# Question abbc9

Nov 2, 2016

A.
P(0) = n!

#### Explanation:

Given a polynomial

$P \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0} = {\sum}_{i = 0}^{n} {a}_{i} {x}^{i}$

we can take its derivative to find

$P ' \left(x\right) = n {a}_{n} {x}^{n - 1} + \left(n - 1\right) {a}_{n - 1} {x}^{n - 2} + \ldots + 2 {a}_{2} x + {a}_{1}$

$= {\sum}_{i = 0}^{n - 1} \left(i + 1\right) a \left(i + 1\right) {x}^{i}$

or evaluate $P \left(x\right)$ at $0$ to get

$P \left(0\right) = {a}_{n} \cdot 0 + {a}_{n - 1} \cdot 0 + \ldots + {a}_{1} \cdot 0 + {a}_{0} = {a}_{0}$

thus we can see that our goal is to find ${a}_{0}$.

As we have $P \left(x\right) - P ' \left(x\right) = {x}^{n}$, we can add $P ' \left(x\right)$ to both sides to get

$P \left(x\right) = {x}^{n} + P ' \left(x\right)$

Substituting in our arbitrary polynomial, we have

${\sum}_{i = 0}^{n} {a}_{i} {x}^{i} = {x}^{n} + {\sum}_{i = 0}^{n - 1} \left(i + 1\right) {a}_{i + 1} {x}^{i}$

As corresponding coefficients must be equal, we can equate the coefficients on each side:

${a}_{i} = \left\{\begin{matrix}1 \mathmr{if} i = n \\ \left(i + 1\right) {a}_{i + 1} \mathmr{if} 0 \le i < n\end{matrix}\right.$

Working our way through the coefficients, starting from ${a}_{n} = 1$, that gives

${a}_{n - 1} = \left(n - 1 + 1\right) \left(1\right) = n$

${a}_{n - 2} = \left(n - 2 + 1\right) \left(n\right) = n \left(n - 1\right)$

${a}_{n - 3} = \left(n - 3 + 1\right) \left(n \left(n - 1\right)\right) = n \left(n - 1\right) \left(n - 2\right)$

At this point we can notice that pattern that each time, we are adding one more term of n!. This can be expressed concisely as a_i = (n!)/(i!)

So, plugging in our target of ${a}_{0}$, we get

a_0 = (n!)/(0!) = (n!)/1 = n!#