I read somewhere that this sum can be written as: $$\sum_{r+s=n}a_rb_s=\sum_{r=0}^na_rb_{n-r}\tag1$$ This means to create all possible orders of $(r,s)$ and add these together.

Now, my question is how do you write this summation in terms of the RHS above:

$$\sum_{j_1+j_2+\cdots+j_m=n}a^{j_1}_1a^{j_2}_2\cdots a_m^{j_m}=?\tag2$$

  • 1
    $\begingroup$ It just means that you sum the value of $a_1^{j_1}\dots$ for all ordered lists $j_1,\dots,j_m$ such that $j_1+\dots+j_m=n$. $\endgroup$ – Benedict Randall Shaw Nov 3 '18 at 13:25
  • $\begingroup$ @BenedictRandallShaw yes, but I want the notation to be (or $(2)$ be written) like the RHS of $(1)$ $\endgroup$ – John Glenn Nov 3 '18 at 13:27
  • $\begingroup$ That's going to take multiple summation signs. Are you OK with that? $\endgroup$ – Arthur Nov 3 '18 at 13:30
  • $\begingroup$ @JohnGlenn Why do you want that? Curiosity? To help program it? For an assignment? Would you allow multiple summation symbols with an ellipsis between them or do you just want a single one (or any fixed number?)? All of this context would help. $\endgroup$ – Mark S. Nov 3 '18 at 13:31
  • $\begingroup$ @Arthur It only requires multiple summation signs if you restrict yourself to using the exponents as the indices of summation. $\endgroup$ – Mark S. Nov 3 '18 at 16:39

For $m=3$,

$$\sum_{r=0}^n\sum_{s=0}^{n-r}a_rb_s c_{n-r-s}.$$

For $m=4$,

$$\sum_{r=0}^n\sum_{s=0}^{n-r}\sum_{t=0}^{n-r-s}a_rb_s c_td_{n-r-s-t}.$$

And so on.

  • $\begingroup$ I would say it's a hypertetrahedron which is the hyperface opposite the origin, one dimension higher than the one you think of. It's of little consequence, though. $\endgroup$ – Arthur Nov 3 '18 at 23:18
  • $\begingroup$ @Arthur: that's right, my bad. $\endgroup$ – Yves Daoust Nov 3 '18 at 23:36

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