# Partial Fraction

1. Feb 5, 2006

### hanson

Hi all!
how would you do this partial fraction problem?
$$\frac{1}{s(s+1)^3(s+2)}$$

The answer is $$\frac{1}{2s}+\frac{1}{2(s+2)}-\frac{1}{s+1}-\frac{1}{(s+1)^3}$$

I know that it can be done by letting
$$\frac{1}{s(s+1)^3(s+2)} = \frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+1}+\frac{D}{(s+1)^2}+\frac{E}{(s+1)^3}$$ and solve for A,B,C,D and E. I tried and it is very tedious. It is easy to find A, B and E but not for the others.

Can anyone tell me a quicker way to do this? Thanks

2. Feb 5, 2006

### cepheid

Staff Emeritus
Sure! Here's a neat trick. Let's derive a formula for the coefficients A, B, C, ... etc.

$$f(s) = \frac{1}{s(s+1)^3(s+2)} = \frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+1}+\frac{D}{( s+1)^2}+\frac{E}{(s+1)^3}$$

Now, to solve for B, what if we started by multiplying everything by s+2?

$$(s+2)f(s) = \frac{(s+2)}{s(s+1)^3(s+2)} = B + (s+2)\left(\frac{A}{s} +\frac{C}{s+1}+\frac{D}{( s+1)^2}+\frac{E}{(s+1)^3}\right)$$

Now, let's evaluate the expression at s = -2:

$$(s+2)f(-2) = \frac{1}{-2(-2+1)^3} = B + (-2+2)\left(\frac{A}{s} +\frac{C}{s+1}+\frac{D}{( s+1)^2}+\frac{E}{(s+1)^3}\right)$$

$$(s+2)f(-2) = \frac{1}{-2(-1)^3} = B + 0$$

$$(s+2)f(-2) = \frac{1}{2} = B$$

I think you can see the general trend. To calculate the coefficient of the partial fraction expansion term having (s-a) in the denominator, multiply the original expression by (s-a), and then evaluate the whole thing at s = a.

3. Feb 5, 2006

### hanson

Thank Cepheid!
But I still meet a problem when finding, say, C.
Following the trend, I should multipy
$$s+1$$ or
$$(s+1)^3$$ to f(s).

But for the former case, the denominator of $$(s+1)f(s)=\frac{1}{s(s+1)^2(s+2)}$$ tends to infinite when s = -1.

For the latter case, the denominator of $$(s+1)^3f(s)=\frac{1}{s(s+2)}$$ is ok when s = -1 but C is now is stuck with $$C(S+1)^2$$.
I still can't get C and D.

Or I missed something? Please point it out.

Last edited: Feb 5, 2006
4. Feb 5, 2006

### VietDao29

No, I don't think you are missing anything. You can just choose randomly 2 more x's, plus it in the expression, and solve for C, and D using A, B, and E.
Or if you don't want to solve equations, you can rearrange it, and solve for C, and D.
$$\frac{1}{s(s + 1)^3 (s + 2)} = \frac{A}{s} + \frac{B}{s + 2} + \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2} + \frac{E}{(s + 1) ^ 3}$$ (1), and you know that:
A = 1 / 2, B = 1 / 2, E = -1, by plugging it in the expression (1), we have:
$$\frac{1}{s(s + 1)^3 (s + 2)} = \frac{1}{2s} + \frac{1}{2(s + 2)} + \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2} - \frac{1}{(s + 1) ^ 3}$$
Isolate the unknows:
$$\Leftrightarrow \frac{1}{s(s + 1) ^ 3 (s + 2)} - \frac{1}{2s} - \frac{1}{2(s + 2)} + \frac{1}{(s + 1) ^ 3} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 - (s + 1) ^ 3 (s + 2) - s(s + 1) ^ 3 + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 - (s + 1) ^ 3 ((s + 2) + s) + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 - 2 (s + 1) ^ 4 + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 (1 - (s + 1) ^ 4) + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 (1 - (s + 1) ^ 2) (1 + (s + 1) ^ 2) + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{2 (1 - (s + 1)) (1 + (s + 1)) (1 + (s + 1) ^ 2) + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{-2s (s + 2) (1 + (s + 1) ^ 2) + 2s(s + 2)}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow \frac{-2s (s + 2) (s + 1) ^ 2}{2s(s + 1) ^ 3 (s + 2)} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$
$$\Leftrightarrow -\frac{1}{(s + 1) ^ 2} = \frac{C}{s + 1} + \frac{D}{(s + 1) ^ 2}$$. From here, one can say that C = 0, and D = -1.
However, I think it's damn long, and solving equations may be faster.
--------------
Now, why don't you do the first way, by plugging 2 more x's in and solve equations, to see if you can arrive at the same answer?
You can go from here, right? :)

Last edited: Feb 5, 2006