In Hearthstone battlegrounds as of patch 18.6.1 the Amalgadon card have the following effect: Battlecry: For each different minion type you have among other minions, Adapt randomly. When playing Amalgadon the number of adapts will be in the range between...

Given the selection of Danish investment funds from Sparindex and Danske Invest, it can be difficult to see how different the funds are. As a measure of similarity, one can calculate the overlap between two funds. The overlap between fund...

In Denmark, all ETFs are taxed yearly by unrealized returns. On the other hand, some Danish investment funds are taxed by realized gain, if they fulfill some criteria of paid out dividends. The dividend paid out will be: earned dividends...

Many brokers (at least in Denmark), have a pricing model of \(q\%\) of the trading amount or minimum \(q\) Euro. Given a fixed amount to invest every month, we might think it is best to accumulate the cash to reach...

Following the source: A. Rizzo, S. Coriani, and K. Ruud, Computational Strategies for Spectroscopy. From Small Molecules to Nano Systems, edited by V. Barone (John Wiley and Sons, 2012) Chap. 2, pp.77–135. It is given in Eq. (2.35) that: \[\varepsilon(\omega)...

In quantum chemstry integrals of the following form are sometimes needed: \[F_x = \left<\phi_i\left|\nabla^x r^{-1}\right|\phi_j\right>\] But these integrals are often not directly available. What instead is available is integrals of the form: \[f_x = \left<\nabla^x\phi_i\left|r^{-1}\right|\phi_j\right>\] \[f_{x,y} = \left<\nabla^x\phi_i\left|r^{-1}\right|\nabla^y\phi_j\right>\] Let us...

Consider the electrostatic potential due to multipoles places at the position of atoms: \[E_{i}=\sum_{j}^{atom}\sum_{n}^{multipole}\frac{\left(-1\right)^{n}}{n!}T_{ij}^{(n)}m_{j}^{(n)}\] If the quantum mechanical electrostatic potential is to be minimized, a cost function can written of the form: \[z=\sum_{i}^{point}\left(V_{i,\mathrm{QM}}-E_{i}\right)^{2}+\sum_{l}^{constraints}\lambda_{l}g_{l}\] Expanding the square: \[z=\sum_{i}^{point}\left(V_{i,\mathrm{QM}}^{2}+E_{i}^{2}-2E_{i}V_{i,\mathrm{QM}}\right)+\sum_{l}^{constraints}\lambda_{l}g_{l}\] It is known...

Consider a function: \[f\left(x_{1}\left(\lambda_{1},\lambda_{2},..,\lambda_{M}\right),x_{2}\left(\lambda_{1},\lambda_{2},..,\lambda_{M}\right),..,x_{N}\left(\lambda_{1},\lambda_{2},..,\lambda_{M}\right)\right)\] The first partial derivative is given as: \[\frac{\partial}{\partial\lambda_{i}}f\left(x_{1},x_{2},..,x_{N}\right)=\frac{\partial f}{\partial x_{1}}\frac{\partial x_{1}}{\partial\lambda_{i}}+\frac{\partial f}{\partial x_{2}}\frac{\partial x_{2}}{\partial\lambda_{i}}+...+\frac{\partial f}{\partial x_{N}}\frac{\partial x_{N}}{\partial\lambda_{i}}\] The above equation can be formulated as: \[\frac{\partial}{\partial\lambda_{i}}f\left(x_{1},x_{2},..,x_{N}\right)=\sum_{k}^{N}\frac{\partial f}{\partial x_{k}}\frac{\partial x_{k}}{\partial\lambda_{i}}\] Now consider the second derivative: \[\frac{\partial^{2}}{\partial\lambda_{j}\partial\lambda_{i}}f\left(x_{1},x_{2},..,x_{N}\right)=\frac{\partial}{\partial\lambda_{j}}\left(\frac{\partial f}{\partial x_{1}}\frac{\partial...