In Saber Elaydi's book "An Introduction to Difference Equations", 3rd ed., Sec. 2.1 (page 59), the discrete analogue of the Fundamental Theorem of Calculus is stated: \begin{equation} \sum_{k=n_0}^{n-1} \Delta x(k) = x(n) - x(n_0) \end{equation} and \begin{equation} \Delta \left(\sum_{k=n_0}^{n-1}x(k) \right) = x(n) \end{equation} where $\Delta x(k) = x(k+1)-x(k)$ is the difference operator. I do not understand the second part. The $\Delta$ operator is linear so why does it not enter the sum? Also, even if the sum is evaluated first, and then the difference is taken, the result is the same as in the first equation. Is there any missing notation on which $x's$ the $\Delta$ should act on?

Thank you in advance for the support.


1 Answer 1


The first equation, when the operator is written inside the sum is meant to be analogous to $$\int^b_a f'(x) dx = f(b) - f(a).$$

In the second equation when the operator is written outside the sum, we are reminded of: $$ \frac{d}{dx} \int^x_a f(t) dt = f(x).$$

While the forward difference operator is indeed linear, we can not simply move the operator inside the summation. This is because in the second equation, the operator is acting on the index $n$, not on $k$ as it does in the first equation. We are considering the sequence $a_n = \displaystyle \sum_{k=n_0}^{n-1} x(k)$ so $$ \begin{equation} \Delta \left(\sum_{k=n_0}^{n-1}x(k) \right) = \Delta (a_n) = a_{n+1} - a_n = \sum_{k=n_0}^{n} x(k) -\sum_{k=n_0}^{n-1} x(k)= x(n) \end{equation} $$

so the second equation is correct as written.

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    $\begingroup$ I see. But isn't this notation misleading? I mean, in continuous calculus you write $d/dx$ to specify the derivative is with respect to $x$. Wouldn't $\Delta_n$ be a more adequate notation? And is this common in difference calculus literature? Anyway, thank you very much for the answer. $\endgroup$ Sep 10, 2012 at 14:00
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    $\begingroup$ @GabrielLandi Indeed perhaps if one is legitimately dealing with several index variables then a notation like the one you propose would have to be taken up. In this case however, after one gets used to it there is little chance for confusion, and there is actually no ambiguity - $n$ is the only legitimate variable inside the brackets, the $k$ is simply a dummy variable for the sum and $n_0$ is understood to be constant. Similarly if someone asked you "What is the derivative of $ \int^x_{x_0} f(t) dt $ ?" you would probably say $f(x)$, since $t$ is just a dummy variable and $x_0$ is constant. $\endgroup$ Sep 10, 2012 at 14:18
  • $\begingroup$ Good Point. :) Once again, thank you. $\endgroup$ Sep 10, 2012 at 23:54

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