Power series representation for $xarctan(x)$. How?












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I am supposed to start with the P.S representation for $dfrac{1}{1-x}$ and then figure out the P.S representation for $xtan^{-1}(x)$. I know the process of making it for just the $tan^{-1}x$. But wouldn't the fact that there is an extra $x$ at the front change everything dramatically? Or am I not getting something?



Can I just do the same process as for $tan^{-1}(x)$ and then just add an extra $x$ after sigma?



Thanks!






calculus power-series







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    $begingroup$
    Yes, you can just multiply. Get $x$ inside the sum and that's it.
    $endgroup$
    – Joaquin San
    Dec 13 '18 at 3:35
















1












$begingroup$


I am supposed to start with the P.S representation for $dfrac{1}{1-x}$ and then figure out the P.S representation for $xtan^{-1}(x)$. I know the process of making it for just the $tan^{-1}x$. But wouldn't the fact that there is an extra $x$ at the front change everything dramatically? Or am I not getting something?



Can I just do the same process as for $tan^{-1}(x)$ and then just add an extra $x$ after sigma?



Thanks!






calculus power-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, you can just multiply. Get $x$ inside the sum and that's it.
    $endgroup$
    – Joaquin San
    Dec 13 '18 at 3:35














1












1








1





$begingroup$


I am supposed to start with the P.S representation for $dfrac{1}{1-x}$ and then figure out the P.S representation for $xtan^{-1}(x)$. I know the process of making it for just the $tan^{-1}x$. But wouldn't the fact that there is an extra $x$ at the front change everything dramatically? Or am I not getting something?



Can I just do the same process as for $tan^{-1}(x)$ and then just add an extra $x$ after sigma?



Thanks!






calculus power-series







share|cite|improve this question











$endgroup$




I am supposed to start with the P.S representation for $dfrac{1}{1-x}$ and then figure out the P.S representation for $xtan^{-1}(x)$. I know the process of making it for just the $tan^{-1}x$. But wouldn't the fact that there is an extra $x$ at the front change everything dramatically? Or am I not getting something?



Can I just do the same process as for $tan^{-1}(x)$ and then just add an extra $x$ after sigma?



Thanks!






calculus power-series




calculus power-series






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edited Dec 13 '18 at 4:06









Nosrati

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26.6k62354










asked Dec 13 '18 at 3:27









GlebGleb

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82








  • 1




    $begingroup$
    Yes, you can just multiply. Get $x$ inside the sum and that's it.
    $endgroup$
    – Joaquin San
    Dec 13 '18 at 3:35














  • 1




    $begingroup$
    Yes, you can just multiply. Get $x$ inside the sum and that's it.
    $endgroup$
    – Joaquin San
    Dec 13 '18 at 3:35








1




1




$begingroup$
Yes, you can just multiply. Get $x$ inside the sum and that's it.
$endgroup$
– Joaquin San
Dec 13 '18 at 3:35




$begingroup$
Yes, you can just multiply. Get $x$ inside the sum and that's it.
$endgroup$
– Joaquin San
Dec 13 '18 at 3:35










3 Answers
3






active

oldest

votes


















1












$begingroup$

Recall that
$$arctan x=int_0^xfrac{dt}{1+t^2}$$
We also know that, for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
So we then have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
So for $|x|<1$,
$$
begin{align}
arctan x=&int_0^xsum_{kgeq0}(-1)^kt^{2k}dt\
=&sum_{kgeq0}(-1)^kint_0^xt^{2k}dt\
=&sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}\
end{align}
$$
Which means that
$$xarctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+2}}{2k+1}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    No, it doesn't change everything dramatically.
    Just multiply each term of power series of $tan^{-1}x$ by $x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
      $endgroup$
      – hardmath
      Dec 13 '18 at 3:42



















