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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] (current)
nikolaj
Line 45: 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 $x$:
  
-$f(x+d)=\sum_{k=0}^\infty\frac{1}{k!}(\frac{\partial^k}{\partial x^k}f(x))\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(d\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$.+$f(x+h)=\sum_{k=0}^\infty\frac{1}{k!}f^{(k)}(x)\cdot((x+h)-x)^k=\sum_{k=0}^\infty\frac{1}{k!}\left(h\frac{\partial}{\partial x}\right)^k f(x)=\exp(h\frac{\partial}{\partial x})\,f(x)$.
  
-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 ===
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