9
$\begingroup$

With the $n$th order derivative ($n$ as a positive integer) of $e^{ax}$ given by $$D^{n}e^{ax}=a^ne^{ax},$$ is the generalized (or fractional) derivative the same? Does it apply for any arbitrary $\alpha$? That is $$D^{\alpha}e^{ax}=a^{\alpha}e^{ax}?$$ For example...$$D^{\frac{1}{2}}e^{2x}=2^{0.5}e^{2x}?$$ Do you know of a study or paper showing how the integer order derivative of an exponential function is generalized into a fractional order?

Thank you.

$\endgroup$
4
  • $\begingroup$ en.wikipedia.org/wiki/Fractional_calculus $\endgroup$
    – guaraqe
    Commented Jun 17, 2014 at 9:15
  • $\begingroup$ There is not the fractional derivative. If I remember correctly there are common definitions out there were not even $D^\alpha \exp(x)=\exp(x)$ for all $\alpha\in\Bbb R$. $\endgroup$
    – M. Winter
    Commented May 28, 2017 at 22:23
  • 3
    $\begingroup$ For all of us superannuated mathematicians with defective vision, it is an unfriendly act to use both $a$ and $\alpha$ in the same formula. $\endgroup$
    – Lubin
    Commented May 28, 2017 at 23:09
  • $\begingroup$ Possible duplicate of Non integer derivative of $1/p(x)$ $\endgroup$
    – Alex M.
    Commented May 29, 2017 at 7:56

3 Answers 3

2
$\begingroup$

The short answer is: it depends. You have different definitions given different starting point $a$

http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html

Look also here for a more extended discussion: http://mathpages.com/home/kmath616/kmath616.htm

$\endgroup$
2
$\begingroup$

Good question. This reminds me that in $11$th grade I gave a student lecture on the anout this using PowerPoint Presentations.

As is usual for Fractional Calculus, there is no such thing as "the Fractional Derivative". For example, Leonhard Euler (one of the first to deal with this) already discovered various possibilities for defining a Fractional Derivative of the exponential function.and both are different.


You have already used Euler's first approach. Simply order the derivatives of the exponential function according to their order of derivative, find a general formula that describes an nth derivative of it and generalize it: $$ \begin{align*} \operatorname{D}^{0}\left[ e^{\lambda \cdot x} \right] &= \lambda^{0} \cdot e^{\lambda \cdot x}\\ \operatorname{D}^{1}\left[ e^{\lambda \cdot x} \right] &= \lambda^{1} \cdot e^{\lambda \cdot x}\\ \operatorname{D}^{2}\left[ e^{\lambda \cdot x} \right] &= \lambda^{2} \cdot e^{\lambda \cdot x}\\ \operatorname{D}^{3}\left[ e^{\lambda \cdot x} \right] &= \lambda^{3} \cdot e^{\lambda \cdot x}\\ \operatorname{D}^{4}\left[ e^{\lambda \cdot x} \right] &= \lambda^{4} \cdot e^{\lambda \cdot x}\\ \operatorname{D}^{5}\left[ e^{\lambda \cdot x} \right] &= \lambda^{5} \cdot e^{\lambda \cdot x}\\ &\cdots\\ \operatorname{D}^{n}\left[ e^{\lambda \cdot x} \right] &= \lambda^{n} \cdot e^{\lambda \cdot x},\, \text{for}\, n \in \mathbb{N_{0}}\\ \\ \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \lambda^{\alpha} \cdot e^{\lambda \cdot x},\, \text{for}\, \alpha \in \mathbb{C}\\ \end{align*} $$


Euler's other approach would be what we would do with the Caputo Derivative: deriving the series expansion around $0$ of the exponential function aka $e^{\lambda \cdot x} = \sum\limits_{k = 0}^{\infty}\left[ \frac{\lambda^{k}}{k!} \cdot x^{k} \right]$: $$ \begin{align*} \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \operatorname{D}^{\alpha}\left[\sum\limits_{k = 0}^{\infty}\left[ \frac{\lambda^{k}}{k!} \cdot x^{k} \right] \right]\\ \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \sum\limits_{k = 0}^{\infty}\left[ \operatorname{D}^{\alpha}\left[ \frac{\lambda^{k}}{k!} \cdot x^{k} \right] \right]\\ \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\lambda^{k}}{k!} \cdot \operatorname{D}^{\alpha}\left[ x^{k} \right] \right]\\ \end{align*} $$

