Løsninger til forløb om opdatering af vægte i en kunstig neuron
\[ \begin{aligned} s^{(1)} & = 0.1 + 0.1 \cdot -1 + 0.1 \cdot 2 = 0.2 \\ \\ o^{(1)} & = \sigma (0.2) \approx 0.5498339973 \\ \\ e^{(1)} & = (1 - o^{(1)})^2 \approx 0.2026494300 \\ \\ \\ s^{(2)} & = 0.5 \\ \\ o^{(2)} & \approx 0.6224593311 \\ \\ e^{(2)} & \approx 0.3874556189 \\ \\ \\ s^{(3)} & = 0.9 \\ \\ o^{(3)} & \approx 0.7109495025 \\ \\ e^{(3)} & = \approx 0.08355019010 \\ \\ \\ E & = \frac {1}{2}(e^{(1)}+e^{(2)}+e^{(3)}) \approx 0.3368276195 \end{aligned} \]
\[ \begin{aligned} E_0^{(1)} & = (t_1-o^{(1)}) \cdot o^{(1)} \cdot (1-o^{(1)}) \cdot 1 \\ & = (1-0.5498339973) \cdot 0.5498339973 \cdot (1-0.5498339973) \cdot 1 \\ & \approx 0.1114235461 \\ \\ E_0^{(2)} & \approx -0.1462802535 \\ \\ E_0^{(3)} & \approx 0.05939996609 \\ \\ \frac{\partial E}{\partial w_0} & = -(E_0^{(1)}+E_0^{(2)}+E_0^{(3)}) \\ & = - (0.1114235461-0.1462802535+0.05939996609) \\ & \approx -0.02454325869 \\ \\ \\ E_1^{(1)} & \approx -0.1114235461 \\ \\ E_1^{(2)} & = 0 \\ \\ E_1^{(3)} & \approx 0.05939996609 \\ \\ \frac{\partial E}{\partial w_1} & \approx 0.05202358001 \\ \\ \\ E_2^{(1)} & \approx 0.2228470922 \\ \\ E_2^{(2)} & \approx -0.5851210140 \\ \\ E_2^{(3)} & \approx 0.4157997626 \\ \\ \frac{\partial E}{\partial w_2} & \approx -0.0535258408 \end{aligned} \]
\[ \begin{aligned} w_0^{(\textrm{ny})} & \approx 0.1 - 0.1 \cdot (-0.02454325869) \approx 0.1024543259 \\ \\ w_1^{(\textrm{ny})} & \approx 0.09479764200 \\ \\ w_2^{(\textrm{ny})} & \approx 0.1053525841 \end{aligned} \]
\[ \begin{aligned} E & \approx 0.3362680478 \\ \\ w_0^{(\textrm{ny})} & \approx 0.1045005370 \\ \\ w_1^{(\textrm{ny})} & \approx 0.08949754490 \\ \\ w_2^{(\textrm{ny})} & \approx 0.1086460194 \end{aligned} \]