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I know how to do these problems but this one is giving me some trouble: Expand into partial fractions


*Should I break it down into:
1) $\frac A{x+1}+\frac B{x-1}+\frac C{x+1}$ or
2) $\frac A{x+1}+\frac B{(x+1)^2}+\frac C{x-1}$

I tried it both ways and came up with different possibilities for the constraints:
If Broken Down Like #1 from Above $$A+B+C=0,\hspace{10pt} 2B = 0,\hspace{10pt} A+B-C=5$$

If Broken Down Like #2 from Above $$A+C=0,\hspace{10pt} B+2C=0,\hspace{10pt} A+B-C=5$$

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You need to go to page 2 of the link @MhenniBenghorbal provided above, to see how to deal with repeated factors in the denominator. –  Harald Hanche-Olsen Nov 26 '12 at 8:15
I haven't gone through the calculation, but something about your first attempt doesn't look right. The expression in 1) is symmetric wrt $A$ and $C$, so the resulting equations should have the same symmetry. In fact your equations, if set up right, should look like three equations in two unknowns, $A+C$ and $B$, and so overdetermined. Which gives you a strong hint that 1) is not going to work. –  Harald Hanche-Olsen Nov 26 '12 at 8:18
The simplest way is to multiply the equation $\dfrac{5}{(x+1)(x^2-1)} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x-1}$ by the denominator $(x+1)(x^2-1) = (x+1)^2(x-1)$ and look at the coefficients of each power of $x$. –  Robert Israel Nov 26 '12 at 8:31
A cleverer way is (1) multiply by $x-1$ and substitute $x=1$, getting the value of $C$. (2) multiply by $(x+1)^2$ and substitute $x=-1$, getting the value of $B$. (3) after plugging in those values for $B$ and $C$, take the difference between the two sides, simplify as see what $A$ must be. –  Robert Israel Nov 26 '12 at 8:34

1 Answer 1

I will add this answer because I think there is a method that is getting lost in mathematical culture. The method is based on Ruffini-Horner algorithm to evaluate polynomials. This is one of the most efficient methods to compute these coefficients. It also allows for a calculation of any partial fraction decomposition in which the denominator splits completely.

First I will give the explanation, which has some length, but then we can see that the actual computation is very very efficient using Ruffini.

Assume that we have $\frac{P(x)}{Q(x)}$ (reduced fraction) and $(x-a)^n$ is the highest power of $(x-a)$ that divides $Q$. We want to write $$\frac{P(x)}{Q(x)}=\frac{A_1}{(x-a)}+\frac{A_2}{(x-a)^2}+\ldots+\frac{A_n}{(x-a)^n}+h(x),$$

where $h$ doesn't have a pole at $x=a$. Let us multiply by $(x-a)^n$. We get


So, computing the coefficients $A_1,\ldots, A_n$ is the same as computing the first few terms of the Taylor expansion of the Rational function $\frac{(x-a)^nP(x)}{Q(x)}$ at the point $x=a$.


Computing the first few terms the Taylor of a rational function:

If we have a rational function $\frac{P(x)}{Q(x)}$ and the numerator and denominator are written in decreasing powers of $(x-a)$ then we get the first few terms of the Taylor expansion of $P(x)/Q(x)$ (very efficiently) by doing long division.

Therefore we only need to see how to (efficiently) write polynomials in powers of $(x-a)$. ..........................

(re)Writing polynomials in powers of $(x-a)$:

Given a polynomial $P(x):=a_nx^n+\ldots+a_0$, say written in powers of $x$. We would like to compute


Notice that $A_0$ is the result of computing $P(a)$. For this computation, one of the most efficient ways is to use [Ruffini-Horner algorithm][1]. The cool thing is that Ruffini doesn't only give you $P(a)$, but the partial computations you do give you the result of $(P(x)-A_0)/(x-a)$. It is clear that $$(P(x)-A_0)/(x-a)=A_1+A_2(x-a)+\ldots+A_n(x-a)^{n-1}.$$

Therefore, what we need to do is to apply again Ruffini to this new polynomial, who's coefficients we already have from the previous Ruffini. Repeating $n$ times you get all the $A_i$'s. ..............................


One of the most efficient ways to compute the partial fraction decomposition is to first do Ruffini enough times (degree of the polynomial) with the numerator and the denominator of the given fraction. This will give you the coefficients for writing the numerator and denominator as powers of $(x-a)$. Then do $n$ steps, where $n$ is the order of the pole $x=a$, of the long division with these two polynomials to get the coefficients you are looking for. DONE! ..............................

PS: We teach in courses what is more easy for the teachers to teach. This is not always what is more easy for the students to learn, or what is more easy for the students to use. Ruffini-Horner is a simple and powerful algorithm with many applications. These applications are not finding its way into the classroom and it is our fault.

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These calculations aren’t in the first-year calculus classroom because they don’t belong there. The mechanics of the traditional method are far more comprehensible, since they’re either already familiar (combining over the least common denominator) or easily justified (equating coefficients), and efficiency is not the goal. –  Brian M. Scott Jul 6 '13 at 7:31
Sr. You are so wrong. All you said is true, but for the professor. It is so easy to tell the students how indeterminate coefficients work and then leave in their hands two steps full of computations: expanding to equate coefficients, and worst of all solving the system. Look back to your exams and see how many you have failed because they didn't know how to solve the system (which is an entirely different skill). Compare with the little addition and multiplication table needed here (math.stackexchange.com/a/437154/26489). This is the post in more detail. –  ABC Jul 6 '13 at 11:13
No, I am not wrong. Understanding is far more important than mere computational efficiency, and the ideas behind the traditional method are far more accessible. Some of them are also important, like the understanding that two polynomials over $\Bbb R$ are identically equal iff they are in fact the same polynomial. And solving small linear systems is a basic operation; students should come to see it as a routine operation. You want to introduce a mysterious black box of a computation for this one special purpose when important general purpose tools will do the job; that’s horrible pædagogy. –  Brian M. Scott Jul 6 '13 at 19:36
HA!!! Sr. It is precisely understanding what makes Ruffini so much better than indeterminate coefficients. It might be a mysterious black-box for you, if you keep your head square like a box. Ruffini and long division are as clear as dividing numbers in decimal notation. You just need to teach it. It is simple ignorance saying that long division and Ruffini are not very general techniques. I have no more interest in explaining more to you. You are part of the problem in North American classrooms. If you were a student I would try harder. I just hope you are not let to confuse more students. –  ABC Jul 6 '13 at 20:17
You’re missing the point completely. Of course division of polynomials is important and rather easy. What is not so easy is explaining why the R-H method works. You appear to have the same fundamental misconception as many students: that it’s enough to know what calculations to perform and how to perform them, with little or no understanding of why they work. In students it’s understandable, since many have been exposed mainly to poor teaching that focusses on computational proficiency rather than on understanding; in teachers in an academic setting it’s unforgivable. –  Brian M. Scott Jul 6 '13 at 21:15

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