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

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?

• This is actually distantly related to my dissertation (it was a Leibniz formula for operators that are formally similar to the derivative). – Matt Samuel Nov 28 '16 at 13:24
• 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)? – Najib Idrissi Nov 28 '16 at 13:29
• If $y$ is a function of $x$, then you might try writing $$\lambda_{x} (y) = \frac{y(x) - y(0)}{x}$$ – AJY Nov 28 '16 at 13:34
• @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. – Matt Samuel Nov 28 '16 at 17:04
• @MattSamuel Should "when for $x = 0$" say "for $x \neq 0$? – AJY Nov 28 '16 at 17:06

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