Example of a function that has fractional derivatives of order less than 1 but not 1 I have recently learned that some fractals can have fractional derivatives of order less than 1, say of 1/2 even if they are not differentiable (have no derivative of order 1).
I wonder if there is a function that has all fractional derivatives up to order 1, but not including 1?
 A: Define the Weirstrass Function by
$$W_{\lambda}^s(t)=\sum_{k=0}^{\infty} {\lambda}^{k \cdot (s-2)} \cdot \sin({\lambda}^k \cdot t)$$
Where $\lambda \gt 1$ , $s \lt 2$ and $t$ is real.
What is the derivative of this expression? Well, if $1 \lt s\lt 2$ it's nowhere differentiable.
Proof
This is actually quite simple. Just differentiate term by term.
$${{dW_{\lambda}^s(t)} \over dt}=\sum_{k=0}^{\infty} {\lambda}^{k \cdot (s-2)} \cdot {\lambda}^k \cdot \cos({\lambda}^k \cdot t)=\sum_{k=0}^{\infty} {\lambda}^{k \cdot (s-1)} \cdot \cos({\lambda}^k \cdot t)$$
Since $s \lt 2$ we know that for $ 1 \lt s \lt 2$ the exponent will range $0$ to $1$. Hence, the series will diverge and thus the derivative can't exist.
Fractional Calculus
You don't really need to do this, but define the fractional derivative of $\sin(a \cdot t)$ of order $\mu$ to be $a^{\mu} \cdot \sin(a \cdot t+{{\mu \cdot \pi} \over 2})$
Taking the fractional derivative of the Weirstrass Function now yields...
$${{d^{\mu} W_{\lambda}^s(t)} \over d^{\mu} t}=\sum_{k=0}^{\infty} {\lambda}^{k \cdot (s-2)} \cdot {\lambda}^{k \cdot \mu} \cdot \sin({\lambda}^k \cdot t +{{\mu \cdot \pi} \over 2})=\sum_{k=0}^{\infty} {\lambda}^{k \cdot (s+\mu-2)} \cdot \sin({\lambda}^k \cdot t +{{\mu \cdot \pi} \over 2})$$
Lets look at exponent term $s+\mu-2$ in more detail. Since the exponent must be less than $0$ to converge that means $\mu \lt 2-s$. Choosing $s=1$, means the the order of the derivative must be less than 1.
Shown below is the fractal generated for $\lambda=2$, $s=1$

Here is the $1/2$ derivative generated for $\lambda=2$, $s=1$, $\mu=1/2$

