equating coefficients in algebraic expansion












1












$begingroup$


If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of



$(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$



$(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$



$(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$



$(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$



Answers given in $n,2n,3n,4n$ formats



What I tried:



$displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$



$displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $



$displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$



How do I find other coefficients? Help me, please.






binomial-theorem







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of



    $(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$



    $(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$



    $(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$



    $(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$



    Answers given in $n,2n,3n,4n$ formats



    What I tried:



    $displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$



    $displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $



    $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$



    How do I find other coefficients? Help me, please.






    binomial-theorem







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of



      $(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$



      $(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$



      $(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$



      $(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$



      Answers given in $n,2n,3n,4n$ formats



      What I tried:



      $displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$



      $displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $



      $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$



      How do I find other coefficients? Help me, please.






      binomial-theorem







      share|cite|improve this question











      $endgroup$




      If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of



      $(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$



      $(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$



      $(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$



      $(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$



      Answers given in $n,2n,3n,4n$ formats



      What I tried:



      $displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$



      $displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $



      $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$



      How do I find other coefficients? Help me, please.






      binomial-theorem




      binomial-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 1 at 5:37









      J. W. Tanner

      4,6441420




      4,6441420










      asked Feb 1 at 5:03









      jackyjacky

      1,344816




      1,344816






















          2 Answers
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          active

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          1












          $begingroup$

          You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but



          $b_{1}=2n$



          $b_{2}=2n^2$



          $b_{3}=frac23(n+2n^3)$



          To go up to $b_{10}$, the series has to be expended up to $x^{10}$.



          By brut force :



          $b_{4}=frac23(2n^2+n^4)$



          $b_{5}=frac{2}{15}(3n+10n^3+2n^5)$



          $b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$



          $b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$



          $b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$



          $b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$



          $b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$



          $frac{3b_{3}-b_{1}}{b_{2}}=2n$



          Then, a supposed recurrence :



          $frac{1b_{2}-0b_{0}}{b_{1}}=n$



          $frac{2b_{4}-1b_{2}}{b_{3}}=n$



          $frac{3b_{6}-2b_{4}}{b_{5}}=n$



          $frac{4b_{8}-3b_{6}}{b_{7}}=n$



          $frac{5b_{10}-4b_{8}}{b_{9}}=n$



          $...$



          $frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.




            We obtain
            begin{align*}
            color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
            &=[x^j](1-x)^{-n}(1+x)^n\
            &=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
            &=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
            &=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
            &,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
            end{align*}




            Comment:




            • In (1) we apply the binomial series expansion


            • In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$


            • In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.


            • In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.







            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              2 Answers
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              active

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              1












              $begingroup$

              You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but



              $b_{1}=2n$



              $b_{2}=2n^2$



              $b_{3}=frac23(n+2n^3)$



              To go up to $b_{10}$, the series has to be expended up to $x^{10}$.



              By brut force :



              $b_{4}=frac23(2n^2+n^4)$



              $b_{5}=frac{2}{15}(3n+10n^3+2n^5)$



              $b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$



              $b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$



              $b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$



              $b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$



              $b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$



              $frac{3b_{3}-b_{1}}{b_{2}}=2n$



              Then, a supposed recurrence :



              $frac{1b_{2}-0b_{0}}{b_{1}}=n$



              $frac{2b_{4}-1b_{2}}{b_{3}}=n$



              $frac{3b_{6}-2b_{4}}{b_{5}}=n$



              $frac{4b_{8}-3b_{6}}{b_{7}}=n$



              $frac{5b_{10}-4b_{8}}{b_{9}}=n$



              $...$



              $frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but



                $b_{1}=2n$



                $b_{2}=2n^2$



                $b_{3}=frac23(n+2n^3)$



                To go up to $b_{10}$, the series has to be expended up to $x^{10}$.



