Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision Last revision Both sides next revision | ||
|
smooth_function_of_a_linear_operator [2015/01/20 13:19] nikolaj |
smooth_function_of_a_linear_operator [2015/02/02 20:18] ben |
||
|---|---|---|---|
| Line 7: | Line 7: | ||
| Let $A(z),B(z)$ be function with expansion around $a,b$, respectively. | Let $A(z),B(z)$ be function with expansion around $a,b$, respectively. | ||
| - | $A(D)\,B(X) | + | $A(D)\,B(X)=\sum_{k=0}^\infty A^{(k)}(a) \dfrac{1}{k!} (D-a)^k \cdot |
| - | =\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 | \sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} (X-b)^j | ||
| $ | $ | ||
| Line 16: | Line 15: | ||
| For $D$ a linear operator with $D^{n+1}X^{n}=0$ in an $\mathbb C$-algebra with unit and infinite sums, we find | 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) | + | $A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} |
| - | =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} | + | |
| \left( | \left( | ||
| \sum_{k=0}^j A^{(k)}(a) \dfrac{1}{k!} (D-a)^k (X-b)^j | \sum_{k=0}^j A^{(k)}(a) \dfrac{1}{k!} (D-a)^k (X-b)^j | ||
| Line 30: | Line 28: | ||
| and find | and find | ||
| - | $A(D)\,B(X) | + | $A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} |
| - | =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} | + | |
| \left( | \left( | ||
| \sum_{k=0}^j A^{(k)}(0) {\large{j \choose k}} Y^k (X-b)^{j-k} | \sum_{k=0}^j A^{(k)}(0) {\large{j \choose k}} Y^k (X-b)^{j-k} | ||
| Line 42: | Line 39: | ||
| and find | and find | ||
| - | $A(D)\,B(X) | + | $A(D)\,B(X) =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} (Y+(X-b))^j=B(Y+X)$. |
| - | =\sum_{j=0}^\infty B^{(j)}(b) \dfrac{1}{j!} (Y+(X-b))^j | + | |
| - | =B(Y+X)$. | + | |
| == Direct prove == | == Direct prove == | ||
| Line 50: | Line 45: | ||
| Note that the last bit can be proven more directly by Taylor expansion of $f(x+h)$ around $x$: | Note that the last bit can be proven more directly by Taylor expansion of $f(x+h)$ around $x$: | ||
| - | $f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$ | + | $f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$. |
| Curiously, note how | Curiously, note how | ||
| - | $\dfrac{\partial}{\partial x}\,f(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}f(x)$ | + | $\dfrac{\partial}{\partial x}\,f(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}f(x)$, |
| - | $f(x)=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}\int_0^x f(y)\,{\mathrm d}y$ | + | $f(x)=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}\int_0^x f(y)\,{\mathrm d}y$. |
| === References === | === References === | ||
