Determine the recurrence formula












4












$begingroup$


$int_{-1}^{1}(1-x^2)^ndx$



I have trouble with finding recurrence formula for this integral. $n$ is natural parameter. I've tried to split up $(1-x^2)^n = (1+x)^n(1-x)^n$ and then to integrate partially, but it only makes things more complicate.



Maybe substitution $x=sint$ can lead to solution?
When I apply it I get:



$int_{-pi/2}^{pi/2}(cost)^{n+1}dt$



What to do next then?






integration definite-integrals recurrence-relations reduction-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
    $endgroup$
    – Sujit Bhattacharyya
    Jan 25 at 11:48


















4












$begingroup$


$int_{-1}^{1}(1-x^2)^ndx$



I have trouble with finding recurrence formula for this integral. $n$ is natural parameter. I've tried to split up $(1-x^2)^n = (1+x)^n(1-x)^n$ and then to integrate partially, but it only makes things more complicate.



Maybe substitution $x=sint$ can lead to solution?
When I apply it I get:



$int_{-pi/2}^{pi/2}(cost)^{n+1}dt$



What to do next then?






integration definite-integrals recurrence-relations reduction-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
    $endgroup$
    – Sujit Bhattacharyya
    Jan 25 at 11:48
















4












4








4


0



$begingroup$


$int_{-1}^{1}(1-x^2)^ndx$



I have trouble with finding recurrence formula for this integral. $n$ is natural parameter. I've tried to split up $(1-x^2)^n = (1+x)^n(1-x)^n$ and then to integrate partially, but it only makes things more complicate.



Maybe substitution $x=sint$ can lead to solution?
When I apply it I get:



$int_{-pi/2}^{pi/2}(cost)^{n+1}dt$



What to do next then?






integration definite-integrals recurrence-relations reduction-formula







share|cite|improve this question











$endgroup$




$int_{-1}^{1}(1-x^2)^ndx$



I have trouble with finding recurrence formula for this integral. $n$ is natural parameter. I've tried to split up $(1-x^2)^n = (1+x)^n(1-x)^n$ and then to integrate partially, but it only makes things more complicate.



Maybe substitution $x=sint$ can lead to solution?
When I apply it I get:



$int_{-pi/2}^{pi/2}(cost)^{n+1}dt$



What to do next then?






integration definite-integrals recurrence-relations reduction-formula




integration definite-integrals recurrence-relations reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 16:58









clathratus

4,8901338




4,8901338










asked Jan 25 at 11:34









Nikola MijuškovićNikola Mijušković

274




274












  • $begingroup$
    how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
    $endgroup$
    – Sujit Bhattacharyya
    Jan 25 at 11:48




















  • $begingroup$
    how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
    $endgroup$
    – Sujit Bhattacharyya
    Jan 25 at 11:48


















$begingroup$
how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
$endgroup$
– Sujit Bhattacharyya
Jan 25 at 11:48






$begingroup$
how about substituting $x=sin theta$? Your case is same as $n=-frac{1}{2}$ in : mathworld.wolfram.com/BetaFunction.html
$endgroup$
– Sujit Bhattacharyya
Jan 25 at 11:48












3 Answers
3






active

oldest

votes


















3












$begingroup$

You can integrate by parts directly to obtain
begin{align}
int limits_{-1}^1 (1-x^2)^n , mathrm{d} x &= left[x (1-x^2)^n right]_{x=-1}^{x=1} + 2n int limits_{-1}^1 x^2 (1-x^2)^{n-1} , mathrm{d} x \
&= 2 n left[int limits_{-1}^1 (1-x^2)^{n-1} , mathrm{d} x - int limits_{-1}^1 (1-x^2)^n , mathrm{d} xright]
end{align}
for $n in mathbb{N}$, which is the recurrence relation you are after.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Not a recurrence, but it seems that
    $$
    I_n = int_{-1}^{1}(1-x^2)^ndx = frac{2^{a_n}}{b_n}
    $$
    where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.