    0












    $begingroup$

    $tan^{-1}x=intfrac1{1+x^2}=intsum_n(-x^2)^n=sum_nint(-x^2)^n=sum_n(-1)^nfrac{x^{2n+1}}{2n+1}$.
    Thus $xcdottan^{-1}x=sum_n(-1)^nfrac{x^{2n+2}}{2n+1}$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Recall that
      $$arctan x=int_0^xfrac{dt}{1+t^2}$$
      We also know that, for $|t|<1$,
      $$frac1t=sum_{kgeq0}(1-t)^k$$
      So we then have that
      $$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
      So for $|x|<1$,
      $$
      begin{align}
      arctan x=&int_0^xsum_{kgeq0}(-1)^kt^{2k}dt\
      =&sum_{kgeq0}(-1)^kint_0^xt^{2k}dt\
      =&sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}\
      end{align}
      $$
      Which means that
      $$xarctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+2}}{2k+1}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Recall that
        $$arctan x=int_0^xfrac{dt}{1+t^2}$$
        We also know that, for $|t|<1$,
        $$frac1t=sum_{kgeq0}(1-t)^k$$
        So we then have that
        $$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
        So for $|x|<1$,
        $$
        begin{align}
        arctan x=&int_0^xsum_{kgeq0}(-1)^kt^{2k}dt\
        =&sum_{kgeq0}(-1)^kint_0^xt^{2k}dt\
        =&sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}\
        end{align}
        $$
        Which means that
        $$xarctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+2}}{2k+1}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Recall that
          $$arctan x=int_0^xfrac{dt}{1+t^2}$$
          We also know that, for $|t|<1$,
          $$frac1t=sum_{kgeq0}(1-t)^k$$
          So we then have that
          $$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
          So for $|x|<1$,
          $$
          begin{align}
          arctan x=&int_0^xsum_{kgeq0}(-1)^kt^{2k}dt\
          =&sum_{kgeq0}(-1)^kint_0^xt^{2k}dt\
          =&sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}\
          end{align}
          $$
          Which means that
          $$xarctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+2}}{2k+1}$$






          share|cite|improve this answer









          $endgroup$



          Recall that
          $$arctan x=int_0^xfrac{dt}{1+t^2}$$
          We also know that, for $|t|<1$,
          $$frac1t=sum_{kgeq0}(1-t)^k$$
          So we then have that
          $$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
          So for $|x|<1$,
          $$
          begin{align}
          arctan x=&int_0^xsum_{kgeq0}(-1)^kt^{2k}dt\
          =&sum_{kgeq0}(-1)^kint_0^xt^{2k}dt\
          =&sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}\
          end{align}
          $$
          Which means that
          $$xarctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+2}}{2k+1}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 3:53









          clathratusclathratus

          4,725337




          4,725337























              0












              $begingroup$

              No, it doesn't change everything dramatically.
              Just multiply each term of power series of $tan^{-1}x$ by $x$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
                $endgroup$
                – hardmath
                Dec 13 '18 at 3:42
















              0












              $begingroup$

              No, it doesn't change everything dramatically.
              Just multiply each term of power series of $tan^{-1}x$ by $x$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
                $endgroup$
                – hardmath
                Dec 13 '18 at 3:42














              0












              0








              0





              $begingroup$

              No, it doesn't change everything dramatically.
              Just multiply each term of power series of $tan^{-1}x$ by $x$.






              share|cite|improve this answer









              $endgroup$



              No, it doesn't change everything dramatically.
              Just multiply each term of power series of $tan^{-1}x$ by $x$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 13 '18 at 3:36









              MartundMartund

              1,667213




              1,667213












              • $begingroup$
                If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
                $endgroup$
                – hardmath
                Dec 13 '18 at 3:42


















              • $begingroup$
                If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
                $endgroup$
                – hardmath
                Dec 13 '18 at 3:42
















              $begingroup$
              If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
              $endgroup$
              – hardmath
              Dec 13 '18 at 3:42




              $begingroup$
              If you want to make an Answer out of this idea (already mooted in the Question and a Comment), why not go beyond the mere claim by presenting a justification?
              $endgroup$
              – hardmath
              Dec 13 '18 at 3:42











              0












              $begingroup$

              $tan^{-1}x=intfrac1{1+x^2}=intsum_n(-x^2)^n=sum_nint(-x^2)^n=sum_n(-1)^nfrac{x^{2n+1}}{2n+1}$.
              Thus $xcdottan^{-1}x=sum_n(-1)^nfrac{x^{2n+2}}{2n+1}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $tan^{-1}x=intfrac1{1+x^2}=intsum_n(-x^2)^n=sum_nint(-x^2)^n=sum_n(-1)^nfrac{x^{2n+1}}{2n+1}$.
                Thus $xcdottan^{-1}x=sum_n(-1)^nfrac{x^{2n+2}}{2n+1}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $tan^{-1}x=intfrac1{1+x^2}=intsum_n(-x^2)^n=sum_nint(-x^2)^n=sum_n(-1)^nfrac{x^{2n+1}}{2n+1}$.
                  Thus $xcdottan^{-1}x=sum_n(-1)^nfrac{x^{2n+2}}{2n+1}$.






                  share|cite|improve this answer









                  $endgroup$



                  $tan^{-1}x=intfrac1{1+x^2}=intsum_n(-x^2)^n=sum_nint(-x^2)^n=sum_n(-1)^nfrac{x^{2n+1}}{2n+1}$.
                  Thus $xcdottan^{-1}x=sum_n(-1)^nfrac{x^{2n+2}}{2n+1}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 3:53









                  Chris CusterChris Custer

                  13.9k3827




                  13.9k3827






























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