By user1091990

2019-03-13 08:21:58 8 Comments

Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$ The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$

And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$

But, if you were to set it up like this: $$\frac{A}{x+1} + \frac{B}{(x^2+3)} + \frac{C}{(x^2+3)^2}$$ You end up with the same answer: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$

I want to know why this is.

I thought that for irreducible quadratic factors, you needed a linear term in the numerator. Is this purely coincidence that it worked with constant terms in the numerator?


@John Coleman 2019-03-13 11:54:42

I agree with the other answers that in some sense it is a coincidence, but I wouldn't say that it is just a coincidence. If this came from a textbook, then the author of the textbook started with the answer, which happens to have constant terms. If the book in question is a calculus textbook and the intention of the problem is to integrate the original function by doing a partial fraction decomposition, then the problem was probably rigged in advance so that neither the algebra nor the subsequent integration are too onerous for a homework problem. These considerations bias the problems, so that solutions with constant numerators appear in textbooks more often than they would otherwise be expected to. If thought of as a problem in reverse-engineering a textbook (or exam) problem, a reasonable heuristic might be to first see if there is a solution with constant terms, and only if and when that heuristic fails, search for a more general solution. A danger of following such a heuristic is that a grader might nevertheless penalize you since most students who use that heuristic don't understand that it is just a heuristic.

@Martin Hansen 2019-03-13 08:37:49

In the first expansion B = D = 0, but with a different but similar question that would not necessarily happen.

I can provide an example easily by starting with values for A = B = C = D = E = 1, working backwards to a new question that will not have the coincidental property that your one has.

Here is the new question:

$$\frac{2x^4+2x^3+11x^2+8x+44}{(x+1)(x^2 +3)^2}$$ The decomposition will, as before, look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$ and with the values substituted in, $$\frac{1}{x+1} + \frac{x+1}{(x^2+3)} + \frac{x+1}{(x^2+3)^2}$$ Robert Z has added a neat answer that goes further and looks at the underlying theory.

Incidentally, the free online Wolfram Alpha is really handy for playing around with these sort of expressions and it'll let you check your answers from partial fraction questions.


@Robert Z 2019-03-13 08:37:42

The five functions $$f_1(x)=\frac{1}{x+1},\; f_2(x)=\frac{x}{x^2+3}, \; f_3(x)=\frac{1}{x^2+3},\; f_4(x)=\frac{x}{(x^2+3)^2}, \; f_5(x)=\frac{1}{(x^2+3)^2}.$$ is a basis which linearly generates all rational functions of the form $$\frac{P(x)}{(x+1)(x^2 +3)^2}$$ where $P$ is ANY polynomial of degree $\leq 4$.

In your case, we simply have that in the vector representation of the given function the coordinates related to $f_2$ and $f_4$ are zero. So, yes, this is a purely coincidence.

@Martin Hansen 2019-03-13 08:50:40

That's a neat and succinct answer, and 'bigger picture' than mine. I like it. It get's my up vote.

@Robert Z 2019-03-13 09:00:04

@MartinHansen Thank you for your nice comment.

@JP McCarthy 2019-03-13 08:30:14

I would consider this a coincidence.

$$\frac{-4x^4-2x^3-26x^2-8x-4\mathbf{5}}{(x+1)(x^2 +3)^2},$$

for example, requires the linear term.

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