Proving $A_1, A_2, \dotsc, A_k$ are real numbers in an arbitrary partial fraction decomposition.

Question

I have two extra credit assignments for my Calculus 2 class and after attempting them for several days and unsuccessfully trying to find any resources on how to do these questions I've decided to ask here.

The first question is as follows:

Let $\alpha$ be an arbitrary real number. Let k be one of the following numbers: $1, 2, 3, \dotsc$. Let $a_{k-1}, a_{k−2}, \dotsc, a_{0}$ be real numbers. Let $p(x)= a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \dotsb + a_{0}x^{0}$. Prove that there exists real numbers $A_{1}, A_{2}, \dotsc, A_{k}$ such that

$$\frac{p(x)}{(x-\alpha)^{k}} = \frac{A_{1}}{(x-\alpha)} + \frac{A_{2}}{(x-\alpha)^{2}} + \dotsb +\frac{A_{k}}{(x-\alpha)^{k}}$$

The second question is as follows:

Let $\beta$ and $\gamma$ be arbitrary real numbers. Let $m$ be one of the following numbers: $1, 2, 3, \dotsc$. Let $b_{2m-1}, b_{2m-2}, \dotsc, b_{0}$ be real numbers. Let $q(x) = b_{2m-1}x^{2m-1} + b_{2m-2}x^{2m-2} +\dotsb +b_{0}x^{0}$. Prove there exist real numbers $B_{1}, C_{1}, B_{2}, C_{2}, \dotsc, B_{m},C{m}$ such that

$$\frac{q(x)}{(x^{2}+\beta x+\gamma)^{m}} = \frac{B_{1}x+C_{1}}{x^{2}+\beta x+\gamma} + \frac{B_{2}x+C_{2}}{(x^{2}+\beta x+\gamma)^{2}} + \dotsb + \frac{B_{m}x+C_{m}}{(x^{2}+\beta x+\gamma)^{m}}$$

EDIT: I must also state that all $\alpha$, $k$, $a_{k-1}$, $\beta$, $\gamma$, $m$, $b_{2m-1}$, etc must remain arbitrary in the solutions and I cannot plug-in any numbers for them and solve for that to prove the statement based on just that number that was plugged-in.

Any help would be extremely helpful in understanding and solving these problems. I just can't seem to find anything in my book or on the web dealing with a problem like these, only the basic equations for a partial fraction.

Answer 1

Hint: For the first problem, if $\alpha=0$ simply expand $\frac{p(x)}{x^k}$ into a sum of fractions. Otherwise, note that $a_{k-1}x^{k-1} = a_{k-1}(x-\alpha)^{k-1} + \text{lower-order terms}$; continuing (technically, using induction), you can write $p(x)$ as a polynomial in $(x-\alpha)$. Now divide both sides by $(x-\alpha)^k$.

The second problem is similar.

Answer 2

For the first one, multiply the equation by $(x-\alpha)^k$. Then $$ p(x) = \sum_{n=0}^{k-1} A_{k-n} (x-\alpha)^{n}. $$ Now you can either expand the brackets and look at the coefficients of $x^r$, or differentiate both sides $r$ times and set $x=\alpha$. The latter gives you a formula for the $A_r$ directly.

The second one is worse because the differentiation trick doesn't apply. Multiply up in the same way, expand the brackets, and show that the system of equations has a unique solution.