Problem source

Let $\mathrm{f}(x)=\sin2x\cos x.$ Find the 1988th derivative of $\mathrm{f}(x).$

Show that the smallest positive value of $x$ for which this derivative
is zero is $\frac{1}{3}\pi+\epsilon,$ where $\epsilon$ is approximately
equal to 
\[
\frac{3^{-1988}\sqrt{3}}{2}.
\]

Solution source

\begin{align*}
&& f(x) &= \sin 2x \cos x \\ 
&&&= \frac12 \l \sin 3x + \sin x \r  \\
\Rightarrow && f^{(1988)}(x) &= \frac12 \l  3^{1988} (-1)^{994} \sin 3x+ (-1)^{994} \sin x \r \\
&&&= \boxed{\frac12 \left (3^{1998} \sin 3x + \sin x \right)} \\
\\
f^{(1988)}(x) = 0: && 0 &= 3^{1988} \sin 3x + \sin x \\
\Rightarrow && 0 &= 3^{1988} ( 3\sin x-4\sin^3 x) + \sin x \\
\Rightarrow && 0 &= \sin x \left (1+3^{1989}-4\cdot 3^{1988}\sin^{2} x \right)
\end{align*}

Since $\sin x$ will first contribute a zero when $x = \frac{\pi}{2}$ we focus on the second bracket, in particular, we need:

\begin{align*}
&& \sin^2 x &= \frac{3}{4} \left ( 1 + \frac{1}{3^{1988}} \right) \\
\Rightarrow && \sin x &= \frac{\sqrt{3}}2 \left (1 + \frac{1}{2 \cdot 3^{1988}} + \cdots \right )
\end{align*}

Since near $\frac{\pi}{3}$, 
\begin{align*}
\sin (\frac{\pi}{3} + \epsilon) &= \sin \frac{\pi}{3} \cos \epsilon + \cos \frac{\pi}{3} \sin \epsilon \\
&\approx \frac{\sqrt{3}}{2} (1-\epsilon^2 + \cdots ) + \frac{1}{2}(\epsilon + \cdots) \\
&= \frac{\sqrt{3}}2 + \frac12 \epsilon + \cdots
\end{align*}

Therefore by comparison we can see that $x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} 3^{-1988}$ will be a very good approximation for the root.