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So the way I was taught to decompose into partial fractions was something like this:

Say I begin with a fraction like $\frac{1}{x^2-1}$ that I want to decompose that into two fractions with linear denominators.

I would do $$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1},$$

which then gives $1=A(x+1)+B(x-1)$.

Then to solve quickly, you can set $x=1$, and solve for $A$, then $x=-1$ and solve for $B$.

My question is, why can you let $x=1$ and $x=-1$? Aren't those excluded values? Isn't the equation $$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$$ only valid for all $x$ other than $1$ and $-1$?

19aksh
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1 Answers1

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Let $A,B$ be unknown constants.

Let $S=\{x\in\mathbb{R}\mid x\ne\pm 1\}$.

Then for all $x\in S$, \begin{align*} &\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}\\[4pt] \implies\;&1=A(x+1)+B(x-1)\\[4pt] \implies\;&A(x+1)+B(x-1)-1=0\\[4pt] \end{align*} Now consider the polynomial function $p:\mathbb{R}\to\mathbb{R}$ given by $$p(x)=A(x+1)+B(x-1)-1$$ We've shown that $p(x)=0$ for all $x\in S$.

But then, since $S$ is infinite, and a nonzero polynomial can't have infinitely many zeros, it follows that $p$ is the zero polynomial.

In particular, we must have $p(1)=0$ and $p(-1)=0$, which justifies the substitutions in question.

quasi
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