Recently, I decided to try and create a formula for the $n$th order derivative of a polynomial, and I believe I succeeded! I tried to do a proof by induction to confirm this for myself, but since I haven't done one before, I'm not really sure if this is correct. Any insight would be greatly appreciated!

So, here's my (attempted) proof:

Let $P(x)$ be a polynomial of degree $s$ with integer coefficients $c_0,c_1,c_2,...c_s$ of the form $$P(x)=c_0+c_1x+c_2x^2+c_3x^3+\cdots+c_sx^s$$

Then, I propose that $$P^{(n)}(x)=\sum_{i=0}^{s-n}c_{n+i}\frac{(n+i)!}{i!}x^i$$

Base case for $n=0$: $$P^{(0)}(x)=\sum_{i=0}^{s}c_{i}x^i$$ $$=c_0+c_1x^1+c_2x^2+c_3x^3+\cdots+c_sx^s$$ $$=P(x)$$

Inductive step:
Assume that for a particular natural number $k$, the case $n=k$ is true (induction hypothesis): $$P^{(k)}(x)=\sum_{i=0}^{s-k}c_{k+i}\frac{(k+i)!}{i!}x^i$$ Expanding the first term gives: $$P^{(k)}(x)=c_kk!+\sum_{i=1}^{s-k}c_{k+i}\frac{(k+i)!}{i!}x^i$$ Differentiating both sides with respect to $x$ gives: $$\frac{d}{dx}P^{(k)}(x)=\frac{d}{dx}(c_kk!+\sum_{i=1}^{s-k}c_{k+i}\frac{(k+i)!}{i!}x^i)$$ $$P^{(k+1)}(x)=\sum_{i=1}^{s-k}c_{k+i}\frac{(k+i)!}{(i-1)!}x^{i-1}$$ The right side can be simplified as: $$\sum_{i=1}^{s-k}c_{k+i}\frac{(k+i)!}{(i-1)!}x^{i-1}=\sum_{i=0}^{s-k-1}c_{k+i+1}\frac{(k+i+1)!}{i!}x^i$$ $$=\sum_{i=0}^{s-(k+1)}c_{(k+1)+i}\frac{((k+1)+i)!}{i!}x^i$$ Equating both sides once more gives: $$P^{(k+1)}(x)=\sum_{i=0}^{s-(k+1)}c_{(k+1)+i}\frac{((k+1)+i)!}{i!}x^i$$ Therefore, the statement is true for $n=k+1$, proving the inductive step.

Since the base case and the inductive step have been proven true, the statement holds for all positive integers $n$.

Hopefully that was all correct! Let me know if I messed up somewhere, it'd be appreciated!

Edit: As Aman Kushwaha described, the inductive step has an assumption that $n\leq s$, and that for $n>s$, we can just define $P^{(n)}(x)=0$.

Edit 2: Ryszard Szwarc showed me that there is actually a formula on Wikipedia analogous to the power rule for regular derivatives, but for fractional derivatives instead. Whereas the normal power rule for the nth order derivative is $$f^{(n)}(x)=\frac{d^n}{dx^n}cx^k=\frac{k!}{k-n}cx^{k-n}$$ the generalized power rule for derivatives of a real number order (excluding the negative integers) is $$f^{(n)}(x)=\frac{d^n}{dx^n}cx^k=\frac{\Gamma (k+1)}{\Gamma (k-n+1)}cx^{k-n}$$ Using this, a new formula that encompasses fractional derivatives can be made that avoids the ambiguity associated with the non integer subscript polynomial coefficients. The only remaining issue is the ambiguity with the upper bound of the summation, however, when analyzing the generalized power rule, it becomes clear that constant terms in the polynomial only evaluate to 0 after each whole number iteration of the differential operator. Therefore the issue can actually be avoided entirely by using the floor function!

So, overall, this is what I came up with: $$P^{(n)}(x)=\sum_{i=0}^{s-\lfloor n \rfloor}c_{\lfloor n \rfloor +i}\frac{\Gamma (\lfloor n \rfloor +i+1)}{\Gamma (\lfloor n \rfloor -n+i+1)}x^{\lfloor n \rfloor -n+i},\: n \neq \mathbb{Z}^-$$

As far as I can tell, it seems to work, but I have absolutely no idea how to prove it! The formula isn't exactly pretty though, so if anyone has any idea how to simplify it, or how to prove it of course, I'd love to hear about it.

  • 1
    $\begingroup$ In the inductive step you have an underlying assumption that $k+1≤s$ or $k<s$. So, the conclusion you can draw from your proof is that $P^{(n)}(x)=\sum_{i=0}^{s-n}c_{n+i}\frac{(n+i)!}{i!}x^i$ for all natural numbers $n<s$ . Now, notice that your proposition also holds true for $n=s$. For $n>s$ you can simply define $P^n(x)=0$, the proof of which would be trivial, though you could use induction again, this time taking the base case to be $n=s+1$. $\endgroup$ May 7, 2022 at 18:41
  • $\begingroup$ @AmanKushwaha Thank you for the advice, that all completely makes sense! I updated the post now to show your suggestions :) $\endgroup$
    – cdog
    May 7, 2022 at 23:08
  • 1
    $\begingroup$ Consult en.m.wikipedia.org/wiki/Fractional_calculus When you scroll down you will find a formula for the fractional derivative of the monomials. $\endgroup$ May 8, 2022 at 6:00
  • $\begingroup$ @RyszardSzwarc Thank you so much for the suggestion! I found the formula and implemented it into my own to make a generalized polynomial derivative formula. The post is now updated to reflect your insight! $\endgroup$
    – cdog
    May 8, 2022 at 17:51


You must log in to answer this question.