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Let $P(x) = a_{n}\cdot x^{n} + a_{n-1}\cdot x^{n-1} + \cdots + a_{0}$ be a polynomial with real coefficients. How can we prove that its $n$th derivative is:$$P^{(n)}(x) = a_{n}n!$$

Gerry
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4 Answers4

4

Hint:

From here observe that $$\frac{dx^m}{d^nx}=\frac{m!}{(m-n)!}x^{m-n}$$ (Proof)

$$\text{ So, }\frac{dx^m}{d^nx}=0 \text{ if }0\le m<n\text { and } \frac{dx^n}{d^nx}=\frac{n!}{(n-n)!}x^{n-n}=n!$$

lab bhattacharjee
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By induction: you have $(a_1 x+a_0)' = a_1\cdot 1!$. Thus, the claim holds for $n = 1$. Assume now that the claim holds for $n-1$ and let us show that it implies the claim for $n$.

Let $$ P_n(x) = \sum_{k=0}^n a_kx^k $$ be any polynomial of the degree $n$, then $$ P_n'(x) = \sum_{k=1}^{n}k\cdot a_{k}x^{k-1} = \sum_{k=0}^{n-1}b_k x^k =: P_{n-1}(x) $$ is a polynomial of the degree $n-1$ with the main term $b_{n-1} = na_{n}$. We know that $$ P^{(n)}_n = P^{(n-1)}_{n-1} = (n-1)!b_{n-1} = n! a_n $$

SBF
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  • How do you know that $P^{(n-1)}{n-1} = (n-1)!b{n-1}$? This is what you want to prove isn't it? – Gerry Mar 04 '13 at 16:01
  • @Gerry: well, that's the whole point of the induction. I should have been more explicit about that, though. Just edited - see 2nd and 3rd sentences of the answer. – SBF Mar 04 '13 at 16:05
  • Ok I see. The part that confused me at the beginning was that $P_{n}^{'}(x) = P_{n-1}(x)$ because when you differentiate the original polynomial the result is not the same $(P)$ polynomial but another, say $S_{n-1}(x)$ right? – Gerry Mar 04 '13 at 16:13
  • @Gerry: of course, I meant $P_n$ and $P_{n-1}$ to bedifferent polynomials (see my definition $...=:P_{n-1}$). – SBF Mar 04 '13 at 16:17
  • You are right. Thanks a lot Ilya! I think I have to approve @lab bhattacharjee's answer 'cause it was a bit faster. I did upvoted you though! – Gerry Mar 04 '13 at 16:19
  • @Gerry: thanks! – SBF Mar 04 '13 at 16:22
1

Let you polynomial be $$p(x)=\sum_{k=0}^n a_kx^k$$

Note that for any polynomial $q$ of degree $m<n$, we have that $$q^{(m)}\equiv 0$$ (i.e. it is the zero polynomial)

Moreover, note that $$p(x)=a_nx^n+q(x)$$ where $\operatorname{deg}(q)\leq n-1$. Then $$p^{(n)}(x)=a_n(x^n)^{(n)}+0=a_n(x^n)^{(n)}$$

So, you can prove by induction two things now

$1.$ If $q$ is a polynomial with degree $m<n$, $q^{(n)}=0$.

$2.$ If $f(x)=x^n$ then $f^{(n)}=n!$

and you'll be done.

Pedro
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The derivative is linear and lower monomials vanish so it suffices to prove $\left(\frac{d}{dx}\right)^n x^n = n!$, we prove a slightly stronger statement by induction:

Theorem $\left(\frac{d}{dx}\right)^k x^n = \frac{n!}{(n-k)!} x^{n-k}$.
proof: $k=0$ is trivial, suppose it's true for $k$ we show it for $k+1$ by differentating RHS once $$\frac{d}{dx} \frac{n!}{(n-k)!} x^{n-k} = \frac{n!}{(n-k-1)!}x^{n-k-1}.$$