High School How Do Discrete Derivatives and Integrals Work in Calculus?

[ForceBoy]

Hello all. I've come across some math which consists of just applying the basic ideas of calculus (derivatives and integrals) onto discrete functions. (The link: http://homepages.math.uic.edu/~kauffman/DCalc.pdf )

The discrete derivative with respect to n is defined as $\Delta_n f(n) = f(n+1) - f(n)$ in the document. The discrete integral is defined as $\sum_{n: a \rightarrow b} f(n) = f(a)+f(a+1)+ ...+f(b-1)$.

It had been a few months since I had read this article so when I started exploring this math, I defined the discrete integral as the regular sum, from a to b ( $\sum_{n= a}^{b} f(n) = f(a)+f(a+1)+ ...+f(b)$ ). For this reason, I found it best to define the discrete integral as $\Delta_n f(n) = f(n) - f(n-1)$
The Fundamental Theorem:

$\Delta_n F(n) = f(n) \leftrightarrow \sum_{n=a}^{b} f(n) = F(b)-F(a-1)$
I did a few problems and everything worked out well when I didn't make any errors. However, just now I was trying to prove

$\sum_{n} a^{n} = \frac{a^{n+1}-1}{a-1}$

but couldn't seem to. My work was as follows:

$\sum_{n} a^{n}$

$u = a^{n}$

$\Delta_u = (a^{n}-a^{n-1})*\Delta_n$

$\Delta_u = a^{n-1}(a-1) * \Delta_n$

$\frac{\Delta_u*a^{1-n}}{a-1} = \Delta_n$

$\frac{a}{a^{n}(a-1)}*\Delta_u = \Delta_n$

So I then make the substitutions

$\sum_{u} u*\frac{a}{a^{n}(a-1)}$

$\sum_{u} u*\frac{a}{u(a-1)}$

$\sum_{u} \frac{a}{(a-1)}$

I can then bring out that fraction

$\frac{a}{(a-1)} \sum_{u} 1$

$***$ $\sum_{k} 1 = k$ $***$

$\frac{a}{(a-1)} u$

$\frac{a}{(a-1)} a^{n}$

And in the end:

$\frac{a^{k+1}}{a-1}$

so

$\sum_{n} a^{n} = \frac{a^{k+1}}{a-1}$

but

$\frac{a^{k+1}}{a-1} \not= \frac{a^{n+1}-1}{a-1}$

Somewhere I've made a mistake. If someone could either point it out for me or guide me to it, it would be greatly appreciated. Also, if my reasoning in someplace isn't clear, please tell me and I'll explain.

 

[tnich]

Your sum index seems to run from 0 to n, or at least that would result in $\sum_{n} a^{n} = \frac{a^{k+1}}{a-1}$. The problem could be in one of two places, 1) your new definition of the discrete integral, or 2) the fact that you have not explicitly defined the limits of your sum.

 

[ForceBoy]

The equation I was solving was meant to be analogous to solving an indeterminate integral. For this reason I never really established my limits. Since you brought this to my attention I decided to evaluate the sum from 0 to n. I then used the equation I had gotten and the fundamental theorem:

$\sum_{k=0}^{n} a^{k} = \frac{a^{n+1}}{a-1} - \frac{a^{(0-1)+1}}{a-1} = \frac{a^{n+1} -1}{a-1}$

So I got what I wanted. Thank you for your time. I now see where the error was. Have a good day!