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Define the left shift operator as follows:

$$\lambda_x y = \frac{y-(x:=0)y}{x}$$

where $(x:= 0)$ means "replace every copy of $x$ by $0$," or equivalently "evaluation at $0$."

For example,

$$\lambda_x(1+3x-x^2) = \frac{1+3x-x^2-(1+3\cdot 0-0^2)}{x} = 3-x.$$

For comparison, lets compute the derivative, too:

$$\partial_x(1+3x-x^2) = 3-2x.$$

So when $\lambda_x$ is applied to a power series, the constant term is dropped, and the whole series shifts to the left; similar to how differentiation works, except that $\partial_x$ also introduces a modification to the coefficients equal to the exponent of the relevant term.

Now it's clear that $\lambda_x$ is linear. And, given that $\lambda_x$ is similar to $\partial_x$, we might expect some kind of product rule to hold. But I've been staring at the expression $$\lambda_x(y\cdot z)$$ for awhile, and I'm not seeing anything.

Question. Does the left shift operator, as defined here, satisfy an analogue of the product rule, and if so, what?

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  • $\begingroup$ This is actually distantly related to my dissertation (it was a Leibniz formula for operators that are formally similar to the derivative). $\endgroup$
    – Matt Samuel
    Nov 28, 2016 at 13:24
  • $\begingroup$ Just to be clear, $\lambda_x y(x) = (y(x) - y(0))/x$? And the "$x$" in $\lambda_x$ does not actually mean that there is an operator $\lambda_x$ for each $x$ (so it should probably be called $\lambda$ or something else instead)? $\endgroup$
    – Najib Idrissi
    Nov 28, 2016 at 13:29
  • $\begingroup$ If $y$ is a function of $x$, then you might try writing $$\lambda_{x} (y) = \frac{y(x) - y(0)}{x}$$ $\endgroup$
    – AJY
    Nov 28, 2016 at 13:34
  • $\begingroup$ @AJY it's only guaranteed to have an associated function when for $x=0$. Otherwise it may not converge. So arguably it doesn't make sense to use function notation. $\endgroup$
    – Matt Samuel
    Nov 28, 2016 at 17:04
  • $\begingroup$ @MattSamuel Should "when for $x = 0$" say "for $x \neq 0$? $\endgroup$
    – AJY
    Nov 28, 2016 at 17:06

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$$f(x)g(x)-f(0)g(0)=(f(x)-f(0))g(x)+f(0)(g(x)-g(0))$$ Thus your left shift operator satisfies the Leibniz formula $$\lambda_x(yz)=(\lambda_xy)z+((x:=0)y)(\lambda_xz)$$

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