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| Both sides previous revision Previous revision | Last revision Both sides next revision | ||
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smooth_function_of_a_linear_operator [2015/01/20 13:19] nikolaj |
smooth_function_of_a_linear_operator [2015/02/02 20:18] ben |
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| 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 == | ||
