# Fractional Calculus on Sinusoidal functions

1. Aug 16, 2011

### TheDestroyer

Hello guys :),

I have a question which I think is very advanced and weird. But I need the answer for some signal analysis purpose.

As we know, the derivative of a sine function, per se, shifts the phase, by Pi/2; i.e.,

f(x) = A sin (w t)
df(x)/dt = A sin (w t + Pi/2) = A cos(w t)

and of course, the integral does the opposite:
Integral(f(x) dx) = A sin(w t - Pi/2) = - A cos(w t).

As we know, there's something called fractional calculus, which involves taking derivates in non-integer orders. So for example the derivative of

g(x) = x

of the order 1/2, would be something like

(d^(1/2) g(x))/(dx^(1/2)) = Gamma(1/2) x^(1/2),

where Gamma(1/2) is the Euler gamma function (I'm not sure whether inside gamma is over 1/2 or 3/2, but whatever).

------------

My question:

So my question is, could you guys help me in finding the phase dependence of sinusoidal functions on non-integer derivatives? so I'm looking for something like:

d^(n) f(x)/dx^n = A sin(w t + k(n)),

where n is a real number, and k(n) is the phase dependence on the derivative order, n, that I'm looking for.

Thank you for any efforts :)

2. Aug 16, 2011

### atomthick

$\frac{d^{\alpha}Asin(t)}{dt} = Asin(t + \alpha\pi/2)$, this fractional derivative is for w=1.

3. Aug 16, 2011

### TheDestroyer

Is it linear like that? I'm surprised! it's so simple! I thought it would be a little but more complicated! The results from the signal shifting I'm doing gives it something like

Phi = Pi/2 * ArcTan(1 - Exp(n))

proportionality between for n=[0,1] with some coefficients, where n is the derivative order...

Any explanation?

4. Aug 16, 2011

### atomthick

Yes is as simple as that. If you believe you have a different shift try the following test:

- compute the half derivative of sin(t), $g(t) = \frac{d^{\frac{1}{2}}sin(t)}{dt}$
- compute the half derivative of g(t), $h(t) = \frac{d^{\frac{1}{2}}g(t)}{dt}$

In the end you should have $h(t) = sin(t+\pi/2)$ hecause h(t) is the first derivative of sin(t)

With the simple formula you get $g(t) = sin(t+\pi/4)$ and $h(t) = sin(t+\pi/4 + \pi/4) = sin(t+\pi/2)$

You can obtain the shift formula by playing with the fractional derivative of the exponential function.

5. Aug 16, 2011

### pwsnafu

Absolutely not!

First, fractional derivatives do not satisfy the index law in general. For example take f(x) = 1 and then $D^{\frac12} \frac{d}{dx} 1 = 0$ but $\frac{d}{dx} D^{\frac12}1 = \frac{d}{dx} \frac{1}{\sqrt{\pi x}} \neq 0$. In general $D^p D^q \neq D^q D^p$.

Secondly, the Riemann-Liouville fractional derivative of the exponential function with base point a=0 leads to the http://en.wikipedia.org/wiki/Mittag-Leffler_function" [Broken], not another exponential.

In fractional calculus you must be explicit in which fractional derivative you use (yes there is more than one) and specify the base point where appropriate.

Last edited by a moderator: May 5, 2017
6. Aug 17, 2011

### TheDestroyer

So what's the correct answer please, if I have to use such a function? is it like the guys have said, concerning this a special simple case, or is it more complicated?

Last edited by a moderator: May 5, 2017
7. Aug 17, 2011

### pwsnafu

I can't answer because you haven't given enough information. You need to specify
1. What the definition of fractional derivative you intend to use,
2. If there is a base point, you need to specify that,
3. The domain of the function you intend to integrate/differentiate.

And yes they are all significant. If you treat sine as a function whose domain is [0,2pi] you will get a different result to sine with domain over all reals, which is different again compared to sine with domain of the unit circle.

You just choose the definition which has the properties you want in the application. The only advice I can give you is read lots of papers in your field. The convention used in one field will differ from others.

8. Aug 17, 2011

### atomthick

For index values a with 0 < a < 1 the fractional derivative is comutative and additive.

9. Aug 17, 2011

### pwsnafu

For Riemann-Liouville yes. Don't know if it is true for all one parameter fractional derivatives though. For two or more parameter operators, obviously not.

10. Aug 17, 2011

### TheDestroyer

Thank you for your replies guys!

I'm not that familiar with fractional derivatives to understand all the terms you mentioned. The thing I can tell you is that the purpose of the derivatives is shifting the phase between -Pi/2 and Pi/2; i.e, the derivative order has to be

-1 <= a <= 1

What's inside the sin is a real number, and I don't really know what basepoint mean. Could you please explain more so that I can provide sufficient information?

11. Aug 17, 2011

### atomthick

I've based my calculations on functions of a special form for which Liouville began his studies on fractional derivatives.

$f(x)=\sum^{\infty}_{n=0}c_{n}e^{a_{n}x}, Re(a_{n}) > 0$

for which the fractional derivatives are

$D^{\alpha}f(x)=\sum^{\infty}_{n=0}c_{n}a_{n}^{\alpha}e^{a_{n}x}, Re(a_{n}) > 0$

You can write $sin(x)=\frac{1}{2i}e^{ix} - \frac{1}{2i}e^{-ix}$ and try to use the above formula to find out the derivative but you have to take a limit case because Re(i) = 0

Here is a link http://www.umw.edu/cas/math/students/documents/damian1.pdf [Broken]

Last edited by a moderator: May 5, 2017
12. Aug 17, 2011

### TheDestroyer

Thanks a lot. I'll try :)

13. Aug 17, 2011

### pwsnafu

In that case, yep atom's answer is what you want.

The most standard definition of fractional calculus is the Riemann-Liouville. Given a function $f:[a,b]\rightarrow$ we define the fractional integral of order $Re \alpha>0$ as
$$D^\alpha_{a+}f(x) := \frac{1}{\Gamma(\alpha)}\int_a^x (x-t)^{\alpha-1} f(t) dt.$$
The base point refers to that a. Different choices of a result in different different answers when you calculate the integral. The Liouville definition above corresponds to $a\rightarrow -\infty$. This is why it is a limit.

14. Aug 18, 2011

### TheDestroyer

Thanks a lot mate :)

I'll go for that as well ^^