- #1

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find the parial fraction decomposition for th erational expression:

__-8x^3-2__

x^2(x-1)^3

- Thread starter TonyC
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- #1

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find the parial fraction decomposition for th erational expression:

x^2(x-1)^3

- #2

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[tex] \frac{A}{x} + \frac{Bx+C}{x^2} + \frac{D}{x-1}+\frac{Ex+F}{(x-1)^2} + \frac{Gx^2+Hx+I}{(x-1)^3} [/tex] if I remember correctly.

From there its just like any other PFD problem.

- #3

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Got it...whiz bang!

- #4

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I came up with -6/x -2/x^2 + 6/x-1 + 4/(x-1)^2 + 10/(x-1)^3

Well?

Well?

- #5

TD

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When the denominator is either lineair (ax+b), or lineair to a certain power (ax+b)^n, the the proposed nominator is just a number.

It's only when the denominator is a quadratic with discriminant < 0 that the nominators is proposed to be lineair. Quadratics in the nominator never appear.

So the proposed partial fractions are:

[tex] \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}+\frac{D}{(x-1)^2} + \frac{E}{(x-1)^3} [/tex]

@Tony: I see a lot of right nominators but many signs are wrong.

- #6

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I'm still a bit foggy:

[tex]\frac{A}{x} + \frac{B}{x^2} - \frac{C}{x-1}-\frac{D}{(x-1)^2} - \frac{E}{(x-1)^3} [/tex]

I am coming up with negative numbers each time I run it.

[tex]\frac{A}{x} + \frac{B}{x^2} - \frac{C}{x-1}-\frac{D}{(x-1)^2} - \frac{E}{(x-1)^3} [/tex]

I am coming up with negative numbers each time I run it.

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- #7

TD

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Have you tried with my proposal of partial fractions?

- #8

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[tex] \frac{A}{x} + \frac{B}{x^2} - \frac{C}{x-1}-\frac{D}{(x-1)^2} - \frac{E}{(x-1)^3} [/tex]

- #9

TD

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What do you mean you run? Negatives? You still have the undetermined coeffecients :S

- #10

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When I said negative, I meant to say minus.

- #11

TD

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[tex]\frac{{x^4 \left( {a + c} \right) + x^3 \left( { - 3a + b - 2c + d} \right) + x^2 \left( {3a - 3b + c - d + e} \right) + x\left( {3b - a} \right) - b}}

{{x^2 \left( {x - 1} \right)^3 }} = \frac{{ - 8x^3 - 2}}

{{x^2 \left( {x - 1} \right)^3 }}[/tex]

This can give you a (not too hard) 5x5 system, or you could simplify your calculations a bit by choosing smart x's, i.e. zero's of the denominator.

- #12

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I am still lost!

- #13

TD

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Well?"

- #14

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This is why I asked for assistance. I am truly lost at this point.

- #15

TD

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You can probably use the same method to solve what I wrote.

- #16

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[tex] \frac{6}{x} + \frac{2}{x^2} - \frac{6}{x-1}-\frac{4}{(x-1)^2} - \frac{10}{(x-1)^3} [/tex]

- #17

HallsofIvy

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You want to find A, B, C, D, E so that

[tex]\frac{-8x^3- 2}{x^2(x-1)^3}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}+ \frac{D}{(x-1)^2}+ \frac{E}{(x-1)^3}[/tex]

TD wrote the left side of that as a single fraction but I think it is easiest to multiply both sides of that by x

[tex]-8x^3- 2[/tex]= Ax(x-1)^3+ B(x-1)^3+ Cx^2(x-1)^2+ Dx^2(x-1)+ Ex^2[/itex]

The easy thing to do is take x= 0 so that

[tex]-2= -B[/tex] or B= 2

and then take x= 1 so that

[tex]-10= E[/tex].

Since those are the only two x values that make things on the right side 0, you now have two different ways to find A, C, D.

(a) Multiply the whole thing out and combine like powers of x

(b) Use simple values for x, like -1, 2, -2, to get three equation for A, C, D.

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