# kongr45gpen/ece-notes

Some corrections

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kongr45gpen committed Oct 2, 2017
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 @@ -5567,10 +5567,7 @@ \subsection{Θεώρημα Cauchy} \end{array} \) \end{enumparen} Τι σχέση έχουν όλα αυτά με \begin{enumparen} \item Μιγαδικές συναρτήσεις; \item Λογισμό II; \end{enumparen} Τι σχέση έχουν όλα αυτά με (1) μιγαδικές συναρτήσεις, (2) λογισμό II; Τι ασκήσεις μπαίνουν στις εξετάσεις; @@ -5624,11 +5621,11 @@ \subsection{Θεώρημα Cauchy} -2u(m\delta,n\delta)-u\left(m\delta,(n-1)\delta\right)}{\delta^2} \end{gather*} \begin{gather*} 0 = u_{xx} + u_{yy} = \frac{ 0 = u_{xx} + u_{yy} = \mathsmaller{\frac{ -4u(m\delta,n\delta) + u\left((m+1)\delta,n\delta\right) +u\left(m\delta,(n+1)\delta\right)+u\left((m-1)\delta,n\delta\right) +u\left(m\delta,(n-1)\delta\right) }{\delta^2} \\ }{\delta^2}} \\ \implies u(m\delta,n\delta) = \frac{ u\left((m+1)\delta,n\delta\right) +u\left((m-1)\delta,n\delta\right) @@ -5762,7 +5759,7 @@ \subsection{Θεώρημα Cauchy} $\left. \begin{array}{l} u_1 = u(a) = ka+\lambda \\ u_1 = u(b) = kb+\lambda u_2 = u(b) = kb+\lambda \end{array} \right| \implies \boxed{u(x,y) = \frac{u_1-u_2}{a-b}x + \underset{u_1}{?} }$ @@ -5796,7 +5793,7 @@ \subsection{Θεώρημα Cauchy} \end{tikzpicture} \begin{gather*} \hat u(,\rho,\theta) = u(\theta) \\ \hat u(\rho,\theta) = u(\theta) \\ D = \left\lbrace (\rho,\theta): \theta_1 < \theta < \theta_2 \right\rbrace \\ D = \left\lbrace z: \theta_1 < \arg(z) < \theta_2 \right\rbrace \\ u(z) = u_1 \text{ όταν } \arg(z) = \theta_1 \\

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