I was studying the Rule 90 cellular automaton and came across a family of polynomials defined by
\begin{equation}
D_n(x)=\begin{cases}
\displaystyle\sum_{k=0}^{m}(-1)^{m+k}\binom{m+k}{m-k}x^{2k}\ , & n=2m\\
\displaystyle\sum_{k=0}^{m}(-1)^{m+k}\binom{m+k+1}{m-k}x^{2k+1}\ , & n=2m+1
\end{cases}\ .
\end{equation}
As visible in the figure they all seem to be bounded by some curve which I tried to find.
I noticed that
\begin{equation}
D_{2m}=\big(D_m+D_{m-1}\big)\big(D_m-D_{m-1}\big)\equiv d_m^{+}d_m^{-}
\end{equation}
and decided to investigate these factors separately. When plotting them, it seemed like the extrema of $d^+$ are very close to the roots of $d^-$ and vice versa. 
This, however, cannot be true as both $d^+$ and $d^-$ are of the same degree so one of the roots of $d^-$ could not correspond to an extremum of $d^+$. Further investigation showed that for larger values of $m$ this root was moving towards $x=2$ for $d^-$ ($x=-2$ for $d^+$) which, together with $d_m^+(x)=(-1)^m d_m^-(-x)$ led me to consider the differential equation
\begin{equation}
-2(x-2)f'(x)=f(-x)
\end{equation}
solved by
\begin{equation}
f(x)=a_0\bigg(1+\frac{x}{4}\ {}_2F_1\Big(\frac{1}{4},\frac{5}{4};\frac{1}{2};\frac{x^2}{4}\Big)+\frac{x^2}{16}\ {}_2F_1\Big(\frac{5}{4},\frac{5}{4};\frac{3}{2};\frac{x^2}{4}\Big)\bigg)
\end{equation}
in terms of the ordinary hypergeometric function ${}_2F_1$ for some constant $a_0$.
For $a_0=\sqrt{2}$, $E(x)\equiv f(x)f(-x)-1$ is an unreasonably good approximation for the enveloping function I was after as can be seen by the figure 
and I don't understand why.
The result I obtained is not exactly the enveloping function I was looking for and the way I obtained it was really just guessing patterns and solving the equations it suggested, hence the term "unreasonable". Can someone explain why it is such a good fit and if there is a more systematic way to find such an envelope?
Update:
I played around with the parameter $a_0$ and I reckon there is a value close to $1.22$ such that a function similar to $E(x)$ (just shifted such that $\tilde E(x=0)=1$) gives the "actual" envelope. Here is an updated figure with $a_0=1.22$
