Index of summation - integer?












1














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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  • 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
    40 mins ago
















1














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










share|cite|improve this 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
    40 mins ago














1












1








1







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 2 hours ago









Christina Daniel

354




354












  • 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
    40 mins ago


















  • 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
    40 mins ago
















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
40 mins ago




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
40 mins ago










6 Answers
6






active

oldest

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4














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
    1 hour ago






  • 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
    1 hour ago












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



















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





























    2














    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





























      2














      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































        1














        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








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          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





















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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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
              1 hour ago






            • 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
              1 hour ago












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
















            4














            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
              1 hour ago






            • 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
              1 hour ago












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














            4












            4








            4






            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 1 hour ago









            Henry

            98.3k475161




            98.3k475161












            • 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
              1 hour ago






            • 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
              1 hour ago












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


















            • 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
              1 hour ago






            • 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
              1 hour ago












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
















            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
            1 hour ago




            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
            1 hour ago




            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
            1 hour ago






            @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
            1 hour ago














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




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











            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 1 hour ago









                pwerth

                1,747411




                1,747411























                    2














                    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


























                      2














                      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
























                        2












                        2








                        2






                        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 1 hour ago









                        orion

                        13k11836




                        13k11836























                            2














                            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














                              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












                                2








                                2






                                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 1 hour ago

























                                answered 1 hour ago









                                Robert Israel

                                318k23208457




                                318k23208457























                                    1














                                    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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                                      1














                                      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












                                        1








                                        1






                                        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






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                                        answered 1 hour ago









                                        Cardioid_Ass_22

                                        236




                                        236




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                                            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 1 hour ago









                                                marty cohen

                                                72.6k549127




                                                72.6k549127






























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