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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.

Note that I read few questions about it on this site eg,. this answer.

Can someone please help me make sense of it? Any help is genuinely appreciated.

Cameron Williams
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  • @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 $x=1$? –  Oct 10 '21 at 14:16
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    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 $x$. – Peter Oct 10 '21 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. – NeuralNetNomad Oct 10 '21 at 14:27
  • @19LadWithMathFire, yeah I found that too. Looking over it. –  Oct 10 '21 at 14:28
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    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. – Calvin Lin Oct 10 '21 at 14:45
  • @CalvinLin, multiplying $x-1$ provides that $x \neq 1$. Otherwise the first equation is undefined when $x=1$ and the second equation is not. –  Oct 10 '21 at 14:47
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    @Retro you have not understood what Calvin correctly said. – ancient mathematician Oct 10 '21 at 15:40
  • @ancientmathematician, would you elaborate what he just said a bit so that it becomes more comprehensible? –  Oct 10 '21 at 16:04
  • Does this answer your question? Is this a valid partial fraction decomposition? – Hans Lundmark Oct 10 '21 at 19:59
  • I would express what he says (I hope not changing it) as: this is an identity of rational functions in $x$. We have that $x-1\ne 1$ since its coefficients $1, -1$ are not $0$. Of course when we apply the various evaluation maps $x\mapsto a$ (for a given $a\in\mathbb{R}$) we have to take care. But you are objecting to multiplying by $x-1$ too soon. – ancient mathematician Oct 11 '21 at 06:32

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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. – Sammy Black Oct 10 '21 at 19:34