What is the limit of $2^ncdot(1-exp(-a/2^n))$ for $a>0$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I am trying to apply the Poisson limit theorem for a random variable that is



$operatornameBin(2^n,(1-exp(-a/2^n))$- distributed. This means I have to calculate $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))$ for some $a>0$ but have no idea how to do this.



The answer should be $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))=a$ but even knowing this I can't come up with a proof.




real-analysis probability limits




share|cite|improve this question





















  • Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
    – Jam
    Aug 2 at 18:08














up vote
0
down vote

favorite












I am trying to apply the Poisson limit theorem for a random variable that is



$operatornameBin(2^n,(1-exp(-a/2^n))$- distributed. This means I have to calculate $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))$ for some $a>0$ but have no idea how to do this.



The answer should be $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))=a$ but even knowing this I can't come up with a proof.




real-analysis probability limits




share|cite|improve this question





















  • Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
    – Jam
    Aug 2 at 18:08












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to apply the Poisson limit theorem for a random variable that is



$operatornameBin(2^n,(1-exp(-a/2^n))$- distributed. This means I have to calculate $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))$ for some $a>0$ but have no idea how to do this.



The answer should be $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))=a$ but even knowing this I can't come up with a proof.




real-analysis probability limits




share|cite|improve this question













I am trying to apply the Poisson limit theorem for a random variable that is



$operatornameBin(2^n,(1-exp(-a/2^n))$- distributed. This means I have to calculate $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))$ for some $a>0$ but have no idea how to do this.



The answer should be $limlimits_n rightarrow infty$ $2^n(1-exp(-a/2^n))=a$ but even knowing this I can't come up with a proof.





real-analysis probability limits





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 18:27









Jam

4,10111230




4,10111230









asked Aug 2 at 17:59







user577146


















  • Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
    – Jam
    Aug 2 at 18:08
















  • Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
    – Jam
    Aug 2 at 18:08















Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
– Jam
Aug 2 at 18:08




Please consider formatting your question with mathjax to make it easier for us to read. math.meta.stackexchange.com/questions/5020/…
– Jam
Aug 2 at 18:08










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Hint: Let $2^n=dfrac1t$ and use L'Hospital rule
$$lim_ntoinfty2^n(1-exp(-a/2^n))=lim_tto0dfrac1-e^-att=a$$






share|cite|improve this answer























  • If $nto infty$ it does not follow that $tto 0$
    – callculus
    Aug 2 at 18:08










  • thanks. I did type it wrongly.
    – user 108128
    Aug 2 at 18:10

















up vote
4
down vote













Recall that as $x=frac a 2^nto 0$



$$e^x=1+x+o(x)$$



therefore



$$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=frac a xleft(1-1+x+o(x)right)=a+o(1)to a$$



or by standard limit as $t to 0 quad frace^t-1tto 1$



$$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=afrac1-e^-xx=afrac1e^xfrace^x-1xto acdot 1cdot 1=a$$






share|cite|improve this answer






























    up vote
    3
    down vote













    Without invoking L'Hospital, it is well-known that
    $$1-mathrm e^usim_0 -u,quadtexthencequad1-mathrm e^-tfrac a2^nsim_ntoinftyfrac a2^n,$$
    so that in the end
    $$2^nBigl(1-mathrm e^-tfrac a2^nBigr)sim_ntoinfty2^nfrac a2^n=a.$$






    share|cite|improve this answer





















      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );








       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870331%2fwhat-is-the-limit-of-2n-cdot1-exp-a-2n-for-a0%23new-answer', 'question_page');

      );

      Post as a guest
















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      Hint: Let $2^n=dfrac1t$ and use L'Hospital rule
      $$lim_ntoinfty2^n(1-exp(-a/2^n))=lim_tto0dfrac1-e^-att=a$$






      share|cite|improve this answer























      • If $nto infty$ it does not follow that $tto 0$
        – callculus
        Aug 2 at 18:08










      • thanks. I did type it wrongly.
        – user 108128
        Aug 2 at 18:10














      up vote
      2
      down vote



      accepted










      Hint: Let $2^n=dfrac1t$ and use L'Hospital rule
      $$lim_ntoinfty2^n(1-exp(-a/2^n))=lim_tto0dfrac1-e^-att=a$$






      share|cite|improve this answer























      • If $nto infty$ it does not follow that $tto 0$
        – callculus
        Aug 2 at 18:08










      • thanks. I did type it wrongly.
        – user 108128
        Aug 2 at 18:10












      up vote
      2
      down vote



      accepted







      up vote
      2
      down vote



      accepted






      Hint: Let $2^n=dfrac1t$ and use L'Hospital rule
      $$lim_ntoinfty2^n(1-exp(-a/2^n))=lim_tto0dfrac1-e^-att=a$$






      share|cite|improve this answer















      Hint: Let $2^n=dfrac1t$ and use L'Hospital rule
      $$lim_ntoinfty2^n(1-exp(-a/2^n))=lim_tto0dfrac1-e^-att=a$$







      share|cite|improve this answer















      share|cite|improve this answer



      share|cite|improve this answer








      edited Aug 2 at 18:09


























      answered Aug 2 at 18:05









      user 108128

      18.8k41544




      18.8k41544











      • If $nto infty$ it does not follow that $tto 0$
        – callculus
        Aug 2 at 18:08










