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Suppose we have:

$ \frac{f(x)}{g(x)h(x)} $

and we want to break it down into;

$ \frac{I(x)}{g(x)} + \frac{J(x)}{h(x)}$

and that;

$deg(f) \leq deg(g)+deg(h)$ , $deg(i) < deg(g)$,

$deg(j) < deg(h)$

What is the general way of doing this?

I don't understand the intuition behind having to express;

$\frac{ax^{2}+bx+c }{(dx+e)(f x^{2}+g) } = \frac{A}{dx+e} + \frac{Bx+C}{fx ^{2}+g }$.

Is there a general rule I'm missing.

Thanks.

Edit: Just to clarify, these are all polynomials.

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First what is the generalized theorem?

Theorem Given three polynomials $P,Q,R$such that $\gcd(P,Q)=1$ there exists $U,V$ such that: $$\frac{R(x)}{P(x)Q(x)}=\frac{U(x)}{P(x)}+\frac{V(x)}{Q(x)} $$ with $\deg(U)+\deg(R)< \deg(P), \deg(V)+\deg(R)<\deg(Q)$

Proof It follows easily from Bézout's identity and extended GCD algorithm

As we observe the condition $P$ and $Q$ being relatively coprime is neccessary for the validity of the theorem because we can find simple examples which dies not satisfy the theorem.

Second how to compute $U$ and $V$?

  • Algebraic method using the extended GCD algorithm. and this method is the general method it works always.
  • Analytic method, for example if we want to find $U$ and we know the roots of $P$ we know that $deg(U)\leq deg(P)$ so we only need to find the values of $U$ in at least $deg(P)$ points this can be done using: $$I(x)=\frac{R(x)}{Q(x)}=U(x)+\frac{V(x)}{Q(x)}P(x) \tag 1$$ It's easy to see that if we evaluate $I$ in the roots of $P$ we can find the values of $U$, if we have multiple roots we evalute the derivatives of $I$ also.

  • There is other methods using for example the roots of $Q$ in the formula $1$ in which case we use limits and evaluating in some remarkable points such as $a,1,-1$ the roots of $P,R,Q£.

Third why we need this decomposition, there is a lot of applications of the decomposition like this in computing integrals and solving differential equations

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As long as the denominator has only simple irreducible factors, it's rather straightforward. Here, for instance, I'll suppose $f,g$ have the same sign, so that $fx^2+g$ is irreducible and it has complex roots. The general strategy is to:

  1. Multiply both sides of equality by $(dx+e)(fx^2+g)$. You obtain a polynomial identity: $$ax^2+bx+c=A(fx^2+g)+(Bx+C)(dx+e).$$
  2. Set $x=-\dfrac ed\,$ and get an equation for $A$.
  3. Set $x=\mathrm i\mkern 1.5mu\sqrt{-\dfrac gf}$. You get an equation with complex coefficients for $d$ and $e$, i.e. two equations with real coefficients.

Also remember you first have to divide (with remainder) the numerator by the denominator, so as to have a proper rational function ($\deg$(numerator) $<\deg$(denominator).

For a proper rational function, the decomposition theorem says they are are sums of proper rational function with primary denominators (=powers of one irreducible polynomial) and $\deg$(numerator) $<\deg($irreducible polynomial). In the case of an irreducible quadratic polynomial, this means the numerator will have degree at most $1$.

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  • $\begingroup$ I understand how to solve for the coefficients, but I do not understand how to choose what the polynomials on the RHS numerators will look like. For example, why is the numerator of the second fraction, $Bx+C$ and not just some number $h$. $\endgroup$ – J.Gudal Apr 10 '15 at 13:20
  • $\begingroup$ Please see my completed answer. $\endgroup$ – Bernard Apr 10 '15 at 13:30
  • $\begingroup$ @J.Gudal: See my answer for precisely how to choose the partial fractions and and roughly why it works. The concepts of dependence and span are crucial in understanding partial fractions completely, so I hope you have some background in linear algebra. $\endgroup$ – user21820 Apr 10 '15 at 13:31
  • $\begingroup$ Ok thanks guys. I appreciate your responses. Haven't yet digested what you've written, but I hope it will clarify my doubts. $\endgroup$ – J.Gudal Apr 10 '15 at 13:34
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Indeed you have asked a good question. The general structure that you are looking for is that the number of unknown coefficients you have is exactly the same as the number of parameters of the same type, in this case 3 coefficients $A,B,C$ for 3 parameters $a,b,c$. It is 'clear' that you need at least the same number, but it is not always true that the same number is enough. For that we need the 'parts' to be independent. This kind of generic consideration turns up everywhere.

Now more specifically, we have 3 parts $\frac{1}{dx+e},\frac{1}{fx^2+g},\frac{x}{fx^2+g}$, and you can check that each is equal to $\frac{p(x)}{(dx+e)(fx^2+g)}$ for some polynomial $p$ with $\deg(p) < 3$. If the 3 corresponding polynomials are linearly independent, we know that we can construct any given polynomial of degree less than $3$ by some linear combination of them. That is what partial fractions is actually accomplishing. How you find the coefficients is another matter, and there are many ways and tricks, but the reason why it is expected to succeed is this.

On the other hand, if the corresponding polynomials are linearly dependent, then linear combinations will not be able to cover all possible polynomials of degree less than $3$. In your example, this will in fact be the case if $(fx^2+g)$ is divisible by $(dx+e)$. In that case, the original denominator is of the form $(dx+e)^2 (fx+g)$, and if $(dx+e),(fx+g)$ are independent, then the set of basic parts we need is $\{ \frac{1}{dx+e},\frac{1}{(dx+e)^2},\frac{1}{fx+g} \}$ or equivalently $\{ \frac{1}{(dx+e)^2},\frac{x}{(dx+e)^2},\frac{1}{fx+g} \}$. But if the original denominator is of the form $(dx+e)^3$, then we need $\{ \frac{1}{dx+e},\frac{1}{(dx+e)^2},\frac{1}{(dx+e)^3} \}$ or $\{ \frac{1}{(dx+e)^3},\frac{x}{(dx+e)^3},\frac{x^2}{(dx+e)^3} \}$. You can check that each of these are indeed linearly independent.

The above generalizes, as long as you group the identical factors in the denominator together as done above. It can be proven (not so simple though) that the set generated in that way will always be independent, and so the original rational function can always be expressed using partial fractions.

Note that the theorem in Elaqqad's answer can be used to prove by induction one of the versions of partial fractions I gave, namely where repeated factors in the denominator give rise to combined parts of the form $\dfrac{A+Bx+...+Cx^{km-1}}{(a+bx+...+cx^m)^k}$. You just factorize the denominator into powers of irreducible factors and apply that theorem repeatedly, pulling one power out at a time. To get the other version, with combined parts of the form $\dfrac{A+Bx+...+Cx^{m-1}}{(a+bx+...+cx^m)}+\dfrac{D+Ex+...+Fx^{m-1}}{(a+bx+...+cx^m)^2}+...+\dfrac{G+Hx+...+Ix^{m-1}}{(a+bx+...+cx^m)^k}$ you just need to check that the span is the same.

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