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Besides e.g. $\sum\limits_{b=c}^d\sum\limits_{a=c}^bf(a,b)=\sum\limits_{a=c}^d\sum\limits_{b=a}^df(a,b)$ , are there any further good formulae about interchange summation signs of multi-summation?

For example,

$\sum\limits_{b=0}^\infty\sum\limits_{a=0}^{kb+n}f(a,b)=\sum\limits_{a=?}^?\sum\limits_{b=?}^?f(?,?)$ , where $k,n\in\mathbb{N}$

$\sum\limits_{c=0}^\infty\sum\limits_{b=0}^\infty\sum\limits_{a=0}^{kb+mc+n}f(a,b,c)=\sum\limits_{a=?}^?\sum\limits_{c=?}^?\sum\limits_{b=?}^?f(?,?,?)$ , where $k,m,n\in\mathbb{N}$

Harry Peter
  • 7,819

1 Answers1

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The more general formula is the following, given two finite sets $A$ and $B$ and an application $S:A\to P(B)$ where $P(B)$ is the set of subsets of of $B$ we have the following result:

$$\sum_{i\in A}\sum_{j\in S(i)} f(i,j)=\sum_{j\in B} \sum_{i\in S^{-1}(j)}f(i,j)$$ with $S^{-1}(j)=\big\{i\in A\Big/j\in S(i)\big\}$

Applications

  • For fixed integers $n,m$, when $A=B=[n,m]$ and $S(i)=[n,i]$, notice that $S^{-1}(j)=[j,n]$ and we have : $$\sum_{i=n}^m\sum_{j=n}^if(i,j)=\sum_{i\in [n,m]}\sum_{j\in [n,i]} f(i,j)=\sum_{j\in [n,m]}\sum_{i\in [j,m]} f(i,j)=\sum_{j=n}^m\sum_{i=j}^m f(i,j)$$
  • For fixed integers $n,m,k$, when $A=[0,n], B=[0,kn+m]$ and $S(i)=[0,ki+m]$ and you can observe that $S^{-1}(j)=\Big[\left \lceil \frac{j-m}{k}\right \rceil,n\Big]$ so we have the following identity: $$\sum_{i=0}^n\sum_{j=0}^{ki+m}f(i,j)=\sum_{j=0}^{kn+m}\sum_{i=\lceil \frac{j-m}{k} \rceil}^{n}f(i,j)$$ now you can tend $n\to \infty$ to solve your first question
  • For the second summation, ($k,m,n,N$ fixed) you can take $A=[0,N]\times[0,N]$ $B=[0,kN+mN+n]$ and $S(b,c)=[0,kb+mc+n]$ and the calculus will follow the same pattern
Elaqqad
  • 13,725