Multiplying summation with same indices and limits












1














What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$



I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.



Any suggestion?






summation summation-method







share|cite|improve this question






















  • These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
    – T. Bongers
    Nov 19 '18 at 0:23










  • k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
    – herb steinberg
    Nov 19 '18 at 0:42










  • As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
    – hakkunamattata
    Nov 19 '18 at 0:54










  • @hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
    – Clement C.
    Nov 19 '18 at 1:29












  • I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
    – Ethan Bolker
    Nov 19 '18 at 1:57
















1














What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$



I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.



Any suggestion?






summation summation-method







share|cite|improve this question






















  • These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
    – T. Bongers
    Nov 19 '18 at 0:23










  • k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
    – herb steinberg
    Nov 19 '18 at 0:42










  • As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
    – hakkunamattata
    Nov 19 '18 at 0:54










  • @hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
    – Clement C.
    Nov 19 '18 at 1:29












  • I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
    – Ethan Bolker
    Nov 19 '18 at 1:57














1












1








1


0





What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$



I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.



Any suggestion?






summation summation-method







share|cite|improve this question













What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$



I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.



Any suggestion?






summation summation-method




summation summation-method






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 '18 at 0:15









hakkunamattata

475




475












  • These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
    – T. Bongers
    Nov 19 '18 at 0:23










  • k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
    – herb steinberg
    Nov 19 '18 at 0:42










  • As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
    – hakkunamattata
    Nov 19 '18 at 0:54










  • @hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
    – Clement C.
    Nov 19 '18 at 1:29












  • I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
    – Ethan Bolker
    Nov 19 '18 at 1:57


















  • These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
    – T. Bongers
    Nov 19 '18 at 0:23










  • k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
    – herb steinberg
    Nov 19 '18 at 0:42










  • As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
    – hakkunamattata
    Nov 19 '18 at 0:54










  • @hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
    – Clement C.
    Nov 19 '18 at 1:29












  • I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
    – Ethan Bolker
    Nov 19 '18 at 1:57
















These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
Nov 19 '18 at 0:23




These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
Nov 19 '18 at 0:23












k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
Nov 19 '18 at 0:42




k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
Nov 19 '18 at 0:42












As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
Nov 19 '18 at 0:54




As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
Nov 19 '18 at 0:54












@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
Nov 19 '18 at 1:29






@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
Nov 19 '18 at 1:29














I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
Nov 19 '18 at 1:57




I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
Nov 19 '18 at 1:57










2 Answers
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oldest

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0














Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$




Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$






share|cite|improve this answer





























    0














    The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.




    The following representations are valid
    begin{align*}
    left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
    left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
    &=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
    end{align*}




    Comment:




    • In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


    • In (2) we denote the index variable of the right-most sum with $j$.




    Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.







    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Let's examine series multiplication first:
      $$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
      $$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
      $$AB=big(a_0+a_1+dotsbig)B$$
      $$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
      $$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
      $$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
      This pattern continues:
      $$AB=a_0B+a_1B+a_2B+a_3B+dots$$
      $$AB=sum_{igeq0}a_iB$$
      Now note the following:
      $$a_iB=a_isum_{kgeq0}b_k$$
      $$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
      $$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
      $$a_iB=sum_{kgeq0}a_ib_k$$
      Plugging in:
      $$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
      Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
      $$
      begin{align}
      AB & = a_0b_0+a_0b_1+a_0b_2+dots \
      & + a_1b_0+a_1b_1+a_1b_2+dots \
      & + a_2b_0+a_2b_1+a_2b_2+dots \
      & +dots
      end{align}
      $$
      Which can be greatly abbreviated:
      $$AB=sum_{i,kin S}a_ib_k$$
      Where
      $$S={xinBbb Z:xgeq0}={0,1,2,dots}$$




      Now we can move onto something more related to your problem:
      $$A=sum_{i=0}^{m}a_i$$
      $$B=sum_{i=0}^{m}b_i$$
      These are just like the case above:
      $$AB=sum_{i,kin S}a_ib_k$$
      Where $$S={xinBbb Z:0leq xleq m}$$
      And of course the fact
      $$(1-A)(1-B)=AB-A-B+1$$
      still holds when $A$ and $B$ are series. But one should note:
      $$-A=-a_0-a_1-a_2-dots$$
      and not $$-A=-a_0+a_1+a_2+dots$$






      share|cite|improve this answer


























        0














        Let's examine series multiplication first:
        $$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
        $$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
        $$AB=big(a_0+a_1+dotsbig)B$$
        $$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
        $$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
        $$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
        This pattern continues:
        $$AB=a_0B+a_1B+a_2B+a_3B+dots$$
        $$AB=sum_{igeq0}a_iB$$
        Now note the following:
        $$a_iB=a_isum_{kgeq0}b_k$$
        $$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
        $$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
        $$a_iB=sum_{kgeq0}a_ib_k$$
        Plugging in:
        $$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
        Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
        $$
        begin{align}
        AB & = a_0b_0+a_0b_1+a_0b_2+dots \
        & + a_1b_0+a_1b_1+a_1b_2+dots \
        & + a_2b_0+a_2b_1+a_2b_2+dots \
        & +dots
        end{align}
        $$
        Which can be greatly abbreviated:
        $$AB=sum_{i,kin S}a_ib_k$$
        Where
        $$S={xinBbb Z:xgeq0}={0,1,2,dots}$$




