0
$\begingroup$

Regarding partial fractions, where rational functions, whose denominator can be factorised into degree 2 polynomials with complex roots such as $\frac{2}{(x+1)(x^2+4)}$, can you evaluate the numerators in the decomposed fraction for the quadratic factor as follows?

$$\frac{2}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{B}{x+2i}+\frac{C}{x-2i}$$

$$A=\lim_{x \to -1} \frac{2}{x^2+4}$$ $$B=\lim_{x \to 2i} \frac{2}{(x+1)(x-2i)}$$ $$C=\lim_{x \to -2i} \frac{2}{(x+1)(x+2i)}$$

$\endgroup$
  • 2
    $\begingroup$ You can, but its better to limit to $$\dfrac A{x+1}+\dfrac{Bx+C}{x^2+4}$$ $\endgroup$ – lab bhattacharjee Jun 5 '17 at 7:23
0
$\begingroup$

I suppose a few typo's in your post.

Starting with $$\frac{2}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{B}{x+2i}+\frac{C}{x-2i}$$ just reduce to same denominator to write $$2=A(x+2i)(x-2i)+B(x+1)(x-2i)+C(x+1)(x+2i)$$ Since this is valid for all $x$, just set successively, as you did, $x=-1$, $x=2i$, $x=-2i$. To me, this looks simpler than using limits but the general underlying idea is good.

$\endgroup$

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.