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г 2010
72
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˲ ֲί ֲ ̲ Ѳ Ҳ
$f(z)=\sum\limits_{k=0}^{\i}f_kz^k$ $a$ $m$ $(z_n)$ -- $S_n(z)=\sum\limits_{k=0}^{n}f_kz^k$, $a$. $|z_n-a|$ $0$.
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