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laplace_transform [2016/03/08 16:17]
nikolaj
laplace_transform [2016/03/08 16:22]
nikolaj
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 === Discussion === === Discussion ===
 == Action on monomials == == Action on monomials ==
-With $u=E/​T\implies dE=Tdu$ and the definition of the [[Gamma function]], we find+In this article we use $\beta=\frac{1}{T}$.  
 + 
 +>todo: clean this $\beta$ vs. $T$ Thing up. Maybe drop $T$ altogether. 
 + 
 +With $u=\beta\,E=E/​T\implies dE=Tdu$ and the definition of the [[Gamma function]], we find
  
 $\int_0^\infty \left(\frac{1}{n!}E^n\right){\mathrm e}^{-E/T}dE = T^{n+1} \frac{1}{n!} \int_0^\infty u^{(n+1)-1}{\mathrm e}^{-u}du = T^{n+1},$ $\int_0^\infty \left(\frac{1}{n!}E^n\right){\mathrm e}^{-E/T}dE = T^{n+1} \frac{1}{n!} \int_0^\infty u^{(n+1)-1}{\mathrm e}^{-u}du = T^{n+1},$
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 $L[f](T) = T\sum_{n=0}^\infty f^{(n)}(0)\,​ T^n$ $L[f](T) = T\sum_{n=0}^\infty f^{(n)}(0)\,​ T^n$
  
-Note how this means $f(E)=1$ is connected to $\dfrac{1}{\beta}$ and $f(E)=\exp(E)$ is connected to $\dfrac{1}{\beta}\dfrac{1}{1-\frac{1}{\beta}}=-\dfrac{1}{1-\beta}$.+Note how this means $f(E)=1$ is connected to $\dfrac{1}{\beta}=T$ and $f(E)=\exp(E)$ is connected to $-\dfrac{1}{1-\beta}=\dfrac{1}{\beta}\dfrac{1}{1-\frac{1}{\beta}}=T\dfrac{1}{1-T}$.
  
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