I'll explain my question with the following example from wikipedia. Suppose, we have a function: $$ f(x)=\frac{1}{x^2+2x-3} $$ Here, the denominator splits into two distinct linear factors: $$ q(x)=x^2+2x-3 = (x+3)(x-1) $$ so we have the partial fraction decomposition $$ f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1} $$ Multiplying through by $x^2 + 2x − 3$, we have the polynomial identity $$ 1=A(x-1)+B(x+3) $$ Substituting $x = −3$ into this equation gives $A = −1/4$, and substituting $x = 1$ gives $B = 1/4$, so that $$ f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right) $$ My doubt: For $x = -3$ and $x = 1$, our function $f(x)$ is undefined. So, is it valid to substitute these values for $x$? If it is, then why? And if it isn't, then why do these values work?

  • $\begingroup$ Just to make my point more clear, please note that when we multiply through $x^2+2x-3$, we'll be multiplying both sides by 0 if $x=-3$ or $x=1$. I really do not see how substituting these particular values for $x$ could be a valid operation. $\endgroup$ – shobhu May 11 '15 at 4:22
  • $\begingroup$ Don't worry ! Your derivation of the parameters $A$ and $B$ is entirely valid and the result is correct. Even if $x = -3$ or $x = 1$ were noy valid points, then still the LHS and RHS should be equal for $x$ very close to these values. In other words, it is impossible that some other solution exists. And your solution is perfect. $\endgroup$ – M. Wind May 11 '15 at 5:08

There really isn't a problem - this is a convenient trick which works. You can solve the identity as well by equating coefficients, and solving linear equations. It can be treated as a polynomial identity and solved in an entirely symbolic way. If there is a solution, it must be this one. It is when you regard the symbolic expression as a function that you get issues about undefined values.

Here is another way of looking at eliminating $A$, which may help. Set $x=1+t$ and $x=1-t$ to obtain:

$$1=At+B(4+t)$$ $$1=-At+B(4-t)$$

Now add these two equations and divide by $2$ to obtain $$1=4B$$

This is exactly what you would have got by putting $x=1$ in the original equation.

  • $\begingroup$ Cool way of eliminating A. How did you come up with $x=1+t$ and $x=1-t$? $\endgroup$ – shobhu May 11 '15 at 15:15
  • $\begingroup$ @shobhu Well they are symmetric around the "problem" point, so the first order term is going to vanish. The $1$ in $1\pm t$ makes the constant term in $A$ vanish and the symmetry deals with the first order term. This is true even in a more complicated case where quadratic terms appear. Equating coefficients of the equivalent of $t$ will eliminate the equivalent of $A$. This process should remind you of differentiation - derivatives do come into the theory when there are double roots in the denominator. It is usually more practical to spot short cuts than to learn a full theoretical solution. $\endgroup$ – Mark Bennet May 11 '15 at 15:27

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