关于拆项,有理函数积分,留数法
在數學上經常會遇到一類問題
那就是拆項
數列求和中有列項相消:
1n(n+1)=1n?1n+1\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}n(n+1)1?=n1??n+11?
高等數學中有有理函數積分:
∫5x?1x2?x?2dx=∫5x?1(x+1)(x?2)dx\int \frac{5x-1}{x^2-x-2} dx=\int \frac{5x-1}{(x+1)(x-2)}dx∫x2?x?25x?1?dx=∫(x+1)(x?2)5x?1?dx
5x?1(x+1)(x?2)=A(x+1)+B(x?2)\frac{5x-1}{(x+1)(x-2)}=\frac{A}{(x+1)} +\frac{B}{(x-2)}(x+1)(x?2)5x?1?=(x+1)A?+(x?2)B?
A(x?2)+B(x+1)=5x?1?{A+B=5?2A+B=?1A(x-2)+B(x+1)=5x-1 \Rightarrow \left \{\begin{array}{c} A+B=5\\-2A+B=-1 \end{array}\right. A(x?2)+B(x+1)=5x?1?{A+B=5?2A+B=?1?
自動控制中有留數法:
F(s)=A(s)(s+p1)(s+p2)?(s+pn)=α1s+p1+α2s+p2+?+αns+pnF(s)=\frac{A(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}= \frac{\alpha_1}{s+p_1}+\frac{\alpha_2}{s+p_2}+\cdots+\frac{\alpha_n}{s+p_n} F(s)=(s+p1?)(s+p2?)?(s+pn?)A(s)?=s+p1?α1??+s+p2?α2??+?+s+pn?αn??
αk為s=?pk處的留數\alpha_k為s=-p_k處的留數αk?為s=?pk?處的留數
αk求解辦法:等式兩邊×(s+pk),再帶入s=?pk\alpha_k求解辦法:等式兩邊×(s+p_k), 再帶入s=-p_kαk?求解辦法:等式兩邊×(s+pk?),再帶入s=?pk?
右邊=αk右邊=\alpha_k右邊=αk?
左邊=[(s+pk)A(s)(s+p1)(s+p2)?(s+pn)]s=?pk左邊=\left[(s+p_k)\frac{A(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}\right]_{s=-p_k}左邊=[(s+pk?)(s+p1?)(s+p2?)?(s+pn?)A(s)?]s=?pk??
即:αk=[(s+pk)A(s)B(s)]s=?pk\alpha_k=\left[(s+p_k)\frac{A(s)}{B(s)}\right]_{s=-p_k}αk?=[(s+pk?)B(s)A(s)?]s=?pk??
總結
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