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smooth_function_of_a_linear_operator [2015/02/02 20:18] ben |
smooth_function_of_a_linear_operator [2015/03/01 15:55] nikolaj |
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| 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 $x$: | ||
| - | $f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$. | + | $f(x+h)=\sum_{k=0}^\infty\frac{1}{k!}f^{(k)}(x)\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(h\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$. |
| - | Curiously, note how | + | Curiously, note how therefore the approximation to a tangent can be expressed as |
| - | $\dfrac{\partial}{\partial x}\,f(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}f(x)$, | + | $\dfrac{f(x+h)-f(x)}{h}=\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}f(x)$ |
| - | + | ||
| - | $f(x)=\lim_{h\to 0}\dfrac{\exp(h\frac{\partial}{\partial x})-1}{h}\int_0^x f(y)\,{\mathrm d}y$. | + | |
| === References === | === References === | ||
