Integer derivatives of $x^n$:
$\partial_x x^n = n x^{n-1}$
$\partial_x^2 x^n = n (n-1) x^{n-2}$
$\partial_x^m x^n = (n-m+1)\cdot(n-m+2)\cdot ... \cdot (n-1) \cdot n \cdot x^{n-m}$
$= \Pi_{j=0}^{m-1} (n-j) x^{n-m}$
$= \frac{n!}{(n-m)!} x^{n-m}$
$= \frac{\Gamma(n+1)}{\Gamma(n-m+1)} x^{n-m}$
With the exception that $\partial_x^m x^n = 0$ for $m > n$.

By this definition, the half derivative of $x^n$ is:
$\partial_x^{1/2} = \frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})} x^{(n-\frac{1}{2})}$

Verifying that it reconstructs the original derivative:
$\partial_x^{1/2} (\partial_x^{1/2} x^n) = \partial_x^{1/2} (\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})} x^{(n-\frac{1}{2})})$
$= \frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})} \frac{\Gamma(n+\frac{1}{2})}{\Gamma(n)} x^{n-1}$
$= \frac{\Gamma(n+1)}{\Gamma(n)} x^{n-1}$
$= n x^{n-1}$

And therefore $\partial_x^{1/2} (\partial_x^{1/2} x^n) = \partial_x x^n$

Now is there a change of variable $\partial_x^{1/2} = \partial_y$ for $y = y(x)$?
$\partial_x^{1/2} x^n = \partial_y x(y)^n$
$\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})} x^{(n-\frac{1}{2})} = \frac{\Gamma(n+1)}{\Gamma(n)} x^{n-1} \frac{d x}{d y}$
$\int d y = \frac{\Gamma(n+\frac{1}{2})}{\Gamma(n)} \int \frac{1}{\sqrt{x}} dx$
$y = 2 \frac{\Gamma(n+\frac{1}{2})}{\Gamma(n)} \sqrt{x} + C$

Now $\partial_x x^n$ equals $\partial_y^2 x^n$.

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Integer derivatives of analytical functions $f(x) = \Sigma_{i=0}^n p_i x^i$
$\partial_x^m f(x) = \Sigma_{i=m}^n \frac{\Gamma(i+1)}{\Gamma(i-m+1)} p_i x^{i-m}$
Notice the lower bound of the sum is incremented to eliminate the derivatives of $x^0$.

Half derivative of $f(x)$
$\partial_x^{1/2} f(x) = \Sigma_{i=0}^n \frac{\Gamma(i+1)}{\Gamma(i+\frac{1}{2})} p_i x^{(i-\frac{1}{2})}$
But what about the lower bound of the sum? Keeping it at $i=0$ means the matching term $x^{(-\frac{1}{2})}$ now approaches $\infty$ for $x \rightarrow 0$

Change of variable:
$\partial_x^{1/2} = \partial_y$ for $y = y(x)$
$\partial_x^{1/2} f(x) = \partial_y f(x)$
$\Sigma_{i=0}^n \frac{\Gamma(i+1)}{\Gamma(i+\frac{1}{2})} p_i x^{(i-\frac{1}{2})} = \Sigma_{i=1}^n \frac{\Gamma(i+1)}{\Gamma(i)} p_i x^{i-1} \frac{dx}{dy}$
$y = \int \frac{ \Sigma_{i=1}^n \frac{\Gamma(i+1)}{\Gamma(i)} p_i x^{i-1} }{ \Sigma_{i=0}^n \frac{\Gamma(i+1)}{\Gamma(i+\frac{1}{2})} p_i x^{(i-\frac{1}{2})} } dx$
Notice the sum range lower bound on the two different polynomials is different.

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Heat Equation:
$\partial_t \phi = \partial_x^2 \phi$

Wave Equation:
$\partial_t^2 \psi = \partial_x^2 \psi$

How to turn the heat equation into a wave equation?
Look for a change of variable $\tau = \tau(t)$
such that the operator exists, $\partial_\tau \phi = \partial_t^{1/2} \phi$
So we end up with $\partial_\tau^2 \phi = \partial_t \phi$

Assume $\phi$ is analytic wrt $t$
So $\phi(x,t) = \Sigma_{i=0}^n p_i(x) t^i$
Next use our previous trick to let $\partial_t^{1/2} \phi = \partial_\tau \phi$ for $\tau = \tau(t)$
Then we come up with the change of variables:
$\tau = \int \frac{ \Sigma_{i=1}^n \frac{\Gamma(i+1)}{\Gamma(i)} p_i(x) t^{i-1} }{ \Sigma_{i=0}^n \frac{\Gamma(i+1)}{\Gamma(i+\frac{1}{2})} p_i(x) t^{(i-\frac{1}{2})} } dt$
And now $\phi(x, \tau)$ is a wave equation, such that:
$\partial_\tau^2 \phi = \partial_x^2 \phi$

Just watch out for $t=0$...