3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.