Generating Function For Catalan Numbers Type SequenceHow do you find generating function?Generating function...

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Generating Function For Catalan Numbers Type Sequence


How do you find generating function?Generating function of shifted Catalan NumbersGenerating function satisfying a second degree equationGenerating Function for 2-Associated Stirling Numbers of the Second KindFinding the generating function for the Catalan number sequenceCatalan numbers generating functionCatalan numbers formula derivationExpress C(x) through A(x) and B(x), where C(x), A(x), B(x), are generating functions of sequences $c_{n}$, $a_{n}$, $b_{n}$ respectivelyFind a generating function and a closed form for the sequence $c_{n+1} = sum_{i=0}^{n}c_i$Generating function for the Catalan numbersSequence $rightarrow$ Generating function simple practice problems













2












$begingroup$


I've been working my way through an old post, but I don't think the solution offered can be correct.



The question is;



Find the generating function (within a choice of sign) for:
$$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\c_0=1, ;c_1=3$$



I think this recurrence relation generates the numbers
$$1, 3, 12, 66, 408, 2712, ...$$



The solution offered is;



Let $$g(x)=sum_{nge 0}c_nx^n$$ be the ordinary generating function for the sequence. Then the standard formula for the Cauchy product of two summations yields



$$big(g(x)big)^2=left(sum_{nge 0}c_nx^nright)^2=sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=frac12sum_{nge 0}c_{n+1}x^n;,$$



and multiplication by $2x$ gives us



$$2xbig(g(x)big)^2=xsum_{nge 0}c_{n+1}x^n=sum_{nge 1}c_nx^n=g(x)-c_0;.$$



This is a quadratic in $g(x)$, so it can straightforwardly be solved for $g(x)$.



From this I've deduced that
$$g(x)=frac{1-sqrt(1-8x)}{4x}$$



I've gone through this proof carefully and can't see an error but I know from Wolfram Alpha that it does not generate the numbers I was expecting. It also makes no use of the fact that $c_1 = 3$ which can't be right.
I think the correct generating function is;
$$GF=frac{1-sqrt(1-8x-8x^2)}{4x}$$
but can't see how to obtain this.
It generates the numbers I was expecting and is in Sloan : https://oeis.org/search?q=1%2C3%2C12%2C66%2C408&language=english&go=Search



FYI : The original post, from 2013, is here : How do you find generating function?



For anyone interested in The Catalan Numbers, this would be a good 'one step on' question so if anyone can spot where the glitch is, I think it would be most useful.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I've been working my way through an old post, but I don't think the solution offered can be correct.



    The question is;



    Find the generating function (within a choice of sign) for:
    $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\c_0=1, ;c_1=3$$



    I think this recurrence relation generates the numbers
    $$1, 3, 12, 66, 408, 2712, ...$$



    The solution offered is;



    Let $$g(x)=sum_{nge 0}c_nx^n$$ be the ordinary generating function for the sequence. Then the standard formula for the Cauchy product of two summations yields



    $$big(g(x)big)^2=left(sum_{nge 0}c_nx^nright)^2=sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=frac12sum_{nge 0}c_{n+1}x^n;,$$



    and multiplication by $2x$ gives us



    $$2xbig(g(x)big)^2=xsum_{nge 0}c_{n+1}x^n=sum_{nge 1}c_nx^n=g(x)-c_0;.$$



    This is a quadratic in $g(x)$, so it can straightforwardly be solved for $g(x)$.



    From this I've deduced that
    $$g(x)=frac{1-sqrt(1-8x)}{4x}$$



    I've gone through this proof carefully and can't see an error but I know from Wolfram Alpha that it does not generate the numbers I was expecting. It also makes no use of the fact that $c_1 = 3$ which can't be right.
    I think the correct generating function is;
    $$GF=frac{1-sqrt(1-8x-8x^2)}{4x}$$
    but can't see how to obtain this.
    It generates the numbers I was expecting and is in Sloan : https://oeis.org/search?q=1%2C3%2C12%2C66%2C408&language=english&go=Search



    FYI : The original post, from 2013, is here : How do you find generating function?



    For anyone interested in The Catalan Numbers, this would be a good 'one step on' question so if anyone can spot where the glitch is, I think it would be most useful.










