# Partial fractions: how can we tell if the numerator should be constant or linear?

For example, if I have

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

How do I know that this can be rewritten as $$f(s) = \frac{As+B}{s^2}+\frac{C}{s+1}$$ as opposed to $$f(s) = \frac{A}{s^2}+\frac{B}{s}+\frac{C}{s+1}$$

In particular, how can I tell when I need to add a linear function or when I should just keep the constant? Why should I write the first equation to solve and not the second?

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In your case, either one will work, they are exactly the same. –  André Nicolas Dec 13 '12 at 2:48
A relevant reference: en.wikipedia.org/wiki/Partial_fraction#General_result –  Jonas Meyer Dec 13 '12 at 2:51

The two are exactly the same, if you interchange $A$ and $B$. Now let's turn to the general case.

Suppose that in the denominator you have, among other things, $(L(s))^4$, where $L(s)$ is a degree $1$ polynomial, like $5s+3$. Assume also that no other term in the denominator is a constant time $L(s)$. Then you will need $$\frac{A_1}{L(s)}+\frac{A_2}{(L(s))^2}+\frac{A_3}{(L(s))^3}+\frac{A_4}{(L(s))^4}.$$

Over the reals, if at the bottom you have $(Q(s))^k$, where $Q(s)$ is a quadratic with no real roots, you will need terms $\dfrac{A_is+B_i}{(Q(s))^i}$, $i$ going $1$ to $k$.

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Because the method of partial fractions says that the form of the denominator dictates the form of the numerators.

In your case, the denominator has factors of $s^2$ and $s+1$. The $s^2$ factor is referred to as a repeated linear factor (think $s^2=s\cdot s$ and $s$ is of course a linear factor since it is of the form $Cs+D$). On the other hand, $s+1$ is a distinct (nonrepeated) linear factor.

Looking in your text, Wikipedia, etc. will show that the method of partial fractions says that when you have a distinct linear factor, the numerator is just a constant. When you have a repeated linear factor that is repeated $k$ times, then you need to "count up" on powers of that factor until you reach its $k$th power. The numerator of each of those terms needs to be a distinct (that's important) constant.

That's why in your case, $$f(s) = \underbrace{\frac{A}{s^2}+\frac{B}{s}}_{\text{repeated linear factor: counting up" on powers of }s\text{ (read from right to left)}}+\underbrace{\frac{C}{s+1}}_\text{distinct linear factor}$$.

Finally, the method of partial fractions also says that if you have an (distinct) irreducible quadratic factor (i.e., it can't be factored over the reals), then you get a general linear function in the numerator rather than just a constant. For example, if you had $${1\over s^2(s+1)(s^2+1)},$$ then the first two parts work like before, but the new factor in the denominator is an irreducible quadratic, so $${1\over s^2(s+1)(s^2+1)}={A\over s}+{B\over s^2}+{C\over s+1}+{Ds+E\over s^2+1}.$$

Most generally, you can keep playing this game by allowing repeated irreducible quadratic factors, e.g. $(s^2+1)^3$, etc. But I think you have enough to go on...

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