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I have found the following differential identity: $$\left(-x^2\frac{d}{dx}\right)^nf(x)=(-1)^n x^{n+1}\frac{d^n}{dx^n}\left(x^{n-1}f(x)\right)$$ I have used it to find an alternative Rodrigues representation for Bessel polynomials.

My proof

we have placed: $D=\frac{d}{dx}$ in order to save space.

We start observing that, for n=1 the following is trivial that:

\begin{equation} \left(x^{2}D\right)=x^{n+1}D^{n}x^{n-1},n=1 \end{equation}

Therefore for n=1 the identity holds. For induction we proof that, if the identity holds for n, then it holds also for n+1. Applying the operator$\left(x^{2}D\right)$

to both members:

\begin{equation} \left(x^{2}D\right)\left(x^{2}D\right)^{n}=\left(x^{2}D\right)\left(x^{n+1}D^{n}x^{n-1}\right) \end{equation}

\begin{equation} \left(x^{2}D\right)^{n+1}=x^{2}Dxx^{n}D^{n}x^{n-1}=x^{2}Dx^{n}xD^{n}x^{n-1} \end{equation}

remembering the following commutation rules:

\begin{equation} Dx^{n}\centerdot-x^{n}D=nx^{n-1} \end{equation}

\begin{equation} D^{n}x\centerdot-xD^{n}=nD^{n-1} \end{equation}

we can write:

\begin{equation} \left(x^{2}D\right)\left(x^{2}D\right)^{n}=x^{2}\left[Dx^{n}\centerdot\right]\left[xD^{n}\right]x^{n-1}=x^{2}\left[x^{n}D+nx^{n-1}\right]\left[D^{n}x\centerdot-nD^{n-1}\right]x^{n-1} \end{equation} \begin{equation} =x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n-1}D^{n}x\centerdot-x^{n}DnD^{n-1}-nx^{n-1}nD^{n-1}\right]x^{n-1} \end{equation}

Now we can show that the term in square parentheses is null, indeed applying above seen commutation rules:

\begin{equation} =x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n-1}xD^{n}+nx^{n-1}nD^{n-1}-nx^{n}D{}^{n}-nx^{n-1}nD^{n-1}\right]x^{n-1}= \end{equation}

\begin{equation} =x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n}D^{n}-nx^{n}D{}^{n}+nx^{n-1}nD^{n-1}-nx^{n-1}nD^{n-1}\right]x^{n-1}=x^{n+2}D^{n+1}x^{n}\centerdot \end{equation}

Therefore:

\begin{equation} \left(x^{2}D\right)^{n+1}=\left(x^{2}D\right)\left(x^{2}D\right)^{n}=x^{n+2}D^{n+1}x^{n}\centerdot \end{equation}

Questions

Is my proof correct?

Is this identity already known? If yes, what are the references?

Where can I find a possibly more elegant proof?

Edit

So, it seems the identity was true, and already known. In "Mathematical Analysis I" (Springer), Zorich V.A, in chapter "The Basic Rules of Differentiation", exercise 3 at page 213, the identity of $D^nf(1/x)=(-1)^nx^{n+1}D^n x^{n-1}f(1/x)$ has to be demonstrated. Unfortunately no hint was given and no proof is sketched, so I remain without a more straightforward proof ...)

giorgiomugnaini
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  • @giorgiomugnaini What you're doing, as far as I can see, is differentiating with respect to the variable $\frac{1}{x}$ where the variable of the function $f$ is $x$. However, that is not so elegant. Instead, it'll be better for you to write the variable of the function as $\frac{1}{x}$ and use the "usual" $\frac{d}{dx}$, However, I think it should remain the same. – Hasan Saad Apr 08 '15 at 08:58
  • @HasanSaad the identity emerges from the attempt to differentiate $(d/d)^n f(1/x)$ in terms of derivatives of $f(x)$. I don't think to return to the form $f(1/x)$. The original problem was to rewrite the standard Rodrigues formula for Bessel polynomials i.e: $D^n exp(1/x)x^{2m}$ in terms of derivatives of $exp(x)x^{-m}$ – giorgiomugnaini Apr 08 '15 at 09:08
  • I really have no idea... I just found it in Zorich in the form I told you about, but still, whichever form you use, the same proof should hold. Also, you might want to note that one can move from one form to the other without much hassle. All that would remain is to take $t=\frac{1}{x}$.. – Hasan Saad Apr 08 '15 at 09:13

1 Answers1

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I think you can write your identity better as your $\frac{d}{d(\frac{1}{x})}$ is undefined. You can write your identity like that:

$\frac{1}{x^{n+1}}f^{(n)}(\frac{1}{x})=(-1)^n\frac{d^n}{dx^n}(x^{n-1}f(\frac{1}{x}))$

To prove it, you can do this, without using the differentiation operator (I hate using it...):

1- The base case where $n=0$ is trivial. Prove that and the case where $n=1$. This should be very easy.

2- Now, start from the right side like this:

$(-1)^{n+1}\frac{d^{n+1}}{dx^{n+1}}(x^{n}f(\frac{1}{x}))=(-1)^{n+1}(\frac{d^n}{dx^n}(x\cdot x^{n-1}f(\frac{1}{x})))'$

Use the General Leibniz Rule and the product, knowing that any $n$th derivative of $x$ where $n\geq2$ is 0 and use the base case you proved, to follow by induction.

QED.

Also, as far as I know, it doesn't have a name. I have seen this in Zorich, and seeing that he didn't mention a name for it though he usually names important theorems in exercises, I believe it doesn't have one.

Hasan Saad
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