View source for Residue Theory/Partial Fractions

This is probably the most basic technique, and doesn't require a lot of theory, mainly just algebraic manipulation.  However, it does have its limitations, namely it really only works with polynomials.  It is more of a cookbook method:  here's the recipe, follow the steps.

Given two polynomial functions <math>P(z)</math> and <math>Q(z)</math>, where the degree of Q is greater than the degree of P, we define another function to be the quotient of the two polynomials:

<math>f(z) = \frac{P(z)}{Q(z)}</math>

And we note that if we factor Q we obtain:

<math>Q(z) = (z-z_0)^{a_0} \cdot (z-z_1)^{a_1} \cdot ...</math>

Then depending on the form of <math>(z-z_0)^{a_0}</math> we can reduce the function into a limited series of simple or m-th order poles as so:

<math>f(z) = \frac{P(z)}{Q(z)} = \frac{A_0(x)(+ B_0)}{(z-z_0)^{a_0}} + \frac{A_0(x)(+ B_0)}{(z-z_0)^{a_0-1}} + ... + \frac{A_1(x)(+ B_1)}{(z-z_1)^{a_1}} + \frac{A_1(x)(+ B_1)}{(z-z_1)^{a_1-1}} + ...</math>

And then the coefficients <math>A_0,...,B_0,...</math> can be solved for and the function can be stated into an explicit form with readable residues.

Of course, this is taught a lot better by example and case.

=Case 1, simple one-order factors=

We begin with the function:

<math>f(z) = \frac{1}{z^2 + 3z + 2}</math>

And note that it can be factored thusly:

<math>f(z) = \frac{1}{z^2 + 3z + 2} = \frac{1}{(z+1)(z+2)}</math>

For this case the correct form we "guess" is like so:

<math>\frac{1}{(z+1)(z+2)} = \frac{A_0}{z+1} + \frac{A_1}{z+2}</math>

The remaining portion is by algebra, we multiply both sides by <math>(z+1)(z+2)</math>:

<math>1 = A_0z + 2A_0 + A_1z + A_1</math>

Which gives us two equations:

<math>0 = A_0z + A_1z</math>

<math>1 = 2A_0 + A_1</math>

Thus:

<math>A_0 = 1</math> and <math>A_1 = -1</math>

And our function can be rewritten as:

<math>f(z) = \frac{1}{z+1} - \frac{1}{z+2}</math>

The remainder  of this section discusses suggests fractional forms that aid in separation, since the actual method and theory hold.

=The Other Two Cases=

Case 1, Unfactorable Terms. In our generic expression, there is A(x)+B term, but really should include an additional polynomial for possible "unfactorable" terms (i.e., terms that can't be factored with only real numbers, although if the terms are factored correctly with imaginary numbers, the method works).  To account for this the "guessed" fraction, include these extra terms. For example,

<math>f(z) = \frac{1}{(z^2+4)(z-1)} = \frac{A_0x + B_0}{z^2+4} + \frac{A_1}{z-1}</math>

With <math>A_0</math>, <math>B_0</math>, and <math>A_1</math> to be solved.

Case 2, Term(s) Raised To A Power, The correct "guess" will include a trailing series of decreasing powers of the factor. For example,

<math>f(z) = \frac{1}{(z+1)^2(z-1)} = \frac{A_0}{(z+1)^2} + \frac{A_1}{z+1} + \frac{A_2}{z-1}</math>

=Remark=

Again, partial fractioning only really works with polynomials, and can be a huge hassle for large denominators.  The next section looks at a more general way of determining the residue.

[[Category:Residue Theory]]