0

Its been years since I solved PF. Now I am having hard time solving this partial fraction

$$ F = \frac{1}{\left( x^2+1\right)^2} $$

I proceeded with(Is this right ?)

$$ \frac{1}{\left( x^2+1\right)^2} = \frac{A}{\left( x+\iota\right)} + \frac{B}{\left( x+\iota\right)^2} + \frac{C}{\left( x-\iota\right)} + \frac{D}{\left( x-\iota\right)^2}$$

I don't get the solution that MATLAB gives:

>> [A,B,C] = residue([1],[1,0,2,0,1])

A =

  -0.0000 - 0.2500i
  -0.2500          
  -0.0000 + 0.2500i
  -0.2500          


B =

  -0.0000 + 1.0000i
  -0.0000 + 1.0000i
  -0.0000 - 1.0000i
  -0.0000 - 1.0000i


C =

 []

Kindly help me solve this PF. Thanks in advance!

moki
  • 3

1 Answers1

0

$$F(x) = \dfrac{1}{\left( x^2+1\right)^2}=\dfrac{1}{\left( x+i\right)^2 \left( x-i\right)^2}=\dfrac{A}{x+i}+\dfrac{B}{\left( x+i\right)^2} +\dfrac{C}{x-i}+\dfrac{D}{\left( x-i\right)^2}= \dfrac{A(x+i)(x-i)^2+B(x-i)^2+C(x-i)(x+i)^2+D(x+i)^2}{\left( x+i\right)^2 \left( x-i\right)^2}$$ Then $$1\equiv {A(x+i)(x-i)^2+B(x-i)^2+C(x-i)(x+i)^2+D(x+i)^2}$$ $$\left.\begin{matrix}{ x^3 \\ x^2\\x=i\\x=-i}\end{matrix}\right| \begin{matrix}{A+C=0 \\ A(-2i+i)+B+C(-i+2i)+D=0\\-4D=1\\-4B=1}\end{matrix}$$ $$\begin{cases} A+C=0 \\ -Ai+B+Ci+D=0\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4} \end{cases}\quad\Rightarrow \quad\begin{cases} A+C=0 \\ (-A+C)i=\dfrac{1}{2}\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4} \end{cases} \quad\Rightarrow \quad\begin{cases} A+C=0 \\ -A+C=-\dfrac{i}{2}\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4} \end{cases}$$ so $$\begin{cases} A=\dfrac{i}{4}, \\ C=-\dfrac{i}{4},\\D=-\dfrac{1}{4},\\B=-\dfrac{1}{4}. \end{cases}$$

M. Strochyk
  • 8,397