Continuous differentiation for polynomials?

Has this concept been explored & if so what name does it go by?

Taking a simple polynomial & its derivatives:

$$y = x^3 + x ^ 2 + x + 1$$ $$\frac{dy}{dx} = 3x^2 + 2x + 1$$ $$\frac{d^2y}{dx^2} = 6x + 2$$ $$\frac{d^3y}{dx^3} = 6$$

At a glance, differentiation seems like a discrete operation. However it looks like it cuold be continuous.

Let's say q is the degree of differentiation, and for normal highschool differentiation $q = 1$, and we now want to try differentiating where $0 < q < 1$.

The process of differentiating a component in a polynomial becomes something like:

$vx^u$ becomes $(v+qu(v-1))x^(u-q)$

When q = 1 we still get the first descrete derivative as $v+1u(v-1) = vu$, while when q = 0 we get the original formula as $v+0u(v-1) = v$, so we have a q representing a continuous variable between the two formulae. Dragging q as a slider on a graph confirms it toggles between the two:

https://www.desmos.com/calculator/eqitd8v37x

I'm interested to know:

• What name does this sort of operation go by?
• Are there smoother or more elegant ways to make make continuous paths between a function and its derivative?
• Is there a more general way which allows q > 1 to represent higher derivatives? (Factorials?)
• Are there more general operations allowing us to make continuous paths between other formulae? (Beyond just weighting & adding their outputs)
• $v+u(v-1)$ is not $uv=u+u(v-1)$. Do you want $(u+qu(v-1))$ as your coefficient? Or, ah, you just want $quv+(1-q)v$ for a smooth path between $v$ and $uv$. – Kevin Arlin Mar 13 '14 at 0:06
• yep just constructing a smooth path. There are probably smoother ways to do it, this is a little klutzy. – Brendan Hill Mar 13 '14 at 3:49

1 Answer

See http://en.wikipedia.org/wiki/Fractional_calculus - this may contain all that you are looking for.

• Exactly what I'm after, thanks! – Brendan Hill Mar 13 '14 at 0:12
• @BrendanHill Sure! – user122283 Mar 13 '14 at 0:13