I have kinda stupid question on the sum I am working with:$$\sum \limits _{k=0}^\infty \frac{z^{2k}}{(2k)!}\prod \limits _{n=1}^k(a+2n-2))=1+a\frac{z^2}{2!}+a(a+2)\frac{z^4}{4!}+\cdots .$$So, all the terms come out correct IF I assume that product in the term $k=0$ is equal to $1$. So, I guess my question is - is it founded to consider the product equal to $1$ if the limits for variable are inconsistent or is this just sloppy math and should be written down some other way?
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ It's reasonable to set the empty product to 1, in the same way it's reasonable to set the empty sum to 0. However, just for clarity, you might want to note that you're doing that. $\endgroup$– Barry CarterAug 14, 2022 at 14:02
-
$\begingroup$ @BarryCarter Makes sense, putting accompanying note seems safe. I was just wondering if professional mathematicians would scoff at this. $\endgroup$– i_prob_should_know_thisAug 14, 2022 at 14:18
-
2$\begingroup$ "if professional mathematicians would scoff at this" Probably, as with many things, it depends on the mathematician, and the field they study. (I was once at a talk where an algebraist had a very hard time understanding a probabilist's use of the phrase "polynomial growth", for example.) Among, say, algebraic and enumerative combinatorialists, no explanation / justification / terminological note would be required. $\endgroup$– JBLAug 14, 2022 at 15:02
-
2$\begingroup$ Wikipedia isn't a source but en.wikipedia.org/wiki/Empty_product suggests 1 is normal $\endgroup$– Barry CarterAug 14, 2022 at 15:40
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The empty product is equal to $1$, in the same way that the empty sum is equal to $0$. Note that you already need this to be true to properly evaluate $0! = 1$ which occurs in the denominator of your series for $k = 0$.
Some people say this is a convention but in my opinion it is a theorem. It is the only result consistent with the very desirable property that a product of products is a product: that is,
$$\left( \prod_{i=1}^n a_i \right) \left( \prod_{i=n+1}^m a_i \right) = \prod_{i=1}^m a_i.$$
Now set $n = 0$. This should be regarded as a generalized form of associativity. For a citation to this argument in the literature you can see Bjorn Poonen's Why all rings should have a $1$ although his argument makes no reference to addition and is really about associativity.
answered Aug 15, 2022 at 11:55