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I am trying to prove the following

Let $F$, $\Phi$ and $\phi$ be arbitrarily differentiable functions of $x$. I have come across the following identity in some literature I am reading which I need help in proving:

$$\frac{1}{\phi^{r-1}}D^{r-1}\left[\frac{(F-\Phi)^r}{\phi}\right]=D_{(r-1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{r}\right],\qquad r=1,2,\ldots$$

where $\displaystyle D^r=\frac{d^r}{dx^r}$ and $$D_{(r)}=\left(D-\frac{\phi'}{\phi}\right)\left(D-2\frac{\phi'}{\phi}\right)\cdots\left(D-r\frac{\phi'}{\phi}\right)$$ and $D_0$ is just the identity.

I can see this for the case $r=2$: $$\begin{align*} \frac{1}{\phi}D\left[\frac{(F-\Phi)^{2}}{\phi}\right]&=\frac{1}{\phi}\left[\frac{\phi D(F-\Phi)^{2}-(F-\Phi)^{2}\phi'}{\phi^{2}}\right]\\ &=\left(D-\frac{\phi'}{\phi}\right)\left[\frac{(F-\Phi)^{2}}{\phi^{2}}\right]\\ &=D_{(1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{2}\right]\end{align*} $$

Its just an application of the quotient rule and some algebra. But how would I prove this for an arbitrary positive integer $r$?

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Is $D^r = \frac{d}{dx}$, or is $D^r$ the composition of $D=\frac{d}{dx}$ with itself $r$ times? –  Arturo Magidin May 15 '11 at 22:38
    
@Arturo Magidin, Typo, I meant, $\displaystyle D^r=\frac{d^r}{dx^r}$ the $r^{th}$ derivative –  aukie May 15 '11 at 22:55
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1 Answer 1

up vote 1 down vote accepted

HINT $\rm\displaystyle\ \ \ \ D + n\ \frac{f'}f\ =\:\ f^{\:-n}\ D\ f^{\:n} $

Below are some papers on general techniques.

Steven Roman, Operational Formulas, Lin. Multilin. Algebra, 1982, 1-20.

L. Carlitz, A theorem on differential operators, Amer. Math. Monthly, 83(5) 1976, 351-354.

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Thankyou very much, that worked a charm. Mind I ask you where you found this identity from? Any references in which I could read more about such things would be appreciated. –  aukie May 16 '11 at 1:01
1  
@aukie I added some references. –  Bill Dubuque May 16 '11 at 1:34
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