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## Generalizations of “standard” calculus

We have the usual analogy between infinitesimal calculus (integrals and derivatives) and finite calculus (sums and forward differences), and also the generalization of infinitesimal calculus to fractional calculus (which allows for real and even complex powers of the differential operator). Have people worked on a "fractional finite" calculus, where instead of a differintegral we had some sort of "differsum"? I don't know much about it, but I was thinking maybe the answer might come from umbral calculus?

To give a motivating example/special case for this question: the Wikipedia article on fractional calculus uses the example of the $\frac{1}{2}$th derivative, which when applied twice gives the standard derivative. What is the operator $D$ on sequences such that, when applied twice, it gives the forward difference of the original sequence?

Also, I have perhaps a related question: The solution to $\frac{d}{dx}f=f$ is $f=e^x$, while the solution to $\Delta f = f$ is $f=2^x$. Is the fact that $e$ is close to 2 a coincidence, or is there something connecting these results? Is there more generally some sort of spectrum of calculi lying between "finite" and "infinitesimal" each with its own "$e$"?

EDIT: After looking around some more I found time scales, which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!

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 @EDIT. If that's what you're after then you might look into what are called "Hybrid Systems". For example, see the work of André Platzer @ symbolaris.com. There is also the literature on Qualitative Reasoning and Qualitative Simulation, for example, the work of Ben Kuipers @ cs.utexas.edu/~kuipers. – Jon Awbrey Dec 25 2009 at 4:50

I don't know if you have seen this but there are papers devoted to "discrete fractional calculus". Like this one for example http://arxiv.org/abs/0911.3370 or http://www.math.u-szeged.hu/ejqtde/sped1/103.pdf . Like in fractional calculus, of course the discrete fractional integral is easier to define than the discrete fractional derivative.

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 Thanks for the links! Looks like exactly what I was thinking of. – Zev Chonoles Dec 23 2009 at 22:02

I can answer your last question, at least. The derivative acts as a shift operator on Taylor series, so the operator $\frac{d}{dx} - 1$ acts as the forward difference on Taylor series. So their eigenvectors are basically the same; the eigenvectors of the derivative are the exponential functions $e^{\lambda x}$, which have Taylor coefficients $a_n = \lambda^n$, and these are the eigenvectors of the forward difference. I'll think about your other questions.

Edit: Here's a possible way to get a notion of "fractional forward difference." If we write the forward difference operator $\Delta f(n) = f(n+1) - f(n)$ as $D - 1$ where $D$ is the shift operator, it follows that

$\displaystyle \Delta^k f(n) = (D - 1)^k f(n) = \sum_{i=0}^{k} {k \choose i} (-1)^{k-i} f(n+i)$

by the binomial theorem. So a possible extension to non-integer values of $k$ is to use the generalized binomial theorem formally in the above expansion. Unfortunately, the above sum is then infinite and therefore not guaranteed to converge. I'm not really sure how useful or interesting this is.

Edit 2: Well, that doesn't work for a stupid reason; $\sqrt{D - 1}$ doesn't have a Taylor series expansion at $0$. However, the backward difference operator $\nabla f(n) = f(n) - f(n-1)$ can be written as $1 - E$ where $E$ is the other shift operator and $(1 - E)^k$ has a Taylor series expansion for every $k \in \mathbb{C}$. (Note in particular the case $k = -1$.)

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Hmm. Why doesn't $\sqrt{D-1}$ have a Taylor series at $0$? As long as you are willing to consider complex numbers (and why wouldn't you?!), $\sqrt{D-1}=i\sqrt{1-E}$ :) – Mariano Suárez-Alvarez Dec 23 2009 at 13:43
The $E$ should be a $D$, of course... (comments should be editable!) – Mariano Suárez-Alvarez Dec 23 2009 at 13:44

Taking the point of view that the binary domain $\mathbb{B} = \lbrace 0, 1 \rbrace$ is more general than the real domain $\mathbb{R}$, and cranking (if you'll excuse the expression) all the analogies in sight, we find ourselves in the wonderland of Differential Logic, where plus and minus are the same operation, playing the devil with our usual suspicions about the diff between differential and integral calculus. So $\exists\sum\operatorname{fun}$ to be had with that.

