Path to proving partial fractions and the fundamental theorem of algebra

Question

As I've learned Calculus, I've tried to follow along with proofs of the rules that I use. In most cases, like say the Power Rule, I'm able to follow along with the proofs using concepts I understand, or things I've already proved like the Binomial Theorem. In other cases, like the Extreme Value Theorem the concepts used to prove it seem to require math that's beyond me, so I save proving them for later.

I recently learned Partial Fraction Decomposition in the context of Integrals, and I became enamored with this technique. I searched for proofs of why it works, and most of them seem to rely on the Fundamental Theorem of Algebra. Most of the proofs of the FTA I've seen rely on either Number Theory or Complex Analysis. I'm interested in proving the FTA regardless of whether I need it to prove partial fractions.

If I want to:

1. Prove the Fundamental Theorem of Algebra
2. Prove why Partial Fractions work

What would be the recommended math path for me following Calc? Would it be to study Number Theory concepts like Euclidean Division and GCD, or to focus on complex numbers to prove the FTA?

Also, am I correct to assume that proving FTA should come before proving Partial Fractions always work because the latter proof will follow easily from the former?

Answer 1

Partial fraction decomposition can be stated more generally (and independently of FTA) as:

Given polynomials $\,P(x), Q(x)\,$ with $\,\gcd\big(P(x),Q(x)\big) = 1\,$ i.e. relatively coprime, there exist $\,A(x), B(x)\,$ with $\,\deg A \lt \deg P\,$ and $\,\deg B \lt \deg Q\,$ such that:

$$ \frac{1}{P(x)Q(x)} = \frac{A(x)}{P(x)}+\frac{B(x)}{Q(x)} $$

This follows directly from Bézout's identity for polynomials, which states that polynomials $\,A,B\,$ exist that satisfy the respective degree constraints such that:

$$ 1 = \gcd\big(P(x),Q(x)\big) = B(x)P(x) + A(x) Q(x) $$

Dividing the equality by $\,P(x)Q(x)\,$ yields the partial fraction decomposition.

In particular:

• for $\,P(x)=x+p_0\,$ this means $\,\frac{1}{(x+p)Q(x)} = \frac{a_0}{x+p_0}+\frac{B(x)}{Q(x)}\,$ for some constant $\,a_0\,$;

• for $\,P(x)=x^2+p_1x+p_0\,$ this means $\,\frac{1}{(x^2+p_1x+p_0)Q(x)} = \frac{a_1 x + a_0}{x^2+p_1x+p_0}+\frac{B(x)}{Q(x)}\,$.

Applying the same steps to $\,\frac{1}{Q(x)}\,$ in the second term gives in the end the familiar decomposition with linear and quadratic denominators - and the existence of such factorization is the (only) part which relies on the FTA.

Answer 2

This can be proved using basic linear algebra. Lets say we want to use partial fraction decomposition on polynomials $\frac{P(x)}{Q(x)}$ with $\deg(P(x)) < \deg(Q(x))$ (if degree were not lesser user polynomial division algorithm to get it in that form). Then consider the vector space over $\mathbb{R}$ $$ \cal{S} =\{\frac{G(x)}{Q(x)}: \deg(G(x)) < \deg(Q(x))\}.$$ If we set $n = \deg(Q(x))$ then $\cal{S}$ has degree $n$. Then note by the fundamental theorem of algebra $Q(x)$ can be decomposed as a product $\prod (x-a_i)(x^2 + b_ix + c_i)$ where note we may have repeated elements in the product. If we group them up, then lets say we have a term of the form $(x-a)^m$, then note that elements of the form $\{\frac{1}{x-a}, \frac{1}{(x-a)^2},\ldots,\frac{1}{(x-a)^m}\}$ are linearly independent. Note that such elements are in $\mathcal{S}$ by noting that $$\frac{1}{(x-a)^s} = \frac{\frac{Q(x)}{(x-a)^s}}{Q(x)}, 1 \leq s \leq m.$$ Similarly if we have a grouping of $\frac{1}{(x^2 + bx + c)^k}$ then we have linearly independent vectors of the form $\{\frac{1}{x^2+bx+c}, \frac{x}{x^2+bx+c}, \ldots, \frac{1}{(x^2 + bx + c)^k}, \frac{x}{(x^2 + bx + c)^k}\}$ where again those vectors are in $\mathcal{S}$. Its not hard to see if we put all those basis elements together, they form a linearly independent set of size $n = \dim\mathcal{S}$ and so form a basis of $\mathcal{S}$. This then becomes the basis of writing $$\frac{P(x)}{Q(x)} = \sum_i\sum_j \frac{A_{i,j}}{(x-a_i)^j} + \sum_t\sum_k(\frac{B_{t,k}}{(x^2+b_tx+c_t)^k} + \frac{C_{t,k}x}{(x^2+b_tx+c_t)^k}).$$