1
$\begingroup$

Prove $D^n(x^n)=n!$ for n=1,2,3...

*note $D^n(x^n)$ is the nth derivative of $x^n$

Ok so I am very familiar with proof by induction and all the preliminary context but I am stuck where i need to prove for k+1. I currently have: $$D^k(x^k)*D^{k+1}(x^{k+1})$$

$$=n!*D^{k+1}(x^{k+1})$$

I am not sure if I am suppose to have multiplication between the terms and I do not know how to handle the $D^{k+1}(x^{k+1})$

2 Answers 2

7

Assuming $D$ is a linear operator:

$$D^{k+1}(x^{k+1})=D^k(D^1(x^{k+1}))$$$$=D^k((k+1)x^k)=(k+1)D^k(x^k)$$$$=(k+1)k!=(k+1)!$$

3

Hint: $D^{k+1}(x^{k+1}) = D(D^k(x^k \cdot x))$.

First apply the $k-$iterated product rule, then differentiate. What happens to the lone $x$ inside?

  • 3
    That is much more complicated than it need to be. Reversing the order is better: $D^{k+1}[x^{k+1}] = D^k[D[x^{k+1}]] = (k+1) D^k[x^k]$. It does require us to prove $D x^n = n x^{n-1}$ though, but I still think this is easier / more clear than applying the $k$-iterated product rule. – 2017-02-01
  • 1
    @Winther If it was a snake, it would have bit me. Woops. – 2017-02-01
  • 0
    Ah yes. Life is truly short, like the life of my ice cream. – 2017-02-02