2
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ 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 '16 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 '16 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 '16 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 '16 at 17:04
  • $\begingroup$ @MattSamuel Should "when for $x = 0$" say "for $x \neq 0$? $\endgroup$ – AJY Nov 28 '16 at 17:06
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.