,b]$ such that $lvert arvert,lvert brvert<2$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
      $endgroup$
      – Omojola Micheal
      Dec 31 '18 at 19:56


















    1












    $begingroup$

    No, your proof surely does not work for all intervals $[a,b]$ of $mathbb R$. It only works for those intervals $[a,b]$ such that $lvert arvert,lvert brvert<2$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
      $endgroup$
      – Omojola Micheal
      Dec 31 '18 at 19:56
















    1












    1








    1





    $begingroup$

    No, your proof surely does not work for all intervals $[a,b]$ of $mathbb R$. It only works for those intervals $[a,b]$ such that $lvert arvert,lvert brvert<2$.






    share|cite|improve this answer









    $endgroup$



    No, your proof surely does not work for all intervals $[a,b]$ of $mathbb R$. It only works for those intervals $[a,b]$ such that $lvert arvert,lvert brvert<2$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 17:28









    José Carlos SantosJosé Carlos Santos

    153k22123225




    153k22123225












    • $begingroup$
      (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
      $endgroup$
      – Omojola Micheal
      Dec 31 '18 at 19:56




















    • $begingroup$
      (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
      $endgroup$
      – Omojola Micheal
      Dec 31 '18 at 19:56


















    $begingroup$
    (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 19:56






    $begingroup$
    (+1) Sorry, I asked a wrong question. I've edited it. So, how do I make it work for all intervals of the kind $[a,b]$? Any hint?
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 19:56













    1












    $begingroup$

    Happy New Year! Credits to Jakobian



    begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq dfrac{1} {k!}left( dfrac{left| xright|} {2}right)^{n+2k}leq dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}.end{align}
    Since,
    begin{align} limlimits_{ktoinfty}left|dfrac{k!} {(k+1)!}left(dfrac{c}{2}right)^{n+2k+2} left(dfrac{2}{c}right)^{n+2k} right|&=limlimits_{ktoinfty}left|dfrac{1} {(k+1)}left(dfrac{c}{2}right)^{2} right|\&=dfrac{c^2}{4}limlimits_{ktoinfty}dfrac{1} {(k+1)} \&=0<1end{align}
    Thus, $sum^{infty}_{k=0} dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
    begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
    converges uniformly on any $[a,b]subseteq Bbb{R}$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Happy New Year! Credits to Jakobian



      begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq dfrac{1} {k!}left( dfrac{left| xright|} {2}right)^{n+2k}leq dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}.end{align}
      Since,
      begin{align} limlimits_{ktoinfty}left|dfrac{k!} {(k+1)!}left(dfrac{c}{2}right)^{n+2k+2} left(dfrac{2}{c}right)^{n+2k} right|&=limlimits_{ktoinfty}left|dfrac{1} {(k+1)}left(dfrac{c}{2}right)^{2} right|\&=dfrac{c^2}{4}limlimits_{ktoinfty}dfrac{1} {(k+1)} \&=0<1end{align}
      Thus, $sum^{infty}_{k=0} dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
      begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
      converges uniformly on any $[a,b]subseteq Bbb{R}$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Happy New Year! Credits to Jakobian



        begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq dfrac{1} {k!}left( dfrac{left| xright|} {2}right)^{n+2k}leq dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}.end{align}
        Since,
        begin{align} limlimits_{ktoinfty}left|dfrac{k!} {(k+1)!}left(dfrac{c}{2}right)^{n+2k+2} left(dfrac{2}{c}right)^{n+2k} right|&=limlimits_{ktoinfty}left|dfrac{1} {(k+1)}left(dfrac{c}{2}right)^{2} right|\&=dfrac{c^2}{4}limlimits_{ktoinfty}dfrac{1} {(k+1)} \&=0<1end{align}
        Thus, $sum^{infty}_{k=0} dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
        begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
        converges uniformly on any $[a,b]subseteq Bbb{R}$






        share|cite|improve this answer











        $endgroup$



        Happy New Year! Credits to Jakobian



        begin{align} left| dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k}right|leq dfrac{1} {k!}left( dfrac{left| xright|} {2}right)^{n+2k}leq dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}.end{align}
        Since,
        begin{align} limlimits_{ktoinfty}left|dfrac{k!} {(k+1)!}left(dfrac{c}{2}right)^{n+2k+2} left(dfrac{2}{c}right)^{n+2k} right|&=limlimits_{ktoinfty}left|dfrac{1} {(k+1)}left(dfrac{c}{2}right)^{2} right|\&=dfrac{c^2}{4}limlimits_{ktoinfty}dfrac{1} {(k+1)} \&=0<1end{align}
        Thus, $sum^{infty}_{k=0} dfrac{1} {k!}left(dfrac{c}{2}right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
        begin{align} B_{n}(x):=sum^{infty}_{k=0} dfrac{(-1)^k}{k!(n+k)!} left(dfrac{x}{2}right)^{n+2k} .end{align}
        converges uniformly on any $[a,b]subseteq Bbb{R}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 0:29

























        answered Dec 31 '18 at 23:54









        Omojola MichealOmojola Micheal

        1,749324




        1,749324






























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