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Compute the derivatives of g(x) = xn f(x)

by RoRi

Assume f is a polynomial with f(0) = 1. Define

    \[ g(x) = x^n f(x). \]

Compute the values of g(0), g'(0), \ldots, g^{(n)}(0).

Assume n is a non-negative integer (otherwise g is undefined at x = 0). Then, we make the following claim:

Claim: The polynomial g(x) has derivatives at 0 given by the following

    \[ g^{(k)}(0) = \begin{cases} 0 & \text{if } 0 \leq k < n \\ n! & \text{if } k =n. \end{cases} \]

Proof. Since f is a polynomial we may write,

    \[ f(x) = a_m x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0. \]

Furthermore, since f(0) = 1 is given we know a_0 = 1. Now, multiplying by x^n we have

    \begin{align*} g(x) &= x^n f(x) = x^n (a_m x^m + \cdots + a_1 x + 1) \\ &= a_m x^{m+n} + \cdots + a_1 x^{n+1} + x^n. \end{align*}

Next, we will use induction to prove that the kth derivative of g(x) for 0 \leq k < n is given by

    \[ g^{(k)}(x) = c_m x^{m+n-k} + c_{m-1} x^{m+n-k-1} + \cdots + c_1 x^{n+1-k} + \frac{n!}{\cdot (n-k)!} x^{n-k},\]

for constants c_1, \ldots, c_m. Since the derivative g'(x) is given by

    \begin{align*} g'(x) &= (m+n) a_m x^{m+n-1} + (m+n-1) a_{m-1} x^{m+n-2} + \cdots + (n+1) a_1 x^n + n x^{n-1} \\ &= b_m x^{m+n-1} + b_{m-1} x^{m+n-2} + \cdots + b_1 x^n + \frac{n!}{(n-1)!}x^{n-1}, \end{align*}

for constants b_1, \ldots, b_m, we see that the formula holds for k = 1. Assume then that it holds for some k,

    \[ g^{(k)}(x) = b_m x^{m+n-k} + b_{m-1} x^{m+n-k-1} + \cdots + b_1 x^{n+1-k} + \frac{n!}{(n-k)!}x^{n-k}. \]

Then, taking the derivative of this we have,

    \begin{align*} g^{(k+1)}(x) &= b_m (m+n-k) x^{m+n-k-1} + b_{m-1} (m+n-k-1) x^{m+n-k-2} + \cdots + b_1 (n+1-k) x^{n+1-k-1} + \frac{n!}{(n-k)!} (n-k) x^{n-k-1)} \\ &= c_m x^{m+n-(k+1)} + c_{m-1} x^{m+n-(k+1)-1} + \cdots + c_1 x^{n+1-(k+1)} + \frac{n!}{(n-(k+1))!} x^{n-(k+1)}. \end{align*}

Hence, the formula holds for all 0 \leq k \leq n. But then, if 0 \leq k < n we have

    \begin{align*} &&g^{(k)}(x) &= b_m x^{m+n-k} + b_{m-1} x^{m+n-k-1} + \cdots + b_1 x^{n+1-k} + \frac{n!}{(n-k)!}x^{n-k} \\ \implies g^{(k)}(0) &= 0. \end{align*}

If k =n then x^{n-k} = x^0 = 1 for all x; hence,

    \[ g^{(n)}(0) &= \frac{n!}{0!} x^0 = n!. \qquad \blacksquare\]

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