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I've been working my way through an old post, but I don't think the solution offered can be correct.



      The question is;



      Find the generating function (within a choice of sign) for:
      $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\c_0=1, ;c_1=3$$



      I think this recurrence relation generates the numbers
      $$1, 3, 12, 66, 408, 2712, ...$$



      The solution offered is;



      Let $$g(x)=sum_{nge 0}c_nx^n$$ be the ordinary generating function for the sequence. Then the standard formula for the Cauchy product of two summations yields



      $$big(g(x)big)^2=left(sum_{nge 0}c_nx^nright)^2=sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=frac12sum_{nge 0}c_{n+1}x^n;,$$



      and multiplication by $2x$ gives us



      $$2xbig(g(x)big)^2=xsum_{nge 0}c_{n+1}x^n=sum_{nge 1}c_nx^n=g(x)-c_0;.$$



      This is a quadratic in $g(x)$, so it can straightforwardly be solved for $g(x)$.



      From this I've deduced that
      $$g(x)=frac{1-sqrt(1-8x)}{4x}$$



      I've gone through this proof carefully and can't see an error but I know from Wolfram Alpha that it does not generate the numbers I was expecting. It also makes no use of the fact that $c_1 = 3$ which can't be right.
      I think the correct generating function is;
      $$GF=frac{1-sqrt(1-8x-8x^2)}{4x}$$
      but can't see how to obtain this.
      It generates the numbers I was expecting and is in Sloan : https://oeis.org/search?q=1%2C3%2C12%2C66%2C408&language=english&go=Search



      FYI : The original post, from 2013, is here : How do you find generating function?



      For anyone interested in The Catalan Numbers, this would be a good 'one step on' question so if anyone can spot where the glitch is, I think it would be most useful.










      share|cite|improve this question









      $endgroup$




      I've been working my way through an old post, but I don't think the solution offered can be correct.



      The question is;



      Find the generating function (within a choice of sign) for:
      $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\c_0=1, ;c_1=3$$



      I think this recurrence relation generates the numbers
      $$1, 3, 12, 66, 408, 2712, ...$$



      The solution offered is;



      Let $$g(x)=sum_{nge 0}c_nx^n$$ be the ordinary generating function for the sequence. Then the standard formula for the Cauchy product of two summations yields



      $$big(g(x)big)^2=left(sum_{nge 0}c_nx^nright)^2=sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=frac12sum_{nge 0}c_{n+1}x^n;,$$



      and multiplication by $2x$ gives us



      $$2xbig(g(x)big)^2=xsum_{nge 0}c_{n+1}x^n=sum_{nge 1}c_nx^n=g(x)-c_0;.$$



      This is a quadratic in $g(x)$, so it can straightforwardly be solved for $g(x)$.



      From this I've deduced that
      $$g(x)=frac{1-sqrt(1-8x)}{4x}$$



      I've gone through this proof carefully and can't see an error but I know from Wolfram Alpha that it does not generate the numbers I was expecting. It also makes no use of the fact that $c_1 = 3$ which can't be right.
      I think the correct generating function is;
      $$GF=frac{1-sqrt(1-8x-8x^2)}{4x}$$
      but can't see how to obtain this.
      It generates the numbers I was expecting and is in Sloan : https://oeis.org/search?q=1%2C3%2C12%2C66%2C408&language=english&go=Search



      FYI : The original post, from 2013, is here : How do you find generating function?



      For anyone interested in The Catalan Numbers, this would be a good 'one step on' question so if anyone can spot where the glitch is, I think it would be most useful.







      sequences-and-series elementary-number-theory generating-functions catalan-numbers cauchy-product






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 17 hours ago









      Martin HansenMartin Hansen

      12711




      12711






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          The trouble seems to be in this step:
          $$
          sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=
          frac12sum_{nge 0}c_{n+1}x^n.tag{*}
          $$

          It is true that
          $$
          sum_{k=0}^nc_kc_{n-k}=frac12 c_{n+1},
          $$

          but only for $nge1$. The constant term in (*) is $c_0^2 = 1$, which is not $frac12 c_1 = frac32$.



