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