Splitting a double summation












0












$begingroup$


I'm trying to figure out double summations.
I was wondering, when trying to simplify them. Can I just follow the same rules as with a standard summation?



I know that I'm allowed to split a summation, like this:



$$sum_{i=1}^{n}(x_{i}+y_{j})=sum_{i=1}^{n}x_{i}+sum_{i=1}^{n}x_{j}$$



But am I allowed to the same with double summations like so?



$$sum_{i=1}^{n}sum_{j=1}^{n}(i-j)=sum_{i=1}^{n}sum_{j=1}^{n}(i)-sum_{i=1}^{n}sum_{j=1}^{n}(j)$$



This would allow me to remove the double summation and turn it into a single one, I believe.



I couldn't find any clear answers anywhere else, so I'm hoping you guys would be able to help me.



Thanks in advance!






statistics summation







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  • 1




    $begingroup$
    Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
    $endgroup$
    – darij grinberg
    Dec 2 '18 at 21:48










  • $begingroup$
    Thanks! Yes I just saw it too, I meant to write $y_{j}$
    $endgroup$
    – Viperdream
    Dec 2 '18 at 21:53
















0












$begingroup$


I'm trying to figure out double summations.
I was wondering, when trying to simplify them. Can I just follow the same rules as with a standard summation?



I know that I'm allowed to split a summation, like this:



$$sum_{i=1}^{n}(x_{i}+y_{j})=sum_{i=1}^{n}x_{i}+sum_{i=1}^{n}x_{j}$$



But am I allowed to the same with double summations like so?



$$sum_{i=1}^{n}sum_{j=1}^{n}(i-j)=sum_{i=1}^{n}sum_{j=1}^{n}(i)-sum_{i=1}^{n}sum_{j=1}^{n}(j)$$



This would allow me to remove the double summation and turn it into a single one, I believe.



I couldn't find any clear answers anywhere else, so I'm hoping you guys would be able to help me.



Thanks in advance!






statistics summation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
    $endgroup$
    – darij grinberg
    Dec 2 '18 at 21:48










  • $begingroup$
    Thanks! Yes I just saw it too, I meant to write $y_{j}$
    $endgroup$
    – Viperdream
    Dec 2 '18 at 21:53














0












0








0





$begingroup$


I'm trying to figure out double summations.
I was wondering, when trying to simplify them. Can I just follow the same rules as with a standard summation?



I know that I'm allowed to split a summation, like this:



$$sum_{i=1}^{n}(x_{i}+y_{j})=sum_{i=1}^{n}x_{i}+sum_{i=1}^{n}x_{j}$$



But am I allowed to the same with double summations like so?



$$sum_{i=1}^{n}sum_{j=1}^{n}(i-j)=sum_{i=1}^{n}sum_{j=1}^{n}(i)-sum_{i=1}^{n}sum_{j=1}^{n}(j)$$



This would allow me to remove the double summation and turn it into a single one, I believe.



I couldn't find any clear answers anywhere else, so I'm hoping you guys would be able to help me.



Thanks in advance!






statistics summation







share|cite|improve this question









$endgroup$




I'm trying to figure out double summations.
I was wondering, when trying to simplify them. Can I just follow the same rules as with a standard summation?



I know that I'm allowed to split a summation, like this:



$$sum_{i=1}^{n}(x_{i}+y_{j})=sum_{i=1}^{n}x_{i}+sum_{i=1}^{n}x_{j}$$



But am I allowed to the same with double summations like so?



$$sum_{i=1}^{n}sum_{j=1}^{n}(i-j)=sum_{i=1}^{n}sum_{j=1}^{n}(i)-sum_{i=1}^{n}sum_{j=1}^{n}(j)$$



This would allow me to remove the double summation and turn it into a single one, I believe.



I couldn't find any clear answers anywhere else, so I'm hoping you guys would be able to help me.



Thanks in advance!






statistics summation




statistics summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 21:43









ViperdreamViperdream

31




31








  • 1




    $begingroup$
    Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
    $endgroup$
    – darij grinberg
    Dec 2 '18 at 21:48










  • $begingroup$
    Thanks! Yes I just saw it too, I meant to write $y_{j}$
    $endgroup$
    – Viperdream
    Dec 2 '18 at 21:53














  • 1




    $begingroup$
    Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
    $endgroup$
    – darij grinberg
    Dec 2 '18 at 21:48










  • $begingroup$
    Thanks! Yes I just saw it too, I meant to write $y_{j}$
    $endgroup$
    – Viperdream
    Dec 2 '18 at 21:53








1




1




$begingroup$
Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
$endgroup$
– darij grinberg
Dec 2 '18 at 21:48




$begingroup$
Yes. Just split the inner sum first and then the outer sum. But your first formula has typos in it; there cannot be any $x_j$ without a $j$.
$endgroup$
– darij grinberg
Dec 2 '18 at 21:48












$begingroup$
Thanks! Yes I just saw it too, I meant to write $y_{j}$
$endgroup$
– Viperdream
Dec 2 '18 at 21:53




$begingroup$
Thanks! Yes I just saw it too, I meant to write $y_{j}$
$endgroup$
– Viperdream
Dec 2 '18 at 21:53










1 Answer
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$begingroup$

First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$,
$$
sum_{k=1}^{n}left[a_{k}+b_{k}right]=sum_{k=1}^{n}a_{k}+sum_{k=1}^{n}b_{k}.
$$
Under the convention $sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$.
Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}geq1$.
Then,
$$
sum_{k=1}^{n_{0}}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}a_{k}+sum_{k=1}^{n_{0}-1}b_{k}=sum_{k=1}^{n_{0}}a_{k}+sum_{k=1}^{n_{0}}b_{k}.
$$
The desired result follows by induction.



