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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 Answers 1

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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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    Good Point. :) Once again, thank you.2012-09-10