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From this paper, in Eq. (4), it is asserted that (LHS being (3) and RHS being (4))

$$ (-1)^{n} z^{n} \frac{d^{n} \operatorname{sinc}(z)}{(z d z)^{n}} = (-1)^{n+1} z^{n+1} \frac{d^{n+1} \cos (z)}{(z d z)^{n+1}} $$

with $\operatorname{sinc}(z) = \sin(z)/z$ and $n \in \mathbb{N}_0$ and $z\in \mathbb{C}$. The notation employed in the cited paper is a bit ambiguous but in other literature (or here on SE), as far as I see it,

$$\frac{d^n}{(zdz)^n} f(z) := \left(\frac{1}{z} \frac{d}{d z}\right)^{n} := \underbrace{ \left(\frac{1}{z} \frac{d}{d z}\right) \left(\frac{1}{z} \frac{d}{d z}\right) \cdots \left(\frac{1}{z} \frac{d}{d z}\right)}_{\text{n times}} f(z) $$

I can't verify this though. When I substitute $n=1$ to test, I get for the LHS: $-\frac{\operatorname{cos}(z)}{z}+\frac{\operatorname{sin}(z)}{z^{2}}$ but for the RHS $z \left(\frac{\sin (z)}{z^2}-\frac{\cos (z)}{z}\right)$. Hence it feels like in the equation above, the $z^{n+1}$ on the RHS should actually be just $z$.

Is this an error in the paper or am I missing something?

ste
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    What is the notation $(zdz)^{n}$? – Golden_Ratio Mar 14 '22 at 14:32
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    I second @Golden_Ratio's clarification request. At first I thought $\frac{d^n}{(zdz)^n}$ would mean $(z^{-1}D)^n$ with $D:=\frac{d}{dz}$ (a helpful abbreviation for a question edit), whence the claim simplifies to$$(z^{-1}D)^n\operatorname{sinc}z=-z(z^{-1}D)^{n+1}\cos z.$$But the $n=0$ case $\operatorname{sinc}z=-D\cos z$ of that is false. – J.G. Mar 14 '22 at 14:48
  • I tried to make it more explicit. I'm not 100% sure this is what it means (could it be that the 'error' lies in the unexplained notation?) but at least fairly certain as I've seen it elsewhere been applied like this. I think of it as an operator that is applied sequentially. – ste Mar 14 '22 at 15:24

1 Answers1

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I think there is a typo and the derivation of the formula is simpler than one thinks.

First note that $$ \frac{1}{z}\frac{d}{dz}\cos(z)=-\mathrm{sinc}(z). $$ Hence defining the operator $$ X:=\frac{1}{z}\frac{d}{dz} $$ we have $$ X\cos=-\mathrm{sinc} $$ and trivially, by repeated application: $$ X^{n}\mathrm{sinc}=-X^{n+1}\cos $$ or, if we want to make it look like the original formula: $$ (-1)^{n}z^{n} \left(\frac{1}{z}\frac{d}{dz}\right)^{n}\mathrm{sinc}(z)=(-1)^{n+1}z^{n}\left(\frac{1}{z}\frac{d}{dz}\right)^{n+1}\cos(z) $$

lcv
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