Series representation of $1/x^2$?












0












$begingroup$


According to WolframAlpha the series representation of $1/x^2$ is



$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$



Can somebody tell me how to prove this result?






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
    $endgroup$
    – TonyK
    Dec 19 '18 at 15:19
















0












$begingroup$


According to WolframAlpha the series representation of $1/x^2$ is



$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$



Can somebody tell me how to prove this result?






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
    $endgroup$
    – TonyK
    Dec 19 '18 at 15:19














0












0








0





$begingroup$


According to WolframAlpha the series representation of $1/x^2$ is



$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$



Can somebody tell me how to prove this result?






sequences-and-series







share|cite|improve this question











$endgroup$




According to WolframAlpha the series representation of $1/x^2$ is



$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$



Can somebody tell me how to prove this result?






sequences-and-series




sequences-and-series






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share|cite|improve this question













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edited Dec 19 '18 at 15:14









amWhy

1




1










asked Dec 19 '18 at 15:09









LightLight

111




111












  • $begingroup$
    This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
    $endgroup$
    – TonyK
    Dec 19 '18 at 15:19


















  • $begingroup$
    This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
    $endgroup$
    – TonyK
    Dec 19 '18 at 15:19
















$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19




$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19










4 Answers
4






active

oldest

votes


















1












$begingroup$

This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.



Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
    $$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
    which is valid for $|y|<1$. Take the geometric series
    $$frac1{1-y}=sum_{n=0}^infty y^n$$
    and either differentiate it or square it.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Hint: start from the geometric series
      $$sum_{n=0}^infty x^n = frac1{1-x}.$$
      Alternatively: Taylor at $x = 1$.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        By the geometric series, you obtain



        $$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$



        You can also consider the Taylor Series, which gets



        $$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$



        and see what happens at $a = 1$...






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.



          Alternatively, you can start from the geometric series
          $$
          frac{1}{1-u} = 1 + u + u^2 + cdots,
          $$
          differentiate term by term with respect to $u$ and substitute $u =1-x$.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.



            Alternatively, you can start from the geometric series
            $$
            frac{1}{1-u} = 1 + u + u^2 + cdots,
            $$
            differentiate term by term with respect to $u$ and substitute $u =1-x$.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.



              Alternatively, you can start from the geometric series
              $$
              frac{1}{1-u} = 1 + u + u^2 + cdots,
              $$
              differentiate term by term with respect to $u$ and substitute $u =1-x$.






              share|cite|improve this answer









              $endgroup$



              This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.



              Alternatively, you can start from the geometric series
              $$
              frac{1}{1-u} = 1 + u + u^2 + cdots,
              $$
              differentiate term by term with respect to $u$ and substitute $u =1-x$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 19 '18 at 15:15









              Ethan BolkerEthan Bolker

              43.4k551116




              43.4k551116























                  4












                  $begingroup$

                  This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
                  $$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
                  which is valid for $|y|<1$. Take the geometric series
                  $$frac1{1-y}=sum_{n=0}^infty y^n$$
                  and either differentiate it or square it.






                  share|cite|improve this answer









                  $endgroup$


















                    4












                    $begingroup$

                    This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
                    $$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
                    which is valid for $|y|<1$. Take the geometric series
                    $$frac1{1-y}=sum_{n=0}^infty y^n$$
                    and either differentiate it or square it.






                    share|cite|improve this answer









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
                      $$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
                      which is valid for $|y|<1$. Take the geometric series
                      $$frac1{1-y}=sum_{n=0}^infty y^n$$
                      and either differentiate it or square it.






                      share|cite|improve this answer









                      $endgroup$



                      This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
                      $$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
                      which is valid for $|y|<1$. Take the geometric series
                      $$frac1{1-y}=sum_{n=0}^infty y^n$$
                      and either differentiate it or square it.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 19 '18 at 15:14









                      Lord Shark the UnknownLord Shark the Unknown

                      104k1160132




                      104k1160132























                          2












                          $begingroup$

                          Hint: start from the geometric series
                          $$sum_{n=0}^infty x^n = frac1{1-x}.$$
                          Alternatively: Taylor at $x = 1$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Hint: start from the geometric series
                            $$sum_{n=0}^infty x^n = frac1{1-x}.$$
                            Alternatively: Taylor at $x = 1$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Hint: start from the geometric series
                              $$sum_{n=0}^infty x^n = frac1{1-x}.$$
                              Alternatively: Taylor at $x = 1$.






                              share|cite|improve this answer









                              $endgroup$



                              Hint: start from the geometric series
                              $$sum_{n=0}^infty x^n = frac1{1-x}.$$
                              Alternatively: Taylor at $x = 1$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 19 '18 at 15:13









                              Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                              34.4k42871




                              34.4k42871























                                  1












                                  $begingroup$

                                  By the geometric series, you obtain



                                  $$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$



                                  You can also consider the Taylor Series, which gets



                                  $$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$



                                  and see what happens at $a = 1$...






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    By the geometric series, you obtain



                                    $$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$



                                    You can also consider the Taylor Series, which gets



                                    $$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$



                                    and see what happens at $a = 1$...






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      By the geometric series, you obtain



                                      $$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$



                                      You can also consider the Taylor Series, which gets



                                      $$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$



                                      and see what happens at $a = 1$...






                                      share|cite|improve this answer









                                      $endgroup$



                                      By the geometric series, you obtain



                                      $$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$



                                      You can also consider the Taylor Series, which gets



                                      $$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$



                                      and see what happens at $a = 1$...







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 19 '18 at 15:29









                                      KM101KM101

                                      6,0251525




                                      6,0251525






























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