Find a polynomial satisfying given conditions

Find a polynomial satisfying
$P'(x) - 3P(x) = 4 - 5x + 3x^2.$

Prove that there is only one such polynomial.
Given a polynomial $Q(x)$ , prove there is exactly one polynomial $P(x)$ such that
$P'(x) - 3P(x) = Q(x).$

Proof. (Finding the polynomial will prove that it is unique since we will not have any choices to make while deriving the polynomial $P(x)$ .) First, we write
$\begin{align*} P(x) = \sum_{k=0}^n c_k x^k && \implies && P'(x) &= \sum_{k=1}^n c_k (k) x^{k-1} \\ &&&&& = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k. \end{align*}$

Thus, we have

$P'(x) - 3P(x) = \sum_{k=0}^{n-1} (k+1)c_{k+1} x^k - \sum_{k=0}^n 3c_k x^k = \sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n.$

Setting this equal to $4 - 5x + 3x^2$ we have

$\sum_{k=0}^{n-1} ((k+1)c_{k+1} - 3c_k)x^k - 3c_n x^n = 4 - 5x + 3x^2.$

But, this implies $n = 2$ and $c_2 = -1$ since $-3c_n x^n$ is the only $x^n$ term on the left (so if $n > 2$ , then we couldn’t have $x^2$ the largest power of $x$ on the right). Therefore, $P(x)$ is a degree polynomial and with $c_2 = -1$ , so we have

$P(x) = c_0 + c_1 x - x^2 \quad \implies \quad P'(x) = c_1 - 2x.$

Hence we have

$\begin{align*} P'(x) - 3P(x) = 4 - 5x + 3x^2 && \implies && c_1 - 2x - 3c_0 - 3c_1 x + 3x^2 &= 4 - 5x + 3x^2 \\ && \implies && (c_1 - 3c_0) - (2+3c_1)x &= 4 - 5x. \end{align*}$

Thus we have the equations

$c_1 - 3c_0 = 4 \quad \text{and} \quad 2+3c_1 = 5.$

These uniquely determine $c_0$ and $c_1$ ,

$c_1 = 1,\quad c_0 = -1.$

Hence, there is a unique $P(x)$ satisfying this equation,

$P(x) = -1 + x - x^2. \qquad \blacksquare$
Proof. Let $Q(x)$ be a given polynomial and suppose there exist two polynomials $P_1 (x)$ and $P_2(x)$ such that
$P'(x) - 3P(x) = Q(x) \quad \text{and} \quad R'(x) - 3R(x) = Q(x).$

This implies

$(P'(x) - R'(x)) - 3(P(x) - R(x)) = 0 \quad \implies \quad (P(x) - R(x))' - 3(P(x) - R(x)) = 0.$

Now, if $P(x) - R(x) \neq 0$ then it is of degree $n$ for some $n \geq 1$ . We know its derivative has degree $n-1$ (Apostol, Page 166). But then, this would imply

$(P(x) - R(x))' - 3(P(x) - R(x))$

has degree $n$ (since the coefficient of $x^n$ in $(P(x) - R(x))'$ is zero since it is degree $n-1$ , and the coefficient of $3(P(x) - R(x))$ is nonzero since it has degree $n$ ). But we know this difference is 0, which means it cannot have degree $n$ for any $n \geq 1$ . Thus, we must have $P(x) - R(x) = 0$ or $P(x) = R(x). \qquad \blacksquare$

One comment

January 10, 2021 at 6:17 pm

Anonymous says:

Not to be a pedant, but for part b you would need to show that at least one solution exists before you can proceed to the step of showing that any two solutions must be equal. Though this could be easily shown like in part a, so if you were like me, you were just too lazy to have to write that all out in LaTeX :)

Stumbling Robot

Find a polynomial satisfying given conditions

Related

One comment

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply