# Shift operator is useful for what?

1. Apr 16, 2014

### Jhenrique

Definition: $f(x+k) = \exp(k \frac{d}{dx}) f(x)$

So I thought, how take advantage this definition? Maybe it be usefull in integration like is the laplace transform. So I tried to integrate the expression

$\int f(x+k) dx = \int \exp(k \frac{d}{dx}) f(x) dx$ that is an integration by parts, so is necessary to know to integrate and/or differentiate $exp(k \frac{d}{dx})$ and I don't know how do it!

I'm in the correct path?

2. Apr 16, 2014

### pasmith

There is no integration by parts. $\exp\left(k \frac{d}{dx}\right)$ is an operator. It doesn't make sense until you apply it to a smooth function. Formally $$\exp\left(k \frac{d}{dx}\right) = \sum_{n=0}^\infty \frac{k^n}{n!}\frac{d^n}{dx^n}$$ and by convention $d^{0}f/dx^{0} = f$. Hence $$\exp\left(k \frac{d}{dx}\right)f = \sum_{n=0}^\infty \frac{k^n}{n!}\frac{d^nf}{dx^n}$$ which is the Taylor series for $f$ near $x$, and is equal to $f(x + k)$ if $f$ is analytic at $x$ and $k$ is within the radius of convergence of the series.

3. Apr 17, 2014

### Jhenrique

What you give me was an analytical definition. I still don't understand which is the use of shift operator...