Formula for the application of a linear differential operator to the product of exponential and polynomial...
$begingroup$
In the context of linear differential equations, I've stumbled upon the following identity for an arbitrary pair of polynomials $P$ and $Q$ with real or complex coefficients:
$$
Pleft(frac{d}{dx}right)bigl(e^{xy}Q(x)bigr)
=sum_{n=0}^inftyfrac{P^{(n)}(y)e^{xy}Q^{(n)}(x)}{n!}
= Qleft(frac{d}{dy}right)bigl(e^{xy}P(y)bigr).
$$
This can be more or less easily checked by using Taylor expansions of $Pbigl(frac{d}{dx}bigr)$ at $y$ and of $Qbigl(frac{d}{dy}bigr)$ at $x$:
$$
Pleft(frac{d}{dx}right)
=sum_{n=0}^inftyfrac{P^{(n)}(y)}{n!}left(frac{d}{dx} - yright)^n,
quad
Qleft(frac{d}{dy}right)
=sum_{n=0}^inftyfrac{Q^{(n)}(x)}{n!}left(frac{d}{dy} - xright)^n.
$$
Is there any easy way to "see" that
$Pbigl(frac{d}{dx}bigr)bigl(e^{xy}Q(x)) = Qbigl(frac{d}{dy}bigr)bigl(e^{xy}P(y)bigr)$
without "getting hands dirty"?
Is this identity a part of some general theory?
It makes me think of Fourier analysis, but I do not know much about it.
ordinary-differential-equations polynomials exponential-function differential-operators
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
$endgroup$
add a comment |
$begingroup$
In the context of linear differential equations, I've stumbled upon the following identity for an arbitrary pair of polynomials $P$ and $Q$ with real or complex coefficients:
$$
Pleft(frac{d}{dx}right)bigl(e^{xy}Q(x)bigr)
=sum_{n=0}^inftyfrac{P^{(n)}(y)e^{xy}Q^{(n)}(x)}{n!}
= Qleft(frac{d}{dy}right)bigl(e^{xy}P(y)bigr).
$$
This can be more or less easily checked by using Taylor expansions of $Pbigl(frac{d}{dx}bigr)$ at $y$ and of $Qbigl(frac{d}{dy}bigr)$ at $x$:
$$
Pleft(frac{d}{dx}right)
=sum_{n=0}^inftyfrac{P^{(n)}(y)}{n!}left(frac{d}{dx} - yright)^n,
quad
Qleft(frac{d}{dy}right)
=sum_{n=0}^inftyfrac{Q^{(n)}(x)}{n!}left(frac{d}{dy} - xright)^n.
$$
Is there any easy way to "see" that
$Pbigl(frac{d}{dx}bigr)bigl(e^{xy}Q(x)) = Qbigl(frac{d}{dy}bigr)bigl(e^{xy}P(y)bigr)$
without "getting hands dirty"?
Is this identity a part of some general theory?
It makes me think of Fourier analysis, but I do not know much about it.
ordinary-differential-equations polynomials exponential-function differential-operators
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
$endgroup$
add a comment |
$begingroup$
In the context of linear differential equations, I've stumbled upon the following identity for an arbitrary pair of polynomials $P$ and $Q$ with real or complex coefficients:
$$
Pleft(frac{d}{dx}right)bigl(e^{xy}Q(x)bigr)
=sum_{n=0}^inftyfrac{P^{(n)}(y)e^{xy}Q^{(n)}(x)}{n!}
= Qleft(frac{d}{dy}right)bigl(e^{xy}P(y)bigr).
$$
This can be more or less easily checked by using Taylor expansions of $Pbigl(frac{d}{dx}bigr)$ at $y$ and of $Qbigl(frac{d}{dy}bigr)$ at $x$:
$$
Pleft(frac{d}{dx}right)
=sum_{n=0}^inftyfrac{P^{(n)}(y)}{n!}left(frac{d}{dx} - yright)^n,
quad
Qleft(frac{d}{dy}right)
=sum_{n=0}^inftyfrac{Q^{(n)}(x)}{n!}left(frac{d}{dy} - xright)^n.
$$
Is there any easy way to "see" that
$Pbigl(frac{d}{dx}bigr)bigl(e^{xy}Q(x)) = Qbigl(frac{d}{dy}bigr)bigl(e^{xy}P(y)bigr)$
without "getting hands dirty"?
Is this identity a part of some general theory?
It makes me think of Fourier analysis, but I do not know much about it.
ordinary-differential-equations polynomials exponential-function differential-operators
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
$endgroup$
In the context of linear differential equations, I've stumbled upon the following identity for an arbitrary pair of polynomials $P$ and $Q$ with real or complex coefficients:
$$
Pleft(frac{d}{dx}right)bigl(e^{xy}Q(x)bigr)
=sum_{n=0}^inftyfrac{P^{(n)}(y)e^{xy}Q^{(n)}(x)}{n!}
= Qleft(frac{d}{dy}right)bigl(e^{xy}P(y)bigr).
$$
This can be more or less easily checked by using Taylor expansions of $Pbigl(frac{d}{dx}bigr)$ at $y$ and of $Qbigl(frac{d}{dy}bigr)$ at $x$:
$$
Pleft(frac{d}{dx}right)
=sum_{n=0}^inftyfrac{P^{(n)}(y)}{n!}left(frac{d}{dx} - yright)^n,
quad
Qleft(frac{d}{dy}right)
=sum_{n=0}^inftyfrac{Q^{(n)}(x)}{n!}left(frac{d}{dy} - xright)^n.
$$
Is there any easy way to "see" that
$Pbigl(frac{d}{dx}bigr)bigl(e^{xy}Q(x)) = Qbigl(frac{d}{dy}bigr)bigl(e^{xy}P(y)bigr)$
without "getting hands dirty"?
Is this identity a part of some general theory?
It makes me think of Fourier analysis, but I do not know much about it.
ordinary-differential-equations polynomials exponential-function differential-operators
ordinary-differential-equations polynomials exponential-function differential-operators
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
edited Dec 3 '18 at 12:56
Alexey
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
asked Dec 3 '18 at 7:53
AlexeyAlexey
756623
756623
add a comment |
add a comment |
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