test (Q16): Difference between revisions

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1 January 1970

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defining formula

\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{*}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}

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Revision as of 15:51, 21 February 2025 Admin (talk \| contribs) Bots, Bureaucrats, Interface administrators, Administrators 572 edits ‎Changed claim: defining formula (P29): \begin{align} \frac{\mathrm d}{\mathrm d z} \psi_\mathbf{g}(z) &= 2\mathrm{i} \pi s_\mathbf{g} \psi_\mathbf{g}(z)+ \frac{\mathrm{i} \pi}{\rho_\mathbf{g}}\sum_{\mathbf{h}\in \Lambda^*_m} U_{\mathbf{g-h}}\psi_\mathbf{h}(z)\\ \quad s_g &= -\frac{g\cdot(2k_0+g)}{2n\cdot(k_0+g)}\end{align} ← Older edit		Revision as of 15:51, 21 February 2025 Admin (talk \| contribs) Bots, Bureaucrats, Interface administrators, Administrators 572 edits ‎Changed claim: defining formula (P29): \begin{align} \frac{\mathrm d}{\mathrm d z} \psi_\mathbf{g}(z) &= 2\mathrm{i} \pi s_\mathbf{g} \psi_\mathbf{g}(z)+ \frac{\mathrm{i} \pi}{\rho_\mathbf{g}}\sum_{\mathbf{h}\in \Lambda^*_m} U_{\mathbf{g-h}}\psi_\mathbf{h}(z)\\ \quad s_g &= -\frac{g\cdot(2k_0+g)}{2n\cdot(k_0+g)}\end{align} Newer edit →
Property / defining formula		Property / defining formula
	$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & a m p; = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$ `\begin{align} \frac{\mathrm d}{\mathrm d z} \psi_\mathbf{g}(z) &= 2\mathrm{i} \pi s_\mathbf{g} \psi_\mathbf{g}(z)+ \frac{\mathrm{i} \pi}{\rho_\mathbf{g}}\sum_{\mathbf{h}\in \Lambda^_m} U_{\mathbf{g-h}}\psi_\mathbf{h}(z)\\ \quad s_g &= -\frac{g\cdot(2k_0+g)}{2n\cdot(k_0+g)}\end{align}`		$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$ `\begin{align} \frac{\mathrm d}{\mathrm d z} \psi_\mathbf{g}(z) &= 2\mathrm{i} \pi s_\mathbf{g} \psi_\mathbf{g}(z)+ \frac{\mathrm{i} \pi}{\rho_\mathbf{g}}\sum_{\mathbf{h}\in \Lambda^_m} U_{\mathbf{g-h}}\psi_\mathbf{h}(z)\\ \quad s_g &= -\frac{g\cdot(2k_0+g)}{2n\cdot(k_0+g)}\end{align}`

test (Q16): Difference between revisions

Admin (talk | contribs)

$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{*}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & a m p; = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$

$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{*}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$

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test (Q16): Difference between revisions

Admin (talk | contribs)

ddzψ𝐠(z)=2iπs𝐠ψ𝐠(z)+iπρ𝐠∑𝐡∈Λm*U𝐠−𝐡ψ𝐡(z)sgamp;=−g⋅(2k0+g)2n⋅(k0+g)

ddzψ𝐠(z)=2iπs𝐠ψ𝐠(z)+iπρ𝐠∑𝐡∈Λm*U𝐠−𝐡ψ𝐡(z)sg=−g⋅(2k0+g)2n⋅(k0+g)

Revision as of 15:51, 21 February 2025

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$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{*}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & a m p; = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$

$\begin{matrix} \frac{d}{d z} ψ_{𝐠} (z) & = 2 i π s_{𝐠} ψ_{𝐠} (z) + \frac{i π}{ρ_{𝐠}} \sum_{𝐡 \in Λ_{m}^{*}} U_{𝐠 - 𝐡} ψ_{𝐡} (z) \\ s_{g} & = - \frac{g \cdot (2 k_{0} + g)}{2 n \cdot (k_{0} + g)} \end{matrix}$