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UnkleRhaukus

  • 3 years ago

Partial-Fraction Decomposition \[\frac1{p^2(p^2-4)}\]

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  1. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\frac1{p^2(p^2-4)}&=\frac1{p^2(p+2)(p-2)}\\ \\&=\frac Ap+\frac B{p^2}+\frac C{p+2}+\frac D{p-2}\\ \\\end{align*}\]

  2. UnkleRhaukus
    • 3 years ago
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    \[1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)\]

  3. Yahoo!
    • 3 years ago
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    Put Values for p and Find A , B , C D

  4. Yahoo!
    • 3 years ago
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    1st put P = 2 and Find D

  5. RolyPoly
    • 3 years ago
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    Covering up! Nice!

  6. UnkleRhaukus
    • 3 years ago
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    \[1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)\] \[\text{for }p=0\\\qquad1=-4B\]\[\qquad B=\frac{-1}4\] \[\text{for }p=2\]\[\qquad1=16D\]\[\qquad D=\frac 1{16}\] \[\text{for }p=-2\]\[\qquad1=-16C\]\[\qquad C=\frac {-1}{16}\] \[\text{for }p=1\]\[\qquad1=3A+3B-C+3D\] \[\qquad1=3A-\frac34+\frac 1{16}+\frac3{16}\]\[\qquad1=3A-\frac34+\frac 4{16}\]\[\qquad\frac32=3A\]\[\qquad A=\frac{1}{2}\] \[\frac1{p^2(p^2-4)}=\frac 1{2p}-\frac 1{4p^2}-\frac 1{16(p+2)}+\frac 1{16(p-2)}\]

  7. UnkleRhaukus
    • 3 years ago
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    i think \(A\) is ment to be zero , , can you see my mistake ?

  8. UnkleRhaukus
    • 3 years ago
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    found it / (them)\[\begin{align*}\text{for }p=1&&1&=-3A-3B-C+3D\\ &&1&=-3A+\frac34+\frac 1{16}+\frac3{16}\\ &&1&=-3A+1\\ \\ &&A&=0\\ \end{align*}\]

  9. UnkleRhaukus
    • 3 years ago
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    \[F(p)=\frac1{p^2(p^2-4)}=-\frac 1{4p^2}-\frac 1{16(p+2)}+\frac 1{16(p-2)}\]

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