# Partial Fractions

• Aug 3rd 2008, 05:19 PM
jalsdn
Partial Fractions
I have 2 questions:

1. What is the reasoning for having to have separate fractions like: A/(x-b) + B/(x-b)^2 + C/(x-b)^3 + D/x to do a partial fraction decomposition of p(x)/x[(x-b)^3] ?

2. [solved] Lets say you have a polynomial P(x)=ax^3 + bx^2 + cx +d and you divide it by x-b and you get a remainder (r). You can now write the polynomial as Q(x)=(x-b)(ex^2 + fx + g + (r/x-b)). Since Q(x)=P(x) they should have the same zeros right? But this isn't always true because that would mean b must be a zero but this is not necessarily true.

Basically, I am saying that according to a theorem--I think it is called the factor theorem--if x-b is factor of P(x) then b is a zero of P(x). But the problem is that you can divide a second degree or greater polynomial P(x) by any polynomial x-b even if b is not a factor of the polynomial, and just write the remainder (r) as (r/x-b) inside the quotient, and those polynomials should be equal but their zeros are not necessarily equal.
• Aug 3rd 2008, 07:16 PM
TKHunny
Quote:

Originally Posted by jalsdn
1. What is the reasoning for having to have separate fractions like: A/(x-b) + B/(x-b)^2 + C/(x-b)^3 + D/x to do a partial fraction decomposition of p(x)/x[(x-b)^3] ?

Three off the top of my head - not necessarily in a priority order.

1) Learn new things. Of course, not all new things are productive.

2) Reasonable technique for finding an antiderivative. Of course, any good computer-based integrator can dust the partial fraction process.

3) You will need it in Complex Analysis. Residue Theory is in your future?
• Aug 4th 2008, 06:54 PM
jalsdn
Ok, I figured out my second question. Its because the two polynomials are not exactly the same, specifically, the one that was divided into pieces is indeterminate at b, since b-b=0 in the denominator making it indeterminate.

But I still don't know the answer to my first question and the response that I got does not make any sense.
• Aug 4th 2008, 07:58 PM
TKHunny
Just a social note...

One really ought to ask a follow-up question before assuming understanding has reached its peak. It would have read more appropriately if you had stated, "...the response that I got does not make any sense TO ME." Sadly, you settled on a more offensive statement.

Just an honest evaluation...

You asked what it was for. I told you a couple of things that are very appropriate. There you have it. Would you have been more satisfied if I had said, "Trust me, you'll need this technique for your mathematical studies down the road. For now, just memorize it." Frankly, I doubt it. "Wait until you are old enough" almost always is viewed as unsatisfactory.

Perhaps this:

1) Is there harm in learning an interesting techinque? Just learn it and muse on its significance.

2) The Ancient Egyptians were quite good at breaking things down to unit fractions (fractions with only unity (1) in the numerator). For example, it is easy enough to see that 1/2 + 1/3 = 5/6, but could you do it the other way? This technique could be related to how they did it! Try 149/308.

3) You may need this technique if your studies take you to Integral Calculus. If you have not seen Integral Calculus, you will not have any reference for what this means or for what value it may provide. It is possible, by the time you reach Integral Calculus, that the technique will have fallen out of favor. You may not encounter it.

4) You may need this technique if your studies take you to Complex Analysis. If you have not seen Complex Analysis, you will not have any reference for what this means or for what value it may provide.
• Aug 4th 2008, 09:32 PM
jalsdn
Hey, sorry if I offended you, this is just getting frustrating. I am actually taking Calculus II right now, and we are doing partial fraction decompositions to help with integration.

My question is not why should I learn this, it is why is it that you need to have separate fractions in your decomposition for a repeated factor of degree N and have separate fractions for 1 through N?

Go to Partial fraction - Wikipedia, the free encyclopedia and click "A repeated first-degree factor in the denominator" to see what I'm talking about.

It says you need to do this but I have not seen an explanation anywhere of why, it is probably some simple obvious answer but I am not very good at algebraic manipulations so after trying a few things it doesn't make sense so I am just asking here.
• Aug 5th 2008, 04:25 AM
TKHunny
No worries. Offense is not really something I do. I have learned to recognize things that others may deem offensive.

It is a simple thing, really. Can it be done?

If the denominator has a quadratic expression, the most the numerator can hold is Ax+B. You must use this if the denominator is NOT a perfect square. You do not have to use it if the denominator IS a perfect square. There is no rule that says you mustn't. It is a simpler expression if you break it up into the pieces. For an antiderivative, the antiderivative is trivial with only a constant in the numerator. It is just a hair trickier with a more complicated exppression.

The real answer? It makes life easier and does not add complexity.