Partial Fraction Decomposition Problem

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  • #1
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[tex]\frac{7}{3s^{2}(3s+1)}[/tex]

Can this be decomposed, and how?
 

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  • #2
symbolipoint
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I'd say, yes; it can be decomposed; without my first relearning the method and trying to decompose to partial fractions. Your denominators might be [tex] \[
3s^2
\]
[/tex] and [tex] \[
3s + 1
\]
[/tex]
 
  • #3
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Yep. I understood that much, but I'm confused about what I can do from there because I end up with:

[tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex]

which would equal out to

[tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex] = [tex]\frac{7}{3s^{2}(3s+1)}[/tex]

then

7 = [tex]A(3s+1)[/tex] + [tex]3Bs^{2}[/tex]

--which is where I'm stuck, because eliminating "s" leaves me with A = 7, which doesn't make sense if I want to separate the fraction.
 
  • #4
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It is important to remember that if one of the factors are squared, then you need to fractions for that expression, i.e. for (7/3)//(s^2(3s+1)), you need to decompose it into a/s AND b/s^2, in addition to c/(3s+1).
 
  • #5
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[tex]\frac{7}{3s^{2}(3s+1)}=\frac{As+B}{3s^{2}}+\frac{C}{3s+1}[/tex]
A=-21
B=7
C=21
 
Last edited:
  • #6
symbolipoint
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You may need to fill in some missing steps, but this seemed to work:
[tex] \[
\begin{array}{l}
\frac{a}{{3s^2 }} + \frac{b}{{3s + 1}} = performSteps = \frac{{(3s + 1)a + 3s^2 b}}{{3s^2 (3s + 1)}} \\
fromOriginalDeno\min ator,\;0s + 7 = 3s^2 b + 3sa + a \\
Answer:\;\;a = 7\quad b = \frac{{ - 7}}{s} \\
\end{array}
\]
[/tex]
 
  • #7
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Hmm, I'll have to try to this in a second.


--wow the method gao xiong did works but I don't recall seeing this in my initial searches through the textbook or online.

Nice method, thanks.



Big-T, I don't understand where the a/s would come from if the highest polynomial is 9s^3 (or the highest term is cubic) when you combine the denominator.
 
Last edited:
  • #8
symbolipoint
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gao_xiong's form makes the best sense, and he obtained constants as answers. He decided that [tex] \[
3s^2
\] [/tex]
is an irreducible quadratic expression and gave a suitable linear binomial expression with it.
 
  • #9
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another problem that's confusing me

I have one more problem.

Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

simplify to something like

[tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]
 
  • #10
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I have one more problem.

Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

simplify to something like

[tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

well first look if [tex]\{x^{2}+13x+38}[/tex] has any roots over reals so the problem could get more simplified, since you could express that to some form of (x+-a)(x+-b) where a,b are the real roots.
then it would be of the form
[tex]\frac{A}{x+-a} + \frac{B}{x+-b}+\frac{C}{x}[/tex]
 
Last edited:
  • #11
HallsofIvy
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[tex]\frac{7}{3s^{2}(3s+1)}[/tex]

Can this be decomposed, and how?

I'd say, yes; it can be decomposed; without my first relearning the method and trying to decompose to partial fractions. Your denominators might be [tex] \[
3s^2
\]
[/tex] and [tex] \[
3s + 1
\]
[/tex]

With that "s2", you are going to need both 1/s and 1/s2.
[tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]
Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.
 
  • #12
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I don't doubt what you're saying, but how do [you] rationalize

a 1/s and 1/s^2 from one 1/s^2. I've never seen that before.
 
Last edited:
  • #13
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With that "s2", you are going to need both 1/s and 1/s2.
[tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]
Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.

You lost the 7 in the original numerator. The correct values are A = -3, B = 1, C = 9.
 

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