Stupid question about partial fraction decomposition:

mAdD INDIAN

Diamond Member
Oct 11, 1999
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I forgot my calc book in my locker and all my friends are at school (I'm sick today so I couldn't make it).

Anyway I'm trying to figure out how to decompose: 2/[x^3 * (x^2+9)]

Thanks. I forgot what the rules where, and when to use Ax+B and such.

Thanks again.
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
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That equation would decompose in to this:

A/x + B/x^2 + C/x^3 + (D*x+E)/(x^2+9)

You should be able to solve the rest from that.
 

mAdD INDIAN

Diamond Member
Oct 11, 1999
7,804
1
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Originally posted by: Kyteland
That equation would decompose in to this:

A/x + B/x^2 + C/x^3 + (D*x+E)/(x^2+9)

You should be able to solve the rest from that.
Thanks. I can solve it from here.

But my question is, why does it decompose to that?

And I already read that link.
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
In partial fraction decomposition, the denominators need to be factors of the overall denominator. So let's take your example of:

x^3*(x^2+9) as the denominator

You can technically decompose it into ones with X^3 and X^2+9 as denonators, but you can further decompose the X^3 into terms of X and X^2. So to get all the possible combinations you need denominators of.

X, X^2, X^3, X^2+9

As for the numerator, it could be any polynomial with a smaller order.
 

mAdD INDIAN

Diamond Member
Oct 11, 1999
7,804
1
0
Originally posted by: TuxDave
In partial fraction decomposition, the denominators need to be factors of the overall denominator. So let's take your example of:

x^3*(x^2+9) as the denominator

You can technically decompose it into ones with X^3 and X^2+9 as denonators, but you can further decompose the X^3 into terms of X and X^2. So to get all the possible combinations you need denominators of.

X, X^2, X^3, X^2+9

As for the numerator, it could be any polynomial with a smaller order.
Ok makes sense. However, if in the numerator polynomial has to be off smaller order, then shouldn't the numerator of x^3 be Cx^2?

I mean to say, why does only the numerator of x^2+9 have to be Dx + E.
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
Originally posted by: mAdD INDIAN
Originally posted by: TuxDave
In partial fraction decomposition, the denominators need to be factors of the overall denominator. So let's take your example of:

x^3*(x^2+9) as the denominator

You can technically decompose it into ones with X^3 and X^2+9 as denonators, but you can further decompose the X^3 into terms of X and X^2. So to get all the possible combinations you need denominators of.

X, X^2, X^3, X^2+9

As for the numerator, it could be any polynomial with a smaller order.
eh... get back to you on it...
I mean to say, why does only the numerator of x^2+9 have to be Dx + E.
lol.. my brain is retarded. Ok, so for x^2+9, it has to be dx+e because that is a polynomial of a lesser order. But x,x^2,and x^3 are all derived from the same factor, x^3. x^3 used to be ax^2+bx+c/x^3 if we didn't break it down. But I guess it's hard to show but you can break ax^2+bx+c/x^3 = d/x+e/x^2+f/x^3... ok, I suck, who's next?

 

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