Is this an empty product?

Question

Assume that we have: $$\Psi_{j,k}=(t_{j+1}-\tau)\times...\times(t_{j+k-1}-\tau)$$ that can be written as:

$$\Pi_{i=j+1}^{j+k-1} (t_i-\tau)$$

If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.

It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write

$$\Psi_{j,1}=(t_{j+1}-\tau)\times(t_{j}-\tau)$$

Is this wrong? I am not familiar with this notation and it may be very trivial.

Answer 1

Read : $\Pi_{i=j+1}^{j+k-1} (t_i-\tau)$ as

"The product of $t_i - \tau$ FROM $j+1$ TO $j+k-1$"

Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.

Answer 2

no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $\psi_{j,1}=(t_{j+1}-\tau)\times (t_j - \tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot. in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$