有理分式的拆分

F(x)=\frac{B(x)}{A(x)}=\frac{b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{0}}{a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}}

要将F(x)拆分成多个分式,首先要保证该分式是真分式,即m<n。如果不是真分式,用多项式除法化为真分式。

假设F(x)已经是真分式,对A(x)进行因式分解,会有几种情况:

A(x)=0全是不同的实根

A(x)=\left(x-p_{1}\right)\left(x-p_{2}\right) \cdots\left(x-p_{n}\right)

则:

F(x)=\frac{c_{1}}{x-p_{1}}+\frac{c_{2}}{x-p_{2}}+\cdots+\frac{c_n}{x-p_{n}}

等式两边同乘x-p_{i},再让x \rightarrow p_{i},右边就剩c_{i}了,即:

c_{i}=\lim _{x \rightarrow p_{i}}\left(x-p_{i}\right) F(x)

例如:

\frac{x+3}{(x-2)(x-3)}=\frac{C_1}{(x-2)}+\frac{C_2}{(x-3)}

C_1 = \lim_{x\rightarrow 2}(x-2)\frac{x+3}{(x-2)(x-3)}=-5

C_2 = \lim_{x\rightarrow 3}(x-3)\frac{x+3}{(x-2)(x-3)}=6

A(x)=0有相同的实根

A(x)=\left(x-p_{1}\right)^m\left(x-p_{2}\right) \left(x-p_{3}\right)

则:

F(x)=\frac{c_{m}}{\left(x-p_{1}\right)^{m}}+\frac{c_{m-1}}{\left(x-p_{1}\right)^{m-1}}+\cdots+\frac{c_{1}}{x-p_{1}}+\frac{c_{m+1}}{x-p_2}+\frac{c_{m+2}}{x-p_3}

c_{m}=\lim _{x \rightarrow p_{1}}\left(x-p_{1}\right)^{m} F(x)

c_{m-1}=\frac{1}{1 !} \lim _{x \rightarrow p_{1}} \left[\left(x-p_{1}\right)^{m} F(x)\right]^{\prime}

c_{m-j}=\frac{1}{j !} \lim _{x \rightarrow p_{1}} \left[\left(x-p_{1}\right)^{m} F(x)\right]^{(j)}

例如:

\frac{4 x^{2}-6 x-1}{(x+1)(2 x-1)^{2}}=\frac{c_2}{(2 x-1)^{2}}+\frac{c_1}{2 x-1}+\frac{c_3}{x+1}

c_{2}=\lim _{x \rightarrow \frac{1}{2}}(2 x-1)^{2} F(x) =-2

c_{1}=\lim _{x \rightarrow \frac{1}{2}}[(2 x-1)^{2} F(x)]^{\prime}=0

c_{3}=\lim _{x \rightarrow -1}(x+1) F(x)=1

A(x)=0有虚根

有虚根,就是有二次因式不能分解。对于高等数学而言,不能分解的二次因式就不分解了,对应项的分子用一次因式,然后用待定系数法做,例如:

\frac{x+3}{(x+2)(x^{2}+2x+2)^2}= \frac{A}{x+2}+\frac{Bx+C}{x^{2}+2x+2}+\frac{Dx+E}{(x^{2}+2x+2)^{2}}

对于工科,运用傅立叶变换或拉普拉斯变换时,虚根按无重根处理:

\frac{x+3}{x^{2}+2x+2} = \frac{x+3}{(x+1-\mathrm{i})(x+1+\mathrm{i})}=\frac{c_{1}}{x+1-\mathrm{i}}+\frac{c_{2}}{x+1+\mathrm{i}}

c_{1}=\lim _{x \rightarrow-1+1}(x+1-\mathrm{i}) \frac{x+3}{(x+1-\mathrm{i})(x+1+\mathrm{i})}=\frac{2+\mathrm{i}}{2 \mathrm{i}}

c_{2}=\lim _{x \rightarrow-1-1}(x+1+i) \frac{x+3}{(x+1-i)(x+1+i)}=\frac{2-i}{-2 i}

posted @ 2021-11-20 16:55:40
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