Empty Range in Multiplication

Question

Why is the following true?

$$\prod_{i=6}^2 3^i = 1$$

I can't quite wrap my head around it. I understand that

$$\prod_{i=1}^k 3^i = \underbrace{3^1 \cdot 3^2 \cdots 3^{k-1} \cdot 3^k}_{3^{\sum_{j=1}^k}}$$

Is it because that sum is now zero and $3^0 = 1$?

Seems odd, where did you see this kind of relation? I think the answer should be $\frac{1}{531~441}$. — Matti P., Oct 24 '18 at 07:22
It was on exercise sheet. It could have also been $3i$, i don't remember. It's been a while — hh32, Oct 24 '18 at 07:24
Well, that's a world of difference there. But I think surely the answer should not be 1 ... — Matti P., Oct 24 '18 at 07:25
Your title suggests that the product actually has no factors. I.e. the product is taken over ${i\in \mathbb{N}|i\geq 6 \text{ and } i\leq2}={}$. However this would also be weird notation. It is however reasonable to define the empty product as 1. — Jürg W. Spaak, Oct 24 '18 at 07:29

score 3 · Accepted Answer · 2018-10-24T08:54:05.763

3

Presumably, the author implicitly used a definition such that

$$\prod_{k=a}^b$$ involves all $k$ in $$a\le k\le b,$$ and if that range is empty, the product is $1$ by default (just as a sum is $0$ by default).

Other authors allow bounds reversal.

edited Oct 24 '18 at 08:54

answered Oct 24 '18 at 08:23

score 0 · Answer 2 · answered Oct 24 '18 at 07:39

0

Well, given a ring $R$. By definition, the empty sum is equal to the zero element $0$ of the ring, and the empty product is equal to the unit element $1$ of the ring.

answered Oct 24 '18 at 07:39

Wuestenfux

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Empty Range in Multiplication

2 Answers2