          Taking this into account, one instead finds
          $$
          g(x)^2 = frac12sum_{nge 0}c_{n+1}x^n - frac12,
          $$

          and the resulting quadratic is
          $$
          2x(g(x))^2 = g(x) - x - 1.
          $$

          One of the solutions of this quadratic is exactly the result you expected.



          By the way, to find the error I found it helpful to write out the low-order terms of $g(x)$ and $g(x)^2$. Those terms are where the complications are typically found in generating function problems.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
            $endgroup$
            – Martin Hansen
            17 hours ago










          • $begingroup$
            I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
            $endgroup$
            – Martin Hansen
            9 hours ago





















          0












          $begingroup$

          Thanks so much for your answer FredH. I agree with what you say.



          Thinking about it some more, and to clarify for anyone looking at this later on, the key step was moving from
          $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\$$
          to
          $$sum_{k=0}^{n}c_k c_{n-k}=frac{c_n+1}{2};;;n=1,2,3,4,dots\$$
          This needed this to be valid for $n=0,1,2,3,4,...$ in order to make the substitution to carry the main thread of the argument onward.
          So, clearly great care must be taken over such extensions to include $n=0$.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            The trouble seems to be in this step:
            $$
            sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=
            frac12sum_{nge 0}c_{n+1}x^n.tag{*}
            $$

            It is true that
            $$
            sum_{k=0}^nc_kc_{n-k}=frac12 c_{n+1},
            $$

            but only for $nge1$. The constant term in (*) is $c_0^2 = 1$, which is not $frac12 c_1 = frac32$.



            Taking this into account, one instead finds
            $$
            g(x)^2 = frac12sum_{nge 0}c_{n+1}x^n - frac12,
            $$

            and the resulting quadratic is
            $$
            2x(g(x))^2 = g(x) - x - 1.
            $$

            One of the solutions of this quadratic is exactly the result you expected.



            By the way, to find the error I found it helpful to write out the low-order terms of $g(x)$ and $g(x)^2$. Those terms are where the complications are typically found in generating function problems.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
              $endgroup$
              – Martin Hansen
              17 hours ago










            • $begingroup$
              I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
              $endgroup$
              – Martin Hansen
              9 hours ago


















            2












            $begingroup$

            The trouble seems to be in this step:
            $$
            sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=
            frac12sum_{nge 0}c_{n+1}x^n.tag{*}
            $$

            It is true that
            $$
            sum_{k=0}^nc_kc_{n-k}=frac12 c_{n+1},
            $$

            but only for $nge1$. The constant term in (*) is $c_0^2 = 1$, which is not $frac12 c_1 = frac32$.



            Taking this into account, one instead finds
            $$
            g(x)^2 = frac12sum_{nge 0}c_{n+1}x^n - frac12,
            $$

            and the resulting quadratic is
            $$
            2x(g(x))^2 = g(x) - x - 1.
            $$

            One of the solutions of this quadratic is exactly the result you expected.



            By the way, to find the error I found it helpful to write out the low-order terms of $g(x)$ and $g(x)^2$. Those terms are where the complications are typically found in generating function problems.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
              $endgroup$
              – Martin Hansen
              17 hours ago










            • $begingroup$
              I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
              $endgroup$
              – Martin Hansen
              9 hours ago
















            2












            2








            2





            $begingroup$

            The trouble seems to be in this step:
            $$
            sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=
            frac12sum_{nge 0}c_{n+1}x^n.tag{*}
            $$

            It is true that
            $$
            sum_{k=0}^nc_kc_{n-k}=frac12 c_{n+1},
            $$

            but only for $nge1$. The constant term in (*) is $c_0^2 = 1$, which is not $frac12 c_1 = frac32$.



            Taking this into account, one instead finds
            $$
            g(x)^2 = frac12sum_{nge 0}c_{n+1}x^n - frac12,
            $$

            and the resulting quadratic is
            $$
            2x(g(x))^2 = g(x) - x - 1.
            $$

            One of the solutions of this quadratic is exactly the result you expected.



            By the way, to find the error I found it helpful to write out the low-order terms of $g(x)$ and $g(x)^2$. Those terms are where the complications are typically found in generating function problems.






            share|cite|improve this answer









            $endgroup$



            The trouble seems to be in this step:
            $$
            sum_{nge 0}left(sum_{k=0}^nc_kc_{n-k}right)x^n=
            frac12sum_{nge 0}c_{n+1}x^n.tag{*}
            $$

            It is true that
            $$
            sum_{k=0}^nc_kc_{n-k}=frac12 c_{n+1},
            $$

            but only for $nge1$. The constant term in (*) is $c_0^2 = 1$, which is not $frac12 c_1 = frac32$.



