User:Skeptical scientist/scratchwork

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It's also incorrect. \sum_{n=0}^\infty a(r)^n is indeed a geometric series, but the sum of this geometric series is not \frac{a}{1-r} by definition, but because of the definition of the limit of a series, and because the partial sums are given by \frac{a(1-r^{n+1})}{1-r}. There is a reason for the middle equality, and it's not the definition of a geometric series, as the poster said, but rather a fundamental property of limits.


2α α


N \int (e^{-(k-k0)^2/(2\Gamma^2) + i (k-k0)x} dk) = e^{-x^2\Gamma^2/2}

{1\over 2} < \left\lfloor \mathrm{mod}\left( f(n) 2^{-100 \lfloor x \rfloor - \mathrm{mod}(\lfloor y \rfloor, 100)},2\right)\right\rfloor


\hat H \psi\left(\mathbf{r}, t\right) = \frac{ih}{2 \pi} \frac{\partial \psi}{\partial t} \left(\mathbf{r}, t\right)