Generalizations of "standard" calculus

Question

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!

Answer 1

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.

Answer 2

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 $k = -1$.)

Answer 3

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 $e^x$.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

Fractional derivative (differintegral) can be defined via Fourier or Laplace transform:

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{s^{q}{\mathcal {L}}[f(t)]\right\}.}$

This way, one can similarly define the fractional discrete derivative (because $\Delta=e^D-1$):

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(e^{i\omega}-1 )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{(e^s-1)^{q}{\mathcal {L}}[f(t)]\right\}.}$

Discrete derivative also can be defined via Newton series:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}f^{(k)}(x).}$

One can replace the derivative with difference operator to get fractional difference:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}\Delta^k f(x).}$

The Fourier transform method can be realized in Mathematica as follows:

FourierTransform[InverseFourierTransform[f[t], t, w] (E^(I w) - 1)^q, 
  w, x] // FullSimplify

(Mathematica realizes a different definition of Fourier transform, so there is a slight difference in formula)

And the Laplace transform method is

LaplaceTransform[InverseLaplaceTransform[f[t], t, w] (1 - E^w)^q, w, 
   x] // FullSimplify

(again, due to specifics of Mathematica, the formula is a bit different to make it more powerful).

The 1/2-th difference of the function $1/x$ is thus $\frac{i \sqrt{\pi } \Gamma \left(x-\frac{1}{2}\right)}{2 \Gamma (x+1)}$

Plot of function $1/x$ (blue), imaginary part of its 1/2-th (yellow) and 1-st (green) finite differences.

For function $f(x)=1/x$ the general expression holds: $\Delta^q [1/x]= \frac{(-1)^q \Gamma (q+1) \Gamma (x-q)}{\Gamma (x+1)}$