How to find a generating function in simple, closed form

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











up vote
0
down vote

favorite












I’m stuck on something in generating functionology. The first problem asks:
Find the ordinary power series generating functions of the sequence in simple closed form for the sequence $a_n = n$. The sequence is defined as $n ≥ 0$.



I figured out how to get to $A(x) = x/((1-x)^2)$. That’s not an issue.



However, the book lists the answer as $(xD)(1/(1-x)) = x/((1-x)^2)$



Where did the D come from? How can I get my answer in terms of D?




generating-functions




share|cite|improve this question




















  • The $D$ is a differential operator.
    – John Wayland Bales
    Aug 8 at 21:17










  • $D$ is $frac ddx$.
    – Lord Shark the Unknown
    Aug 8 at 21:34














up vote
0
down vote

favorite












I’m stuck on something in generating functionology. The first problem asks:
Find the ordinary power series generating functions of the sequence in simple closed form for the sequence $a_n = n$. The sequence is defined as $n ≥ 0$.



I figured out how to get to $A(x) = x/((1-x)^2)$. That’s not an issue.



However, the book lists the answer as $(xD)(1/(1-x)) = x/((1-x)^2)$



Where did the D come from? How can I get my answer in terms of D?




generating-functions




share|cite|improve this question




















  • The $D$ is a differential operator.
    – John Wayland Bales
    Aug 8 at 21:17










  • $D$ is $frac ddx$.
    – Lord Shark the Unknown
    Aug 8 at 21:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I’m stuck on something in generating functionology. The first problem asks:
Find the ordinary power series generating functions of the sequence in simple closed form for the sequence $a_n = n$. The sequence is defined as $n ≥ 0$.



I figured out how to get to $A(x) = x/((1-x)^2)$. That’s not an issue.



However, the book lists the answer as $(xD)(1/(1-x)) = x/((1-x)^2)$



Where did the D come from? How can I get my answer in terms of D?




generating-functions




share|cite|improve this question












I’m stuck on something in generating functionology. The first problem asks:
Find the ordinary power series generating functions of the sequence in simple closed form for the sequence $a_n = n$. The sequence is defined as $n ≥ 0$.



I figured out how to get to $A(x) = x/((1-x)^2)$. That’s not an issue.



However, the book lists the answer as $(xD)(1/(1-x)) = x/((1-x)^2)$



Where did the D come from? How can I get my answer in terms of D?





generating-functions





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 8 at 21:13









user582707

31




31











  • The $D$ is a differential operator.
    – John Wayland Bales
    Aug 8 at 21:17










  • $D$ is $frac ddx$.
    – Lord Shark the Unknown
    Aug 8 at 21:34
















  • The $D$ is a differential operator.
    – John Wayland Bales
    Aug 8 at 21:17










  • $D$ is $frac ddx$.
    – Lord Shark the Unknown
    Aug 8 at 21:34















The $D$ is a differential operator.
– John Wayland Bales
Aug 8 at 21:17




The $D$ is a differential operator.
– John Wayland Bales
Aug 8 at 21:17












$D$ is $frac ddx$.
– Lord Shark the Unknown
Aug 8 at 21:34




$D$ is $frac ddx$.
– Lord Shark the Unknown
Aug 8 at 21:34










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So
$$
(xD)frac11-x=xleft(frac11-xright)'=xtimesfrac11-x^2=fracx1-x^2.
$$
Note that if
$$
B(x)=sum_ngeq 0 b_n z^n
$$
then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^infty$.






share|cite|improve this answer



























    up vote
    0
    down vote













    begineqnarray
    A(x)&=&0+1cdot x+2cdot x^2+3cdot x^3+cdots\
    &=&x(1+2x+3x^2+cdots)\
    &=&xfracddx(1+x+x^2+x^3+cdots)\
    &=&xfracddxleft(frac11-xright)\
    &=&fracx(1-x)^2
    endeqnarray






    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%2f2876599%2fhow-to-find-a-generating-function-in-simple-closed-form%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote



      accepted










      You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So
      $$
      (xD)frac11-x=xleft(frac11-xright)'=xtimesfrac11-x^2=fracx1-x^2.
      $$
      Note that if
      $$
      B(x)=sum_ngeq 0 b_n z^n
      $$
      then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^infty$.






      share|cite|improve this answer
























        up vote
        0
        down vote



        accepted










        You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So
        $$
        (xD)frac11-x=xleft(frac11-xright)'=xtimesfrac11-x^2=fracx1-x^2.
        $$
        Note that if
        $$
        B(x)=sum_ngeq 0 b_n z^n
        $$
        then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^infty$.






        share|cite|improve this answer






















          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So
          $$
          (xD)frac11-x=xleft(frac11-xright)'=xtimesfrac11-x^2=fracx1-x^2.
          $$
          Note that if
          $$
          B(x)=sum_ngeq 0 b_n z^n
          $$
          then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^infty$.






          share|cite|improve this answer












          You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So
          $$
          (xD)frac11-x=xleft(frac11-xright)'=xtimesfrac11-x^2=fracx1-x^2.
          $$
          Note that if
          $$
          B(x)=sum_ngeq 0 b_n z^n
          $$
          then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 21:19









          Foobaz John

          18.1k41245




          18.1k41245




















              up vote
              0
              down vote













              begineqnarray
              A(x)&=&0+1cdot x+2cdot x^2+3cdot x^3+cdots\
              &=&x(1+2x+3x^2+cdots)\
              &=&xfracddx(1+x+x^2+x^3+cdots)\
              &=&xfracddxleft(frac11-xright)\
              &=&fracx(1-x)^2
              endeqnarray






              share|cite|improve this answer
























                up vote
                0
                down vote













                begineqnarray
                A(x)&=&0+1cdot x+2cdot x^2+3cdot x^3+cdots\
                &=&x(1+2x+3x^2+cdots)\
                &=&xfracddx(1+x+x^2+x^3+cdots)\
                &=&xfracddxleft(frac11-xright)\
                &=&fracx(1-x)^2
                endeqnarray






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  begineqnarray
                  A(x)&=&0+1cdot x+2cdot x^2+3cdot x^3+cdots\
                  &=&x(1+2x+3x^2+cdots)\
                  &=&xfracddx(1+x+x^2+x^3+cdots)\
                  &=&xfracddxleft(frac11-xright)\
                  &=&fracx(1-x)^2
                  endeqnarray






                  share|cite|improve this answer












                  begineqnarray
                  A(x)&=&0+1cdot x+2cdot x^2+3cdot x^3+cdots\
                  &=&x(1+2x+3x^2+cdots)\
                  &=&xfracddx(1+x+x^2+x^3+cdots)\
                  &=&xfracddxleft(frac11-xright)\
                  &=&fracx(1-x)^2
                  endeqnarray







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 8 at 21:38









                  John Wayland Bales

                  13.1k21137




                  13.1k21137






















                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2876599%2fhow-to-find-a-generating-function-in-simple-closed-form%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      -

                      這個網誌中的熱門文章

                      How to combine Bézier curves to a surface?

                      圖片
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
                      閱讀完整內容

                      Mutual Information Always Non-negative

                      圖片
                      Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
                      閱讀完整內容

                      Why am i infinitely getting the same tweet with the Twitter Search API?

                      圖片
                      up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
                      閱讀完整內容