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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 $\beta=\frac{1}{T}$. |
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+ | >todo: clean this $\beta$ vs. $T$ Thing up. Maybe drop $T$ altogether. | ||
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+ | 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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