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I would appreciate help showing $e^{D}(f(x)) = f(x+1)$

Where $D$ is the linear operator $D: \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ where (in the context where this statement arose) $x \in \mathbb{N}$;

$f(x) \mapsto \frac{d}{dx} f(x)$

By the Taylor series expansion $e^{D} = \sum_{n=0}^{\infty} \frac{D^n}{n!}$

$(1)$ Then $e^{D} (f(x)) = f(x) + f'(x) + \frac{f''(x)}{2!} +\dots$

I would appreciate help showing that the above line $(1)$ is equal to $f(x +1)$

This is what I have tried, but it feels forced:

for a Taylor series representation of $f(x +1)$ near $x$ I could write

$f(x+1) = f(x) + f'(x)(x+1 - x) + \frac{f''(x)}{2!}(x+1 -x)^2 \dots$

This then is equal to the RHS of line $(1)$.

Could this line of thinking be correct? Thanks very much.

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    $e^D$ is a linear operator, so it suffices to see what it does on a basis. A convenient basis for $\mathbb{C}[x]$ is formed by the monomials. – Daniel Fischer Oct 10 '13 at 22:22
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    Checking on a basis may be illuminating for other purposes, but if you want to understand why the formula is true, and are performing computations on a basis, something is wrong, because it is just a restatement of Taylor's formula (which does not refer to a basis) in different notation. All the computations have already been performed in deriving the Taylor formula, it is a matter of repackaging the end result in the language of operators. @DanielFischer – zyx Oct 10 '13 at 22:50

2 Answers2

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Use the binomial theorem. $$ e^D x^n = \sum_{k=0}^\infty \frac{D^k}{k!} x^n. $$ All of the terms after $k=n$ are $0$, i.e. $D^k x^n=0$ if $k>n$, so this is \begin{align} \sum_{k=0}^n \frac{D^k}{k!} x^n & = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} x^{n-k} \\[6pt] & = \sum_{k=0}^n \binom nk x^{n-k} \\[6pt] & = (x+1)^n.\quad\longleftarrow\text{By the binomial theorem.} \end{align}

Michael Hardy
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    Exercise: Show that every shift-equivariant linear operator on polynomials in one variable is a power series in $D$. Hint: Shift equivariance is how the binomial theorem gets involved. And the hint is most of the proof. Corollary: Every shift-equivariant linear operator on polynomials in one variable maps every polynomial to a polynomial of equal or lesser degree. – Michael Hardy Oct 10 '13 at 22:40
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The Taylor series is $$ f(x+h) = \displaystyle \sum_{n \geq 0} \frac{f^{(n)}(x)}{n!}h^n $$ and here $e^A$ is nothing more than a notation for the power series $\sum A^n/n!$. The series is a well-defined operator for any operator $A$ that is "locally nilpotent" (only a finite number of terms are nonzero when applied to any $f$). Using $(hD)^n = h^n D^n$ it is a matter of comparing definitions to see that the operator that takes input any polynomial $g(x)$ and outputs $g(x+h)$ is $e^{hD}$.

That is all for polynomials, in characteristic $0$.

zyx
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