0
$\begingroup$

I have a function $f(x)$ that I want to sum in two separate ways:

  • across integer values of $x\geqslant0$
  • across all real values of $x\geqslant0$

I am interested in the notation for both situations. Is it legitimate to say something like

$$\sum_{x \in \mathbb{Z}\geqslant0} f(x)$$

and $$\sum_{x \in \mathbb{R}\geqslant0} f(x)$$

I realise that this second example is also equivalent to a partial integral, but since the expression isn't algebraically integrable, I want to explore alternative notations.

$\endgroup$
  • $\begingroup$ Both are not really sums, since you can only sum finitely many summands. For the first I can guess a definition as the limit of $\sum_{x=0}^n f(x)$ as $n\to\infty$. For the second you have to give a definition. $\endgroup$ – Christoph Sep 19 '18 at 15:42
  • $\begingroup$ OK, understood. Thanks. $\endgroup$ – Richard Burke-Ward Sep 19 '18 at 15:53
0
$\begingroup$

The first sum doesn't really make sense in the way you phrased it (but can be rephrased to make sense). The second sum doesn't even make sense, unless you redefine what it means to take the summation.

The way summation is defined on a finite set $S$ like below$$\sum\limits_{x\in S} f(x)$$ is to first order the set with some bijective map from the set of numbers from $1$ to $n$, then evaluate the summation as $$f(x_1)+f(x_2)+...+f(x_n)$$Now, for countably infinite sets, we can generalize this procedure. We begin by creating a bijective map from the natural numbers to the set $S$, and then, we compute the sum $$f(x_1)+f(x_2)+..+f(x_n)$$ for all $n$. After doing so, we can take the limit as $n\to\infty$.

As such, your first summation doesn't make much sense, unless we create a mapping between your set $\mathbb{N}\cup\{0\}$ and $\mathbb{N}$, which is relatively trivial to do. Your second summation is not well defined, as there exists no bijective mapping between the set and the natural numbers.

If you seek to redefine what the summation means, be my guest, but make sure to be rigorous and precise when dealing with concepts as such.

$\endgroup$

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.