# 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.

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}}\{e^{ax}(D+a)^{k}y\}\\&{}=e^{ax}{\frac {d}{dx}}\{(D+a)^{k}y\}+ae^{ax}\{(D+a)^{k}y\}\\&{}=e^{ax}\left\{\left({\frac {d}{dx}}+a\right)(D+a)^{k}y\right\}\\&{}=e^{ax}(D+a)^{k+1}y.\end{aligned}}}

This completes the proof.

The shift theorem applied equally well to inverse operators:

${\displaystyle {\frac {1}{P(D)}}(e^{ax}y)=e^{ax}{\frac {1}{P(D+a)}}y.}$

There is a similar version of the shift theorem for Laplace transforms (${\displaystyle t):

${\displaystyle {\mathcal {L}}(e^{at}f(t))={\mathcal {L}}(f(t-a)).}$