
Is there such a thing as half a derivative? - ColinWright
http://www.askamathematician.com/2015/05/q-is-there-such-a-thing-as-half-a-derivative/
======
KeytarHero
If you have a bit of signal processing background, fractional derivatives
actually make perfect sense in the frequency domain.

A derivative is equivalent to a highpass filtering with a slope of +20
dB/decade (or about +6 dB/octave), and an integral a lowpass filter of -20
dB/dec. So if you filter something by +10 dB/dec, you get half a derivative.

~~~
mgraczyk
I assume that this is what the author alluded to when he said "One way [to
define non-integer derivatives] is to use Fourier Transforms."

To be more specific, if we write the Fourier transform of a function as
FT(x(t)), then FT(d/dt x(t)) = j _2_ pi _f_ FT(x(t)).

In fact, this equality holds for higher order derivatives:

FT(d^n/dt^n x(t)) = (j _2_ pi _f)^n_ FT(x(t))

Where d^n/dt^n is the nth derivative.

We naturally extend this definition to "1/2 derivatives" the same way we often
extend integer valued functions to take rational arguments: we plug in a
rational and see what happens:

FT(d^(1/2)/dt^(1/2) x(t)) = (j _2_ pi _f)^(1 /2) _ FT(x(t)) = sqrt(j _2_ pi
_f)_ FT(x(t))

Then we take the inverse Fourier transform to find the half-derivative, which
is what we were originally looking for:

d^(1/2)/dt^(1/2) x(t) = FT^-1 (sqrt(j _2_ pi _f)_ FT(x(t)))

------
leni536
You can treat derivation as a linear operator on vector-space of functions,
let's name it D. Now second and third derivatives as operators are quite
simple since they are just D^2 and D^3. The question is if you can well define
sqrt(D), or saying that if there is one and only one operator D_half such that
D_half^2 is D (existence and unicity). It breaks at unicity since there are
multiple half derivative constructions (frequency domain approach, Taylor-
series approach,...).

On the other hand sqrt is well defined for positive real numbers (however not
well defined for general complex numbers). There is a similar definition only
for positive operators on Hilbert spaces (you pick the positive operator from
all the possible square roots). However derivation is not a positive operator
on the most used Hilbert-spaces.

~~~
ngoldbaum
At some point in my college Linear Algebra class I realized that the
exponential function is an eigenvector of the derivative operator. Hilbert
spaces are fun.

~~~
c_lebesgue
And the Gaussian distribution is an eigenvector of the Fourier transform.

Not surprisingly both of these facts correlate with how fundamental is the
exponents function to solving linear differential equations, and how
fundamental is the Gaussian distribution in probability theory.

------
ssivark
Intuitive essence of fractional derivatives: Interpret the n-th derivative as
the coefficient of \epsilon^n in a "generalized" Taylor expansion of
f(x+\epsilon). If the "function" behaves weirdly enough to not have a nice
Taylor expansion with integer powers, then you can essentially extract the
behaviour under increments by \epsilon. HTH. (IANAMathematician :-)

~~~
mdevere
Is the author's definition consistent with that?

~~~
yablak
From the functional form of the author's definition, it looks like yes.
However, to connect the two definitions probably requires using Cauchy's
integral formula
([http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula](http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula))
and/or the complex method of residues, along with some line integral tricks
([http://en.wikipedia.org/wiki/Methods_of_contour_integration](http://en.wikipedia.org/wiki/Methods_of_contour_integration)).

------
functional_test
Wikipedia also has a good discussion of the basics as well as some
generalizations:

[http://en.m.wikipedia.org/wiki/Fractional_derivative](http://en.m.wikipedia.org/wiki/Fractional_derivative)

Turns out there are even some applications, albeit rather esoteric ones.

------
yablak
I remember rediscovering fractional derivatives in college after learning
about the Laplace transform derivative formula. I even called them fractional
derivatives; and have a maple notebook somewhere in which I identified some
fractional derivatives for various smooth functions. An interesting fact is
that:

sin^(a)(x) = sin(x+a*pi/2)

(perhaps with some normalizing factor in front).

thus d/dx sin(x) = cos(x), etc.

I found the symmetry of this to be really beautiful.

I was very proud of my accomplishment until I googled the term and realized
someone had beat me to it by ~50-100 years :)

~~~
eridal
Bright minds think alike!

------
one-more-minute
A while back I realised you could do a similar thing by just working out the
nth derivative analytically and chucking a real number in. For example, the
nth derivative of x^m is

    
    
        (x^(m - n) m!)/(m - n)!
    

You can gradually vary n from 0 to 1 to see the smooth change. It doesn't
always behave how you'd expect it to.

Same for e^ix:

    
    
        i^n e^(i x)
    

Which means you can partially differentiate sin, cos etc. (In fact, the power
derivative lets you derive any function via its taylor series).

What do you think the gradient of sin looks like if you take it _gradually_
and animate it? You might expect it to slowly move to the left and become cos.
It does, but it also does a barrel roll around the real axis at the same time.
Kinda neat – I wish there was an easy way to put a demo up.

I tried to figure out some physical applications for this but I've fallen
short so far – would be interested to know if there are any.

