Confusion Regarding Half-Derivatives

Question

When I attempt to find the half-derivative for the function $e^x$ using the Riemann–Liouville Fractional Integral, I end up with $e^x$ as expected. However, whenever I expand $e^x$ using a Maclaurin Series and then take the half-derivative of the resulting summation, I end up with a different result. Can anyone please explain this incongruity? And if you could, can you please inform me on how to use the half-derivative on MacLaurin Series while ensuring my answer stays the same as if I had used the Riemann–Liouville Fractional Integral? Thank you for your help!

Answer 1

Let $f(x) = e^{x}$. The Caputo fractional derivative of order $\alpha \in (0,1]$, with lower limit $a = 0$, is defined as

\begin{equation} {}^C D_x^\alpha f(x) = \frac{1}{\Gamma(1-\alpha)} \int_0^x \frac{f'(t)}{(x-t)^{\alpha}} \, dt \end{equation}

Since $f'(t)=e^{t}$, we replace:

\begin{equation} {}^C D_x^\alpha e^x = \frac{1}{\Gamma(1-\alpha)} \int_0^x \frac{e^t}{(x-t)^\alpha} \, dt \end{equation}

Let $u=x-t$, so $t=x-u$, and $dt =-du$. Changing the integration limits:

\begin{equation} {}^C D_x^\alpha e^x = \frac{1}{\Gamma(1-\alpha)} \int_0^x \frac{e^{x-u}}{u^\alpha} \, du = \frac{e^x}{\Gamma(1-\alpha)} \int_0^x \frac{e^{-u}}{u^\alpha} \, du \end{equation}

This integral is known as the lower incomplete gamma function: \begin{equation} \gamma(1-\alpha, x) = \int_0^x u^{-\alpha} e^{-u} \, du \end{equation}

From this follows,

\begin{equation} \boxed{{}^C D_x^\alpha e^x = \frac{e^x}{\Gamma(1-\alpha)} \, \gamma(1-\alpha, x)} \end{equation}

The ordinary Taylor series allows us to reconstruct an analytic function over its domain of convergence by knowing only a countable set of its integer order derivatives evaluated at a single point. This powerful result forms the basis for much of classical analysis and numerical methods. Now since \begin{equation} e^x =\sum_{k=0}^{+\infty} \frac{x^k}{k!} \end{equation} We can compute the Caputo fractional derivative as \begin{equation} {}^C D_x^\alpha e^x ={}^C D_x^\alpha \sum_{k=0}^{+\infty} \frac{x^k}{k!} =\sum_{k=0}^{+\infty} \frac{{}^C D_x^\alpha x^k}{k!}=\sum_{k=0}^{+\infty} \frac{x^{k-\alpha} \Gamma(k+1)}{k! \Gamma(1+k-\alpha)} \end{equation} Because k is a positive integer or zero the above expression reduces to \begin{equation} {}^C D_x^\alpha e^x =\sum_{k=0}^{+\infty} \frac{x^{k-\alpha}}{\Gamma(1+k-\alpha)} \end{equation} When applying the recurrence relation for the lower incomplete gamma function (see enter link description here) we have \begin{equation} \boxed{{}^C D_x^\alpha e^x =\sum_{k=0}^{+\infty} \frac{x^{k-\alpha}}{\Gamma(1+k-\alpha)}= \frac{e^x \gamma(-\alpha,x)}{\Gamma(-\alpha)}} \end{equation} which is different from the previous boxed expression, except when $\alpha \to 1$. In classical infinitesimal calculus, differentiation is a local operation: the derivative at a point depends only on the behavior of the function in an infinitesimally small neighborhood around that point. As a result, term-by-term differentiation of a power series is valid, since each term can be treated independently without reference to the function's global behavior.

In contrast, fractional derivatives are inherently nonlocal. For instance, the Caputo fractional derivative of a function depends on the function’s values over an entire interval. This nonlocality implies that term-by-term differentiation of a series expansion is generally not valid unless strong convergence conditions are satisfied (such as uniform convergence of the derivative series and appropriate bounds on its terms). Simply ignoring the memory encoded in the function's past values leads to incorrect results.

Extending the concept of differentiation to the fractional case significantly broadens the scope of function representations. The nonlocal nature of fractional derivatives means they often encode more information about the global behavior of a function—such as long-range dependencies or cumulative effects—than their integer-order counterparts. This makes them particularly valuable for modeling systems with memory, hereditary properties, or anomalous diffusion. This insight motivates the development of the fractional Taylor series, in which a function is expanded in terms of its fractional derivatives rather than just integer-order ones. A generalized Taylor expansion using the Caputo derivative of order $\alpha \in (0,1]$ can be written as:

\begin{equation} f(x)=\sum_{j=0}^{+\infty} \frac{(x-a)^{j \alpha}}{\Gamma(1+j \alpha)} {}^C D_{a}^{j \alpha}f(a) \end{equation} where $\Gamma(\cdot)$ denotes the Gamma function, and $\alpha \in (0,1]$.