Question #abbc9

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Answer 1 · 2016-11-02T02:02:48

A.
$P (0) = n!$ $P(0) = n!$

Given a polynomial

$P(x) = a_nx^n + a_(n-1)x^(n-1)+...+a_1x+a_0 = sum_(i=0)^na_ix^i$

we can take its derivative to find

$P'(x) = na_nx^(n-1)+(n-1)a_(n-1)x^(n-2)+...+2a_2x+a_1$

$= sum_(i=0)^(n-1)(i+1)a(i+1)x^i$

or evaluate $P(x)$ at $0$ to get

$P(0) = a_n*0+a_(n-1)*0+...+a_1*0+a_0 = a_0$

thus we can see that our goal is to find $a_0$ .

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

$P(x) = x^n + P'(x)$

Substituting in our arbitrary polynomial, we have

$sum_(i=0)^na_ix^i = x^n+sum_(i=0)^(n-1)(i+1)a_(i+1)x^i$

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

$a_i = {(1 if i=n),((i+1)a_(i+1) if 0 <= i < n):}$

Working our way through the coefficients, starting from $a_n = 1$ , that gives

$a_(n-1) = (n-1+1)(1) = n$

$a_(n-2) = (n-2+1)(n) = n(n-1)$

$a_(n-3) = (n-3+1)(n(n-1)) = n(n-1)(n-2)$

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

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