The Fractional Derivative for the monomial he had also derived from Euler in a similar way as for the exponential function. Simply order the derivatives of the monomial according to their order of derivative, find a general formula that describes an nth derivative of it and generalize it: $$ \begin{align*} \operatorname{D}^{0}\left( x^{m} \right) &= x^{m}\\ \operatorname{D}^{1}\left( x^{m} \right) &= m \cdot x^{m - 1}\\ \operatorname{D}^{2}\left( x^{m} \right) &= m \cdot \left( m - 1 \right) \cdot x^{m - 2}\\ \operatorname{D}^{3}\left( x^{m} \right) &= m \cdot \left( m - 1 \right) \cdot \left( m - 2 \right) \cdot x^{m - 3}\\ \operatorname{D}^{4}\left( x^{m} \right) &= m \cdot \left( m - 1 \right) \cdot \left( m - 3 \right) \cdot x^{m - 4}\\ \operatorname{D}^{5}\left( x^{m} \right) &= m \cdot \left( m - 1 \right) \cdot \left( m - 4 \right) \cdot x^{m - 5}\\ &\cdots\\ \operatorname{D}^{n}\left( x^{m} \right) &= m \cdot \left( m - 1 \right) \cdots \left( m - n + 1 \right) \cdot x^{m - n},\, &\text{for}\, n \in \mathbb{N_{0}}\\ \operatorname{D}^{n}\left( x^{m} \right) &= \frac{m!}{\left( m - n \right)!} \cdot x^{m - n},\, &\text{for}\, n \in \mathbb{N_{0}}\\ \operatorname{D}^{n}\left( x^{m} \right) &= \frac{\Gamma\left( m + 1 \right)}{\Gamma\left( m - n + 1 \right)} \cdot x^{m - n},\, &\text{for}\, n \in \mathbb{C}\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ is the Complete Gamma Function.

This allows us to continue the above formula: $$ \begin{align*} \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\lambda^{k}}{k!} \cdot \operatorname{D}^{\alpha}\left[ x^{k} \right] \right]\\ \operatorname{D}^{\alpha}\left[ e^{\lambda \cdot x} \right] &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\lambda^{k}}{k!} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} $$


Another cool approach is via Euler's Formula $e^{x \cdot i} = \cos\left( x \right) + \sin\left( x \right) \cdot i$ where $i^{2} = -1$ and a generalized formula for the $n$-th derivative of the $\sin$ and $\cos$. Simply order the derivatives of the $\sin$ or $\cos$ according to their order of derivative, find a general formula that describes an nth derivative of it and generalize it: $$ \begin{align*} \operatorname{D}^{1}\left[ \sin\left( x \right) \right] &= +\sin\left( x \right)\\ \operatorname{D}^{1}\left[ \sin\left( x \right) \right] &= +\cos\left( x \right)\\ \operatorname{D}^{2}\left[ \sin\left( x \right) \right] &= -\sin\left( x \right)\\ \operatorname{D}^{3}\left[ \sin\left( x \right) \right] &= -\cos\left( x \right)\\ \operatorname{D}^{4}\left[ \sin\left( x \right) \right] &= +\sin\left( x \right)\\ \operatorname{D}^{5}\left[ \sin\left( x \right) \right] &= +\cos\left( x \right)\\ &\cdots\\ \operatorname{D}^{n}\left[ \sin\left( x \right) \right] &= \sin\left( x + \frac{n \cdot \pi}{2} \right),\, \text{for}\, n \in \mathbb{N_{0}}\\ \\ \operatorname{D}^{n}\left[ \sin\left( x \right) \right] &= \sin\left( x + \frac{n \cdot \pi}{2} \right),\, \text{for}\, n \in \mathbb{C}\\ \end{align*} $$

So follows: $$ \begin{align*} \operatorname{D}^{n}\left[ e^{x \cdot i} \right] &= \operatorname{D}^{n}\left[ \cos\left( x \right) + \sin\left( x \right) \cdot i \right]\\ \operatorname{D}^{n}\left[ e^{x \cdot i} \right] &= \operatorname{D}^{n}\left[ \cos\left( x \right) \right] + \operatorname{D}^{n}\left[ \sin\left( x \right) \cdot i \right]\\ \operatorname{D}^{n}\left[ e^{x \cdot i} \right] &= \operatorname{D}^{n}\left[ \cos\left( x \right) \right] + \operatorname{D}^{n}\left[ \sin\left( x \right) \right] \cdot i\\ \operatorname{D}^{n}\left[ e^{x \cdot i} \right] &= \cos\left( x + \frac{n \cdot \pi}{2} \right) + \sin\left( x + \frac{n \cdot \pi}{2} \right) \cdot i\\ \operatorname{D}^{n}\left[ e^{x \cdot i} \right] &= e^{\left( x + \frac{n \cdot \pi}{2} \right) \cdot i}\\ \end{align*} $$

$\endgroup$
0
$\begingroup$

I would explain my intuition. If you think of operators and its eigenvector, given that the eigenvector of $D$ is $e^x$, and given the fact that a polynomial $P$ applied to an operator $A$ acting on a eigenvector $v$, gives the polynomial of eigenvalue $a$ times the eigenvector like:

$$P(A)v=P(a)v$$

Then if $De^{ax}=ae^{ax}\Rightarrow D^{k}e^{ax}=a^ke^{ax}\quad\forall k\in\Bbb N$. Now the problem is that the square function can't be expanded in a Taylor series arround $0$, so you are forced to expand the polynomial around an arbitrary múltiple of the identity operator, which is kind of weird. The fact above behave well with functions that can be expanded as a Taylor series arround $0$, like $\exp, \cos, \sin$, etc. Interestly by itself, $e^D$ is the shift operator, and $\cos(D), \sin(D)$ can be expressed as linear combination of shifts operator by imaginary unity and I think are related to Hermite polynomials but I haven't prooved myself.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .