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

$\frac5{(x+1)(x^2-1)}$

*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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    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.2012-11-26

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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

$(x-a)^n\frac{P(x)}{Q(x)}=A_1(x-a)^{n-1}+A_2(x-a)^{n-2}+\ldots+A_n+(x-a)^nh(x),$

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

$P(x)=A_0+A_1(x-a)+\ldots+A_n(x-a)^n.$

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. ..............................

Summarizing:

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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    Don't understand me wrong. $I$ndeterminate coefficients should be mentioned in class. Of course it should. And the 'cover up' short-cut too. But giving them as THE method to get partial fractions is just a malpractice.2013-07-07