Index of summation - integer? [on hold]












0














Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?










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put on hold as off-topic by mrtaurho, Dietrich Burde, amWhy, jgon, user21820 22 hours ago


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  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, jgon, user21820

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  • Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
    – user3482749
    yesterday
















0














Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?










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put on hold as off-topic by mrtaurho, Dietrich Burde, amWhy, jgon, user21820 22 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, jgon, user21820

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
    – user3482749
    yesterday














0












0








0







Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?










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Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?







summation






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asked yesterday









Christina Daniel

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put on hold as off-topic by mrtaurho, Dietrich Burde, amWhy, jgon, user21820 22 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, jgon, user21820

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by mrtaurho, Dietrich Burde, amWhy, jgon, user21820 22 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, jgon, user21820

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
    – user3482749
    yesterday


















  • Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
    – user3482749
    yesterday
















Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
– user3482749
yesterday




Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them.
– user3482749
yesterday










6 Answers
6






active

oldest

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5














You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index



Here is an example:



$$sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k = frac{zeta(k-1)}{zeta(k)}$$ where $frac a b$ is a rational expressed in lowest terms and real $k gt 2$






share|cite|improve this answer





















  • I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
    – marty cohen
    yesterday






  • 1




    @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
    – Henry
    yesterday












  • Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
    – marty cohen
    yesterday










  • "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
    – JiK
    yesterday



















3














Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing



$$sum_{i=0}^N f(x_i)$$
you can simply write
$$sum_{xin A} f(x)$$
where $A$ is the set over which summation is done. This is very common simplification of notation.



What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.






share|cite|improve this answer





























    3














    The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
    $$sum_{k=1}^{n}x_{k}$$
    You can also think of this as
    $$sum_{xin S}x$$
    where $S={x_{1},...,x_{n}}$.
    However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
    $$sum_{xin [0,1]}x$$
    The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
    $$sum_{xin S}x<infty$$
    then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.






    share|cite|improve this answer





























      3














      Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as



      $$ sum_{alpha in A} f(alpha) $$



      where $A$ is some set (not necessarily of integers).



      It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(alpha)$ are $0$.






      share|cite|improve this answer































        2














        Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.



        Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).



        An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy

        (see https://planetmath.org/indexingset)



        Also, it's really often inconvenient to limit ourselves to plain old integer indexing.
        More abstract notation is generally more useful.

        Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like



        $sum_{p:p=prime,p<4000} p$






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        New contributor




        Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        • 1




          For anyone who's wondering, there are 550 primes under 4000.
          – Dan
          yesterday



















        0














        As a more practical answer,
        if you want the
        index of summation
        to increment by a constant
        non-integer value,
        you could write



        $sum_{x=0 by .01}^1 f(x)
        $
        .



        However,
        both symbolically and numerically
        I would write this as
        $sum_{n=0}^{100} f(n/100)
        $
        .



        The numerical reason is that
        if the increment is not
        representable exactly
        ($0.01$ is not in binary floating point),
        the loop may execute
        an extra or one less time
        than expected.






        share|cite|improve this answer




























          6 Answers
          6






          active

          oldest

          votes








          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5














          You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index



          Here is an example:



          $$sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k = frac{zeta(k-1)}{zeta(k)}$$ where $frac a b$ is a rational expressed in lowest terms and real $k gt 2$






          share|cite|improve this answer





















          • I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
            – marty cohen
            yesterday






          • 1




            @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
            – Henry
            yesterday












          • Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
            – marty cohen
            yesterday










          • "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
            – JiK
            yesterday
















          5














          You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index



          Here is an example:



          $$sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k = frac{zeta(k-1)}{zeta(k)}$$ where $frac a b$ is a rational expressed in lowest terms and real $k gt 2$






          share|cite|improve this answer





















          • I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
            – marty cohen
            yesterday






          • 1




            @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
            – Henry
            yesterday












          • Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
            – marty cohen
            yesterday










          • "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
            – JiK
            yesterday














          5












          5








          5






          You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index



          Here is an example:



          $$sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k = frac{zeta(k-1)}{zeta(k)}$$ where $frac a b$ is a rational expressed in lowest terms and real $k gt 2$






          share|cite|improve this answer












          You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index



          Here is an example:



          $$sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k = frac{zeta(k-1)}{zeta(k)}$$ where $frac a b$ is a rational expressed in lowest terms and real $k gt 2$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Henry

          98.3k475162




          98.3k475162












          • I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
            – marty cohen
            yesterday






          • 1




            @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
            – Henry
            yesterday












          • Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
            – marty cohen
            yesterday










          • "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
            – JiK
            yesterday


















          • I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
            – marty cohen
            yesterday






          • 1




            @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
            – Henry
            yesterday












          • Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
            – marty cohen
            yesterday










          • "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
            – JiK
            yesterday
















          I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
          – marty cohen
          yesterday




          I get $zeta(k-1)$ for the sum. Where is my mistake? $begin{array}\ sum_{frac{a}{b} in mathbb{Q} cap (0,1]} left(frac{1}{b}right)^k &=sum_{n=1}^{infty} sum_{j=1}^n (1/n)^k\ &=sum_{n=1}^{infty} (1/n)^ksum_{j=1}^n 1\ &=sum_{n=1}^{infty} (1/n)^k n\ &=sum_{n=1}^{infty} (1/n)^{k-1}\ &=zeta(k-1)\ end{array} $
          – marty cohen
          yesterday




          1




          1




          @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
          – Henry
          yesterday






          @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $frac14$ and $frac34$, as $frac24=frac12$ and $frac44=frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$
          – Henry
          yesterday














          Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
          – marty cohen
          yesterday




          Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $phi(n)$. Thanks.
          – marty cohen
          yesterday












          "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
          – JiK
          yesterday




          "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms.
          – JiK
          yesterday











          3














          Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing



          $$sum_{i=0}^N f(x_i)$$
          you can simply write
          $$sum_{xin A} f(x)$$
          where $A$ is the set over which summation is done. This is very common simplification of notation.



          What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.






          share|cite|improve this answer


























            3














            Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing



            $$sum_{i=0}^N f(x_i)$$
            you can simply write
            $$sum_{xin A} f(x)$$
            where $A$ is the set over which summation is done. This is very common simplification of notation.



            What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.






            share|cite|improve this answer
























              3












              3








              3






              Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing



              $$sum_{i=0}^N f(x_i)$$
              you can simply write
              $$sum_{xin A} f(x)$$
              where $A$ is the set over which summation is done. This is very common simplification of notation.



              What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.






              share|cite|improve this answer












              Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing



              $$sum_{i=0}^N f(x_i)$$
              you can simply write
              $$sum_{xin A} f(x)$$
              where $A$ is the set over which summation is done. This is very common simplification of notation.



              What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered yesterday









              orion

              13k11836




              13k11836























                  3














                  The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
                  $$sum_{k=1}^{n}x_{k}$$
                  You can also think of this as
                  $$sum_{xin S}x$$
                  where $S={x_{1},...,x_{n}}$.
                  However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
                  $$sum_{xin [0,1]}x$$
                  The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
                  $$sum_{xin S}x<infty$$
                  then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.






                  share|cite|improve this answer


























                    3














                    The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
                    $$sum_{k=1}^{n}x_{k}$$
                    You can also think of this as
                    $$sum_{xin S}x$$
                    where $S={x_{1},...,x_{n}}$.
                    However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
                    $$sum_{xin [0,1]}x$$
                    The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
                    $$sum_{xin S}x<infty$$
                    then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.






                    share|cite|improve this answer
























                      3












                      3








                      3






                      The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
                      $$sum_{k=1}^{n}x_{k}$$
                      You can also think of this as
                      $$sum_{xin S}x$$
                      where $S={x_{1},...,x_{n}}$.
                      However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
                      $$sum_{xin [0,1]}x$$
                      The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
                      $$sum_{xin S}x<infty$$
                      then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.






                      share|cite|improve this answer












                      The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write
                      $$sum_{k=1}^{n}x_{k}$$
                      You can also think of this as
                      $$sum_{xin S}x$$
                      where $S={x_{1},...,x_{n}}$.
                      However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as
                      $$sum_{xin [0,1]}x$$
                      The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum
                      $$sum_{xin S}x<infty$$
                      then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      pwerth

                      1,757411




                      1,757411























                          3














                          Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as



                          $$ sum_{alpha in A} f(alpha) $$



                          where $A$ is some set (not necessarily of integers).



                          It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(alpha)$ are $0$.






                          share|cite|improve this answer




























                            3














                            Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as



                            $$ sum_{alpha in A} f(alpha) $$



                            where $A$ is some set (not necessarily of integers).



                            It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(alpha)$ are $0$.






                            share|cite|improve this answer


























                              3












                              3








                              3






                              Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as



                              $$ sum_{alpha in A} f(alpha) $$



                              where $A$ is some set (not necessarily of integers).



                              It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(alpha)$ are $0$.






                              share|cite|improve this answer














                              Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as



                              $$ sum_{alpha in A} f(alpha) $$



                              where $A$ is some set (not necessarily of integers).



                              It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(alpha)$ are $0$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited yesterday

























                              answered yesterday









                              Robert Israel

                              318k23208457




                              318k23208457























                                  2














                                  Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.



                                  Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).



                                  An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy

                                  (see https://planetmath.org/indexingset)



                                  Also, it's really often inconvenient to limit ourselves to plain old integer indexing.
                                  More abstract notation is generally more useful.

