Show that for f(x) = e^(-x²): 1° f^((n))(0) = -2

Show that for $f(x) = e^{-x^{2}}$: 1° $f^{(n)}(0) = -2\bigl(n-1\bigr)f^{(n-2)}(0)$; 2° $f^{(2n)}(0) = \bigl(-2\bigr)^{n}\bigl(2n-1\bigr)!!$.

Step 1: The differential equation satisfied by $f$

Direct differentiation gives $f'(x) = -2x\,e^{-x^{2}} = -2x\,f(x)$, that is

\[ f'+2xf = 0. \tag{1} \]

This is the key: instead of differentiating $e^{-x^{2}}$ many times, differentiate the equation.

Step 2: Differentiate (1) exactly $n-1$ times

The first term becomes $f^{(n)}$. For the second, Leibniz applied to the product $x\cdot f$ terminates after two terms, because $x'' = 0$: $\bigl(xf\bigr)^{(n-1)} = x\,f^{(n-1)}+\bigl(n-1\bigr)f^{(n-2)}$.

Hence \[f^{(n)}+2x\,f^{(n-1)}+2\bigl(n-1\bigr)f^{(n-2)} = 0. \tag{2}\]

Step 3: Set $x = 0$

The middle term carries the factor $x$ and disappears: $f^{(n)}(0) = -2\bigl(n-1\bigr)f^{(n-2)}(0)$.

which proves part 1°.

Step 4: The starting values

$f$ is even, so every odd-order derivative is an odd function and therefore vanishes at the origin: $f^{(2k-1)}(0) = 0 \quad\bigl(k\ge1\bigr)$.

For the even orders the recurrence starts from

\[ f^{(0)}(0) = 1, \qquad f^{(2)}(0) = -2\cdot1\cdot f^{(0)}(0) = -2. \]

Step 5: Iterate the recurrence

With $n = 2m$ the recurrence reads $f^{(2m)}(0) = -2\bigl(2m-1\bigr)f^{(2m-2)}(0)$. Applying it $m$ times down to $f^{(0)}(0) = 1$ produces $m$ factors of $-2$ and the descending odd numbers $2m-1,\,2m-3,\,\ldots,\,3,\,1$:

\[ f^{(2m)}(0) = \bigl(-2\bigr)^{m}\bigl(2m-1\bigr)\bigl(2m-3\bigr)\cdots3\cdot1 = \bigl(-2\bigr)^{m}\bigl(2m-1\bigr)!!, \]

which is part 2°.

Step 6: Verification

\[\begin{array}{c|c|c} n & \bigl(-2\bigr)^{n}\bigl(2n-1\bigr)!! & \text{from the recurrence} \\ \hline 0 & 1 & f(0) = 1 \\ 1 & -2\cdot1 = -2 & -2\cdot1\cdot1 = -2 \\ 2 & 4\cdot3 = 12 & -2\cdot3\cdot(-2) = 12 \\ 3 & -8\cdot15 = -120 & -2\cdot5\cdot12 = -120 \end{array}\]

\[\text{Series: } e^{-x^{2}} = 1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{6}+\cdots, \text{ so } f^{(6)}(0) = 6!\left(-\frac16\right) = -120 \;\;\checkmark\]

\[\text{Convention } \bigl(-1\bigr)!! = 1 \text{ makes the case } n = 0 \text{ read correctly} \;\;\checkmark\]