            Taking this into account, one instead finds
            $$
            g(x)^2 = frac12sum_{nge 0}c_{n+1}x^n - frac12,
            $$

            and the resulting quadratic is
            $$
            2x(g(x))^2 = g(x) - x - 1.
            $$

            One of the solutions of this quadratic is exactly the result you expected.



            By the way, to find the error I found it helpful to write out the low-order terms of $g(x)$ and $g(x)^2$. Those terms are where the complications are typically found in generating function problems.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 17 hours ago









            FredHFredH

            1,073612




            1,073612












            • $begingroup$
              Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
              $endgroup$
              – Martin Hansen
              17 hours ago










            • $begingroup$
              I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
              $endgroup$
              – Martin Hansen
              9 hours ago




















            • $begingroup$
              Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
              $endgroup$
              – Martin Hansen
              17 hours ago










            • $begingroup$
              I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
              $endgroup$
              – Martin Hansen
              9 hours ago


















            $begingroup$
            Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
            $endgroup$
            – Martin Hansen
            17 hours ago




            $begingroup$
            Thanks for a quick response; I'm going to look at it very carefully after work finishes later, (following your advice), but it's the sort of small but very annoying error that I knew had to be in there somewhere.
            $endgroup$
            – Martin Hansen
            17 hours ago












            $begingroup$
            I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
            $endgroup$
            – Martin Hansen
            9 hours ago






            $begingroup$
            I've gone over this and am thoroughly satisfied with the argument now. I've added a note to the old post to point it here, and have also added a small further clarification here of exactly why the original six year old post derived a wrong answer.
            $endgroup$
            – Martin Hansen
            9 hours ago













            0












            $begingroup$

            Thanks so much for your answer FredH. I agree with what you say.



            Thinking about it some more, and to clarify for anyone looking at this later on, the key step was moving from
            $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\$$
            to
            $$sum_{k=0}^{n}c_k c_{n-k}=frac{c_n+1}{2};;;n=1,2,3,4,dots\$$
            This needed this to be valid for $n=0,1,2,3,4,...$ in order to make the substitution to carry the main thread of the argument onward.
            So, clearly great care must be taken over such extensions to include $n=0$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Thanks so much for your answer FredH. I agree with what you say.



              Thinking about it some more, and to clarify for anyone looking at this later on, the key step was moving from
              $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\$$
              to
              $$sum_{k=0}^{n}c_k c_{n-k}=frac{c_n+1}{2};;;n=1,2,3,4,dots\$$
              This needed this to be valid for $n=0,1,2,3,4,...$ in order to make the substitution to carry the main thread of the argument onward.
              So, clearly great care must be taken over such extensions to include $n=0$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Thanks so much for your answer FredH. I agree with what you say.



                Thinking about it some more, and to clarify for anyone looking at this later on, the key step was moving from
                $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\$$
                to
                $$sum_{k=0}^{n}c_k c_{n-k}=frac{c_n+1}{2};;;n=1,2,3,4,dots\$$
                This needed this to be valid for $n=0,1,2,3,4,...$ in order to make the substitution to carry the main thread of the argument onward.
                So, clearly great care must be taken over such extensions to include $n=0$.






                share|cite|improve this answer









                $endgroup$



                Thanks so much for your answer FredH. I agree with what you say.



                Thinking about it some more, and to clarify for anyone looking at this later on, the key step was moving from
                $$c_{n+1} = 2sum_{k=0}^{n}c_k c_{n-k},;;;n=1,2,3,4,dots\$$
                to
                $$sum_{k=0}^{n}c_k c_{n-k}=frac{c_n+1}{2};;;n=1,2,3,4,dots\$$
                This needed this to be valid for $n=0,1,2,3,4,...$ in order to make the substitution to carry the main thread of the argument onward.
                So, clearly great care must be taken over such extensions to include $n=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 9 hours ago









                Martin HansenMartin Hansen

                12711




                12711






























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