Help needed concerning discrete calculus

In summary, the conversation discusses the concept of discrete calculus, including the definitions of discrete derivative and integral. The participants also explore the Fundamental Theorem and its application to solving an indeterminate integral. However, a mistake is found in the solution and is eventually corrected with the help of a fellow participant.
  • #1
ForceBoy
47
6
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.
 
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  • #2
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.
 
  • #3
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!
 

1. What is discrete calculus?

Discrete calculus is a branch of mathematics that deals with the study of discrete structures and their properties. It involves the use of discrete variables and functions, as opposed to continuous ones, to model and solve problems.

2. How is discrete calculus different from traditional calculus?

Traditional calculus deals with continuous functions and their derivatives and integrals, while discrete calculus deals with discrete functions and their differences and sums. Discrete calculus is also more focused on solving problems in discrete structures such as graphs, networks, and sequences.

3. What are some real-world applications of discrete calculus?

Discrete calculus has many applications in computer science, engineering, and physics. It is used in the analysis of algorithms, optimization problems, and data compression, among others. It is also used in modeling and analyzing discrete systems, such as electrical circuits and computer networks.

4. Is discrete calculus difficult to learn?

Like any branch of mathematics, discrete calculus can be challenging to learn, especially for those who are not familiar with discrete structures and their properties. However, with proper guidance and practice, it can be mastered by anyone with a strong foundation in calculus and algebra.

5. How can I improve my understanding of discrete calculus?

To improve your understanding of discrete calculus, it is essential to have a solid understanding of traditional calculus and algebra. Additionally, practicing solving problems and working through examples can help improve your skills. Seeking help from a tutor or joining a study group can also be beneficial.

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