      • thanks. I did type it wrongly.
        – user 108128
        Aug 2 at 18:10
















      • If $nto infty$ it does not follow that $tto 0$
        – callculus
        Aug 2 at 18:08










      • thanks. I did type it wrongly.
        – user 108128
        Aug 2 at 18:10















      If $nto infty$ it does not follow that $tto 0$
      – callculus
      Aug 2 at 18:08




      If $nto infty$ it does not follow that $tto 0$
      – callculus
      Aug 2 at 18:08












      thanks. I did type it wrongly.
      – user 108128
      Aug 2 at 18:10




      thanks. I did type it wrongly.
      – user 108128
      Aug 2 at 18:10










      up vote
      4
      down vote













      Recall that as $x=frac a 2^nto 0$



      $$e^x=1+x+o(x)$$



      therefore



      $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=frac a xleft(1-1+x+o(x)right)=a+o(1)to a$$



      or by standard limit as $t to 0 quad frace^t-1tto 1$



      $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=afrac1-e^-xx=afrac1e^xfrace^x-1xto acdot 1cdot 1=a$$






      share|cite|improve this answer



























        up vote
        4
        down vote













        Recall that as $x=frac a 2^nto 0$



        $$e^x=1+x+o(x)$$



        therefore



        $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=frac a xleft(1-1+x+o(x)right)=a+o(1)to a$$



        or by standard limit as $t to 0 quad frace^t-1tto 1$



        $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=afrac1-e^-xx=afrac1e^xfrace^x-1xto acdot 1cdot 1=a$$






        share|cite|improve this answer

























          up vote
          4
          down vote










          up vote
          4
          down vote









          Recall that as $x=frac a 2^nto 0$



          $$e^x=1+x+o(x)$$



          therefore



          $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=frac a xleft(1-1+x+o(x)right)=a+o(1)to a$$



          or by standard limit as $t to 0 quad frace^t-1tto 1$



          $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=afrac1-e^-xx=afrac1e^xfrace^x-1xto acdot 1cdot 1=a$$






          share|cite|improve this answer















          Recall that as $x=frac a 2^nto 0$



          $$e^x=1+x+o(x)$$



          therefore



          $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=frac a xleft(1-1+x+o(x)right)=a+o(1)to a$$



          or by standard limit as $t to 0 quad frace^t-1tto 1$



          $$2^nleft(1-e^-frac a 2^nright)=frac a xleft(1-e^-xright)=afrac1-e^-xx=afrac1e^xfrace^x-1xto acdot 1cdot 1=a$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 2 at 19:46


























          answered Aug 2 at 18:03









          gimusi

          63.8k73480




          63.8k73480




















              up vote
              3
              down vote













              Without invoking L'Hospital, it is well-known that
              $$1-mathrm e^usim_0 -u,quadtexthencequad1-mathrm e^-tfrac a2^nsim_ntoinftyfrac a2^n,$$
              so that in the end
              $$2^nBigl(1-mathrm e^-tfrac a2^nBigr)sim_ntoinfty2^nfrac a2^n=a.$$






              share|cite|improve this answer

























                up vote
                3
                down vote













                Without invoking L'Hospital, it is well-known that
                $$1-mathrm e^usim_0 -u,quadtexthencequad1-mathrm e^-tfrac a2^nsim_ntoinftyfrac a2^n,$$
                so that in the end
                $$2^nBigl(1-mathrm e^-tfrac a2^nBigr)sim_ntoinfty2^nfrac a2^n=a.$$






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Without invoking L'Hospital, it is well-known that
                  $$1-mathrm e^usim_0 -u,quadtexthencequad1-mathrm e^-tfrac a2^nsim_ntoinftyfrac a2^n,$$
                  so that in the end
                  $$2^nBigl(1-mathrm e^-tfrac a2^nBigr)sim_ntoinfty2^nfrac a2^n=a.$$






                  share|cite|improve this answer













                  Without invoking L'Hospital, it is well-known that
                  $$1-mathrm e^usim_0 -u,quadtexthencequad1-mathrm e^-tfrac a2^nsim_ntoinftyfrac a2^n,$$
                  so that in the end
                  $$2^nBigl(1-mathrm e^-tfrac a2^nBigr)sim_ntoinfty2^nfrac a2^n=a.$$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 2 at 18:26









                  Bernard

                  110k635102




                  110k635102






















                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870331%2fwhat-is-the-limit-of-2n-cdot1-exp-a-2n-for-a0%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more