How is the Partial Fraction Series Used in Reducing Polynomials?

[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.

I had two questions:

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 A_n in this case.

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

Any clarification would be great!

 

[pwsnafu]

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.

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.

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.

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.

 

[MathewsMD]

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

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.

 

[pwsnafu]

Also, are A and A' related in any way?

They are not. Just notation.

If we have $\frac {1}{x^2}$ why is it written as $\frac {A}{x} + \frac {B}{x^2}$ instead of $\frac {mx + b}{x^2}$?

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

Having a proof or an explanation why the partial fraction series was formulated as stated above would be helpful.

What specifically are you asking?

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.

If you are asking for existence then a proof sketch is on the Wikipedia page.

 

[MathewsMD]

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 realized that $\frac{x}{(x+1)^2}$ is also a poor example because you can integrate it directly.
What specifically are you asking?

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.

If you are asking for existence then a proof sketch is on the Wikipedia page.

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.

 

[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.

 

[MathewsMD]

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.

Thank you!