                                  Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like



                                  $sum_{p:p=prime,p<4000} p$






                                  share|cite|improve this answer








                                  New contributor




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                                  • 1




                                    For anyone who's wondering, there are 550 primes under 4000.
                                    – Dan
                                    yesterday
















                                  2














                                  Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.



                                  Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).



                                  An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy

                                  (see https://planetmath.org/indexingset)



                                  Also, it's really often inconvenient to limit ourselves to plain old integer indexing.
                                  More abstract notation is generally more useful.

                                  Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like



                                  $sum_{p:p=prime,p<4000} p$






                                  share|cite|improve this answer








                                  New contributor




                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.














                                  • 1




                                    For anyone who's wondering, there are 550 primes under 4000.
                                    – Dan
                                    yesterday














                                  2












                                  2








                                  2






                                  Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.



                                  Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).



                                  An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy

                                  (see https://planetmath.org/indexingset)



                                  Also, it's really often inconvenient to limit ourselves to plain old integer indexing.
                                  More abstract notation is generally more useful.

                                  Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like



                                  $sum_{p:p=prime,p<4000} p$






                                  share|cite|improve this answer








                                  New contributor




                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.



                                  Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).



                                  An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy

                                  (see https://planetmath.org/indexingset)



                                  Also, it's really often inconvenient to limit ourselves to plain old integer indexing.
                                  More abstract notation is generally more useful.

                                  Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like



                                  $sum_{p:p=prime,p<4000} p$







                                  share|cite|improve this answer








                                  New contributor




                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  share|cite|improve this answer



                                  share|cite|improve this answer






                                  New contributor




                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  answered yesterday









                                  Cardioid_Ass_22

                                  336




                                  336




                                  New contributor




                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                  New contributor





                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  Cardioid_Ass_22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                  • 1




                                    For anyone who's wondering, there are 550 primes under 4000.
                                    – Dan
                                    yesterday














                                  • 1




                                    For anyone who's wondering, there are 550 primes under 4000.
                                    – Dan
                                    yesterday








                                  1




                                  1




                                  For anyone who's wondering, there are 550 primes under 4000.
                                  – Dan
                                  yesterday




                                  For anyone who's wondering, there are 550 primes under 4000.
                                  – Dan
                                  yesterday











                                  0














                                  As a more practical answer,
                                  if you want the
                                  index of summation
                                  to increment by a constant
                                  non-integer value,
                                  you could write



                                  $sum_{x=0 by .01}^1 f(x)
                                  $
                                  .



                                  However,
                                  both symbolically and numerically
                                  I would write this as
                                  $sum_{n=0}^{100} f(n/100)
                                  $
                                  .



                                  The numerical reason is that
                                  if the increment is not
                                  representable exactly
                                  ($0.01$ is not in binary floating point),
                                  the loop may execute
                                  an extra or one less time
                                  than expected.






                                  share|cite|improve this answer


























                                    0














                                    As a more practical answer,
                                    if you want the
                                    index of summation
                                    to increment by a constant
                                    non-integer value,
                                    you could write



                                    $sum_{x=0 by .01}^1 f(x)
                                    $
                                    .



                                    However,
                                    both symbolically and numerically
                                    I would write this as
                                    $sum_{n=0}^{100} f(n/100)
                                    $
                                    .



                                    The numerical reason is that
                                    if the increment is not
                                    representable exactly
                                    ($0.01$ is not in binary floating point),
                                    the loop may execute
                                    an extra or one less time
                                    than expected.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      As a more practical answer,
                                      if you want the
                                      index of summation
                                      to increment by a constant
                                      non-integer value,
                                      you could write



                                      $sum_{x=0 by .01}^1 f(x)
                                      $
                                      .



                                      However,
                                      both symbolically and numerically
                                      I would write this as
                                      $sum_{n=0}^{100} f(n/100)
                                      $
                                      .



                                      The numerical reason is that
                                      if the increment is not
                                      representable exactly
                                      ($0.01$ is not in binary floating point),
                                      the loop may execute
                                      an extra or one less time
                                      than expected.






                                      share|cite|improve this answer












                                      As a more practical answer,
                                      if you want the
                                      index of summation
                                      to increment by a constant
                                      non-integer value,
                                      you could write



                                      $sum_{x=0 by .01}^1 f(x)
                                      $
                                      .



                                      However,
                                      both symbolically and numerically
                                      I would write this as
                                      $sum_{n=0}^{100} f(n/100)
                                      $
                                      .



                                      The numerical reason is that
                                      if the increment is not
                                      representable exactly
                                      ($0.01$ is not in binary floating point),
                                      the loop may execute
                                      an extra or one less time
                                      than expected.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered yesterday









                                      marty cohen

                                      72.7k549128




                                      72.7k549128















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