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smooth_function_of_a_linear_operator [2015/01/20 13:18] nikolaj |
smooth_function_of_a_linear_operator [2015/01/20 13:19] nikolaj |
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| =B(Y+X)$. | =B(Y+X)$. | ||
| - | Note that the last bit can be proven more directly. Taylor expansion of $f(x+h)$ around $x$ gives | + | == Direct prove == |
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| + | Note that the last bit can be proven more directly by Taylor expansion of $f(x+h)$ around $x$: | ||
| $f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$ | $f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$ | ||
