Smooth function of a linear operator



Let $A(z),B(z)$ be function with expansion around $a,b$, respectively.

$A(D)\,B(X)=\sum_{k=0}^\infty A^{(k)}(a) \dfrac{1}{k!} (D-a)^k \cdot \sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} (X-b)^j $

Truncation at $k=j$

For $D$ a linear operator with $D^{n+1}X^{n}=0$ in an $\mathbb C$-algebra with unit and infinite sums, we find

$A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} \left( \sum_{k=0}^j A^{(k)}(a) \dfrac{1}{k!} (D-a)^k (X-b)^j \right)$.

(here I say $\mathbb C$-algebra and interpret $a$ as $a$ times the unit. I haven't checked out what the minimal requirements really are, yet)

Differential operators

For $a=0$ and $D=Y\frac{\partial}{\partial X}$, we use ${j \choose k}=\frac{1}{k!}\frac{j!}{(j-k)!}$ and find

$A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} \left( \sum_{k=0}^j A^{(k)}(0) {\large{j \choose k}} Y^k (X-b)^{j-k} \right)$.

Shift operator

For $A=\exp$, i.e. $A^{(k)}(0)=1$, we use $(u+v)^j=\sum_{k=0}^j {j \choose k} u^k v^{j-k}$ and find

$A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} (Y+(X-b))^j=B(Y+X)$.

Direct prove

Note that the last bit can be proven more directly by Taylor expansion of $f(x+h)$ around $x$:

$f(x+h)=\sum_{k=0}^\infty\frac{1}{k!}f^{(k)}(x)\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(h\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$.

Curiously, note how therefore the approximation to a tangent can be expressed as

$\dfrac{f(x+h)-f(x)}{h}=\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}f(x)$


Wikipedia: Shift operator

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