View Single Post
Old 2005-01-17, 19:45   #5
jinydu
 
jinydu's Avatar
 
Dec 2003
Hopefully Near M48

2×3×293 Posts
Default

This is the second part of the post.

Note that the above process applies only when you want to stay in the real numbers. But the original rational function contains complex coefficients (or when you enjoy working with complex numbers, and your teacher allows it), you can actually go a step further.

Maybe you're wondering: Why stop at quadratic factors? After all, Gauss proved that an nth degree polynomial always has n roots. On top of that, there's an easy way to find the two roots of any quadratic: Just use the Quadratic Formula. The answer is that the Quadratic Formula sometimes gives roots that involve the square root of a negative number, which is not allowed in the real number system. But this can also happen with higher degree polynomials, so why give the quadratic a special status? The answer is that with higher degree polynomials, its always possible to split up the polynomial further using only real numbers (although it might not be possible using functions you know about, see my comments on step 2). If the polynomial has an odd-number degree (3, 5, 7, 9, etc.), and all of its coefficients are real, its not hard to show that it will always have at least one real root. We can then factor out this root (neglecting the difficulty of actually finding this root, see comments on step 2), reducing the polynomial to an even-degree polynomial. It can also be shown that an even-degree polynomial can be written as a product of quadratic factors (again, neglecting the difficulty I pointed out for step 2), which may or may not be further factorable into linear factors using only real numbers.

But as I've said before, if you use complex numbers, you can go a step further. If you have the quadratic factor (x^2 + ax + b), it is always possible to find the roots of that factor if you use complex numbers. Then, the factorization is just (x - r(1))*(x - r(2)), where r(1) and r(2) are the (possibly complex) roots of the quadratic factor.

In some sense then, the problem actually becomes simpler because you don't have to worry about quadratic factors any more. Also, all the numerators of your sum are now just constants, instead of some being linear expressions with two constants each. Just keep in mind that the final expression may contain complex numbers, even if your original problem only had real numbes. Here's a simple example:

Find the partial fraction decomposition of 3/(x^2 + 1)

In the complex numbers, x^2 + 1 factors into (x + i)(x - i)

3/(x^2 + 1) = A/(x + i) + B/(x - i)

Multiplying both sides by (x + i)*(x - i):

3 = A*(x - i) + B*(x + i)

Using the method of step 5b, we can let x = i:

3 = A*(i - i) + B*(i + i)

3 = A*(0) + B*(2i)

3 = (2i)*B

B = 3/(2i) = 3/2 * 1/i

Since 1/i = -i:

B = -3i/2

Back in our equation: 3 = A*(x - i) + B*(x + i), we can expand the right-hand side:

3 = Ax - Ai + Bx + Bi

3 = (A + B)x + (-Ai + Bi)

Since the x coefficient on the left-hand side is 0 (i.e. its not there):

A + B = 0

A - 3i/2 = 0

A = 3i/2

Therefore, our partial fraction decomposition is:

3/(x^2 + 1) = (3i/2)/(x + i) - (3i/2)/(x - i)

If you want only one fraction bar in each term, you can simplify this further:

3/(x^2 + 1) = 3i/(2x + 2i) - 3i/(2x - 2i), and you're done!

Notice that if you were limited to real numbers, you wouldn't be able to do anything with the original fraction, 3/(x^2 + 1), since x^2 + 1 = 0 has no real number solution.
jinydu is offline   Reply With Quote