4

Let $$ f(n,x) = \frac{d^n}{dx^n} \cos(x) = \cos(\frac{\pi n}{2}) \: \cos(x) - \sin(\frac{\pi n}{2}) \: \sin(x)$$

Can one take the derivative with respect to $n$ so that one has $$ \frac{\partial}{\partial n} f(n,x) = -\frac{\pi}{2} \left(\sin(\frac{\pi n}{2}) \: \cos(x) + \cos(\frac{\pi n}{2}) \: \sin(x) \right) $$

Is this just nonsense?

DecarbonatedOdes
  • 968
  • There is such a thing as fractional derivatives, but it is non-trivial. However, generally speaking the integers can't be treated like real numbers when it comes to things involving limits (like derivatives). But, the answer below from Kenny Lau gives a workaround, though this changes the meaning of $n$ and you can no longer write $\frac{d^n}{dx^n}$. – Randall Nov 15 '21 at 17:40

1 Answers1

8

It is important to understand that functions, at least in rigorous mathematics, require a domain and a codomain, a fact which the Leibniz notation unfortuantely obscures.

A priori, your $f$ is a function $\Bbb N \times \Bbb R \to \Bbb R$, so you can't take partial derivative w.r.t. the first variable.

However, your formula showed that you can extend $f$ "naturally" to become a function $\Bbb R \times \Bbb R \to \Bbb R$, and now you can take partial derivative w.r.t. the first variable.

Now it is very very important to note that the extended function is not the same as $f$, because they have different domains; however, not every situation requires a separate notation like $\overline{f}$ for the extended function, so sometimes "abuse of notation" allows us to re-use the same variable $f$ for the extended function, when the context is clear enough.

So yes, after you extended the function $f$ using the formula, then you can take partial derivative.

Kenny Lau
  • 25,049
  • 1
    Very clear and concise! Thank you – DecarbonatedOdes Nov 15 '21 at 17:40
  • 3
    Given the title of the question, it should be perhaps emphasized that the extension function is no longer a derivative of $f$, but just a function in two variables where the first one was extended from naturals to reals. Otherwise put, $\frac{\partial}{\partial n} f(n,x)$ can be legitimized by extension, but $\frac{\partial}{\partial n} f^{(n)}(x)$ cannot. – dxiv Nov 15 '21 at 17:45
  • I have seen before that there is some theory of fractional derivatives---for $n$ rational, does the fractional derivative agree with simply generalizing the OP's formula in the "obvious" way? – user7530 Nov 15 '21 at 17:58