        Now we can move onto something more related to your problem:
        $$A=sum_{i=0}^{m}a_i$$
        $$B=sum_{i=0}^{m}b_i$$
        These are just like the case above:
        $$AB=sum_{i,kin S}a_ib_k$$
        Where $$S={xinBbb Z:0leq xleq m}$$
        And of course the fact
        $$(1-A)(1-B)=AB-A-B+1$$
        still holds when $A$ and $B$ are series. But one should note:
        $$-A=-a_0-a_1-a_2-dots$$
        and not $$-A=-a_0+a_1+a_2+dots$$






        share|cite|improve this answer
























          0












          0








          0






          Let's examine series multiplication first:
          $$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
          $$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
          $$AB=big(a_0+a_1+dotsbig)B$$
          $$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
          $$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
          $$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
          This pattern continues:
          $$AB=a_0B+a_1B+a_2B+a_3B+dots$$
          $$AB=sum_{igeq0}a_iB$$
          Now note the following:
          $$a_iB=a_isum_{kgeq0}b_k$$
          $$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
          $$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
          $$a_iB=sum_{kgeq0}a_ib_k$$
          Plugging in:
          $$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
          Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
          $$
          begin{align}
          AB & = a_0b_0+a_0b_1+a_0b_2+dots \
          & + a_1b_0+a_1b_1+a_1b_2+dots \
          & + a_2b_0+a_2b_1+a_2b_2+dots \
          & +dots
          end{align}
          $$
          Which can be greatly abbreviated:
          $$AB=sum_{i,kin S}a_ib_k$$
          Where
          $$S={xinBbb Z:xgeq0}={0,1,2,dots}$$




          Now we can move onto something more related to your problem:
          $$A=sum_{i=0}^{m}a_i$$
          $$B=sum_{i=0}^{m}b_i$$
          These are just like the case above:
          $$AB=sum_{i,kin S}a_ib_k$$
          Where $$S={xinBbb Z:0leq xleq m}$$
          And of course the fact
          $$(1-A)(1-B)=AB-A-B+1$$
          still holds when $A$ and $B$ are series. But one should note:
          $$-A=-a_0-a_1-a_2-dots$$
          and not $$-A=-a_0+a_1+a_2+dots$$






          share|cite|improve this answer












          Let's examine series multiplication first:
          $$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
          $$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
          $$AB=big(a_0+a_1+dotsbig)B$$
          $$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
          $$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
          $$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
          This pattern continues:
          $$AB=a_0B+a_1B+a_2B+a_3B+dots$$
          $$AB=sum_{igeq0}a_iB$$
          Now note the following:
          $$a_iB=a_isum_{kgeq0}b_k$$
          $$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
          $$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
          $$a_iB=sum_{kgeq0}a_ib_k$$
          Plugging in:
          $$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
          Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
          $$
          begin{align}
          AB & = a_0b_0+a_0b_1+a_0b_2+dots \
          & + a_1b_0+a_1b_1+a_1b_2+dots \
          & + a_2b_0+a_2b_1+a_2b_2+dots \
          & +dots
          end{align}
          $$
          Which can be greatly abbreviated:
          $$AB=sum_{i,kin S}a_ib_k$$
          Where
          $$S={xinBbb Z:xgeq0}={0,1,2,dots}$$




          Now we can move onto something more related to your problem:
          $$A=sum_{i=0}^{m}a_i$$
          $$B=sum_{i=0}^{m}b_i$$
          These are just like the case above:
          $$AB=sum_{i,kin S}a_ib_k$$
          Where $$S={xinBbb Z:0leq xleq m}$$
          And of course the fact
          $$(1-A)(1-B)=AB-A-B+1$$
          still holds when $A$ and $B$ are series. But one should note:
          $$-A=-a_0-a_1-a_2-dots$$
          and not $$-A=-a_0+a_1+a_2+dots$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 '18 at 1:52









          clathratus

          3,126331




          3,126331























              0














              The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.




              The following representations are valid
              begin{align*}
              left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
              left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
              &=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
              end{align*}




              Comment:




              • In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


              • In (2) we denote the index variable of the right-most sum with $j$.




              Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.







              share|cite|improve this answer




























                0














                The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.




                The following representations are valid
                begin{align*}
                left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
                left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
                &=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
                end{align*}




                Comment:




                • In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


                • In (2) we denote the index variable of the right-most sum with $j$.




                Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.







                share|cite|improve this answer


























                  0












                  0








                  0






                  The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.




                  The following representations are valid
                  begin{align*}
                  left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
                  left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
                  &=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
                  end{align*}




                  Comment:




                  • In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


                  • In (2) we denote the index variable of the right-most sum with $j$.




                  Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.







                  share|cite|improve this answer














                  The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.




                  The following representations are valid
                  begin{align*}
                  left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
                  left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
                  &=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
                  end{align*}




                  Comment:




                  • In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.


                  • In (2) we denote the index variable of the right-most sum with $j$.




                  Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 19 '18 at 16:38

























                  answered Nov 19 '18 at 12:52









                  Markus Scheuer

                  60k455143




                  60k455143






























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