Anyway, there were a number of interesting little problems that presented themselves near the end of the semester. One fellow student asked if we could consider partial fraction decomposition as a linear operation. The question was never adequately answered in class (merely due to a misunderstanding, $M^2$ still rocks!), but I think I've worked out one decent version of it.

First, to review, partial fraction decomposition is the process of turning a rational fraction into a sum of simpler, partial, rational fractions. That is, moving from

$\begin{eqnarray} \frac{4x+6}{x^3+6x^2+11x+6} \end{eqnarray}$

to

$\begin{eqnarray} \frac{1}{x+1}+\frac{2}{x+2}-\frac{3}{x+3} \end{eqnarray}$

In this example, we are using a relatively straightforward, easily factorable, denominator with a degree greater than that of the numerator. Honestly, I haven't, at this point, worked out the details for some of the other types of partial fraction decomposition.

The above decomposition is typically worked out by first factoring the denominator into distinct linear factors, setting up a sum of those factors with variables in the numerators, and multiplying by a common denominator. The result is an equation with the original numerator on the left and polynomial expressions on the right.

$\begin{eqnarray} \frac{4x+6}{x^3+6x^2+11x+6} & = & \frac{4x+6}{(x+1)(x+2)(x+3)}\\ & = & \frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\\ 4x+6 & = & A(x+2)(x+3)+B(x+1)(x+3)\\ & & +C(x+1)(x+2)\end{eqnarray}$

At this stage, we expand all the expressions on the right, regroup like terms and extract a set of equations representing the coefficients of like terms as follows

$\begin{eqnarray} 4x+6 & = & A(x^2+5x+6)+B(x^2+4x+3)+C(x^2+3x+2)\\ 4x+6 & = & (A+B+C)x^2+(5A+4B+3C)x+(6A+3B+2C)\\\end{eqnarray}$

$\begin{eqnarray} A+B+C & = & 0\\ 5A+4B+3C & = & 4\\ 6A+3B+2C & = & 6\end{eqnarray}$

Now we just solve the system of equations using any method we like to find the three values to use as numerators in the decomposition. It is easy to see a linear solution. We have a system of three equations with three unknowns. We could easily solve this using an augmented coefficient matrix and row reduction.

But there is a better way. Back up to this step.

$\begin{eqnarray} 4x+6=A(x^2+5x+6)+B(x^2+4x+3)+C(x^2+3x+2)\end{eqnarray}$

If you think about this for a minute, this looks rather like a representation of one expression, the polynomial on the left, in terms of a different set of polynomials on the right. And in fact, it is. Also note that the polynomials on the right are the expansions of each multiplicative combination (less one) of the linear factors of our original denominator.

We can look at this as a coordinate change operation. The expression on the right is a coordinate vector of the expression on the left in some basis other than the standard basis for $P$ (polynomials). If you don't see it yet, you will, so follow along.

One way to represent polynomials is as a vector whose elements represent the various terms in a standard polynomial. This is a basic linear transformation often used in linear algebra texts. It looks like this.

$\begin{eqnarray} & T : P^2\rightarrow\mathcal{R}^3\\ & T(f(x)) = \left[ \begin{array}{c} a_1\\ a_2\\ a_3 \end{array}\right]\end{eqnarray}$

where

$\begin{eqnarray} f(x) = a_1 + a_2x + a_3x^2\end{eqnarray}$

Using this common transformation, we can rewrite each polynomial in our equation above as a vector in $\mathcal{R}^3$.

$\begin{eqnarray} 4x+6 & = & \left[ \begin{array}{c} 6\\4\\0 \end{array} \right] \\x^2+5x+6 & = & \left[ \begin{array}{c} 6\\5\\1 \end{array} \right] \\x^2+4x+3 & = & \left[ \begin{array}{c} 3\\4\\1 \end{array} \right] \\x^2+3x+2 & = & \left[ \begin{array}{c} 2\\3\\1 \end{array} \right]\end{eqnarray}$

And now it is quite clear that this is a basis change, expressing a vector in terms of a different basis.

$\begin{eqnarray} \left[ \begin{array}{c} 6\\4\\0 \end{array} \right] = A\left[ \begin{array}{c} 6\\5\\1 \end{array} \right] + B\left[ \begin{array}{c} 3\\4\\1 \end{array} \right] + C\left[ \begin{array}{c} 2\\3\\1 \end{array} \right]\end{eqnarray}$

We can develop a coordinate change matrix $[T]_{\mathcal{B}}$ as the inverse of the concatenation of the vectors on the righthand side. In coordinate changes, we use the equation $[T]_{\mathcal{B}}\vec{v_{\mathcal{B}}}=\vec{v}$. With this we can easily solve this partial fraction decomposition thus

$\begin{eqnarray}& [T]_\mathcal{B} = \left[ \begin{array}{c c c} 6 & 3 & 2\\ 5 & 4 & 3\\ 1 & 1 & 1 \end{array} \right]^{-1} = \left[ \begin{array}{c c c} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\\ -1 & 2 & -4\\ \frac{1}{2} & -\frac{3}{2} & \frac{9}{2} \end{array} \right] \\& \vec{v_{\mathcal{B}}} = \left[ \begin{array}{c} 6\\4\\0 \end{array} \right] \\& [T]_\mathcal{B}\vec{v_{\mathcal{B}}} = \left[ \begin{array}{c c c} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\\ -1 & 2 & -4\\ \frac{1}{2} & -\frac{3}{2} & \frac{9}{2} \end{array} \right] \left[ \begin{array}{c} 6\\4\\0 \end{array} \right] = \left[ \begin{array}{c} 1\\2\\-3 \end{array} \right] = \left[ \begin{array}{c} A\\B\\C \end{array} \right]\end{eqnarray}$

and use the $A$, $B$, $C$ result to complete our decomposition as

before

$\begin{eqnarray} \frac{4x+6}{x^3+6x^2+11x+6} & = & \frac{1}{x+1}+\frac{2}{x+2}+\frac{-3}{x+3}\\\end{eqnarray}$

Well, this is certainly a lot of work for a single application. But, using $[T]_\mathcal{B}$, we can perform any partial fraction decomposition for a rational fraction with this denominator by simple matrix multiplication. In other words, we can repeat this operation, using a different numerator, easily and quickly by plugging in the vector representation of the new numerator above. This should work for any numerator with a degree equal or less than that of the denominator.

Likewise, being an invertible operation, we can add any decomposed fractions with these factors as denominators without the tedious polynomial expansion commonly used. We simply multiply $[T]_{\mathcal{B}}^{-1}$ by the vector representation of $A$, $B$, $C$ and it will spit out the vector form of the numerator. Cool!

This technique really only works, at this point, for rational fractions with easily factored denominators. It may be extendable for use in other cases, but I haven't done the work yet to investigate how easy that might be.

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