How to prove this equation($x^n \frac{{\rm d}^n}{{\rm d}x^n}=\cdots$)?

Question

How to prove $x^n \frac{{\rm d}^n}{{\rm d}x^n}=x\frac{{\rm d}}{{\rm d}x}\Big(x\frac{{\rm d}}{{\rm d}x}-1\Big)\cdot\Big(x\frac{{\rm d}}{{\rm d}x}-n+1\Big)$? F. H. Jackson call this equation is Boole's equation, but when I Google "Boole's equation", I can't get any useful information!

Answer 1

Hint. Use Leibnitz' rule for a higher derivative of a product: $$\frac{d^m}{dx^m}(uv)=\sum_{k=0}^m \binom mk\frac{d^ku}{dx^k}\frac{d^{m-k}v}{dx^{m-k}}\ .$$ If $u=x$ this reduces to $$\frac{d^m}{dx^m}(xv)=x\frac{d^mv}{dx^m}+m\frac{d^{m-1}v}{dx^{m-1}}\ .$$ Your result can now be proved by induction: assume the formula for $n-1$ and apply it to $$x\frac{df}{dx}-(n-1)f\ .$$ Good luck!