Induction proof for stirling of first kind.












2












$begingroup$


I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



I've started like this:



Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.



Now we assume $forall n in mathbb{N} $ with $ n geq 2$
that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.



It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.



Now let $n rightarrow n+1$.



$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.



If we insert our assumption, that leaves us with



$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.



If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.



The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?






induction







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    2












    $begingroup$


    I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



    I've started like this:



    Base case: Let $n$ be $n=2$, then per defintion
    $s_{n,0} = 0$.
    Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.



    Now we assume $forall n in mathbb{N} $ with $ n geq 2$
    that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



    We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.



    It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.



    Now let $n rightarrow n+1$.



    $s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.



    If we insert our assumption, that leaves us with



    $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.



    If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
    $frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.



    The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?






    induction







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



      I've started like this:



      Base case: Let $n$ be $n=2$, then per defintion
      $s_{n,0} = 0$.
      Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.



      Now we assume $forall n in mathbb{N} $ with $ n geq 2$
      that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



      We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.



      It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.



      Now let $n rightarrow n+1$.



      $s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.



      If we insert our assumption, that leaves us with



      $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.



      If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
      $frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.



      The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?






      induction







      share|cite|improve this question









      $endgroup$




      I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



      I've started like this:



      Base case: Let $n$ be $n=2$, then per defintion
      $s_{n,0} = 0$.
      Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.



      Now we assume $forall n in mathbb{N} $ with $ n geq 2$
      that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.



      We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.



      It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.



      Now let $n rightarrow n+1$.



      $s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.



      If we insert our assumption, that leaves us with



      $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.



      If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
      $frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.



      The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?






      induction




      induction






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      asked Jan 17 at 13:50









      D idsea JD idsea J

      325




      325






















          2 Answers
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          $begingroup$

          $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.



          $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.



          Substitute this in the prior equation we get



          $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.



          which simplifies to



          $s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.



          Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you clarify what you mean by identity(2)?
            $endgroup$
            – D idsea J
            Jan 17 at 15:24






          • 1




            $begingroup$
            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
            $endgroup$
            – Satish Ramanathan
            Jan 17 at 15:26










          • $begingroup$
            Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
            $endgroup$
            – D idsea J
            Jan 17 at 15:32












          • $begingroup$
            It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
            $endgroup$
            – D idsea J
            Jan 17 at 15:43










          • $begingroup$
            Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
            $endgroup$
            – D idsea J
            Jan 18 at 8:05



















          3












          $begingroup$

          Induction step:



          $begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
          & =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
          & =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
          & =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
          & =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
          end{aligned}
          $



          In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.



            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.



            Substitute this in the prior equation we get



            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.



            which simplifies to



            $s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.



            Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you clarify what you mean by identity(2)?
              $endgroup$
              – D idsea J
              Jan 17 at 15:24






            • 1




              $begingroup$
              $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
              $endgroup$
              – Satish Ramanathan
              Jan 17 at 15:26










            • $begingroup$
              Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
              $endgroup$
              – D idsea J
              Jan 17 at 15:32












            • $begingroup$
              It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
              $endgroup$
              – D idsea J
              Jan 17 at 15:43










            • $begingroup$
              Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
              $endgroup$
              – D idsea J
              Jan 18 at 8:05
















            3












            $begingroup$

            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.



            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.



            Substitute this in the prior equation we get



            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.



            which simplifies to



            $s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.



            Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you clarify what you mean by identity(2)?
              $endgroup$
              – D idsea J
              Jan 17 at 15:24






            • 1




              $begingroup$
              $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
              $endgroup$
              – Satish Ramanathan
              Jan 17 at 15:26










            • $begingroup$
              Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
              $endgroup$
              – D idsea J
              Jan 17 at 15:32












            • $begingroup$
              It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
              $endgroup$
              – D idsea J
              Jan 17 at 15:43










            • $begingroup$
              Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
              $endgroup$
              – D idsea J
              Jan 18 at 8:05














            3












            3








            3





            $begingroup$

            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.



            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.



            Substitute this in the prior equation we get



            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.



            which simplifies to



            $s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.



            Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.






            share|cite|improve this answer









            $endgroup$



            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.



            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.



            Substitute this in the prior equation we get



            $s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.



            which simplifies to



            $s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.



            Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 17 at 14:57









            Satish RamanathanSatish Ramanathan

            9,71031323




            9,71031323












            • $begingroup$
              Could you clarify what you mean by identity(2)?
              $endgroup$
              – D idsea J
              Jan 17 at 15:24






            • 1




              $begingroup$
              $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
              $endgroup$
              – Satish Ramanathan
              Jan 17 at 15:26










            • $begingroup$
              Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
              $endgroup$
              – D idsea J
              Jan 17 at 15:32












            • $begingroup$
              It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
              $endgroup$
              – D idsea J
              Jan 17 at 15:43










            • $begingroup$
              Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
              $endgroup$
              – D idsea J
              Jan 18 at 8:05


















            • $begingroup$
              Could you clarify what you mean by identity(2)?
              $endgroup$
              – D idsea J
              Jan 17 at 15:24






            • 1




              $begingroup$
              $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
              $endgroup$
              – Satish Ramanathan
              Jan 17 at 15:26










            • $begingroup$
              Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
              $endgroup$
              – D idsea J
              Jan 17 at 15:32












            • $begingroup$
              It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
              $endgroup$
              – D idsea J
              Jan 17 at 15:43










            • $begingroup$
              Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
              $endgroup$
              – D idsea J
              Jan 18 at 8:05
















            $begingroup$
            Could you clarify what you mean by identity(2)?
            $endgroup$
            – D idsea J
            Jan 17 at 15:24




            $begingroup$
            Could you clarify what you mean by identity(2)?
            $endgroup$
            – D idsea J
            Jan 17 at 15:24




            1




            1




            $begingroup$
            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
            $endgroup$
            – Satish Ramanathan
            Jan 17 at 15:26




            $begingroup$
            $s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
            $endgroup$
            – Satish Ramanathan
            Jan 17 at 15:26












            $begingroup$
            Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
            $endgroup$
            – D idsea J
            Jan 17 at 15:32






            $begingroup$
            Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
            $endgroup$
            – D idsea J
            Jan 17 at 15:32














            $begingroup$
            It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
            $endgroup$
            – D idsea J
            Jan 17 at 15:43




            $begingroup$
            It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
            $endgroup$
            – D idsea J
            Jan 17 at 15:43












            $begingroup$
            Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
            $endgroup$
            – D idsea J
            Jan 18 at 8:05




            $begingroup$
            Wolfram states, that the special case for stirling first kind s(n,n-1) is -(n choose 2). So who's right now o_O
            $endgroup$
            – D idsea J
            Jan 18 at 8:05











            3












            $begingroup$

            Induction step:



            $begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
            & =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
            & =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
            & =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
            & =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
            end{aligned}
            $



            In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Induction step:



              $begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
              & =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
              & =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
              & =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
              & =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
              end{aligned}
              $



              In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Induction step:



                $begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
                & =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
                & =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
                & =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
                & =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
                end{aligned}
                $



                In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.






                share|cite|improve this answer









                $endgroup$



                Induction step:



                $begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
                & =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
                & =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
                & =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
                & =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
                end{aligned}
                $



                In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 15:04









                drhabdrhab

                99.8k544130




                99.8k544130






























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