# Limitations of fractional derivative approximation with Taylor series

I was playing around with the concept of fraction derivatives, and came across some base functions for which it is defined, namely power and exponential functions

$$\left(\frac{d}{dt}\right)^\alpha t^k = \frac{\Gamma(k+1)}{\Gamma(k-\alpha+1)} t^{k-\alpha}, \quad k\geq0 \tag{1}$$

$$\left(\frac{d}{dt}\right)^\alpha e^{kt} = k^\alpha e^{kt} = e^{kt + \alpha \log(k)}. \tag{2}$$

I was wondering what the fractional derivative would be of $\sin(\omega t)$ and initially though that if I would calculate its Taylor series at $t=0$, I then could use (1) to find its fraction derivative. But later I came across (2) and realized that in this case it can be found much easier. Namely $\sin(\omega t)$ can be written as

$$\sin(\omega t) = \frac{1}{2i} \left(e^{i\omega t} - e^{-i\omega t}\right), \tag{3}$$

thus the fractional derivative can be found with (2)

$$\left(\frac{d}{dt}\right)^\alpha \sin(\omega t) = \frac{1}{2i} \left(e^{i\omega t + \alpha \log(i\omega)} - e^{-i\omega t + \alpha \log(-i\omega)}\right) = \omega^\alpha \sin\left(\omega t + \alpha \frac{\pi}{2}\right). \tag{4}$$

But if I compare this with the result I get when using the Taylor series I get very wrong results for $t<0$ and for $t$ slightly larger than $0$ I get a transition towards the correct result. For example here are the results I get when $\alpha=\frac 12$, $\omega=1$ and the Taylor series is approximated with 50 terms:

Could it be that (1) is only true for $t>0$ and that taking the factional derivative of a Taylor series might not have a convergence near the point at which the Taylor series is constructed, in my case $t=0$?

The fractional derivative of $e^{kt}$ is not what you supposed.

From the definition based on the Riemann-Liouville operator : $$\frac{d^\nu}{dt^\nu}f(t)=\frac{1}{\Gamma(-\nu)}\int_0^t \frac{f(x) dx}{(t-x)^{\nu+1}}$$ The fractional derivative of the exponential function is : $$\frac{d^\nu}{dt^\nu}e^{bt}=b^\nu e^{bt} \left( 1-\frac{\Gamma(-\nu,bt)}{\Gamma(b+1-\nu)} \right)$$ In the incomplete form : $\frac{d^\nu}{dt}e^{bt}=b^\nu e^{bt}$ , the term due to the lower bound of the integral is neglected, which causes the discrepancy that you observed.

In fact, your equation corresponds to the Weyl’s opetrator in which the lower bound of the integral is $-\infty$ instead of $0$. In “time mode” (used in electrotechnics), this corresponds to the steady-state after the fading of the terms due to a start at finite time.

All this is explained in more details in pages 5-6 in the paper : https://fr.scribd.com/doc/71923015/The-Phasance-Concept

The fractional derivatives of the sine and cosine functions are :

$$\frac{d^\nu}{d t^\nu} \cos(\omega t) =\omega^\nu \cos(\omega t+\frac{\pi}{2}\nu) - \frac{\omega^\nu }{\Gamma(-\nu)} \left( \cos(\omega t)C(\omega t , -\nu) + \sin(\omega t)S(\omega t , -\nu) \right)$$

$$\frac{d^\nu}{d t^\nu} \sin(\omega t) =\omega^\nu \sin(\omega t+\frac{\pi}{2}\nu) - \frac{\omega^\nu }{\Gamma(-\nu)} \left( \sin(\omega t)C(\omega t , -\nu) - \cos(\omega t)S(\omega t , -\nu) \right)$$

In $t>0$. The functions $S$ and $C$ are the Generalized Fresnel integrals : $$S(\theta,a)=\int_\theta^\infty x^{a-1}\sin(x)dx$$ $$C(\theta,a)=\int_\theta^\infty x^{a-1}\cos(x)dx$$ $a=-\nu$ ; $\theta=\omega t$ ; $\theta>0$.

• I have not found a good simple numerical approximation of $S(\theta,a)$ and $C(\theta,a)$ in order to compared it to my other results, but are the polynomial results correct? – Kwin van der Veen Oct 29 '15 at 20:11
• What do you call ''polynomial results" ? – JJacquelin Oct 29 '15 at 20:56
• The Taylor polynomial/equation $(1)$. – Kwin van der Veen Oct 29 '15 at 22:20
• The power series can be used (with condition of convergence). Your result for the fractionnal derivative of $t^k$ is correct. – JJacquelin Oct 31 '15 at 4:42