    Also,
    $$
    frac{1}{(1 - x)^{3/2}} = sum_{n=0}^{infty} frac{2}{I_n} x^n
    $$



    WA gives
    $$
    I_n = frac{sqrt{pi} Gamma(n + 1)}{Gamma(n +frac32)}
    = frac{2^{n+1} n!}{(2n+1)!!}
    $$






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      On the path of Nikola Mijušković...



      begin{align}I_n&=int_{-1}^1 (1-x^2)^n, dx\&=2int_{0}^1 (1-x^2)^n, dxend{align}



      Perform the change of variable $x=sin t$,



      begin{align}I_n&=2int_{0}^{frac{pi}{2}} cos^{2n+1}t, dt\
      &=2W_{2n+1}
      end{align}



      $W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )



      It is known that, for $ngeq 0$,



      begin{align}W_1&=1\
      W_{n+2}&=frac{n+1}{n+2}W_nend{align}



      Therefore,



      begin{align}I_{n+1}&=2W_{2n+3}\
      &=frac{2(2n+2)}{2n+3}W_{2n+1}\
      &=frac{2(n+1)}{2n+3}I_{n}
      end{align}






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        You can integrate by parts directly to obtain
        begin{align}
        int limits_{-1}^1 (1-x^2)^n , mathrm{d} x &= left[x (1-x^2)^n right]_{x=-1}^{x=1} + 2n int limits_{-1}^1 x^2 (1-x^2)^{n-1} , mathrm{d} x \
        &= 2 n left[int limits_{-1}^1 (1-x^2)^{n-1} , mathrm{d} x - int limits_{-1}^1 (1-x^2)^n , mathrm{d} xright]
        end{align}
        for $n in mathbb{N}$, which is the recurrence relation you are after.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          You can integrate by parts directly to obtain
          begin{align}
          int limits_{-1}^1 (1-x^2)^n , mathrm{d} x &= left[x (1-x^2)^n right]_{x=-1}^{x=1} + 2n int limits_{-1}^1 x^2 (1-x^2)^{n-1} , mathrm{d} x \
          &= 2 n left[int limits_{-1}^1 (1-x^2)^{n-1} , mathrm{d} x - int limits_{-1}^1 (1-x^2)^n , mathrm{d} xright]
          end{align}
          for $n in mathbb{N}$, which is the recurrence relation you are after.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            You can integrate by parts directly to obtain
            begin{align}
            int limits_{-1}^1 (1-x^2)^n , mathrm{d} x &= left[x (1-x^2)^n right]_{x=-1}^{x=1} + 2n int limits_{-1}^1 x^2 (1-x^2)^{n-1} , mathrm{d} x \
            &= 2 n left[int limits_{-1}^1 (1-x^2)^{n-1} , mathrm{d} x - int limits_{-1}^1 (1-x^2)^n , mathrm{d} xright]
            end{align}
            for $n in mathbb{N}$, which is the recurrence relation you are after.






            share|cite|improve this answer









            $endgroup$



            You can integrate by parts directly to obtain
            begin{align}
            int limits_{-1}^1 (1-x^2)^n , mathrm{d} x &= left[x (1-x^2)^n right]_{x=-1}^{x=1} + 2n int limits_{-1}^1 x^2 (1-x^2)^{n-1} , mathrm{d} x \
            &= 2 n left[int limits_{-1}^1 (1-x^2)^{n-1} , mathrm{d} x - int limits_{-1}^1 (1-x^2)^n , mathrm{d} xright]
            end{align}
            for $n in mathbb{N}$, which is the recurrence relation you are after.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 at 12:06









            ComplexYetTrivialComplexYetTrivial

            4,8232631




            4,8232631























                1












                $begingroup$

                Not a recurrence, but it seems that
                $$
                I_n = int_{-1}^{1}(1-x^2)^ndx = frac{2^{a_n}}{b_n}
                $$
                where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.



                Also,
                $$
                frac{1}{(1 - x)^{3/2}} = sum_{n=0}^{infty} frac{2}{I_n} x^n
                $$



                WA gives
                $$
                I_n = frac{sqrt{pi} Gamma(n + 1)}{Gamma(n +frac32)}
                = frac{2^{n+1} n!}{(2n+1)!!}
                $$






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Not a recurrence, but it seems that
                  $$
                  I_n = int_{-1}^{1}(1-x^2)^ndx = frac{2^{a_n}}{b_n}
                  $$
                  where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.



                  Also,
                  $$
                  frac{1}{(1 - x)^{3/2}} = sum_{n=0}^{infty} frac{2}{I_n} x^n
                  $$



                  WA gives
                  $$
                  I_n = frac{sqrt{pi} Gamma(n + 1)}{Gamma(n +frac32)}
                  = frac{2^{n+1} n!}{(2n+1)!!}
                  $$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Not a recurrence, but it seems that
                    $$
                    I_n = int_{-1}^{1}(1-x^2)^ndx = frac{2^{a_n}}{b_n}
                    $$
                    where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.



