Derive some properties of the product of ex with a polynomial

Let

$f(x) = e^x p(x) \qquad \text{where} \qquad p(x) = c_0 + c_1 x + c_2x^2.$

Prove that
$f^{(n)} (0) = c_0 + nc_1 + n(n-1)c_2$

where $f^{(n)}$ denotes the $n$ th derivative of $f$ .
Do part (a) in the case that $p(x)$ is a cubic polynomial.
Find a similar formula and prove it in the case that $p(x)$ is a polynomial of degree $m$ .

For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if $h(x) = f(x)g(x)$ then the $n$ th derivative $h^{(n)}(x)$ is given by

$h^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} f^{(k)} (x) g^{(n-k)}(x).$

So, in the case at hand we have $f(x) = e^x p(x)$ and so

$f^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(x) e^x.$

(Since the $(n-k)$ th derivative of $e^x$ is still $e^x$ for all $n$ and $k$ .)

Proof. From the formula above we have
$f^{(n)}(0) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0).$

But, since $p(x)$ is a quadratic polynomial we have

$\begin{align*} p(x) &= c_0 + c_1 x + c_2 x^2 \\[9pt] p'(x) &= c_1 + 2c_2 x \\[9pt] p''(x) &= 2c_2 \\[9pt] p^{(k)} (x) &= 0 & \text{for all } k \geq 3. \end{align*}$

Hence, we have

$\begin{align*} f^{(n)}(0) &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) \\[9pt] &= c_0 + n c_1 + \frac{n(n-1)}{2} 2c_2 \\[9pt] &= c_0 + n c_1 + n(n-1) c_2. \qquad \blacksquare \end{align*}$
If $p(x)$ is a cubic polynomial we may write,
$p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3.$

Claim: If $f(x) = e^x p(x)$ then

$f^{(n)}(0) = c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3.$

Proof. We follow the exact same procedure as part (a) except now we have the derivatives of $p(x)$ given by

$\begin{align*} p(x) &= c_0 + c_1 x + c_2x^2 + c_3x^3 \\[9pt] p'(x) &= c_1 + 2c_2 x + 3c_3 x^2 \\[9pt] p''(x) &= 2c_2 + 6c_3 x \\[9pt] p^{(3)}(x) &= 6c_3 \\[9pt] p^{(k)}(x) &= 0 & \text{for all } k \geq 4. \end{align*}$

Therefore, we now have

$\begin{align*} f^{(n)}(0) &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 \\[9pt] &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) \\[9pt] &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) + \binom{n}{3} p'''(0) \\[9pt] &= c_0 + nc_1 + \frac{n(n-1)}{2} 2c_2 + \frac{n(n-1)(n-2)}{6} 6c_3 \\[9pt] &= c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3. \qquad \blacksquare \end{align*}$
Claim: Let $p(x)$ be a polynomial of degree $m$ ,
$p(x) = \sum_{k=0}^m c_k x^k.$

Let $f(x) = e^x p(x)$ . Then,

$f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k.$

Proof. Using Leibniz’s formula again, we have

$f^{(n)}(0) = \sum_{k=0}^m \binom{n}{k} p^{(k)}(0).$

But for the degree $m$ polynomial $p(x)$ , we know $p^{(k)}(0) = k!c_k$ if $0 \leq k \leq m$ and $p^{(k)} (0) = 0$ for all $k > m$ . Hence, we have

$f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k. \qquad \blacksquare$