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UnkleRhaukus
Group Title
PartialFraction Decomposition
\[\frac1{p^2(p^24)}\]
 one year ago
 one year ago
UnkleRhaukus Group Title
PartialFraction Decomposition \[\frac1{p^2(p^24)}\]
 one year ago
 one year ago

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\begin{align*}\frac1{p^2(p^24)}&=\frac1{p^2(p+2)(p2)}\\ \\&=\frac Ap+\frac B{p^2}+\frac C{p+2}+\frac D{p2}\\ \\\end{align*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[1=Ap(p+2)(p2)+B(p+2)(p2)+Cp^2(p2)+Dp^2(p+2)\]
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Put Values for p and Find A , B , C D
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
1st put P = 2 and Find D
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Covering up! Nice!
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[1=Ap(p+2)(p2)+B(p+2)(p2)+Cp^2(p2)+Dp^2(p+2)\] \[\text{for }p=0\\\qquad1=4B\]\[\qquad B=\frac{1}4\] \[\text{for }p=2\]\[\qquad1=16D\]\[\qquad D=\frac 1{16}\] \[\text{for }p=2\]\[\qquad1=16C\]\[\qquad C=\frac {1}{16}\] \[\text{for }p=1\]\[\qquad1=3A+3BC+3D\] \[\qquad1=3A\frac34+\frac 1{16}+\frac3{16}\]\[\qquad1=3A\frac34+\frac 4{16}\]\[\qquad\frac32=3A\]\[\qquad A=\frac{1}{2}\] \[\frac1{p^2(p^24)}=\frac 1{2p}\frac 1{4p^2}\frac 1{16(p+2)}+\frac 1{16(p2)}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i think \(A\) is ment to be zero , , can you see my mistake ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
found it / (them)\[\begin{align*}\text{for }p=1&&1&=3A3BC+3D\\ &&1&=3A+\frac34+\frac 1{16}+\frac3{16}\\ &&1&=3A+1\\ \\ &&A&=0\\ \end{align*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[F(p)=\frac1{p^2(p^24)}=\frac 1{4p^2}\frac 1{16(p+2)}+\frac 1{16(p2)}\]
 one year ago
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