                By brut force :



                $b_{4}=frac23(2n^2+n^4)$



                $b_{5}=frac{2}{15}(3n+10n^3+2n^5)$



                $b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$



                $b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$



                $b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$



                $b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$



                $b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$



                $frac{3b_{3}-b_{1}}{b_{2}}=2n$



                Then, a supposed recurrence :



                $frac{1b_{2}-0b_{0}}{b_{1}}=n$



                $frac{2b_{4}-1b_{2}}{b_{3}}=n$



                $frac{3b_{6}-2b_{4}}{b_{5}}=n$



                $frac{4b_{8}-3b_{6}}{b_{7}}=n$



                $frac{5b_{10}-4b_{8}}{b_{9}}=n$



                $...$



                $frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but



                  $b_{1}=2n$



                  $b_{2}=2n^2$



                  $b_{3}=frac23(n+2n^3)$



                  To go up to $b_{10}$, the series has to be expended up to $x^{10}$.



                  By brut force :



                  $b_{4}=frac23(2n^2+n^4)$



                  $b_{5}=frac{2}{15}(3n+10n^3+2n^5)$



                  $b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$



                  $b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$



                  $b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$



                  $b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$



                  $b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$



                  $frac{3b_{3}-b_{1}}{b_{2}}=2n$



                  Then, a supposed recurrence :



                  $frac{1b_{2}-0b_{0}}{b_{1}}=n$



                  $frac{2b_{4}-1b_{2}}{b_{3}}=n$



                  $frac{3b_{6}-2b_{4}}{b_{5}}=n$



                  $frac{4b_{8}-3b_{6}}{b_{7}}=n$



                  $frac{5b_{10}-4b_{8}}{b_{9}}=n$



                  $...$



                  $frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.






                  share|cite|improve this answer









                  $endgroup$



                  You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but



                  $b_{1}=2n$



                  $b_{2}=2n^2$



                  $b_{3}=frac23(n+2n^3)$



                  To go up to $b_{10}$, the series has to be expended up to $x^{10}$.



                  By brut force :



                  $b_{4}=frac23(2n^2+n^4)$



                  $b_{5}=frac{2}{15}(3n+10n^3+2n^5)$



                  $b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$



                  $b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$



                  $b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$



                  $b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$



                  $b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$



                  $frac{3b_{3}-b_{1}}{b_{2}}=2n$



                  Then, a supposed recurrence :



                  $frac{1b_{2}-0b_{0}}{b_{1}}=n$



                  $frac{2b_{4}-1b_{2}}{b_{3}}=n$



                  $frac{3b_{6}-2b_{4}}{b_{5}}=n$



                  $frac{4b_{8}-3b_{6}}{b_{7}}=n$



                  $frac{5b_{10}-4b_{8}}{b_{9}}=n$



                  $...$



                  $frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 7:27









                  JJacquelinJJacquelin

                  45.6k21857




                  45.6k21857























                      1












                      $begingroup$

                      Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.




                      We obtain
                      begin{align*}
                      color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
                      &=[x^j](1-x)^{-n}(1+x)^n\
                      &=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
                      &=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
                      &=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
                      &,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
                      end{align*}




                      Comment:




                      • In (1) we apply the binomial series expansion


                      • In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$


                      • In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.


                      • In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.







                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.




                        We obtain
                        begin{align*}
                        color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
                        &=[x^j](1-x)^{-n}(1+x)^n\
                        &=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
                        &=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
                        &=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
                        &,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
                        end{align*}




                        Comment:




                        • In (1) we apply the binomial series expansion


                        • In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$


                        • In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.


                        • In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.







                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.




                          We obtain
                          begin{align*}
                          color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
                          &=[x^j](1-x)^{-n}(1+x)^n\
                          &=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
                          &=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
                          &=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
                          &,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
                          end{align*}




                          Comment:




                          • In (1) we apply the binomial series expansion


                          • In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$


                          • In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.


                          • In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.







                          share|cite|improve this answer









                          $endgroup$



                          Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.




                          We obtain
                          begin{align*}
                          color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
                          &=[x^j](1-x)^{-n}(1+x)^n\
                          &=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
                          &=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
                          &=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
                          &,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
                          end{align*}




                          Comment:




                          • In (1) we apply the binomial series expansion


                          • In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$


                          • In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.


                          • In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.








                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 1 at 20:42









                          Markus ScheuerMarkus Scheuer

                          64.1k460152




                          64.1k460152






























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                              Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

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                              28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
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                              Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

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                              up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
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                              A Topological Invariant for $pi_3(U(n))$

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                              3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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