easier way to decompose fraction into partial fraction

Question

It is a question in a test, and I couldn't manage to complete it. Given a complex fraction $\frac{1}{(z-1)^3(z+1)^3}$, we know that it can be decompose into $\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}$, and then multiplying both sides by $(z-1)^3(z+1)^3$ and then substitute some values to find those $A $, $B$ and $C$.

I have two questions.

$1.$ Must those $A $, $B$ and $C$ be constants? Or some of them can be first order polynomial?

$2.$ How to decompose it faster? It is nearly impossible to substitute values six times and solve the linear equations. One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Answer 1

Sorry. The correct form of above formula is $$A_k=\frac{1}{k!}\Big\{(z-\alpha)^nf(z)\Big\}^{(k)}_{z=\alpha}$$ that $n$ is greatest power of $z-\alpha$ in function $f(z)$ and $(k)$ is derivative order calculate in $z=\alpha$. For example if $\displaystyle f(z)=\frac{1}{(z-1)^3(z+1)^3}$ then $\displaystyle A_2$ in $\displaystyle \frac{A_2}{(z-1)^2}$ will be $$A_2=\frac{1}{2!}\Big\{(z-1)^3f(z)\Big\}^{(2)}_{z=1}=\frac12\Big\{\frac{6}{(z+1)^5}\Big\}_{z=1}=\frac{3}{16}$$

Answer 2

In this particular problem you can simplify the system by letting $u=z-1$, so that your decomposition becomes

$$\frac{A_1}{u^3}+\frac{B_1}{u^2}+\frac{C_1}u+\frac{A_2}{(u+2)^3}+\frac{B_2}{(u+2)^2}+\frac{C_2}{(u+2)}\;,$$

leading to the equation

$$A_1(u+2)^3+B_1u(u+2)^3+C_1u^2(u+2)^3+A_2u^3+B_2u^3(u+2)+C_2u^3(u+2)^2=1\;.$$

Now the terms can be multiplied out mentally:

$$\begin{array}{ccc} &&&&A_1u^3&+&6A_1u^2&+&12A_1u&+&8A_1\\ &&B_1u^4&+&6B_1u^3&+&12B_1u^2&+&8B_1u\\ C_1u^5&+&6C_1u^4&+&12C_1u^3&+&8C_1u^2\\ &&&&A_2u^3\\ &&B_2u^4&+&2B_2u^3\\ C_2u^5&+&4C_2u^4&+&4C_2u^3 \end{array}$$

The system of equations is then

$$\left\{\begin{align*} C_1+C_2&=0\\ B_1+6C_1+B_2+4C_2&=0\\ A_1+6B_1+12C_1+A_2+2B_2+4C_2&=0\\ 6A_1+12B_1+8C_1&=0\\ 12A_1+8B_1&=0\\ 8A_1&=1\;. \end{align*}\right.$$

This is quickly solved from the bottom up for $A_1,B_1$, and $C_1$:

$$\begin{align*} &A_1=\frac18\;;\\\\ &2B_1=-3A_1=-\frac38,B_1=-\frac3{16}\;;\\\\ &4C_1=-6B_1-3A_1=\frac98-\frac38=\frac34,C_1=\frac3{16}\;. \end{align*}$$

The remaining three are quickly found from the top down:

$$\begin{align*} &C_2=-C_1=-\frac3{16}\;;\\\\ &B_2=-B_1-6C_1-4C_2=\frac3{16}-2C_1=\frac3{16}-\frac6{16}=-\frac3{16}\;;\\\\ &A_2=-A_1-(6B_1+12C_1+2B_2+4C_2)=-\frac18\;, \end{align*}$$

since the quantity in parentheses on the last line is easily seen to be $0$. In short, $A_1=\frac18$, $A_2=-\frac18$, $B_1=B_2=C_2=-\frac3{16}$, and $C_1=\frac3{16}$. There’s still a fair bit of computation, but it’s a lot less than with the original variable, and it’s simple enough that I can do it about as fast as I can write.

Answer 3

1. Must those A, B and C be constants? Or some of them can be first order polynomial?

$$\frac{1}{(z-1)^3(z+1)^3}=\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}.$$ $$\tag{1}$$

Since the denominator of the left hand side of $(1)$ has real roots only, all $A_i,B_i$, $i=1,2$, on the right hand side are real constants.