------
chestervonwinch
In a relatedly similar way, you can define fractional B-splines. The nth order
B-spline is the n+1-fold convolution of a box function. n+1 convolutions turn
into raising the fourier transform of the box function to the n+1 power in the
fourier domain. This generalizes to non-integer powers. Then you just invert.
Cool stuff :)

------
mdevere
So, if you plot a three dimensional graph of y = f^(N)(x) (axes: x, y, N), I
wonder what it would look like, for, say, f(x) = x. Anyone handy with Matlab?

~~~
yk
Not too interesting ( a is the derivative axis and x is x):

[https://i.imgur.com/mFEGQ3d.gif](https://i.imgur.com/mFEGQ3d.gif)

Basically it just interpolates between two lines, one with slope 1 and one
with slope 0.

------
bkcooper
The Wikipedia article on fractional calculus mentions that fractional
derivatives are not local in the same way that integer derivatives are. Is
that right? That seems profoundly weird in the context of differential
equations.

~~~
danbruc
Because the local behavior is already completely captured by integer order
derivatives there is no information remaining that could go into fractional
order derivatives. You could maybe just make fractional order derivatives
interpolate between neighboring integer order derivatives but I guess that may
cause some problems and a non-local definition just makes more sense.

------
4ad
A more in depth and more complete treatment:
[http://mathpages.com/home/kmath616/kmath616.htm](http://mathpages.com/home/kmath616/kmath616.htm)

------
amelius
Next question: is there such a thing as a derivative of code?

~~~
teraflop
There is such a thing as a derivative of a context-free language, which turns
out to be useful for writing parsers: [http://matt.might.net/articles/parsing-
with-derivatives/](http://matt.might.net/articles/parsing-with-derivatives/)

~~~
wetmore
This derivative (the Brzozowski derivative) is in fact applicable to any
language, in the sense of a set of words over some alphabet. The language
doesn't need to be context-free.

------
rawnlq
Also see
[http://en.wikipedia.org/wiki/Hausdorff_dimension](http://en.wikipedia.org/wiki/Hausdorff_dimension)
to see how you can have fractional dimensions

~~~
Scene_Cast2
Fractional function application (called "Fractional iteration") is also pretty
neat.

[http://en.wikipedia.org/wiki/Iterated_function#Fractional_it...](http://en.wikipedia.org/wiki/Iterated_function#Fractional_iterates_and_flows.2C_and_negative_iterates)

------
mancmatt
From my masters project covering this I recommend [http://www.amazon.com/The-
Fractional-Calculus-Applications-D...](http://www.amazon.com/The-Fractional-
Calculus-Applications-Differentiation/dp/0486450015)

ended up coding up something similar but less general than Podlubny's
[http://www.mathworks.com/matlabcentral/fileexchange/36570-ma...](http://www.mathworks.com/matlabcentral/fileexchange/36570-matrix-
approach-to-distributed-order-odes-and-pdes)

------
arh68
So instead of swapping (N-1)! out for gamma(N), can you swap in any other
continuous extension to the factorial function, or does it _have_ to be gamma?

I just read about Hadamard's & Luschny's gammas/factorial extensions [1]:
would those not work out?

[1]
[http://www.luschny.de/math/factorial/hadamard/HadamardsGamma...](http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html)

~~~
yablak
Probably not. gamma(N) is the single mostly-analytic function for which
gamma(x+1) = x*gamma(x). This is a requirement.

~~~
tanderson92
> single mostly-analytic function

I think by "mostly-analytic" here you mean meromorphic.

One really cool thing about the gamma function is it is not the only
meromorphic function which satisfies that recurrence relation on the integers!
E.g.: gamma(x) + sin(x*pi) is meromorphic and agrees with the factorial.

~~~
Chinjut
gamma(x) + sin(x * pi) agrees with the factorial [shifted by one, because the
gamma function has that stupid shift in its definition for no good reason] on
integer arguments, but does not satisfy the recurrence relation on non-integer
arguments.

That having been said, one can devise infinitely many analytic functions which
do satisfy the recurrence relation in general. See my other comment.

~~~
tanderson92
Yes indeed, that's why I only said they agreed "on the integers". Your other
example is deviously clever, nicely done.

------
xiaq
I wonder why Fourier and Laplace transforms were dismissed very early on? It
is very tempting to generalize the the differentiation property to non-integer
n and do a inverse transform to get any non-integer derivative. For instance,
I would imagine that f^(1/2) = L^{-1}[ sqrt(s) L[f] ]. Is there a particular
difficulty in doing the integral?

------
D_Alex
Let's turn it up a notch... can you have i-th order derivative (or integral)?

~~~
oofabz
It looks like you can. Most fractional derivative solutions on the Wikipedia
page use the gamma function, which is defined for complex numbers.

------
mdevere
Author seems pretty certain there are no applications and that his/her
definition is just arbitrary but I wonder... ?

~~~
KenoFischer
In quantum field theory you frequently do integrals in 4-ϵ as a way of
regularizing divergent integrals.

------
daveloyall
From the domain name, I expected this to be in English. It's not.

~~~
bitcrusher
From the top of the page:

"There is! For readers not already familiar with first year calculus, this
post will be a lot of non-sense."

FWIW, it is a pretty straight forward explanation of 'half-integrals'... which
by induction means there are 'half-derivatives'.

~~~
Lawtonfogle
It isn't nonsense, but I still couldn't follow all of the steps even with an
okay level of calculus knowledge (multivariable but not PDEs).

