Show that for f(x) = x²e^(-x/a), f^((n))(0) = (n (n-1)

Show that for $f(x) = x^{2}e^{-x/a}$, \[ f^{(n)}(0) = \frac{n\bigl(n-1\bigr)\bigl(-1\bigr)^{n}}{a^{\,n-2}}. \].

Step 1: Why the Leibniz sum is short

Put $u = x^{2}$ and $v = e^{-x/a}$. Then

\[ u = x^{2}, \qquad u' = 2x, \qquad u'' = 2, \qquad u^{(k)} = 0 \ \ \bigl(k\ge3\bigr), \]

so of the $n+1$ terms in the Leibniz formula only three can be non-zero: \[f^{(n)}(x) = \binom{n}{0}x^{2}v^{(n)}+\binom{n}{1}2x\,v^{(n-1)}+\binom{n}{2}\cdot2\cdot v^{(n-2)}. \tag{1}\]

Step 2: The derivatives of the exponential

\[ v^{(j)}(x) = \left(-\frac1a\right)^{j}e^{-x/a} \;\Longrightarrow\; v^{(j)}(0) = \frac{\bigl(-1\bigr)^{j}}{a^{j}}. \]

Step 3: Evaluate (1) at $x = 0$

The first term carries the factor $x^{2}$ and the second the factor $2x$; both vanish at the origin. Only the third survives: \[f^{(n)}(0) = \binom{n}{2}\cdot2\cdot v^{(n-2)}(0) = \frac{n\bigl(n-1\bigr)}{2}\cdot2\cdot\frac{\bigl(-1\bigr)^{\,n-2}}{a^{\,n-2}}.\]

Step 4: Tidy the sign

Since $\bigl(-1\bigr)^{\,n-2} = \bigl(-1\bigr)^{n}\bigl(-1\bigr)^{-2} = \bigl(-1\bigr)^{n}$: $f^{(n)}(0) = \frac{n\bigl(n-1\bigr)\bigl(-1\bigr)^{n}}{a^{\,n-2}}$.

which is the required formula.

Step 5: Verification

\[n = 2:\ \frac{2\cdot1\cdot1}{a^{0}} = 2; \qquad n = 3:\ \frac{3\cdot2\cdot(-1)}{a} = -\frac6a; \qquad n = 4:\ \frac{4\cdot3}{a^{2}} = \frac{12}{a^{2}} \;\;\checkmark\]

\[\text{Series check: } x^{2}e^{-x/a} = x^{2}-\frac{x^{3}}{a}+\frac{x^{4}}{2a^{2}}-\cdots, \text{ so } f^{(4)}(0) = 4!\cdot\frac{1}{2a^{2}} = \frac{12}{a^{2}} \;\;\checkmark\]

\[n = 0 \text{ and } n = 1 \text{ give } 0, \text{ matching } f(0) = f'(0) = 0 \;\;\checkmark\]