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1) First, why are powers of linear factors repeated?

For example, $(x+1)^2$ gets $\frac{A}{(x+1)}$ and $\frac{B}{(x+1)^2}$

2) Why does a quad factor like $x^2+1$ get a term $\frac{Cx+d}{x^2+1}$

Looking for an example that shows WHY you need a linear term above the quadratic, and WHY you need to repeat the factors.

JackOfAll
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  • Powers of any factor get repeated, not just linear. The numerator of each term will be a polynomial of smaller degree than the denominator, however until further examination we don't know ahead of time what degree it is, so we use a polynomial of degree one less with unknown coefficients which may or may not turn out to be zero after solving. – JMoravitz Jan 04 '16 at 20:42
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    You can theoretically break out $x^2+1=(x+i)(x-i)$ where $i$ is the imaginary number, and then when you add $\frac{a}{x+i}+\frac{b}{x-i}$ you get something in the form $\frac{cx+d}{x^2+1}$. – Thomas Andrews Jan 04 '16 at 20:50
  • I'm not sure that it's illuminating, but consider applying partial fractions to, e.g., $\frac{1}{(x + 1)^2}$ itself: Cross-multiplying shows that $\frac{1}{(x + 1)^2}$ cannot coincide with $\frac{A}{x + 1}$ for any $A$, so using only that term cannot suffice. – Travis Willse Jan 04 '16 at 21:33
  • By the way, $\frac1{x^2}=\lim_{h\to 0}\frac1{x(x+h)}$ by direct substitution. Performing partial fractions, we get:$$\frac1{x^2}=\lim_{h\to0}\left(\frac1h\right) \frac1x+\left(-\frac1h\right)\frac1{x+h}$$ – Akiva Weinberger Jan 07 '16 at 01:21
  • See a similar question here (asked a few days after yours). – Akiva Weinberger Jan 07 '16 at 21:36
  • One shouldn't think of $(x - 1)^2$ as a quadratic, but rather as a power of a linear expression. One needs to write terms like $$\frac{A x + B}{C x^2 + D x + E}$$ when the quadratic in the denominator is irreducible, i.., when $D^2 - 4 C E < 0$. Incidentally, the decomposition of $\frac{1}{(x - 1) (x + 1)^2}$ as presently (hand)written looks good to me. Also, you can carry out this procedure in Maple with convert(1 / ((x - 1) * (x + 1)^2), parfrac);. – Travis Willse Jan 08 '16 at 00:35

4 Answers4

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We have to assume that the numerator is of order at most one less than the order of the denominator, since if the order was higher, long division could be performed to reduce it, leaving a remainder whose order is again at most one less than the denominator.

As for the repeated factors, theoretically we should put $$\frac {Ax+B}{(x-1)^2}$$

However,$$\frac {Ax+B}{(x-1)^2}=\frac{Ax-A}{(x-1)^2}+\frac{-A+B}{(x-1)^2}$$ and hence we write $$\frac{A}{(x-1)}+\frac{B}{(x-1)^2}$$

David Quinn
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  • this process assumes that the original expression being decomposed is proper? – JackOfAll Jan 04 '16 at 21:12
  • Ok, since this process supposes the original expression being decomposed is proper, then the A needs to be one degree less. That way, when it gets multiplied going forwards for the LCD, it will get the other linear factor. If it was already degree = 1, then it would now be degree = 2, which no longer proper. Yea? – JackOfAll Jan 04 '16 at 21:21
  • The point is that we want to make the numerators as simple as possible so that we can do things like integrate, binomially expand etc. more easily. Likewise, rather than write $$\frac{Ax^2+Bx+C}{(x-1)^3}$$ we would write $$\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$$ because we know it can be written this way – David Quinn Jan 04 '16 at 21:37
  • You can check which of your decompositions is correct by putting the fractions back together again and seeing if you get the original expression – David Quinn Jan 06 '16 at 21:39
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This results from the theory of finitely generated modules over PIDs (the ring of polynomials $K[x]$ over a field $K$ is a PID).

The general result is this: any power $P(x)^m$ of an irreducible polynomial $P(x)$ in a rational function has a contribution $$\frac{a_1(x)}{P(x)}+\dots+\frac{a_m(x)}{P(x)^m}$$ in the partial fractions decomposition, where $\deg a_i(x)<\deg P(x)$ for all $i$.

Bernard
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    @JackOfAll Bernard's explanation is actually very efficient, but of course it requires knowing a little ring theory, which I suppose is not common when one encounters the Method of Partial Fractions for the first time. – Travis Willse Jan 04 '16 at 21:30
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    @JackOfAll: I do not give any details on how it is proved. I only mentioned where it stems from, and gave the general form of the decomposition, which answers your questions, I think, and makes you in a position to guess what happens in variant cases. – Bernard Jan 04 '16 at 21:40
  • Actually you managed to find values for $A$ and $B$, specifying values for $x$, but didn't write the linear system equivalent to $A(x+1)^2+B(x-2)=1$. If you write this system, you'll see it is inconsistent. Also, if you replace $x=0$ with $x=-1$, you'll get another value for $B$. – Bernard Jan 06 '16 at 21:48
  • Want to do what? On the other hand, the expression does have something to do with your work as you explained it. I just mistyped $B(x-2)$ instead of $B(x-1)$. – Bernard Jan 09 '16 at 20:06
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Here are some trivial counterexamples that illustrate why we must consider terms of the given form:

  1. Consider the rational expression $$\frac{1}{x^2 (x + 1)} .$$ If we did not require terms with lower powers in the denominator, applying partial fractions to this expression would give entail solving $$\frac{1}{x^2 (x + 1)} = \frac{A}{x^2} + \frac{B}{x + 1}$$ for some constant $A$. Cross-multiplying gives $1 = A(x + 1) + B x^2$, and comparing like coefficients yields a contradiction, and hence there is no such decomposition as written.

  2. Similarly, consider the rational expression $$\frac{x}{x^2 + 1} .$$ If we did not require a linear term in the numerator of expressions with irreducible quadratics in the denominator, applying partial fractions to this expression would entail solving $$\frac{x}{x^2 + 1} = \frac{A}{x^2 + 1} .$$ Again, there is no solution $A$.

Edit In the handwritten computations in the edit to the original question, the computations in the candidate decomposition with only two terms is incorrect: Writing $$\frac{1}{(x - 1) (x + 1)^2} = \frac{A}{x - 1} + \frac{B}{(x + 1)^2}$$ and cross-multiplying gives $$1 = A(x + 1)^2 + B(x - 1) = A x^2 + (2 A + B) x + (A - B) .$$ Then, comparing like coefficients gives the system $$ \left\{\begin{array}{rcl} A &=& 0\\ 2 A + B &=& 0\\ A - B &=& 1 \end{array}\right. $$ but this system has no solution, so yet again there is no such decomposition. (One can verify that something has gone wrong in the handwritten solution by combining the expression into a single ratio of polynomials.)

Travis Willse
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  • They examples are certainly sensical, and it's not a "self-referential moot point": They are minimal counterexamples that illustrates the necessity of precisely the kinds of terms that you question. If you don't find these examples satisfactory, you can replace them with, for example, $\frac{1}{x^2 (x + 1)}$ and $\frac{1}{(x^2 + 1)(x + 1)}$, which lead similarly to contradictions. – Travis Willse Jan 06 '16 at 23:16
  • As for your decomposition of $\frac{1}{x^2}$, if we set $C := A + B$, then your candidate decomposition is $\frac{C}{x}$, which is exactly the one I describe (with the name of the constant changed). – Travis Willse Jan 06 '16 at 23:17
  • Reading the handwritten solution, I see now that by "linear factors repeated" you're questioning including fractions with lower powers of the linear factors, not the higher ones. In that case, one does need an example like $\frac{1}{x^2 (x + 1)}$ to see the necessity for including all appropriate linear powers. – Travis Willse Jan 06 '16 at 23:32
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  1. Powers of linear factors in the denominator of a rational function indicate poles of higher order. Any rewriting of the original rational function as a sum of "simpler" rational functions (which is what partial fraction decomposition really is) would need to have a pole of the same order somewhere on the right hand side.

If you don't know what poles are, consider multiplying both sides with $(x+1)$ (no square). Then the left hand side is still unbounded near $x=-1,$ so the right hand side cannot have a finite limit there.

  1. One way to look at this is to note that $x^2+1$ is even, and there is no hope of rewriting a general function (odd+even parts) as a linear combination of even functions.

(update after clarification of question)

If you "arbitrarily" put a rational function like $\frac1{(x+1)^2(x-1)}$ equal to a linear combination of simple rational functions of your own choice, like in your first example, then you will obtain values for the coefficients but these are "necessary" and not "sufficient". If you work the result from your first example backwards then you obtain the rational function $\frac{x-2}{4(x-1)(x+1)^2}$ which is different from the original.

The general setting for this kind of trick is Bezout's theorem for polynomials: if $P(x)$ and $Q(x)$ have no common factors then there exist polynomials $A(x)$ and $B(x)$ such that $A(x)P(x)+B(x)Q(x)$ is the constant $1.$ Applying this to $P(x)=(x+1)^2$ and $Q(x)=x-1$ we see that we can write $\frac1{(x+1)^2(x-1)}=\frac{AP+BQ}{PQ}$ as a linear combination of $\frac1{(x+1)^2}$ and $\frac1{(x-1)}$ but the coefficients are polynomials the degrees of which can be chosen to be smaller than the degrees of the numerators, but not necessarily smaller than that. So instead of repeating the factor $(x+1)$ in the first degree you could also write the decomposition as

$$\frac1{(x+1)^2(x-1)}=\frac E{x-1}+\frac{Fx+G}{(x+1)^2}$$

As it turns out, the more usual form

$$\frac1{(x+1)^2(x-1)}=\frac A{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$

is "equivalent" to this in terms of getting a determined system of linear equations, but more "useful" when finding primitive functions (indefinite integrals).

Justpassingby
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