Problem source

Let $$ {\rm S}_n(x)=\mathrm{e}^{x^3}{{\d^n}\over{\d x^n}}{(\mathrm{e}^{-x^3})}.$$
Show that ${\rm S}_2(x)=9x^4-6x$ and find ${\rm S}_3(x)$. Prove by induction on $n$ that    ${\rm S}_n(x)$ is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient  of the highest power of $x$.
Show also that if $\displaystyle \frac{\d S_n}{\d x}=0$ for  some value $a$ of $x$, then $S_n(a)S_{n+1}(a)\le0$.

Solution source

\begin{align*}
&& S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\
&&&=  e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\
&&&=  e^{x^3}  \left [e^{-x^3}(9x^4-6x) \right] \\
&&&=9x^4-6x \\
\\
&& S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\
&&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\
&&&= e^{x^3}  e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\
&&&= -27x^6 +54x^3-6
\end{align*}

Claim: $S_n$ is a polynomial of degree $2n$ with leading coefficient $(-3)^n$.

Proof: Clearly this is true for $n = 1, 2, 3$ by demonstration. Suppose it is true for some $n = k$, then

\begin{align*}
&& S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\
&& (-3)^kx^{2k} +\cdots &=  e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right]  \\
\Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left (  (-3)^kx^{2k} +\cdots\right) \\
\Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left (  (-3)^kx^{2k} +\cdots\right) + e^{-x^3}  S_k'(x) \\
&&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\
\Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)
\end{align*}

And therefore $S_{k+1}$ is a polynomial degree $2(k+1)$ with leading coefficient $(-3)^{k+1}$ so by induction it's true for all $n$.

If $S'_n(a) = 0$ then $S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0$