Меню Главная Случайная статья Настройки Участник:DataDrafter Материал из https://ru.wikipedia.org ${\displaystyle {\overline {B_{0}}}={\frac {\mu _{0}I}{\pi R}}({\frac {\sqrt {R^{2}-a^{2}}}{R}}+\arcsin {\frac {a}{R}}).}$ ${\displaystyle B={\frac {4\pi IN\mu _{1}\mu _{2}}{c(2d\mu _{1}\mu _{2}+{\frac {L}{2}}-d)(\mu _{1}+\mu _{2}))}}.}$ ${\displaystyle k_{1},k_{2}}$ ${\displaystyle sgn(k_{1})=1,sgn(k_{2})=-1,|k_{1}|<|k_{2}|}$ к примеру k1=1, k2=-20. ${\displaystyle k_{1}}$ ${\displaystyle k_{2}}$ ${\displaystyle H=\prod _{i=1}^{n}(1+H_{i})-1}$ ${\displaystyle H=\prod _{i=1}^{n}(1+H_{i})-1}$ ${\displaystyle H_{i}}$ ${\displaystyle H_{0}=(1.065)^{n}-1}$ ${\displaystyle Expected=Positive/(H+1)}$ ${\displaystyle Expected=(1,04121^{n})\cdot x\cdot k_{1}+(n\cdot 0.0001)\cdot (0.9389^{n})\cdot x\cdot k_{2}}$ ${\displaystyle Positive=(0.9999^{n})\cdot ((1.109^{n})\cdot x)\cdot k_{1}+(1-0.9999^{n})\cdot k_{2}\cdot x}$ ${\displaystyle Positive=(0.99999^{n})\cdot ((1.126^{n})\cdot x)\cdot k_{1}+(n\cdot 0.00001)\cdot k_{2}\cdot x}$ ${\displaystyle Expected=(1.0572^{n})\cdot x\cdot k_{1}+(1-0.99999^{n})\cdot (0.9389^{n})\cdot x\cdot k_{2};}$ ${\displaystyle Positive=(0.9^{n})\cdot ((1.23^{n})\cdot x)\cdot k_{1}+(1-0.9^{n})\cdot k_{2}\cdot x}$ ${\displaystyle Expected=Positive\cdot (0.9389^{n})}$ ${\displaystyle Expected=(0.9^{n})\cdot ((1.23^{n})\cdot x)\cdot k_{1}+(1-0.9^{n})\cdot k_{2}\cdot (0.9389^{n})}$ ${\displaystyle }$ ${\displaystyle E_{b}=(1,04121^{n})\cdot x\cdot k_{1}+(n\cdot 0.0001)\cdot (0.9389^{n})\cdot x\cdot k_{2}}$ ${\displaystyle E_{n}=(1.0572^{n})\cdot x\cdot k_{1}+(1-0.99999^{n})\cdot (0.9389^{n})\cdot x\cdot k_{2};}$ ${\displaystyle E_{m}=(0.9^{n})\cdot ((1.23^{n})\cdot x)\cdot k_{1}+(1-0.9^{n})\cdot k_{2}\cdot (0.9389^{n})}$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$