Is this sum less when it contains less positive terms?












0












$begingroup$


Problem



Let ${X_n}$ be a sequence of real numbers defined by $$x_n=
left{
begin{array}{ccrcl}
{displaystylefrac{1}{2k}} & mbox{if} &
{displaystyle n} & {displaystyle =} &
{displaystyle 2k - 1}
\[1mm]
{displaystylefrac{1}{2k - 1}} & mbox{if} &
{displaystyle n} & {displaystyle =} &
{displaystyle 2k}
end{array}right.
\epsilon_n=[frac{1}{2}]^n\
f(x)=sum_{n:x_n<x}epsilon_n
$$



(A)Compute f(1), f(1/2)



(B) Determine the set of discontinuities for the function



Solution



So with mathematica I computed $x_n={frac{1}{2},1,frac{1}{4},frac{1}{3},frac{1}{6},frac{1}{5},frac{1}{8},frac{1}{7}...}$



So, $f(1)=frac{1}{2}+sum_{n:x_n<1}frac{1}{2}^2=frac{1}{2}+frac{1}{1-1/2}=frac{1}{2}+2=frac{3}{2}$



$f(frac{1}{2})=sum_{n:x_n<1/2}frac{1}{2}^2=frac{1}{1-1/2}=2$



Yet I noticed that the sum of all terms$=f(infty)=$ $sum_0^{infty}frac{1}{2}^2=frac{1}{1-1/2}=2$



Why is $f(1)$ greater than the sum of all terms, when all terms are positive? Shouldn't $f(1)<f(infty)$



For the points of discontinuity, I think the sequence should be continuous everywhere.



Choose $delta$ such that $delta=x_n-x_{n-3},$ Then $forall epsilon > 0$, $forall yin B_{delta}(x_n),|x_n-y|<delta$, $|f(x_n)-f(y)|=|f(x_n)-f(x_n)|<epsilon$






real-analysis







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Problem



    Let ${X_n}$ be a sequence of real numbers defined by $$x_n=
    left{
    begin{array}{ccrcl}
    {displaystylefrac{1}{2k}} & mbox{if} &
    {displaystyle n} & {displaystyle =} &
    {displaystyle 2k - 1}
    \[1mm]
    {displaystylefrac{1}{2k - 1}} & mbox{if} &
    {displaystyle n} & {displaystyle =} &
    {displaystyle 2k}
    end{array}right.
    \epsilon_n=[frac{1}{2}]^n\
    f(x)=sum_{n:x_n<x}epsilon_n
    $$



    (A)Compute f(1), f(1/2)



    (B) Determine the set of discontinuities for the function



    Solution



    So with mathematica I computed $x_n={frac{1}{2},1,frac{1}{4},frac{1}{3},frac{1}{6},frac{1}{5},frac{1}{8},frac{1}{7}...}$



    So, $f(1)=frac{1}{2}+sum_{n:x_n<1}frac{1}{2}^2=frac{1}{2}+frac{1}{1-1/2}=frac{1}{2}+2=frac{3}{2}$



    $f(frac{1}{2})=sum_{n:x_n<1/2}frac{1}{2}^2=frac{1}{1-1/2}=2$



    Yet I noticed that the sum of all terms$=f(infty)=$ $sum_0^{infty}frac{1}{2}^2=frac{1}{1-1/2}=2$



    Why is $f(1)$ greater than the sum of all terms, when all terms are positive? Shouldn't $f(1)<f(infty)$



    For the points of discontinuity, I think the sequence should be continuous everywhere.



