复变函数证明,Q是n阶多项式,有不同的n个解 a1,a2,a3.an ,P是小于n阶多项式,证明P(z)/Q(z)=P(a1)/Q'(a1)(z-a1)+P(a2)/Q'(a2)(z-a2)+P(a3)/Q'(a3)(z-a3)+.P(an)/Q'(an)(z-an)
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![复变函数证明,Q是n阶多项式,有不同的n个解 a1,a2,a3.an ,P是小于n阶多项式,证明P(z)/Q(z)=P(a1)/Q'(a1)(z-a1)+P(a2)/Q'(a2)(z-a2)+P(a3)/Q'(a3)(z-a3)+.P(an)/Q'(an)(z-an)](/uploads/image/z/7259200-16-0.jpg?t=%E5%A4%8D%E5%8F%98%E5%87%BD%E6%95%B0%E8%AF%81%E6%98%8E%2CQ%E6%98%AFn%E9%98%B6%E5%A4%9A%E9%A1%B9%E5%BC%8F%2C%E6%9C%89%E4%B8%8D%E5%90%8C%E7%9A%84n%E4%B8%AA%E8%A7%A3+a1%2Ca2%2Ca3.an+%2CP%E6%98%AF%E5%B0%8F%E4%BA%8En%E9%98%B6%E5%A4%9A%E9%A1%B9%E5%BC%8F%2C%E8%AF%81%E6%98%8EP%28z%29%2FQ%EF%BC%88z%29%3DP%28a1%29%2FQ%27%28a1%29%28z-a1%29%2BP%28a2%29%2FQ%27%28a2%29%28z-a2%29%2BP%28a3%29%2FQ%27%28a3%29%28z-a3%29%2B.P%28an%29%2FQ%27%28an%29%28z-an%29)
复变函数证明,Q是n阶多项式,有不同的n个解 a1,a2,a3.an ,P是小于n阶多项式,证明P(z)/Q(z)=P(a1)/Q'(a1)(z-a1)+P(a2)/Q'(a2)(z-a2)+P(a3)/Q'(a3)(z-a3)+.P(an)/Q'(an)(z-an)
复变函数证明,
Q是n阶多项式,有不同的n个解 a1,a2,a3.an ,P是小于n阶多项式,证明P(z)/Q(z)=P(a1)/Q'(a1)(z-a1)+P(a2)/Q'(a2)(z-a2)+P(a3)/Q'(a3)(z-a3)+.P(an)/Q'(an)(z-an)
复变函数证明,Q是n阶多项式,有不同的n个解 a1,a2,a3.an ,P是小于n阶多项式,证明P(z)/Q(z)=P(a1)/Q'(a1)(z-a1)+P(a2)/Q'(a2)(z-a2)+P(a3)/Q'(a3)(z-a3)+.P(an)/Q'(an)(z-an)
显然Q(z)=A(z-a1)(z-a2)...(z-an),A是其最高次项系数.
按照求导的乘法规则,有
Q'(z)=A(z-a2)...(z-an) +
A(z-a1)(z-a3)...(z-an) + ...+
A(z-a1)(z-a2)...(z-a_{n-1})
所以 Q'(a1) = A(a1-a2)(a1-a3)...(a1-an)
Q'(a2) = A(a2-a1)(a2-a3)...(a2-an)
...
对要证明的式子两边乘以Q(z)
得到P(z) = P(a1)A(z-a2)(z-a3)...(z-an)/Q'(a1) + ...
两边代入 a1,左边=P(a1)
右边=P(a1)A(a1-a2)*...*(a1-an)/Q'(a1)=P(a1)
同样可以验证对于n个互不相同的a1,...,an,两边都相等.
而注意两边都是不超过n-1阶的多项式,所以两边恒等.