# Does this summation index make any sense?

From my textbook, I have this summation:

$$y_f(k) = \sum_{\tau = k_0}^{k-1}a^{k-1-\tau}g(\tau)$$

So far so good. But then there is a "change of variable" $\tau = \theta - m$ and the summation becomes:

$$y_f(k) = \sum_{\theta - m = k_0}^{k-1}a^{k-1-(\theta - m)}g(\theta - m)$$

Notice the index $\theta - m$. How the index of a summation can be a substraction? Does this make any sense to you?

• It's perfectly legit. You can also write it as $\theta = m + k_0$, or $\theta = k^*$, where $k^* = m+k_0$. Mar 21 '13 at 21:19
• I would say it looks odd. Now it is not clear what is the variable that is changing, is it $m$ or $\theta$ or in this case both. It is best to make a basic change of variables, affecting the look of the upper limit as well. There are cases with multiple conditions on the terms of the sum where folks put some symbol $C$ under $\sum$ sign to indicate those collective conditions. But this case is not complicated enough to warrant that. Mar 21 '13 at 21:21
• If it's legit, how am I supposed to interpret it? Mar 21 '13 at 21:27

Just think of it as $\displaystyle\sum_{\theta=k_0+m}^{k-1}$, with the same summand, interpreted as usual.
The only way this can make a problem is if $k_0+m$ is greater than $k-1$. In general, you can think of $\displaystyle\sum_{\theta=a}^b$ as the summation over all $\theta$ satisfying $a\leq \theta \leq b$. If $b<a$, however, there are no such $\theta$, in which case we are left with the "empty sum," which is defined to be $0$.
Remark. The only other way I could see interpreting this would be as a sum over all pairs $(\theta,m)\in \mathbb{Z}^2$ satisfying $k_0 \leq \theta-m \leq k-1$, but this is obviously an infinite sum, which is doubtfully what the context calls for. Besides, $\theta$ and $m$ are never separated in the summand, so it stands to reason they want a single sum (derived from the single sum in the previous line) with "variable" $\theta-m$, interpreted as above.