Sigma notation for sum of $ln(x)^2$ from $2$ to $20$ with steps of $0.5$












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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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    6












    $begingroup$


    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











    $endgroup$















      6












      6








      6


      2



      $begingroup$


      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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      edited Feb 11 at 15:37









      user21820

      39k543153




      39k543153










      asked Feb 10 at 19:26









      H.LinkhornH.Linkhorn

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      415113






















          4 Answers
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          active

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          19












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Arbitrary finite set (or else invoke topology and convergence conditions).
            $endgroup$
            – Marc van Leeuwen
            Feb 11 at 10:17






          • 1




            $begingroup$
            You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
            $endgroup$
            – Jam
            Feb 11 at 10:20



















          15












          $begingroup$

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






          share|cite|improve this answer









          $endgroup$





















            10












            $begingroup$

            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











            $endgroup$





















              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









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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                19












                $begingroup$

                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









                $endgroup$













                • $begingroup$
                  Arbitrary finite set (or else invoke topology and convergence conditions).
                  $endgroup$
                  – Marc van Leeuwen
                  Feb 11 at 10:17






                • 1




                  $begingroup$
                  You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                  $endgroup$
                  – Jam
                  Feb 11 at 10:20
















                19












                $begingroup$

                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









                $endgroup$













                • $begingroup$
                  Arbitrary finite set (or else invoke topology and convergence conditions).
                  $endgroup$
                  – Marc van Leeuwen
                  Feb 11 at 10:17






                • 1




                  $begingroup$
                  You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                  $endgroup$
                  – Jam
                  Feb 11 at 10:20














                19












                19








                19





                $begingroup$

                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 Feb 10 at 22:38









                YiFanYiFan

                4,0101627




                4,0101627












                • $begingroup$
                  Arbitrary finite set (or else invoke topology and convergence conditions).
                  $endgroup$
                  – Marc van Leeuwen
                  Feb 11 at 10:17






                • 1




                  $begingroup$
                  You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                  $endgroup$
                  – Jam
                  Feb 11 at 10:20


















                • $begingroup$
                  Arbitrary finite set (or else invoke topology and convergence conditions).
                  $endgroup$
                  – Marc van Leeuwen
                  Feb 11 at 10:17






                • 1




                  $begingroup$
                  You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                  $endgroup$
                  – Jam
                  Feb 11 at 10:20
















                $begingroup$
                Arbitrary finite set (or else invoke topology and convergence conditions).
                $endgroup$
                – Marc van Leeuwen
                Feb 11 at 10:17




                $begingroup$
                Arbitrary finite set (or else invoke topology and convergence conditions).
                $endgroup$
                – Marc van Leeuwen
                Feb 11 at 10:17




                1




                1




                $begingroup$
                You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                $endgroup$
                – Jam
                Feb 11 at 10:20




                $begingroup$
                You could also index over the set $S$, to get $sum_{i=0}^nf(s_i)$, where $S={s_0,s_1,ldots,s_n}$.
                $endgroup$
                – Jam
                Feb 11 at 10:20











                15












                $begingroup$

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






                share|cite|improve this answer









                $endgroup$


















                  15












                  $begingroup$

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






                  share|cite|improve this answer









                  $endgroup$
















                    15












                    15








                    15





                    $begingroup$

                    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 Feb 10 at 19:29









                    JimmyK4542JimmyK4542

                    41.1k245106




                    41.1k245106























                        10












                        $begingroup$

                        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











                        $endgroup$


















                          10












                          $begingroup$

                          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











                          $endgroup$
















                            10












                            10








                            10





                            $begingroup$

                            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 Feb 10 at 19:37

























                            answered Feb 10 at 19:32









                            KM101KM101

                            6,0251525




                            6,0251525























                                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 Feb 10 at 19:30









                                    PeterPeter

                                    47.6k1039131




                                    47.6k1039131






























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