                    Also,
                    $$
                    frac{1}{(1 - x)^{3/2}} = sum_{n=0}^{infty} frac{2}{I_n} x^n
                    $$



                    WA gives
                    $$
                    I_n = frac{sqrt{pi} Gamma(n + 1)}{Gamma(n +frac32)}
                    = frac{2^{n+1} n!}{(2n+1)!!}
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    Not a recurrence, but it seems that
                    $$
                    I_n = int_{-1}^{1}(1-x^2)^ndx = frac{2^{a_n}}{b_n}
                    $$
                    where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.



                    Also,
                    $$
                    frac{1}{(1 - x)^{3/2}} = sum_{n=0}^{infty} frac{2}{I_n} x^n
                    $$



                    WA gives
                    $$
                    I_n = frac{sqrt{pi} Gamma(n + 1)}{Gamma(n +frac32)}
                    = frac{2^{n+1} n!}{(2n+1)!!}
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 25 at 12:14

























                    answered Jan 25 at 11:51









                    lhflhf

                    166k10171398




                    166k10171398























                        1












                        $begingroup$

                        On the path of Nikola Mijušković...



                        begin{align}I_n&=int_{-1}^1 (1-x^2)^n, dx\&=2int_{0}^1 (1-x^2)^n, dxend{align}



                        Perform the change of variable $x=sin t$,



                        begin{align}I_n&=2int_{0}^{frac{pi}{2}} cos^{2n+1}t, dt\
                        &=2W_{2n+1}
                        end{align}



                        $W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )



                        It is known that, for $ngeq 0$,



                        begin{align}W_1&=1\
                        W_{n+2}&=frac{n+1}{n+2}W_nend{align}



                        Therefore,



                        begin{align}I_{n+1}&=2W_{2n+3}\
                        &=frac{2(2n+2)}{2n+3}W_{2n+1}\
                        &=frac{2(n+1)}{2n+3}I_{n}
                        end{align}






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          On the path of Nikola Mijušković...



                          begin{align}I_n&=int_{-1}^1 (1-x^2)^n, dx\&=2int_{0}^1 (1-x^2)^n, dxend{align}



                          Perform the change of variable $x=sin t$,



                          begin{align}I_n&=2int_{0}^{frac{pi}{2}} cos^{2n+1}t, dt\
                          &=2W_{2n+1}
                          end{align}



                          $W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )



                          It is known that, for $ngeq 0$,



                          begin{align}W_1&=1\
                          W_{n+2}&=frac{n+1}{n+2}W_nend{align}



                          Therefore,



                          begin{align}I_{n+1}&=2W_{2n+3}\
                          &=frac{2(2n+2)}{2n+3}W_{2n+1}\
                          &=frac{2(n+1)}{2n+3}I_{n}
                          end{align}






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            On the path of Nikola Mijušković...



                            begin{align}I_n&=int_{-1}^1 (1-x^2)^n, dx\&=2int_{0}^1 (1-x^2)^n, dxend{align}



                            Perform the change of variable $x=sin t$,



                            begin{align}I_n&=2int_{0}^{frac{pi}{2}} cos^{2n+1}t, dt\
                            &=2W_{2n+1}
                            end{align}



                            $W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )



                            It is known that, for $ngeq 0$,



                            begin{align}W_1&=1\
                            W_{n+2}&=frac{n+1}{n+2}W_nend{align}



                            Therefore,



                            begin{align}I_{n+1}&=2W_{2n+3}\
                            &=frac{2(2n+2)}{2n+3}W_{2n+1}\
                            &=frac{2(n+1)}{2n+3}I_{n}
                            end{align}






                            share|cite|improve this answer









                            $endgroup$



                            On the path of Nikola Mijušković...



                            begin{align}I_n&=int_{-1}^1 (1-x^2)^n, dx\&=2int_{0}^1 (1-x^2)^n, dxend{align}



                            Perform the change of variable $x=sin t$,



                            begin{align}I_n&=2int_{0}^{frac{pi}{2}} cos^{2n+1}t, dt\
                            &=2W_{2n+1}
                            end{align}



                            $W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )



                            It is known that, for $ngeq 0$,



                            begin{align}W_1&=1\
                            W_{n+2}&=frac{n+1}{n+2}W_nend{align}



                            Therefore,



                            begin{align}I_{n+1}&=2W_{2n+3}\
                            &=frac{2(2n+2)}{2n+3}W_{2n+1}\
                            &=frac{2(n+1)}{2n+3}I_{n}
                            end{align}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 25 at 20:55









                            FDPFDP

                            6,06211829




                            6,06211829






























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