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Shift operator is useful for what?

  1. Apr 16, 2014 #1
    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. jcsd
  3. Apr 16, 2014 #2

    pasmith

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    Homework Helper

    There is no integration by parts. [itex]\exp\left(k \frac{d}{dx}\right)[/itex] is an operator. It doesn't make sense until you apply it to a smooth function. Formally [tex]\exp\left(k \frac{d}{dx}\right) = \sum_{n=0}^\infty \frac{k^n}{n!}\frac{d^n}{dx^n}[/tex] and by convention [itex]d^{0}f/dx^{0} = f[/itex]. Hence [tex]
    \exp\left(k \frac{d}{dx}\right)f = \sum_{n=0}^\infty \frac{k^n}{n!}\frac{d^nf}{dx^n}[/tex] which is the Taylor series for [itex]f[/itex] near [itex]x[/itex], and is equal to [itex]f(x + k)[/itex] if [itex]f[/itex] is analytic at [itex]x[/itex] and [itex]k[/itex] is within the radius of convergence of the series.
     
  4. Apr 17, 2014 #3
    What you give me was an analytical definition. I still don't understand which is the use of shift operator...
     
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