Calculate limit of $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$












0














Suppose when $xrightarrow 0$,$f(x) sim x$, $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$
Want to prove: $lim_{n to infty}{x_n}=a(a>0).$



Also,
I don't understand what $sim$ mean?






real-analysis







share|cite|improve this question






















  • I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
    – Sangchul Lee
    Nov 20 at 17:52


















0














Suppose when $xrightarrow 0$,$f(x) sim x$, $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$
Want to prove: $lim_{n to infty}{x_n}=a(a>0).$



Also,
I don't understand what $sim$ mean?






real-analysis







share|cite|improve this question






















  • I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
    – Sangchul Lee
    Nov 20 at 17:52
















0












0








0







Suppose when $xrightarrow 0$,$f(x) sim x$, $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$
Want to prove: $lim_{n to infty}{x_n}=a(a>0).$



Also,
I don't understand what $sim$ mean?






real-analysis







share|cite|improve this question













Suppose when $xrightarrow 0$,$f(x) sim x$, $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$
Want to prove: $lim_{n to infty}{x_n}=a(a>0).$



Also,
I don't understand what $sim$ mean?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











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asked Nov 20 at 17:36









mathnoob

1,794422




1,794422












  • I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
    – Sangchul Lee
    Nov 20 at 17:52




















  • I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
    – Sangchul Lee
    Nov 20 at 17:52


















I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
– Sangchul Lee
Nov 20 at 17:52






I suspect that $f(x) sim x$ means $f(x)/x to 1$ as $xto 0$. Assuming so, the rationale for $x_n to a$ is that $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$, which converges to $int_{0}^{1} 2ax , dx = a$. But the implication $f(x) sim x$ $Rightarrow$ $x_n sim sum_{i=1}^{n} frac{2i-1}{n^2}a$ deserves justification as it does not directly follow from the definition of $sim$. User Mostafa Ayaz's answer addresses this.
– Sangchul Lee
Nov 20 at 17:52












2 Answers
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1














$f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$






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    0














    Hint: Suppose $f(x)=x$, then we see that
    begin{align}
    x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
    =& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
    end{align}
    as $nrightarrow infty$.



    Also, $f(x)sim x$ as $x rightarrow 0$ means
    begin{align}
    lim_{xrightarrow 0}frac{f(x)}{x}=1
    end{align}
    or more usefully
    begin{align}
    f(x)=x+text{ something small and goes to 0}
    end{align}
    when $x$ is close to $0$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1














      $f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$






      share|cite|improve this answer


























        1














        $f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$






        share|cite|improve this answer
























          1












          1








          1






          $f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$






          share|cite|improve this answer












          $f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 17:51









          Mostafa Ayaz

          13.7k3836




          13.7k3836























              0














              Hint: Suppose $f(x)=x$, then we see that
              begin{align}
              x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
              =& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
              end{align}
              as $nrightarrow infty$.



              Also, $f(x)sim x$ as $x rightarrow 0$ means
              begin{align}
              lim_{xrightarrow 0}frac{f(x)}{x}=1
              end{align}
              or more usefully
              begin{align}
              f(x)=x+text{ something small and goes to 0}
              end{align}
              when $x$ is close to $0$.






              share|cite|improve this answer


























                0














                Hint: Suppose $f(x)=x$, then we see that
                begin{align}
                x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
                =& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
                end{align}
                as $nrightarrow infty$.



                Also, $f(x)sim x$ as $x rightarrow 0$ means
                begin{align}
                lim_{xrightarrow 0}frac{f(x)}{x}=1
                end{align}
                or more usefully
                begin{align}
                f(x)=x+text{ something small and goes to 0}
                end{align}
                when $x$ is close to $0$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Hint: Suppose $f(x)=x$, then we see that
                  begin{align}
                  x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
                  =& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
                  end{align}
                  as $nrightarrow infty$.



                  Also, $f(x)sim x$ as $x rightarrow 0$ means
                  begin{align}
                  lim_{xrightarrow 0}frac{f(x)}{x}=1
                  end{align}
                  or more usefully
                  begin{align}
                  f(x)=x+text{ something small and goes to 0}
                  end{align}
                  when $x$ is close to $0$.






                  share|cite|improve this answer












                  Hint: Suppose $f(x)=x$, then we see that
                  begin{align}
                  x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
                  =& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
                  end{align}
                  as $nrightarrow infty$.



                  Also, $f(x)sim x$ as $x rightarrow 0$ means
                  begin{align}
                  lim_{xrightarrow 0}frac{f(x)}{x}=1
                  end{align}
                  or more usefully
                  begin{align}
                  f(x)=x+text{ something small and goes to 0}
                  end{align}
                  when $x$ is close to $0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 20 at 17:49









                  Jacky Chong

                  17.6k21128




                  17.6k21128






























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