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Fractions are normally decomposed by finding the numerator that goes with each individual factor of the denominator of the original fraction. However, when a denominator factor is repeated, the solution algorithm involves finding numerators for ascending powers of that factor. So there appears to be a difference in how we treat repeated and non-repeated factors.

I thought it might be insightful to work out a couple of examples with almost-repeated factors. We can find, either trivially or via partial fraction decomposition, that $$\frac{x + 5}{x^2} = \frac{1}{x} + \frac{5}{x^2}$$ Compare this to the decomposition of a similar fraction $$\frac{x + 5}{(x + \varepsilon)(x)} = \frac{\varepsilon - 5}{\varepsilon(x + \varepsilon)} + \frac{5}{\varepsilon x}$$ where $\varepsilon$ is an arbitrarily small constant. To me, it's not particularly clear that the right-hand side (RHS) of each equation relates to that of the other. We can't set $\varepsilon$ to $0$ in the second RHS, but even using limits to let $\varepsilon$ approach $0$ doesn't yield the first RHS, as I would have guessed. Furthermore, the solution process does not harmonize the degree disparity between the top and bottom RHS; the top has the form $\frac{degree 0}{degree 1} + \frac{degree 0}{degree 2}$, while the bottom has the form $\frac{degree 0}{degree 1} + \frac{degree 0}{degree 1}$.

A less trivial example, comparing $$\frac{x^2 + 1}{(x^2)(x + 3)} = -\frac{1}{9x} + \frac{1}{3x^2} + \frac{10}{9(x + 3)}$$ to $$\frac{x^2 + 1}{(x + \varepsilon)(x)(x + 3)} = \frac{1}{3\varepsilon x} + \frac{\varepsilon^2 + 1}{\varepsilon(\varepsilon - 3)(x + \varepsilon)} + \frac{10}{3(3 - \varepsilon)(x + 3)}$$ reveals a similar absence of clarity. Confusingly, one gets the urge to let some of the $\varepsilon$ approach $0$ while letting others approach $3$, but even arbitrarily allowing this much freedom won't allow us to reach the top RHS in this pair.

Why is this analysis not providing more insight into the repeated factor rule for partial fraction decomposition, and will some other analysis in a similar vein work better?

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  • $\begingroup$ Just because $a = b+c$ and $a = l+r$ does not mean that $l \to b$ and $r \to c$ when some parameter is varied. $\endgroup$
    – Math Lover
    Commented Jul 25, 2018 at 19:38

2 Answers 2

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In your first example, Maclaurin expansion with respect to the variable $\epsilon$ gives (after a short calculation) $$ \frac{\epsilon-5}{\epsilon (x+\epsilon)} = -\frac{5}{\epsilon x} + \left(\frac{1}{x} + \frac{5}{x^2} \right) + O(\epsilon). $$ Cancel the $5/\epsilon x$ terms and then let $\epsilon \to 0$.

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Note that a rational function $R(x) = \frac{f(x)}{q(x)}$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,

$$\frac{x+5}{x(x+e)} = \frac{x}{x(x+e)}+\frac{5}{x(x+e)} = \frac{1}{x+e}+\frac{5}{x(x+e)}. \tag{1}$$ Obviously, $$\lim_{e \to 0} \frac{1}{x+e}+\frac{5}{x(x+e)} = \frac{1}{x}+ \frac{5}{x^2}.\tag{2}$$ But $(1)$ might not be preferred for carrying out integration of $\frac{x+5}{x(x+e)}$.

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