Now, you should be able to use the above result to prove that for any sequences $(a_{kell})$ and $(b_{kell})$ and positive integers $n$ and $m$,
$$
sum_{k=1}^{n}sum_{ell=1}^{m}left[a_{kell}+b_{kell}right]=sum_{k=1}^{n}sum_{ell=1}^{m}a_{kell}+sum_{k=1}^{n}sum_{ell=1}^{m}b_{kell}.
$$






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    0












    $begingroup$

    First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$,
    $$
    sum_{k=1}^{n}left[a_{k}+b_{k}right]=sum_{k=1}^{n}a_{k}+sum_{k=1}^{n}b_{k}.
    $$
    Under the convention $sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$.
    Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}geq1$.
    Then,
    $$
    sum_{k=1}^{n_{0}}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}a_{k}+sum_{k=1}^{n_{0}-1}b_{k}=sum_{k=1}^{n_{0}}a_{k}+sum_{k=1}^{n_{0}}b_{k}.
    $$
    The desired result follows by induction.



    Now, you should be able to use the above result to prove that for any sequences $(a_{kell})$ and $(b_{kell})$ and positive integers $n$ and $m$,
    $$
    sum_{k=1}^{n}sum_{ell=1}^{m}left[a_{kell}+b_{kell}right]=sum_{k=1}^{n}sum_{ell=1}^{m}a_{kell}+sum_{k=1}^{n}sum_{ell=1}^{m}b_{kell}.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$,
      $$
      sum_{k=1}^{n}left[a_{k}+b_{k}right]=sum_{k=1}^{n}a_{k}+sum_{k=1}^{n}b_{k}.
      $$
      Under the convention $sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$.
      Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}geq1$.
      Then,
      $$
      sum_{k=1}^{n_{0}}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}a_{k}+sum_{k=1}^{n_{0}-1}b_{k}=sum_{k=1}^{n_{0}}a_{k}+sum_{k=1}^{n_{0}}b_{k}.
      $$
      The desired result follows by induction.



      Now, you should be able to use the above result to prove that for any sequences $(a_{kell})$ and $(b_{kell})$ and positive integers $n$ and $m$,
      $$
      sum_{k=1}^{n}sum_{ell=1}^{m}left[a_{kell}+b_{kell}right]=sum_{k=1}^{n}sum_{ell=1}^{m}a_{kell}+sum_{k=1}^{n}sum_{ell=1}^{m}b_{kell}.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$,
        $$
        sum_{k=1}^{n}left[a_{k}+b_{k}right]=sum_{k=1}^{n}a_{k}+sum_{k=1}^{n}b_{k}.
        $$
        Under the convention $sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$.
        Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}geq1$.
        Then,
        $$
        sum_{k=1}^{n_{0}}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}a_{k}+sum_{k=1}^{n_{0}-1}b_{k}=sum_{k=1}^{n_{0}}a_{k}+sum_{k=1}^{n_{0}}b_{k}.
        $$
        The desired result follows by induction.



        Now, you should be able to use the above result to prove that for any sequences $(a_{kell})$ and $(b_{kell})$ and positive integers $n$ and $m$,
        $$
        sum_{k=1}^{n}sum_{ell=1}^{m}left[a_{kell}+b_{kell}right]=sum_{k=1}^{n}sum_{ell=1}^{m}a_{kell}+sum_{k=1}^{n}sum_{ell=1}^{m}b_{kell}.
        $$






        share|cite|improve this answer









        $endgroup$



        First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$,
        $$
        sum_{k=1}^{n}left[a_{k}+b_{k}right]=sum_{k=1}^{n}a_{k}+sum_{k=1}^{n}b_{k}.
        $$
        Under the convention $sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$.
        Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}geq1$.
        Then,
        $$
        sum_{k=1}^{n_{0}}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}left[a_{k}+b_{k}right]=a_{n_{0}}+b_{n_{0}}+sum_{k=1}^{n_{0}-1}a_{k}+sum_{k=1}^{n_{0}-1}b_{k}=sum_{k=1}^{n_{0}}a_{k}+sum_{k=1}^{n_{0}}b_{k}.
        $$
        The desired result follows by induction.



        Now, you should be able to use the above result to prove that for any sequences $(a_{kell})$ and $(b_{kell})$ and positive integers $n$ and $m$,
        $$
        sum_{k=1}^{n}sum_{ell=1}^{m}left[a_{kell}+b_{kell}right]=sum_{k=1}^{n}sum_{ell=1}^{m}a_{kell}+sum_{k=1}^{n}sum_{ell=1}^{m}b_{kell}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 22:44









        parsiadparsiad

        17k32353




        17k32353






























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