Why do we add extra coefficients for repeated terms when we do partial fraction decomposition?

Question

I've seen how to do partial fraction decomposition but couldn't really see or understand why we add more coefficients. For example $$\frac{x^2+3}{(x-1)(x-2)^2} = \frac{A_1}{x-1} + \frac{B_1}{x-2} + \frac{B_2}{(x-2)^2} $$ I really don't see where the extra term for $x-2$ comes from So I would like to see the derivation of the partial fraction decomposition thing.

Answer 1

It's actually the $B_1$ term that is "extra".

Consider what would happen if you only had the $A_1$ and $B_1$ terms: Their sum would have the denominator $(x-1)(x-2)$, which is not what we see on the left-hand side. So to make the left-hand side denominator $(x-1)(x-2)^2$, we need a term on the right-hand side that has $(x-2)^2$ as denominator.

This can be handled two ways: Either have a term $$ \frac{B_1x+B_2}{(x-2)^2} $$ or have two terms $$ \frac{B_1}{x-2} + \frac{B_2}{(x-2)^2} $$ You won't get the exact same $B_i$'s from these two approaches, but they are closely related.

Answer 2

The given rational function has a pole of order $1$ at $x_1=1$ and a pole of order $2$ at $x_2=2$.

A rational function $f$ with a pole of order $r\geq1$ at a point $x_0$ has a Laurent expansion there of the form $$f(x)=\sum_{k=1}^r{B_k\over (x-x_0)^k}+\sum_{j=0}^\infty G_j(x-x_0)^j\ .$$ The $B$-terms are the bad terms, and the $G$-terms are the good terms in this expansion. The partial fraction expansion of $f$ collects all the bad terms $f$ has at all its poles. The function $f$ then is the sum of these bad terms and a rational function $p$ which is good at all points in ${\mathbb C}$. This $p$ has to be a polynomial.

Answer 3

Just check what happens when you exclude any of the three terms -- you get a sum whose denominator is different from the original denominator, and this can never give what we want. This is the way to verify that this wouldn't work otherwise.

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To see why directly, you need to understand what we do when we split a fraction into simpler parts.

Suppose we have the fraction $$\frac{x-a}{(x-b)(x-c)},$$ where $b\ne c,$ then we want to split it into two parts, and from our understanding of the reverse process, we know that one part will have one of the factors in the denominator as its own denominator. Also, since the factor is linear, then its numerator can only be some nonzero constant. So we want to have one part of the form $$\frac{A}{x-b},$$ so that now the remainder is of the form $$\frac{x-a-A(x-c)}{(x-b)(x-c)}.$$ However, we want to get rid of the already used factor in the denominator in the second part, so that things simplify. So we want to choose the constant $A$ so that $x-b$ divides into $x-a-A(x-c)$ without remainder. That is, we want $b-a-A(b-c)$ to vanish. And we can always find such an $A$ since $b\ne c.$

Now consider the contrary case. That is, we want to split $$\frac{x-a}{(x-b)^2}$$ into simpler parts. Then the above method will fail if applied as before since $0A=b-a$ has no solution -- except if $b=a,$ in which case we can choose $A$ as we please. Thus we write the numerator as $x-b+x-a-(x-b),$ (this is the simplest way to proceed -- one may use any constant multiple of $x-b$ as well), so that the fraction now becomes $$\frac{1}{x-b}+\frac{B}{(x-b)^2},$$ where $B=b-a,$ and this couldn't be simpler. So you can see why in such decompositions the repeated factors always have to have constants in the numerator even when the denominator is a polynomial of higher degree, albeit a perfect power -- it is by the nature of the decomposition.

I only used the simplest nontrivial case for simplicity of exposition. Hopefully, you can see how to extend this to arbitrary rational functions.

Answer 4

Because it's so much easier to compute $\displaystyle \int \dfrac{a}{(x-b)^n}$.

Answer 5

The possible non-trivial factors of the denominator are $(x-1)$, $(x-2)$, $(x-1)(x-2)$ and $(x-2)^2$. However any term in $\frac1{(x-1)(x-2)}$ can itself be decomposed using partial fractions so it doesn't appear in the result.