    Choose $delta$ such that $delta=x_n-x_{n-3},$ Then $forall epsilon > 0$, $forall yin B_{delta}(x_n),|x_n-y|<delta$, $|f(x_n)-f(y)|=|f(x_n)-f(x_n)|<epsilon$






    real-analysis







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Problem



      Let ${X_n}$ be a sequence of real numbers defined by $$x_n=
      left{
      begin{array}{ccrcl}
      {displaystylefrac{1}{2k}} & mbox{if} &
      {displaystyle n} & {displaystyle =} &
      {displaystyle 2k - 1}
      \[1mm]
      {displaystylefrac{1}{2k - 1}} & mbox{if} &
      {displaystyle n} & {displaystyle =} &
      {displaystyle 2k}
      end{array}right.
      \epsilon_n=[frac{1}{2}]^n\
      f(x)=sum_{n:x_n<x}epsilon_n
      $$



      (A)Compute f(1), f(1/2)



      (B) Determine the set of discontinuities for the function



      Solution



      So with mathematica I computed $x_n={frac{1}{2},1,frac{1}{4},frac{1}{3},frac{1}{6},frac{1}{5},frac{1}{8},frac{1}{7}...}$



      So, $f(1)=frac{1}{2}+sum_{n:x_n<1}frac{1}{2}^2=frac{1}{2}+frac{1}{1-1/2}=frac{1}{2}+2=frac{3}{2}$



      $f(frac{1}{2})=sum_{n:x_n<1/2}frac{1}{2}^2=frac{1}{1-1/2}=2$



      Yet I noticed that the sum of all terms$=f(infty)=$ $sum_0^{infty}frac{1}{2}^2=frac{1}{1-1/2}=2$



      Why is $f(1)$ greater than the sum of all terms, when all terms are positive? Shouldn't $f(1)<f(infty)$



      For the points of discontinuity, I think the sequence should be continuous everywhere.



      Choose $delta$ such that $delta=x_n-x_{n-3},$ Then $forall epsilon > 0$, $forall yin B_{delta}(x_n),|x_n-y|<delta$, $|f(x_n)-f(y)|=|f(x_n)-f(x_n)|<epsilon$






      real-analysis







      share|cite|improve this question











      $endgroup$




      Problem



      Let ${X_n}$ be a sequence of real numbers defined by $$x_n=
      left{
      begin{array}{ccrcl}
      {displaystylefrac{1}{2k}} & mbox{if} &
      {displaystyle n} & {displaystyle =} &
      {displaystyle 2k - 1}
      \[1mm]
      {displaystylefrac{1}{2k - 1}} & mbox{if} &
      {displaystyle n} & {displaystyle =} &
      {displaystyle 2k}
      end{array}right.
      \epsilon_n=[frac{1}{2}]^n\
      f(x)=sum_{n:x_n<x}epsilon_n
      $$



      (A)Compute f(1), f(1/2)



      (B) Determine the set of discontinuities for the function



      Solution



      So with mathematica I computed $x_n={frac{1}{2},1,frac{1}{4},frac{1}{3},frac{1}{6},frac{1}{5},frac{1}{8},frac{1}{7}...}$



      So, $f(1)=frac{1}{2}+sum_{n:x_n<1}frac{1}{2}^2=frac{1}{2}+frac{1}{1-1/2}=frac{1}{2}+2=frac{3}{2}$



      $f(frac{1}{2})=sum_{n:x_n<1/2}frac{1}{2}^2=frac{1}{1-1/2}=2$



      Yet I noticed that the sum of all terms$=f(infty)=$ $sum_0^{infty}frac{1}{2}^2=frac{1}{1-1/2}=2$



      Why is $f(1)$ greater than the sum of all terms, when all terms are positive? Shouldn't $f(1)<f(infty)$



      For the points of discontinuity, I think the sequence should be continuous everywhere.



