4
$\begingroup$

If I have a sum or product whose upper index is less than its start index, how is this interpreted? For example: $$\sum_{k=2}^0a_k,\qquad \prod_{k=3}^1b_k$$

I want to say that they are equal to the empty sum and empty product, respectively, but I don't know.

(This question arises from seeking shortened forms for denoting some nested series/sequences, where the upper index of the inner sum/product is the variable for the outer sum/product.)

Example for why I was wondering: $$\sum_{n=0}^\infty\left[\frac{\prod_{k=0}^{n-1}\left(4k-1\right)}{(2n+1)!}\right]$$ Note that, for the first case of $n=0$, the product is in a situation like I describe.

$\endgroup$
  • $\begingroup$ Whenever I encountered these they represented 0 and 1 respectively, but there may be exceptions. What particular formula is this about ? $\endgroup$ – justt May 2 '13 at 22:30
  • $\begingroup$ @justt It isn't so much for a particular formula, but rather for when I write down nested sums/products. I'll edit a more concrete example into the question. $\endgroup$ – apnorton May 2 '13 at 22:33
  • $\begingroup$ @justt see example now... $\endgroup$ – apnorton May 2 '13 at 22:36
  • 2
    $\begingroup$ Does anybody have an example of $\sum_{k=a}^b$ or $\prod_{k=a}^b$ with $b<a-1$ from real life math literature? I think the case $b=a-1$ (meaning $\sum=0, \prod=1$) is quite common, but I wonder if the different possible interpretations for "even worse" cases are in practical use at all. $\endgroup$ – Hagen von Eitzen May 2 '13 at 22:44
  • $\begingroup$ @anorton The empty product in your example is common (and definitely equals $1$), not the product at the beginning of your question. $\endgroup$ – Did May 2 '13 at 22:46
2
$\begingroup$

A common interpretation of these somewhat unorthodox summation signs is based on the observation that $\sum\limits_{k=i}^{n+1}a_k=a_{n+1}+\sum\limits_{k=i}^{n}a_k$ for every $n\geqslant i$. Extending this, one should expect (and indeed this is the most widely used convention) that $\sum\limits_{k=i}^{i-1}a_k=0$ for every $i$ and every sequence $(a_n)$. Likewise, $\prod\limits_{k=i}^{i-1}b_k=1$ for every $i$ and every sequence $(b_n)$.

The summation $\sum\limits_{k=2}^{0}$ goes one step further but logic would suggest that $\sum\limits_{k=2}^{0}a_k=-a_1$, although I must confess having never met such uses.

Likewise, one should probably consider that $\prod\limits_{k=3}^{1}b_k=1/b_2$ for every nonzero $b_1$, $b_2$, $b_3$, as can be deduced from the identity $\prod\limits_{k=3}^{1}b_k\cdot\prod\limits_{k=2}^{3}b_k=\prod\limits_{k=3}^{3}b_k=b_3$ and from the fact that $\prod\limits_{k=2}^{3}b_k=b_2b_3$.

$\endgroup$
0
$\begingroup$

What you are really doing with $\sum$ when you see $$ \sum_{k\in\mathcal{A}}a_{k} $$ is adding every element in $\mathcal{A}$ to the additive identity $0$. Likewise, when you see $$ \prod_{k\in\mathcal{A}}a_{k}, $$ you are multiplying every element in $\mathcal{A}$ to the multiplicative identity $1$.

Note: the notation with a subscript $k=a$ and superscript $k=b$ is just sugar for $\mathcal{A}=\left\{a,a+1,\ldots,b\right\}$ whenever $a\geq b$ and $\mathcal{A}=\emptyset$ otherwise.

$\endgroup$
  • 2
    $\begingroup$ It depends. Some authors prefer to have $\sum_{k=a}^{b-1}+\sum_{k=b}^{c-1}$ to always eequal $\sum_{k=a}^{c-1}$ $\endgroup$ – Hagen von Eitzen May 2 '13 at 22:40
  • $\begingroup$ Your definition and that presented by Hagen are both valid, and one shall premise in which of the two "environment" is going to work. The question of the post refers to the second $\endgroup$ – G Cab Aug 31 '16 at 9:42

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.