Shift theorem

Shift theorem In mathematics, the (exponential) shift theorem is a theorem about polynomial differential operators (D-operators) and exponential functions. It permits one to eliminate, in certain cases, the exponential from under the D-operators.

Contents 1 Statement 2 Related 3 Examples 4 Notes 5 References Statement The theorem states that, if P(D) is a polynomial D-operator, then, for any sufficiently differentiable function y, {displaystyle P(D)(e^{ax}y)equiv e^{ax}P(D+a)y.} To prove the result, proceed by induction. Note that only the special case {displaystyle P(D)=D^{n}} needs to be proved, since the general result then follows by linearity of D-operators.

The result is clearly true for n = 1 since {displaystyle D(e^{ax}y)=e^{ax}(D+a)y.} Now suppose the result true for n = k, that is, {displaystyle D^{k}(e^{ax}y)=e^{ax}(D+a)^{k}y.} Then, {displaystyle {begin{aligned}D^{k+1}(e^{ax}y)&equiv {frac {d}{dx}}left{e^{ax}left(D+aright)^{k}yright}\&{}=e^{ax}{frac {d}{dx}}left{left(D+aright)^{k}yright}+ae^{ax}left{left(D+aright)^{k}yright}\&{}=e^{ax}left{left({frac {d}{dx}}+aright)left(D+aright)^{k}yright}\&{}=e^{ax}(D+a)^{k+1}y.end{aligned}}} This completes the proof.

The shift theorem can be applied equally well to inverse operators: {displaystyle {frac {1}{P(D)}}(e^{ax}y)=e^{ax}{frac {1}{P(D+a)}}y.} Related There is a similar version of the shift theorem for Laplace transforms ( {displaystyle t

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