      Choose $delta$ such that $delta=x_n-x_{n-3},$ Then $forall epsilon > 0$, $forall yin B_{delta}(x_n),|x_n-y|<delta$, $|f(x_n)-f(y)|=|f(x_n)-f(x_n)|<epsilon$






      real-analysis




      real-analysis






      share|cite|improve this question















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      edited Dec 12 '18 at 5:23









      Felix Marin

      67.8k7107142




      67.8k7107142










      asked Dec 12 '18 at 5:00









      FrankFrank

      16210




      16210






















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          For calculating $f(1)$, see that ${ninmathbb{N}:x_n<x}={ninmathbb{N}:nneq2}$. $$f(1)=sum_{substack{n:x_n<x\n neq2}}varepsilon^n=dfrac12+sum_{n=3}^inftyleft(dfrac12right)^n=dfrac34$$
          Similarly $$fleft(dfrac12right)=sum_{n=3}^inftyleft(dfrac12right)^n=dfrac14$$



          So $f(x)=2>f(1)::forall:x>1$.



          $f(x)$ is discontinuous at every $x_n$. For see, $displaystylelim_{xto1^-}f(x)=dfrac34neqlim_{xto1^+}f(x)=2$



          enter image description here






          share|cite|improve this answer











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            $begingroup$

            For calculating $f(1)$, see that ${ninmathbb{N}:x_n<x}={ninmathbb{N}:nneq2}$. $$f(1)=sum_{substack{n:x_n<x\n neq2}}varepsilon^n=dfrac12+sum_{n=3}^inftyleft(dfrac12right)^n=dfrac34$$
            Similarly $$fleft(dfrac12right)=sum_{n=3}^inftyleft(dfrac12right)^n=dfrac14$$



            So $f(x)=2>f(1)::forall:x>1$.



            $f(x)$ is discontinuous at every $x_n$. For see, $displaystylelim_{xto1^-}f(x)=dfrac34neqlim_{xto1^+}f(x)=2$



            enter image description here






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              For calculating $f(1)$, see that ${ninmathbb{N}:x_n<x}={ninmathbb{N}:nneq2}$. $$f(1)=sum_{substack{n:x_n<x\n neq2}}varepsilon^n=dfrac12+sum_{n=3}^inftyleft(dfrac12right)^n=dfrac34$$
              Similarly $$fleft(dfrac12right)=sum_{n=3}^inftyleft(dfrac12right)^n=dfrac14$$



              So $f(x)=2>f(1)::forall:x>1$.



              $f(x)$ is discontinuous at every $x_n$. For see, $displaystylelim_{xto1^-}f(x)=dfrac34neqlim_{xto1^+}f(x)=2$



              enter image description here






              share|cite|improve this answer











              $endgroup$
















                0












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                0





                $begingroup$

                For calculating $f(1)$, see that ${ninmathbb{N}:x_n<x}={ninmathbb{N}:nneq2}$. $$f(1)=sum_{substack{n:x_n<x\n neq2}}varepsilon^n=dfrac12+sum_{n=3}^inftyleft(dfrac12right)^n=dfrac34$$
                Similarly $$fleft(dfrac12right)=sum_{n=3}^inftyleft(dfrac12right)^n=dfrac14$$



                So $f(x)=2>f(1)::forall:x>1$.



                $f(x)$ is discontinuous at every $x_n$. For see, $displaystylelim_{xto1^-}f(x)=dfrac34neqlim_{xto1^+}f(x)=2$



                enter image description here






                share|cite|improve this answer











                $endgroup$



                For calculating $f(1)$, see that ${ninmathbb{N}:x_n<x}={ninmathbb{N}:nneq2}$. $$f(1)=sum_{substack{n:x_n<x\n neq2}}varepsilon^n=dfrac12+sum_{n=3}^inftyleft(dfrac12right)^n=dfrac34$$
                Similarly $$fleft(dfrac12right)=sum_{n=3}^inftyleft(dfrac12right)^n=dfrac14$$



                So $f(x)=2>f(1)::forall:x>1$.



                $f(x)$ is discontinuous at every $x_n$. For see, $displaystylelim_{xto1^-}f(x)=dfrac34neqlim_{xto1^+}f(x)=2$



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 18 '18 at 10:12

























                answered Dec 12 '18 at 5:42









                Yadati KiranYadati Kiran

                1,773619




                1,773619






























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