Shift operator is useful for what?

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SUMMARY

The discussion centers on the use of the shift operator, defined as f(x+k) = exp(k (d/dx)) f(x), and its application in integration. The user initially attempts to integrate the expression ∫ f(x+k) dx but is corrected by another participant who clarifies that exp(k (d/dx)) is an operator that must be applied to a smooth function. The formal definition of the operator is provided, indicating that it represents the Taylor series expansion of f around x, equating to f(x + k) if f is analytic. The discussion concludes with a request for practical applications of the shift operator.

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  • Understanding of differential operators, specifically d/dx
  • Familiarity with Taylor series and their convergence
  • Basic knowledge of integration techniques
  • Concept of analytic functions in calculus
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  • Explore the application of the shift operator in solving differential equations
  • Learn about the properties and applications of the Taylor series
  • Investigate the relationship between the shift operator and Laplace transforms
  • Study the implications of analytic functions in complex analysis
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Mathematicians, physicists, and engineers interested in advanced calculus, particularly those working with differential equations and series expansions.

Jhenrique
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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?
 
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Jhenrique said:
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?

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.
 
pasmith said:
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.

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

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