Closed form of a power series solution to a differential equation












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Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?






real-analysis sequences-and-series power-series







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    $begingroup$


    Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?






    real-analysis sequences-and-series power-series







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?






      real-analysis sequences-and-series power-series







      share|cite|improve this question









      $endgroup$




      Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?






      real-analysis sequences-and-series power-series




      real-analysis sequences-and-series power-series






      share|cite|improve this question













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      asked Jan 31 at 18:38









      Sei SakataSei Sakata

      10810




      10810






















          2 Answers
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          $begingroup$

          You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:



          $$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$



          So we get



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



          Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.



          EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html






          share|cite|improve this answer











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            $begingroup$

            Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
            $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$






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              2 Answers
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              active

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              2 Answers
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              active

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              5












              $begingroup$

              You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:



              $$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$



              So we get



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



              Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.



              EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:



                $$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$



                So we get



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



                Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.



                EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:



                  $$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$



                  So we get



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



                  Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.



                  EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html






                  share|cite|improve this answer











                  $endgroup$



                  You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:



                  $$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$



                  So we get



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



                  Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.



                  EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 31 at 18:59

























                  answered Jan 31 at 18:48









                  flawrflawr

                  11.5k32446




                  11.5k32446























                      2












                      $begingroup$

                      Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
                      $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
                        $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
                          $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$






                          share|cite|improve this answer









                          $endgroup$



                          Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
                          $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 31 at 18:50









                          zipirovichzipirovich

                          11.3k11631




                          11.3k11631






























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