0
$\begingroup$

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:

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

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.