The Philosophy of Repeat Linear Fractions?

sponsoredwalk

I've been thinking about partial fractions & I've realised I don't have any justification for most of it, it comes out of nowhere & I've just memorized something that works, I'm trying to stop that :pac:

[latex] \frac{P(x)}{(x - 1)(x + 2)^3} \ = \ \frac{A}{x - 1} \ + \ \frac{B}{x + 2} \ + \ \frac{C}{(x + 2)^2} \ + \ \frac{D}{(x + 2)^3} [/latex]

I can understand this in a sense, but;

[latex] \frac{P(x)}{(ax^2 + bx + c)^3(x - 1)(x + 2)^3} \ = \frac{A_1x + B_1}{ax^2 + bx + c}\ \ + \ \frac{A_2x + B_2}{(ax^2 + bx + c)^2}\ \ + \ \frac{A_3x + B_3}{(ax^2 + bx + c)^3}\ \ + \ \frac{C}{x - 1} \ + \ \frac{D}{x + 2} \ + \ \frac{E}{(x + 2)^2} \ + \ \frac{F}{(x + 2)^3} [/latex]

The only justification I have for this is the idea of standard factoring, i.e. when you get a common denominator it will be the highest power in the denominator, but that does nothing to explain the numerator's weird shape...

Did someone just magically invent this out of thin air, & then say for equations that have quartic iin the denominator - the magical shape I place in the numerator is

[latex]A_1x^3 + B_1 x^2 + C_1 x + D[/latex]

How are you supposed to generalize this idea? There is no way I could magically know how to do this for weirder shaped equations...

Edit: Had another question here but I figured it out

Michael Collins

Good question! I remember when I was learning partical fraction expansions I felt I needed a justification of this too.

To get a complete understanding of why this is the case you need to know some abstract algebra (rings, ideals etc) . There is a partial fraction decomposition theorem. I looked at it at the time and think I got most of it - but I can't remember any of it now (it's so annoying when this happens!). But I say this because if I got it, it can't have been that hard. So if you want a full justification I'd say take that route.

I can give you a handwaving reason of why the denominator and numerator are the way they are. Maybe it will help.

A repeated factor in the denominator:

[latex]
\displaystyle \frac{P_{\leq 1}[x]}{(x-p)^2} = \frac{A}{x-p} + \frac{B}{(x-p)^2}
[/latex]

where [latex] P_{\leq 1}[x] [/latex] is any real polynomial of order 1 or less and p is any real number.

Let's think about going the opposite way to decomposition, i.e. combining fractions, going from the RHS to the LHS. Clearly we end up with the LHS above, whether we include the first term (in A) on the RHS or not. But, since any decomposition must hold over all x in R, it is unique, so there will only be one possible decomposition for each rational function on the LHS, hence we need to make it as general as it needs to be, as some functions on the LHS will require the term in A to exist. Which rational functions require this? Functions where the numerator [latex] P[x] [/latex] has exactly degree 1. The term in B can only return real numbers. Hopefully this takes care of the repeated factor case.



An irreducible factor in the denominator:

[latex]
\displaystyle \frac{P[x]}{ax^2+bx+c} = \frac{A}{ax^2+bx+c}
[/latex]

This statement is valid if the polynomial [latex] P[x] [/latex] in the LHS numerator is of zero order, i.e. a constant. If it's of order 1 however, we have a problem analogus to the previous case. We require the LHS to equal to RHS for all real x, but when [latex] P[x] [/latex] is a 1st order polynomial, its value will change with x, whereas A on the RHS wont, hence the formula will only be true for a single x - not what we want. So again, we need to make the RHS numerator as general as it needs to be. So we have

[latex]
\displaystyle \frac{P[x]}{ax^2+bx+c} = \frac{Ax + B}{ax^2+bx+c}
[/latex]



These obviously expand likewise to higher orders. Take the last example, we can expand that to:

[latex]
\displaystyle \frac{P[x]}{ax^3+bx^2+cx +d} = \frac{Ax^2 + Bx + C}{ax^3+bx^2+cx +d}
[/latex]

because this is the most general form.

Note that the order of [latex] P[x] [/latex] is always less than that of its denominator, hence the most general numerator on the RHS is one degree less too. If the order of [latex] P[x] [/latex] is greater than or equal to the LHS denominator, you must divide the denominator into the numerator and get a quotient element, i.e. an additive polynomial appears on the RHS (of zero order if the degress are equal) before you continue the standard partial fraction expansion procedure.

sponsoredwalk

Michael Collins wrote: »

Hopefully this takes care of the repeated factor case.

I'm a little confused, I get that the degree of the numerator should be less than the denominator, not rigorously though - only because the option is there to divide the rational function out. If I attempt a decomposition this way I get wrong answers, I know to divide to reduce the fraction but I just don't know why it goes wrong if you don't divide it out...

