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# 第 5 课 补充讲义 | ||
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## 商多项式 | ||
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商多项式 | ||
目的:不告知 $f(X)$ 是什么的情况下,证明 $y=f(z)$。 | ||
方法: | ||
- $f(z)=y$ 保证了 $f(X)-y$ 在 $X=z$ 处为零,因此 $X-z$ 是它的一个因子。 | ||
- 将 $f(X)-y$ 除以 $X-z$ ,结果是一个次低一阶的多项式 $q(X)$ 。 | ||
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举一个具体例子来帮助理解: | ||
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假设 $f(X) = X^2 + 2X + 1$ ,并且我们知道 $f(1) = 4$ ,也就是 $y = 4$ 。那么: | ||
$f(X) - y = (X^2 + 2X + 1) - 4 = X^2 + 2X - 3$ | ||
现在我们可以验证 $f(X) - y$ 在 $X = 1$ 处是否为零: | ||
$f(1) - y = (1^2 + 2 \cdot 1 - 3) = 0$ | ||
所以 $X = 1$ 是 $f(X) - y$ 的一个根。 | ||
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因此,我们可以将 $f(X) - y$ 分解为: | ||
$f(X) - y = (X - 1)(X + 3)$ | ||
这里的商多项式 $q(X) = X + 3$ 。 | ||
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## 验证着双线性映射的推导 | ||
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验证者 $\mathcal{V}$ 检查 $e\left(C-[y]_{1}, [1]_2 \right) \stackrel{?}{=} e\left(\pi,[\tau]_{2}-[z]_{2}\right)$ . | ||
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$e\left(C-[y]_{1}, [1]_2\right) \stackrel{?}{=} e\left(\pi,[\tau]_{2}-[z]_{2}\right)$ | ||
$e\left(\textcolor{blue}{[f(\tau]_1}-[y]_{1}, [1]_2\right) \stackrel{?}{=} e\left(\textcolor{blue}{[q(\tau)]_1},[\tau]_{2}-[z]_{2}\right)$ | ||
$e\left(\textcolor{blue}{[f(\tau)-y]_{1}}, [1]_2\right) \stackrel{?}{=} e\left([q(\tau)]_1,\textcolor{blue}{[\tau - z]_{2}}\right)$ | ||
$f(\tau)-y \stackrel{?}{=} q(\tau) \cdot (\tau - z)$ | ||
$f(\tau)-y \stackrel{?}{=} \textcolor{blue}{\frac{f(\tau)-y}{\tau-z}} \cdot (\tau - z)$ |