What would it even mean to consider a "half" derivative?

  • Context: Undergrad 
  • Thread starter Thread starter Ethan Singer
  • Start date Start date
  • Tags Tags
    Derivative even Mean
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
Ethan Singer
Messages
19
Reaction score
1
Traditionally, derivatives are taught as a function that have... "Whole" transitions. Take the following example:

If you have the function f(x) = x^2, we find that f'(x) = 2x, and that f''(x) = 2. In other words, it has a first and a second derivative.

But what would it even mean to take a rational number as a derivative? I mean... If there is a half derivative for the function of x, you would first imagine instinctively that it's value would between x^2 and 2x, but beyond that I wouldn't know how to imagine what it would be, or what it would imply.

Is there a derivative for every real number? Imaginary? Are there Partial Half derivatives? What utility do they have? By that I mean, under what circumstance could we use a half-derivative in Physics?
 
Physics news on Phys.org
Ethan Singer said:
Traditionally, derivatives are taught as a function that have... "Whole" transitions. Take the following example:

If you have the function f(x) = x^2, we find that f'(x) = 2x, and that f''(x) = 2. In other words, it has a first and a second derivative.

But what would it even mean to take a rational number as a derivative? I mean... If there is a half derivative for the function of x, you would first imagine instinctively that it's value would between x^2 and 2x, but beyond that I wouldn't know how to imagine what it would be, or what it would imply.

Is there a derivative for every real number? Imaginary? Are there Partial Half derivatives? What utility do they have? By that I mean, under what circumstance could we use a half-derivative in Physics?
I have never heard of such a thing and I doubt it would make sense. Perhaps if we consider the differentiation as an operator and then expand its rational powers in a series, there is a small chance to even define it properly in some range. But more likely is that we run into problems to define convergence or something like ##D^nD^{\frac{1}{n}}=D## which I would expect to be required.
 
For sufficiently well-behaved functions, you can use the Fourier transform. Differentiation in the "time domain" corresponds to multiplication by the function ##(\omega \mapsto i \omega)## in the "frequency domain" and that can be generalised to ##(\omega \mapsto (i \omega) ^\alpha)##. It's not something I've ever looked at in detail, so I'm not sure what "sufficiently well-behaved" really means in this context.

(For A-thead-level readers: more generally, for any operator for which the spectral theorem applies, you can use the ##L^2## representation of the operator.)
 
this was done by riemann already in complex analysis. the secret is to represent the nth derivative of a holomorphic function, as a Cauchy integral of an expression involving an (n+1)st power. Then since fractional powers make sense, the analogous Cauchy integral formula with a fractional power should be a fractional derivative. thus the one half derivative of f(z) at z=a, should be the integral of f(z)/(z-a)^3/2. Or maybe you also need to know the value of the factorial function at rational numbers, i.e. use the gamma function. in fact that is perhaps the deeper part. I.e. once you can define the factorial function at fractions you get fractional derivatives. You can check this by looking up the cauchy integral formula for a derivative and see if it has a factorial in it. yes it does, so you need to know what (1/2)! means for this question. i.e. gamma(3/2).
 
  • Like
Likes   Reactions: WWGD