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→Анализ сепараторов
<tex>x \le \bigg(\dfrac{e^4(f+2)^2n}{4j}\bigg)^{2//\varepsilon_Ff}\bigg(\dfrac{2ej}{fn}\bigg)^{(f-2)/f} </tex>
<tex>t = /\varepsilon_F(f-2)/2 </tex> <tex> x^{f/(f-2)} \le 0.24\Bigg(\dfrac{e^5(f+2)^2}{0.48f}\bigg(\dfrac{ej}{0.12fn}\bigg)^{t-1}\Bigg)^{1/t} \\ \le 0.24\Bigg(3e^5f\bigg(\dfrac{e\delta_F}{0.12}\bigg)^{t-1}\Bigg)^{1/t} </tex> <tex> t-1 \ge \dfrac{\ln(3e^5f)}{\ln(0.12/e\delta_F)} </tex> <tex> x^{f/(f-2)} \le 0.24 </tex> <tex>x <0.32 </tex> <tex> 1.025 \sum\limits_{i\ge1} 0.32^i </tex>
== Доказательство ==