If the denominator has complex roots as e.g. in $(2z+1)/((z-1)(z^2+z+1))$, then the expansion into partial fractions would be of the form

$$\frac{2z+1}{(z-1)(z^2+z+1)}=\frac{A}{z-1}+\frac{Bz+C}{z^2+z+1},\tag{A}$$

where $A,B,C$ are real constants, that is, the factor $(z^2+z+1)$ of the denominator generates the partial fraction $\frac{Bz+C}{z^2+z+1}$, whose numerator is a first degree polynomial.

2. How to decompose it faster? (...) One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Yes, it is, because each differentiation decreases the degree of the current equation by one.

Another special method, addapted from the Residue method, consists of multiplying the given equation by a factor of the denominator and applying adequate limits, which we illustrate as follows, without reducing the speed of the computation very much though.

1. As an illustration, we can start by multiplying $(1)$ by the factor $(z-1)^3$ and let $z$ tend to the root $1$ of this factor: \begin{eqnarray*} \lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}} &=&\lim_{z\rightarrow 1}A_{1}+\left( z-1\right) B_{1}+\left( z-1\right) ^{2}C_{1} \\ &&+\lim_{z\rightarrow 1}\frac{\left( z-1\right) ^{3}A_{2}}{\left( z+1\right) ^{3}}+\frac{\left( z-1\right) ^{3}B_{2}}{\left( z+1\right) ^{2}}+\frac{ \left( z-1\right) ^{3}C_{2}}{z+1}. \end{eqnarray*} On the left hand side $\lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}}=\frac{1}{8}$ . Since the right hand side tends to $A_{1}$, we find $A_{1}=1/8$.
2. Similarly multiplying by $(z+1)^{3}$ and letting $z$ tend to the root $-1$, we get on the left hand side $\lim_{z\rightarrow -1}\frac{1}{(z-1)^{3}}=- \frac{1}{8}$ and $A_{2}$ on the right hand side. So $A_{2}=-\frac{1}{8}$. Multiplying both side by $z$ and letting $z$ tend to $\infty $, yields \begin{equation*} 0=C_{1}+C_{2}. \end{equation*}
3. To reduce the degree of the denominator we can move $\frac{A_{1}}{\left( z-1\right) ^{3}}+\frac{A_{2}}{\left( z+1\right) ^{3}}$ to the left hand side (described here) \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}-\frac{1}{8\left( z-1\right) ^{3}}+ \frac{1}{8\left( z+1\right) ^{3}}=\frac{8-(z+1)^{3}+\left( z-1\right) ^{3}}{ 8\left( z-1\right) ^{3}(z+1)^{3}}=\frac{-6\left( z-1\right) \left( z+1\right) }{8\left( z-1\right) ^{3}(z+1)^{3}}=-\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}. \end{equation*} Consequently \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=\frac{B_{1}}{\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}+\frac{B_{2}}{\left( z+1\right) ^{2}}- \frac{C_{1}}{z+1}.\tag{2} \end{equation*}
4. We can proceed likewise to above, multiplying by $\left( z+1\right) ^{2}$ and let $z\rightarrow -1$ \begin{eqnarray*} \lim_{z\rightarrow -1}-\frac{3}{4\left( z-1\right) ^{2}} &=&\lim_{z \rightarrow -1}\frac{B_{1}\left( z+1\right) ^{2}}{\left( z-1\right) ^{2}}+ \frac{C_{1}\left( z+1\right) ^{2}}{z-1}+B_{2}-\frac{C_{1}\left( z+1\right) ^{2}}{z+1} \\ -\frac{3}{16} &=&0+B_{2}. \end{eqnarray*}
5. Similarly, if we multiply by $\left( z-1\right) ^{2}$ and let $z\rightarrow 1 $, we find $B_{1}=-\frac{3}{16}=B_{2}$ \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=-\frac{3}{16\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{ C_{1}}{z+1}. \end{equation*}
6. Setting e.g. $z=0$ on both sides, and solving for $C_{1}$ we finally get $C_{1}= \frac{3}{16}=-C_{2}$. Thus \begin{align} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}&=\frac{1}{8\left( z-1\right) ^{3}}- \frac{3}{16\left( z-1\right) ^{2}}+\frac{3}{16\left( z-1\right) }\\&-\frac{1}{ 8\left( z+1\right) ^{3}}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{3}{ 16\left( z+1\right) }.\tag{3} \end{align}