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laplace_transform [2016/03/08 16:17]
nikolaj
laplace_transform [2016/03/08 16:19]
nikolaj
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 or, in Terms of $T:​=\frac{1}{\beta}$ or, in Terms of $T:​=\frac{1}{\beta}$
  
-$L[f](T) = \beta\sum_{n=0}^\infty f^{(n)}(0) T^n$+$L[f](T) = T\sum_{n=0}^\infty f^{(n)}(0)\, T^n$
  
-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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