# Partial fraction series

1. Jan 24, 2014

### MathewsMD

http://en.wikipedia.org/wiki/Partial_fraction_decomposition

In general, if you have a proper rational function, then:

if $R(x) = \frac {P(x)}{Q(x)}$ and $Q(x) = (mx + b)^n ... (ax^2 + bx + c)^p$ where $Q(x)$ is composed of distinct linear powers and/or distinct irreducible quadratic powers.

1) Can we actually reduce every polynomial of powers greater than 2 to either a distinct linear or quadratic power? I may not be thinking correctly, but does anyone know of a proof they could link me to? For a an expression like $x^{12} - πx^3 - 542.43x + 21$ I am having trouble simplifying this to an irreducible quadratic or linear factor.

2) Why can we show that $R(x) = \frac {A}{mx+b} + \frac {B}{(mx+b)^2}... \frac {C}{(mx+b)^2} + \frac {Lx + D}{ax^2 + bx + c} + \frac {Mx + E}{(ax^2 + bx + c)^2}...+\frac {Nx + F}{(ax^2 + bx + c)^p}$

For each distinct linear and irreducible quadratic power, why can we not just show this as:

$R(x) = \frac {A}{(mx+b)^n} + \frac {Lx + D}{(ax^2 + bx + c)^p}$

I know the above is incorrect, but I'm wondering how it was proved that we must add a power to the denominator for each subsequent fraction in the series.

Also, I've seen it written as $A_1 + A_2 +....+ A_n$ instead of $A + B +....+ C$ and I was wondering if there's any relationship between A1 and An in this case.

I've looked online but have not found the proof(s) that I'm looking for.

Any clarification would be great!

Last edited: Jan 24, 2014
2. Jan 24, 2014

### pwsnafu

Very easy. The fundamental theorem of algebra shows that the only irreducible polynomials over C are linear. But if the polynomial has real coefficients then the roots must occur in conjugate pairs, hence we are left with quadratics.

Notice that $\frac {A}{(mx+b)^n} + \frac {Lx + D}{(ax^2 + bx + c)^p} \neq \frac {A}{(mx+b)^n} + \frac {A'}{(mx+b)^{n-1}} + \frac {Lx + D}{(ax^2 + bx + c)^p}$ and its easy to see that you get different rational functions. In other words, the left hand side doesn't cover all cases.

Last edited: Jan 24, 2014
3. Jan 24, 2014

### MathewsMD

Okay, I just went to check the theorem again and that question's been answered.

Regarding my second question, I am still confused. I realize my statement is not correct, I was just trying to ask why exactly the partial fractions can be written in the general form I stated above with subsequently increasing powers. Also, are A and A' related in any way?

If we have $\frac {1}{x^2}$ why is it written as $\frac {A}{x} + \frac {B}{x^2}$ instead of $\frac {mx + b}{x^2}$? Having a proof or an explanation why the partial fraction series was formulated as stated above would be helpful. Thank you for referring me to the fundamental theorem of algebra.

4. Jan 25, 2014

### pwsnafu

They are not. Just notation.

It's neither. $\frac{1}{x^2}$ is the partial fraction decomposition of itself.

As to your question, it's easier to work with when integrating. The point of partial fractions is a divide and conquer approach to finding antiderivatives of rational functions. So instead of trying to integrate $\frac{x}{(x+1)^2}$ in one step, you consider $\frac{1}{x-1}-\frac{1}{(x-1)^2}$ in two steps. $\frac{1}{x^2}$ is a bad example.

Edit: I just realised that $\frac{x}{(x+1)^2}$ is also a poor example because you can integrate it directly.

It should be obvious that if $\frac{x}{(x+1)^2} = \frac{Ax+b}{(x+1)^2}$ then A=1 and B=0, hence you haven't made the problem easier so you need to consider lower powers. That was the question in the first post.

5. Jan 25, 2014

### MathewsMD

My question is why we generalize the theorem as:

$\frac{x}{(x+1)^2} = \frac{Ax+b}{(x+1)}+ \frac{Mx+c}{(x+1)^2}$ instead of just $\frac{x}{(x+1)^2} = \frac{Ax+b}{(x+1)^2}$. I undertand in this case that is easy to just let A = 1 and B = 0, but in the general case of $f(x) = \frac {P(x)}{(Q(x))^n} = \frac {Ax + b}{Q(x)} + \frac {Mx + c}{(Q(x))^2} + .... + \frac {Ex + d}{(Q(x))^n}$

I just don't understand why it's necessary when writing the general form of a partial fraction to include a series where the denominators begin from i = 1 to i = n depending on the power (n) of the irreducible quadratic or linear function.

6. Jan 25, 2014

### pwsnafu

This is probably best illustrated by an example. I'm just going to assume $Q(x) = x+1$ and we'll consider successive n. Changing Q doesn't change the argument, it makes it less clear.

Let n=2. The numerator has to have degree less the denominator so it is at most linear. We have
$$\frac{Ax+B}{(x+1)^2} = \frac{a}{x+1}+\frac{b}{(x+1)^2} = \frac{ax + (a+b)}{(x+1)^2}$$
Now, your question was "why can't we just toss the first term?" Well, equating coefficients we see that $a$ is completely determined by $A$, i.e. $a = 0 \iff A = 0$. So we can toss the first term exactly when we are presented a problem where numerator is a constant.

Let n=3. The numerator is at most quadratic. We have:
$$\frac{Ax^2+Bx+C}{(x+1)^3} = \frac{a}{x+1}+\frac{b}{(x+1)^2}+\frac{c}{(x+1)^3} = \frac{ax^2 + (2a+b)x + (a+b+c)}{(x+1)^3}$$
As before, we can toss the first term if and only if A=0. The middle term is more interesting. Because we know $a = A$, we have $b = 0 \iff B = 2A$. So we can toss the middle term only for problems when B is twice A.

And this argument works for n=4 and so on. In general we have an expression
$$\frac{A_{n-1}x^{n-1}+A_{n-2}x^{n-2}+\ldots+A_0}{(x+1)^n} = \frac{ax^{n-1}+(na+b)x^{n-2}+\ldots}{(x+1)^n}$$
and you can see again $a=0 \iff A_{n-1}=0$ and $b=0 \iff A_{n-2}=nA_{n-1}$ et cetera.

Anyway, we write the expression in our form because the P(x) we get at the start is arbitrary. We don't know if $A=0$ or if $B=2A$ when we start.

Last edited: Jan 25, 2014
7. Jan 25, 2014

### MathewsMD

Thank you!!!!