test (Q16): Difference between revisions
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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} |
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} |
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Property / defining formula | Property / defining formula | ||
\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{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} |
Latest revision as of 15:54, 21 February 2025
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English | test |
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1 January 1970
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