Sigma notation for non-integer steps












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Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from 2 to 20 with steps of 0.5, is there a way I could write this in sigma notation or some other form of notation.






summation notation







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    $begingroup$


    Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from 2 to 20 with steps of 0.5, is there a way I could write this in sigma notation or some other form of notation.






    summation notation







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      4


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      $begingroup$


      Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from 2 to 20 with steps of 0.5, is there a way I could write this in sigma notation or some other form of notation.






      summation notation







      share|cite|improve this question









      $endgroup$




      Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from 2 to 20 with steps of 0.5, is there a way I could write this in sigma notation or some other form of notation.






      summation notation




      summation notation






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      asked 6 hours ago









      H.LinkhornH.Linkhorn

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          4 Answers
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          Does $$sum_{n = 0}^{36}lnleft(2+dfrac{1}{2}nright)^2$$ satisfy your requirements?






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            $begingroup$

            You could change $ln(x)^2$ into $lnleft(frac{x}{2}right)^2$ to achieve the steps of $0.5$ in this case. You want $frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limits{x = 4}^{40}lnleft(frac{x}{2}right)^2$.






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              $begingroup$

              In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
              $$sum_{sin S}f(s).$$
              In particular you might have $S={2,2.5,3,dots,20}$.






              share|cite|improve this answer









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                1












                $begingroup$

                Just use $frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.






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                  4 Answers
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                  4 Answers
                  4






                  active

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                  active

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                  6












                  $begingroup$

                  Does $$sum_{n = 0}^{36}lnleft(2+dfrac{1}{2}nright)^2$$ satisfy your requirements?






                  share|cite|improve this answer









                  $endgroup$


















                    6












                    $begingroup$

                    Does $$sum_{n = 0}^{36}lnleft(2+dfrac{1}{2}nright)^2$$ satisfy your requirements?






                    share|cite|improve this answer









                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      Does $$sum_{n = 0}^{36}lnleft(2+dfrac{1}{2}nright)^2$$ satisfy your requirements?






                      share|cite|improve this answer









                      $endgroup$



                      Does $$sum_{n = 0}^{36}lnleft(2+dfrac{1}{2}nright)^2$$ satisfy your requirements?







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 6 hours ago









                      JimmyK4542JimmyK4542

                      41k245105




                      41k245105























                          4












                          $begingroup$

                          You could change $ln(x)^2$ into $lnleft(frac{x}{2}right)^2$ to achieve the steps of $0.5$ in this case. You want $frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limits{x = 4}^{40}lnleft(frac{x}{2}right)^2$.






                          share|cite|improve this answer











                          $endgroup$


















                            4












                            $begingroup$

                            You could change $ln(x)^2$ into $lnleft(frac{x}{2}right)^2$ to achieve the steps of $0.5$ in this case. You want $frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limits{x = 4}^{40}lnleft(frac{x}{2}right)^2$.






                            share|cite|improve this answer











                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$

                              You could change $ln(x)^2$ into $lnleft(frac{x}{2}right)^2$ to achieve the steps of $0.5$ in this case. You want $frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limits{x = 4}^{40}lnleft(frac{x}{2}right)^2$.






                              share|cite|improve this answer











                              $endgroup$



                              You could change $ln(x)^2$ into $lnleft(frac{x}{2}right)^2$ to achieve the steps of $0.5$ in this case. You want $frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limits{x = 4}^{40}lnleft(frac{x}{2}right)^2$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 6 hours ago

























                              answered 6 hours ago









                              KM101KM101

                              5,9651524




                              5,9651524























                                  2












                                  $begingroup$

                                  In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
                                  $$sum_{sin S}f(s).$$
                                  In particular you might have $S={2,2.5,3,dots,20}$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
                                    $$sum_{sin S}f(s).$$
                                    In particular you might have $S={2,2.5,3,dots,20}$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
                                      $$sum_{sin S}f(s).$$
                                      In particular you might have $S={2,2.5,3,dots,20}$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
                                      $$sum_{sin S}f(s).$$
                                      In particular you might have $S={2,2.5,3,dots,20}$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 3 hours ago









                                      YiFanYiFan

                                      3,5511525




                                      3,5511525























                                          1












                                          $begingroup$

                                          Just use $frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            1












                                            $begingroup$

                                            Just use $frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              1












                                              1








                                              1





                                              $begingroup$

                                              Just use $frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Just use $frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 6 hours ago









                                              PeterPeter

                                              47.5k1039129




                                              47.5k1039129






























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