Why is $\frac{d^m}{d(-x)^m}=(-1)^m \frac{d^m}{dx^m}$

Question

Im not a mathematician so dont judge me with my following "proof". I want to show that a Legendre polynomial is odd if its index is odd, and even if its index is even. This in order to prove that $P_1$ and $(P_m)^2$ are orthogonal over the interval $(-1,1)$.

Basically I've said that Legendre polynomials satisfy the equation

$$P_m(u)=\frac{1}{2^m m!}\frac{d^m}{du^m}(u^2-1)^m$$

And if you plug in $-x$ and conveniently say that $\frac{d^m}{d(-x)^m}=(-1)^m \frac{d^m}{dx^m}$ You'll arrive at the relation that a L.P. is odd if its index is odd and even if its index is even.

Now I happen to hate this "proof" because I only accept $\frac{d}{dx}$ as quotient when the derivative is of the first order. Higher order ones are less like quotients in my view. Here I've clearly used that $d(-x)^m=(-1)^mdx^m$ as if I were manipulating some regular number.

This is longer than I wanted it to be, but what proves that relation then?

Thanks.

Answer 1

HINT Use induction. We have $$\dfrac{d^{m+1}}{d^{m+1}(-x)} = \dfrac{d}{d(-x)} \left(\dfrac{d^{m}}{d^{m}(-x)}\right) = -\dfrac{d}{dx} \left(\dfrac{d^{m}}{d^{m}(-x)}\right)$$

Answer 2

The derivative can be thought of as $$\frac{d^n}{dx^n}(f(x))=\underbrace{\frac{d}{dx}\big(\frac{d}{dx}\big(\cdots}_{n\text{ times}}f(x)\big)\cdots\big).$$ Now, if you make the change of variables $x\mapsto -x$, you get the coefficient $(-1)^n$ that you're looking for in your situation.