Partial Fractions

  • Thread starter Nubcakes
  • Start date
  • #1
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I've been working with Laplace Transforms and integration ALOT lately. Many times I windup having to use partial fractions to solve the problem and frankly my algebra skills just aren't up to the task.

Take this fraction for example;
partfrac01Q.gif


I know 3 ways to do it... 1 of the ways doesn't work unless there is a repeated root. Another uses complex numbers which is absurdly messy. And finally, the last method I know rarely seems to work... or I just don't know what I am doing!

Using the repeated root to have a setup like this is easy;
partfrac01T1.gif


This way is pretty clean cut for most applications, but you need a repeated root for it to work.



I find the usage of complex numbers to work the best;
partfrac01T3.gif


But, my professors HATE me when I do this. It tends to make them dock extra points on petty mistakes, so if at all possible I want to avoiding using this method.



Finally, this is the method I want to learn to use better;
partfrac01T2.gif


I can OCCASIONALLY get this method to work when I set V = to i. I sometimes end up with an equation at the end like this;
# - #i = B(V) + Ci

And that is pretty easy to solve for the variables B and C. As you can see though, it just doesn't work here, or I am worse than I thought at algebra!


Thank you for your time!~
 

Answers and Replies

  • #2
196
1
Using the repeated root to have a setup like this is easy;
partfrac01T1.gif


This way is pretty clean cut for most applications, but you need a repeated root for it to work.

Well (v2+9) is not equal to (v-3)(v-3)
 
  • #3
196
1
Finally, this is the method I want to learn to use better;
partfrac01T2.gif


I can OCCASIONALLY get this method to work when I set V = to i. I sometimes end up with an equation at the end like this;
# - #i = B(V) + Ci

And that is pretty easy to solve for the variables B and C. As you can see though, it just doesn't work here, or I am worse than I thought at algebra!


Thank you for your time!~

In this method use (Bv+C) in the numerator of the second half in RHS.
Then compare the coefficients.
 
Last edited:
  • #4
365
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[tex]\frac{4v+8}{(v-1)(v^2+9)}=\frac{A}{v-1}+\frac{Bv+C}{v^2+9}[/tex]

[tex]4v+8=A(v^2+9)+(Bv+C)(v-1)[/tex]

[tex]0*v^2+4v+8=v^2(A+B)+v(-B+C)+9A-C[/tex]

Now solve for A,B,C and substitute in the first equation.

Regards.
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
5,002
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[tex]\frac{4v+8}{(v-1)(v^2+9)}=\frac{A}{v-1}+\frac{Bv+C}{v^2+9}[/tex]

[tex]4v+8=A(v^2+9)+(Bv+C)(v-1)[/tex]

[tex]0*v^2+4v+8=v^2(A+B)+v(-B+C)+9A-C[/tex]

Now solve for A,B,C and substitute in the first equation.

Regards.

Another way to solve for the coefficients is to pick values of v the make the expression simple and plug them in. In this case. choosing [itex]v=1[/itex] would make the 2nd term on the RHS of [tex]4v+8=A(v^2+9)+(Bv+C)(v-1)[/tex] zero and allow you to easily solve for A. Similarily, choosing [itex]v=3i[/itex] would make the 1st term on the RHS zero, and you could then easily solve for B and C.
 
  • #6
1,851
7
You can actually avoid having to solve any equations at all. Simply expand the function around all poles, keeping only the singular terms, and then add up all the expansions. The fact that the sum of the expansions is equal to the function is then a simple consequence of Liouville's theorem.
 

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