As for a numerator of degree zero, I've tried partial fractions & the only possibility I've found is to arrive with your original rational function that you've started with, so unless there are special cases I get that for now.

When I put a degree one function in the numerator and some crazy thing in the denominator, I can rationalize things by thinking about it this way;

[latex] \frac{P(x)}{(x - 1)(x + 2)^2} \ = \ \frac{1}{x + 2}(\frac{P(x)}{(x - 1)(x + 2)}) [/latex]
then you can use partial fractions inside the brackets;

[latex] \frac{1}{x + 2}(\frac{P}{(x - 1)} + \frac{Q}{(x + 2}) = \frac{P}{(x - 1)(x + 2)} + \frac{Q}{(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} \ + \ \frac{Q}{(x + 2)^2} [/latex]

and then use them again on the P term to break it into A & B.

That explains it perfectly, (I found that on a forum online, not in my book :mad:, but it's great). The numerator having degree one will mean it is included in the decomposed fractions just by the multiplication.

I can sort of see what you mean about
[latex] \frac{P(x)}{ax^2 + bx + c} = \frac{A}{ax^2 + bx + c}[/latex]
being valid only for degree zero because if include a B or C rational etc... & you do the algebra it just wont be included, but it doesn't explain the form Ax + B in the way the above derivation does, you know what I mean?

I can understand the repeating fractions for any equation in the denominator by the above method, but I can't rigorously understand the numerator other than the generality of it in staying a degree below the denominator. Perhaps there is no further explanation than that...

equivariant

This is a really good question! I have always meant to work this out properly for myself but never done it. Anyway, here are some thoughts that might help (or might not).

I think that it may come down to the fact that the polymials in one variable over a field form a Euclidean domain. In other words, the euclidean algorithm works just like it does with the integers.

(Aside: If this is gobbledygook here is quick explanation: Suppose that p(x) and q(x) are polynomials. Then there are polynomials a(x) and r(x) such that [latex] q(x) = a(x) p(x) +r(x)[/latex] and such that [latex] deg(r) < deg(p)[/latex]. In other words we can divide p into q and be left with a remainder that has degree strictly smaller than p.)

OK, back to partial fractions. Suppose that we have a rational function [latex] \frac{A(x)}{f(x)g(x)}[/latex]. We may as well assume that deg(A) < deg f + deg g - otherwise do a long division to make that the case. Now consider the equation
[latex] \frac{A(x)}{f(x)g(x)} = \frac{B(x)}{f(x)} +\frac{C(x)}{g(x)}[/latex] This is equivalent to (drop the x from the notation for convenience)
[latex] A = Bg+Cf [/latex]
Now, if f and g are relatively prime (i.e. have no common factors), then the Euclidean algorithm, just like in the case of the integers, says that we can find B and C that make this last equation true.

Now, we have to be a bit careful, because we don't initially know anything about the degrees of B and C. However, since we know that deg A < deg f +deg g, we can probably arrange it so that deg B < deg f and deg C < deg g. This bit might require a bit of work to prove, but should be OK.

Ok, so that guarantees that we can split up our rational function into partial fractions where all the denominators are relatively prime. Now, we have to deal with the repeated factor case. In other words, what do we do if, say [latex] f(x) = p(x)^2q(x)[/latex]. Let me think about that and get back to you. I think that Michael Collins may already have explained that case anyway.

equivariant

By the way, the whole business of having to divide out first so that the numerator has degree less than the denominator is a bit of a red herring. Nothing really goes wrong if you don't do that.

sponsoredwalk

equivariant wrote: »

By the way, the whole business of having to divide out first so that the numerator has degree less than the denominator is a bit of a red herring. Nothing really goes wrong if you don't do that.

[latex]\frac{8x^3 - 4x + 7}{(x - 1)(x + 2)} \ = \ \frac{A}{x - 1} \ + \ \frac{B}{x + 2}[/latex]

[latex]8x^3 - 4x + 7\ = \ A(x + 2)\ + \ B{(x - 1)[/latex]

[latex]A \ = \ \frac{11}{3}, \ B \ = \ \frac{49}{3}[/latex]

[latex]\frac{8x^3 - 4x + 7}{(x - 1)(x + 2)} \ = \ \frac{\frac{11}{3}}{x - 1} \ + \ \frac{\frac{49}{3}}{x + 2}[/latex]

[latex] \frac{\frac{11}{3}}{x - 1} \ + \ \frac{\frac{49}{3}}{x + 2} \ = \ \frac{\frac{11}{3}x \ + \ \frac{22}{3} \ + \ \frac{49}{3}x \ - \ \frac{49}{3}}{(x - 1)(x + 2)} \ = \ \frac{\frac{60}{3}x \ - \ \frac{47}{3}}{(x - 1)(x + 2)}[/latex]

I actually just wrote this whole thing out & have just realised that I'm supposed to put [latex]Ax^2 + Bx + C[/latex] into the numerator or something.

