# Prove that the rational function $f(x)/g(x)$ has a partial fraction decomposition in the case when $g(x)$ factors into distinct linear factors.

To be honest i don't even know where to start. I thought about using diagonalizable matrices and characteristic polynomials, but the class hasn't gotten there yet so there should be a way to solve this with simpler stuff. Any ideas?

• I see an $n$. It must be induction. Perhaps divide $f$ by the product of the first $n-1$ factors of $g$? Feb 9, 2020 at 7:44
• I'm doing it by induction, but the other way around. I showed that it works for n=2, and am showing that if you start with the partial fraction decomposition and reassemble it, you'll end up with a n-1 degree polynomial on the numerator. I'm unsure if that's a valid approach but I think it should be? Feb 9, 2020 at 22:22
• It sounds to me that you are doing it backwards, you start with a numerator of degree $n-1$ rather than end up with one. Feb 10, 2020 at 7:39

We proceed by strong induction. Given that $$n = 1,$$ we have that $$f(x) = a$$ for some real number $$a$$ so that $$\frac{f(x)}{g(x)} = \frac a {x - c_1},$$ as desired. We will assume inductively that for any polynomial $$f(x)$$ of degree $$\leq n - 1,$$ we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_n}{x - c_n}$$ for some real numbers $$a_1, \dots, a_n.$$ Consider the polynomial $$f(x) = b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0$$ of degree $$\leq n.$$ We have that $$\frac{f(x)}{g(x)} = \frac{b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)(x - c_{n + 1})} = \frac 1 {x - c_{n + 1}} \biggl(\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} + \frac{b_{n - 1} x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)} \biggr).$$ Given that $$b_n = 0,$$ the expression in the parentheses is the ratio of a polynomial of degree $$\leq n - 1$$ and a polynomial of degree $$n,$$ hence by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(\frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr) = \frac{d_1}{(x - c_1)(x - c_{n + 1})} + \cdots \frac{d_n}{(x - c_n)(x - c_{n + 1})}$$ for some real numbers $$d_1, \dots, d_n.$$ Once again, by strong induction, we can have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $$a_1, \dots, a_{n + 1}.$$ (Each of the terms has its own partial fraction decomposition.)
Consider the case that $$b_n \neq 0.$$ Observe that $$(x - c_1) \cdots (x - c_n)$$ is a monic polynomial of degree $$n$$ and $$b_n x^n$$ is a polynomial of degree $$n,$$ hence by the Division Algorithm, we have that $$\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} = b_n + \frac{r(x)}{(x - c_1) \cdots (x - c_n)}$$ for some polynomial $$r(x)$$ of degree $$\leq n - 1.$$ Consequently, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(b_n + \frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr)$$ for some real numbers $$d_1, \dots, d_n.$$ Like before, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $$a_1, \dots, a_{n + 1}.$$ Our proof is complete by strong induction. QED.