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-1: Very little of this makes any sense. To say that "X is more general than Y" means that Y can be recovered as a special case of X; this is not the case for B and R (or R and B!). Moreover, differentiation and integration are distinct operations even in characteristic 2. – Pete L. Clark Dec 23 2009 at 20:50
I was going to put "generalization" on the list of overloaded words, since it has many senses, some of them neatly opposed, but "genericity" is already there, so I felt that was covered. If you regard B as included in R, then, sure, it looks more special. But if you count the axioms that define B and R, then more axioms means more special. When we use binary domains as targets of maps f : X → B, they are really just devices for drawing distinctions in X, and distinctions are basic to all structure. It's not that ∂ = ∫, only that the degeneracies of char 2 force new definitions on us. – Jon Awbrey Dec 24 2009 at 2:12
Jon, Pete clearly stated that neither R nor B is "more general than" the other, which the second sentence of your comment seems to ignore. Your subsequent sentence makes no sense relevant to your argument: "...more axioms means more special." "More" can only make sense in this setting if one system is obtained from the other only by adding axioms, and taking none away. The problem here isn't that generalization is overloaded, it is that you are using it incorrectly. – Jonas Meyer Dec 24 2009 at 22:03
A clear statement of opinion is a good thing, but not therefore a theorem. There is an abstract concept that covers B and R and any space A that we use to form a realm of types like X → A, A → X, A^k, A^k → A, and so on. There are analogies or morphisms between corresponding types, say, between R^k → R and B^k → B, and there is a clear sense, deriving from actual applications, in saying that the types built on R are more concrete and more specialized than the corresponding types built on B. – Jon Awbrey Dec 25 2009 at 3:31
I agree with your first sentence. Happy holidays! – Jonas Meyer Dec 25 2009 at 8:30

Part of what makes the exponential function so special is that it's the eigenfunction of a differential operator. When we solve the equation $\frac{df}{dx} = E f$ we are saying that f is an eigenfunction of $\frac{d}{dx}$ with eigenvalue E. The spectrum, in this case, is the whole real line.

If we solve the equation f(n+1) - f(n) = k f(n), with f(0) = 1. Then we have f(n+1) = (k+1)f(n), so $f(n) = (k+1)^n$. In your case, let $k = 2$.

One more example to think about is the quantum harmonic oscillator which looks for eigenfunctions of $H = \frac{d^2}{dx^2} - x^2$. However, we get the factorization $\left( \frac{d}{dx} - x \right)\left(\frac{d}{dx} + x \right) = a_- a_+$. The eigenfunction of $a_-$ is $e^{-x^2/2}$ and I will leave it as an exercise to show $H[e^{-x^2/2}] = 0$ and that $(a_+)^k[ e^{-x^2/2}]$ is an eigenfunction of H for all k.

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This could be an answer for your last question: if you define $\frac{\Delta_t}{t}$ as $\frac{f(x+t)-f(x)}{t}$ for $t>0$ and you solve $\frac{\Delta_t}{t}(f)=f$ you obtain the solution $$f(x)=\left((1+t)^{\frac{1}{t}}\right)^x.$$ So for the operator $\Delta_1(f)=f(x+1)-f(x)$ the solution is $2^x$, and as $t\rightarrow 0$ the solution tends to $e^x$.

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I realize this question was posted a while ago, but I would like to make a note on your edit stating,

After looking around some more I found time scales, which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!

Time scale calculus is fairly new development. It came about in 1989 in Hilger's Ph.D. thesis (he initially called it a measure chain). At the time of writing his thesis, mathematicians did not embrace his idea. Then, in the early 90s, the ideas were brought to the U.S. and now there are papers being written about time scale all over the world. However, I suppose there are still some mathematicians who are not studying/teaching time scale.

If you are looking for resources on this, you might consider taking a look at [1].

Furthermore, I see above that you are trying to define the exponential function for "discrete(difference) calculus" (if I am understanding your posting correctly).

Note that for the continuous case, we have that $(e^{at})^{'}=ae^{at}$

Now, consider $\Delta (1+ \alpha)^{t} = (1+ \alpha)^{t+1} - (1 + \alpha)^{t}= (1+\alpha)(1+\alpha)^{t}- (1+\alpha)^{t} = ((1+\alpha)-1)(1+\alpha)^{t}=\alpha(1+\alpha)^{t}$.

Thus, the discrete exponential should actually be $(1+ \alpha)^{t}$.

[2] is a good resource for discrete calculus.

Also, [1] discusses exponentials in time scale. There is a delta and nabla exponential form. You may consider checking that out.

Now, to your fractional calculus question, I believe I recall my professor mentioning some problem with fractional calculus and time scale. I cannot remember what it was exactly, but it was an open problem regarding fractional calculus. (Note: There are many open problems in time scale.)

I hope this helps!

[1] Bohner, M. & Peterson, A. (2001). Dynamic Equations on Time Scales: An Introduction with Applicaitons. Boston, MA: Birkhäuser.

[2] Kelley, W. & Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.

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Just to draw the attention of the curious reader to take a look at the book titled "Discrete Calculus by Analogy" witten by Izadi, Aliev, and Bagirev to see the differences as well as similirities between Discrete Calculs and Continous one.

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