# Skepticism concerning Heaviside's "Cover-up Method" for $\textbf{partial fraction decomposition}$

I was reading this paper from MIT and it introduces Heaviside’s Cover-up Method for partial fraction decomposition. In that paper in Example $$1$$ it solves a problem using that method and just when explaining why it works (on the same page-1) it says-

Why does the method work? The reason is simple. The “right” way to determine $$A$$ from equation $$(1)$$ would be to multiply both sides by $$(x −1)$$ ; this would give $$\frac{x − 7}{ ~~~~~~~~~(x + 2)} = A + \frac{B}{ x + 2} (x − 1) ~~~~~~~~\qquad(4)$$

Now if we substitute $$x = 1$$, what we get is exactly equation $$(2)$$, since the term on the right disappears.

Which seems absurd to me since multiplying both sides by $$x-1$$ should render that $$x \neq 1$$ otherwise it would mean $$\frac{0}{0}$$ is equal to $$1$$ because we could've written $$A$$ as such $$\frac{A\cdot(x-1)}{x-1}$$ and substituting by $$x = 1$$ would give us $$\frac{A\cdot 0}{0}$$. I looked over other places too where this method is used but those more or less follows the same way.

• @Peter, imagine I didn't cancel $(x-1)$ and it becomes $$\frac{x − 7}{ ~~~~~~~~~(x + 2)} = \frac{A \cdot (x-1)}{x-1} + \frac{B}{ x + 2} (x − 1)$$ Now where does that leaves us is we substitute $x=1$?
– user964242
Oct 10, 2021 at 14:16
• What we actually do is computing $A$ for $x\rightarrow 1$. This limit exists since every $x$ in a small enough neighbourhood of $1$ can be inserted and the value we get for $A$ depends continously on $x$. Oct 10, 2021 at 14:23
• In case you're asking how this avoids division by zero or multiplication of the form 0/0, maybe this would help link. Oct 10, 2021 at 14:27
• What you have is an algebraic identity that you're trying to match coefficients (as opposed to solving an equation). The identity holds on all points, and so you can still multiply by $x-1$ and consider what happens. Oct 10, 2021 at 14:45
• @Retro you have not understood what Calvin correctly said. Oct 10, 2021 at 15:40

I've at last found what I was looking for and contributed the answer.

$$A(x+2)+B(x-1)$$ is a polynomial in $$x$$, as is $$x-7$$. I wanted values of $$A$$ and $$B$$ that make these two polynomials equal for all real numbers except $$1$$ and $$-2$$. But polynomials are continuous and so two polynomials that agree at infinitely many real numbers are necessarily identically equal and will therefore agree at all real numbers. In particular, $$A(x+2)+B(x-1)$$ will hold at $$x=1$$ and $$x=-2$$ if it holds at all other integers, so we can substitute $$x=1$$ and $$x=-2$$ as shortcuts to finding the correct values of $$A$$ and $$B$$.
• In fact, two polynomials (in a single variable $x$, with coefficients in $\mathbb{R}$) of degree $d$ that agree at just $d+1$ points are necessarily identical. Oct 10, 2021 at 19:34