I'm even more confused now...

In my original post I've got [latex]A_1 x + B_1[/latex] on top of a quadratic irreducible which is definition yet if the numerator's degree is greater than that of the denominator am I to put some extra cubic or quadratic terms in the A and B numerators to make everything work?

Btw: the Euclidian algorithm explanation has me lost :P

If

q(x) = a(x)p(x) + r(x)

and you divide p into q, I mean I don't get it, q is
defined to be a(x)p(x) + r(x), it's like your saying in
ƒ(x) = 3x² + 2 we'll divide x² into ƒ,

[latex] \frac{f(x)}{x^2} \ = \ 3 \ + \ \frac{2}{x^2} [/latex]

but if you rewrite f(x) in the fraction as 3x² + 2 you're back
to where you started, i.e. mathematical circularity.

That's how I understand it, probably completely missed the point :pac:

equivariant

sponsoredwalk wrote: »

[latex]\frac{8x^3 - 4x + 7}{(x - 1)(x + 2)} \ = \ \frac{A}{x - 1} \ + \ \frac{B}{x + 2}[/latex]

[latex]8x^3 - 4x + 7\ = \ A(x + 2)\ + \ B{(x - 1)[/latex]

[latex]A \ = \ \frac{11}{3}, \ B \ = \ \frac{49}{3}[/latex]

[latex]\frac{8x^3 - 4x + 7}{(x - 1)(x + 2)} \ = \ \frac{\frac{11}{3}}{x - 1} \ + \ \frac{\frac{49}{3}}{x + 2}[/latex]

[latex] \frac{\frac{11}{3}}{x - 1} \ + \ \frac{\frac{49}{3}}{x + 2} \ = \ \frac{\frac{11}{3}x \ + \ \frac{22}{3} \ + \ \frac{49}{3}x \ - \ \frac{49}{3}}{(x - 1)(x + 2)} \ = \ \frac{\frac{60}{3}x \ - \ \frac{47}{3}}{(x - 1)(x + 2)}[/latex]

I actually just wrote this whole thing out & have just realised that I'm supposed to put [latex]Ax^2 + Bx + C[/latex] into the numerator or something.

I'm even more confused now...

In my original post I've got [latex]A_1 x + B_1[/latex] on top of a quadratic irreducible which is definition yet if the numerator's degree is greater than that of the denominator am I to put some extra cubic or quadratic terms in the A and B numerators to make everything work?

In this case you could decompose as follows

[latex]
\frac{8x^3-4x+7}{(x-1)(x+2) = \frac{Ax^2+Bx+C}{x-1} +\frac{D}{x+2}[/latex]

or you could decompose as follows
[latex]\frac{8x^3-4x+7}{(x-1)(x+2) = \frac{A}{x-1} +\frac{Bx^2+Cx+D}{x+2}[/latex]. There are other possibilities as well.

Anyway, I probably should not have bothered mentioning this as it is not really important and distracts from the main issue which is the Euclidean property of polynomials in one variable.

Btw: the Euclidian algorithm explanation has me lost :P

If

q(x) = a(x)p(x) + r(x)

and you divide p into q, I mean I don't get it, q is
defined to be a(x)p(x) + r(x), it's like your saying in
ƒ(x) = 3x² + 2 we'll divide x² into ƒ,

[latex] \frac{f(x)}{x^2} \ = \ 3 \ + \ \frac{2}{x^2} [/latex]

but if you rewrite f(x) in the fraction as 3x² + 2 you're back
to where you started, i.e. mathematical circularity.

That's how I understand it, probably completely missed the point :pac:

q is not defined to be ap+r. On the contrary assume that p and q are given, then the Euclidean property says that there exists some a and some r, with deg(r) < deg(p) such that q = ap+r. Think of the Euclidean property as formalising the idea of long division of polynomials. It says that if we divide q by p, then the remainder has degree strictly less than p.

Example: suppose [latex] q(x) = x^4++x+1[/latex] and [latex]p(x) = x^2+1[/latex] Then by using long division of polynomials we see that [latex] q(x) = p(x)(x^2-1) + (-x+2)[/latex] In other words, the remainder when q is divided by p is -x+2. Note that in this case the reminder has degree 1 (which of course is smaller than the degree of p).

equivariant

Damn! Latex gets messed up when I try to edit a post - sorry.

Anyway, forget the whole business about the degree of the numerator not having to be less than the denominator - I think I just unnecessarily confused the situation by mentioning that.