Limit of a series of difference quotients minus derivatives of functions












0












$begingroup$


Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval



Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
f_n(x)
=
{
dfrac{n}{(2n-1)^{a+x}}
-
dfrac{n}{(2n)^{a+x}}
}
$$
and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
g_n(x)
=
dfrac
{d}{dx}
f_n(x)
=
dfrac{n cdot ln(2n)}{(2n)^{a+x}}
-
dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
$$
Let $h : I to mathbb{R}$ be the funtion
$$
h(x)=
sum_{n=1}^infty
left|
dfrac{f_n(x)-f_n(0)}{x}
-
g_n(0)
right|^2
$$
My question is if it is true that
$$
lim_{x to 0^+}
h(x)=0
$$
Thanks.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval



    Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
    $$
    f_n(x)
    =
    {
    dfrac{n}{(2n-1)^{a+x}}
    -
    dfrac{n}{(2n)^{a+x}}
    }
    $$
    and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
    $$
    g_n(x)
    =
    dfrac
    {d}{dx}
    f_n(x)
    =
    dfrac{n cdot ln(2n)}{(2n)^{a+x}}
    -
    dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
    $$
    Let $h : I to mathbb{R}$ be the funtion
    $$
    h(x)=
    sum_{n=1}^infty
    left|
    dfrac{f_n(x)-f_n(0)}{x}
    -
    g_n(0)
    right|^2
    $$
    My question is if it is true that
    $$
    lim_{x to 0^+}
    h(x)=0
    $$
    Thanks.






    real-analysis sequences-and-series limits







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval



      Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
      $$
      f_n(x)
      =
      {
      dfrac{n}{(2n-1)^{a+x}}
      -
      dfrac{n}{(2n)^{a+x}}
      }
      $$
      and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
      $$
      g_n(x)
      =
      dfrac
      {d}{dx}
      f_n(x)
      =
      dfrac{n cdot ln(2n)}{(2n)^{a+x}}
      -
      dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
      $$
      Let $h : I to mathbb{R}$ be the funtion
      $$
      h(x)=
      sum_{n=1}^infty
      left|
      dfrac{f_n(x)-f_n(0)}{x}
      -
      g_n(0)
      right|^2
      $$
      My question is if it is true that
      $$
      lim_{x to 0^+}
      h(x)=0
      $$
      Thanks.






      real-analysis sequences-and-series limits







      share|cite|improve this question











      $endgroup$




      Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval



      Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
      $$
      f_n(x)
      =
      {
      dfrac{n}{(2n-1)^{a+x}}
      -
      dfrac{n}{(2n)^{a+x}}
      }
      $$
      and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
      $$
      g_n(x)
      =
      dfrac
      {d}{dx}
      f_n(x)
      =
      dfrac{n cdot ln(2n)}{(2n)^{a+x}}
      -
      dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
      $$
      Let $h : I to mathbb{R}$ be the funtion
      $$
      h(x)=
      sum_{n=1}^infty
      left|
      dfrac{f_n(x)-f_n(0)}{x}
      -
      g_n(0)
      right|^2
      $$
      My question is if it is true that
      $$
      lim_{x to 0^+}
      h(x)=0
      $$
      Thanks.






      real-analysis sequences-and-series limits




      real-analysis sequences-and-series limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 12:59







      Matey Math

















      asked Dec 14 '18 at 10:29









      Matey MathMatey Math

      850514




      850514






















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          $begingroup$

          Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.



          By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
          or
          $$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
          $$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$



          where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.






          share|cite|improve this answer









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            $begingroup$

            Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.



            By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
            or
            $$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
            $$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$



            where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.



              By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
              or
              $$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
              $$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$



              where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.



                By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
                or
                $$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
                $$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$



                where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.






                share|cite|improve this answer









                $endgroup$



                Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.



                By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
                or
                $$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
                $$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$



                where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 17:08









                Sameer BahetiSameer Baheti

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