function with zero first to n'th derivative at end points As an extension of my earlier question, 
It is required to find f(x) with following properties,
$$ f(0) = 0 \hspace{1cm}f(1) = 1 \\
f'(0) = 0 \hspace{1cm}f'(1) = 0 \\
f''(0) = 0 \hspace{1cm}f''(1) = 0 \\
f'''(0) = 0 \hspace{1cm}f'''(1) = 0 \\
\dots \\
f^{(n)}(0) = 0 \hspace{1cm}f^{(n)}(1) = 0 \\
$$
and most important condition being $f'(\xi) \ge = 0 \hspace{.5cm}\forall \xi \in (0,1)$. Here prime denotes derivative of the function.
I have a solution for n = 1 and n= 2 as follows respectively(I don't assume that they are unique),
\begin{eqnarray}
f_1(x) &=&  \frac{1}{2} \left[1- \cos(\pi x)\right] \\
f_2(x) &=& x - \frac{\sin(2\pi x)}{2\pi}
\end{eqnarray}
With graphs as follwing, 

It is clear from the graph that with increasing n, the solution would approach step function with jump at $x = 0.5$. I will be happy if someone can point out solution for n = 3 or higher degrees. Thanks for the attention
 A: We know that
$$1 = (x+(1-x))^{2n+1} = \sum_{i=0}^{2n+1} {2n+1 \choose i} x^i(1-x)^{2n+1-i}$$
This can be separated into two symmetrical pieces:
$$1 = f(x) + f(1-x)$$
$$f(x) =\sum_{i=n+1}^{2n+1} {2n+1 \choose i} x^i(1-x)^{2n+1-i}$$
Since $f(x)$ has a factor of $x^{n+1}$ it has $n$ zero derivatives at $x=0$. Symmetry ensures the same derivatives are zero at $x=1$. And the end points are correct too. 
A: Do not know general solution for n, but n=3,4 has following solution(out of many possible),
\begin{eqnarray}
f_3(x) &=& \frac{1}{12\pi} \left[ -8\sin(2\pi x)+ \sin(4\pi x) + 12\pi x\right] \\
f_4(x) &=& \frac{1}{60\pi} \left[ -45\sin(2\pi x) + 9\sin(6 \pi x) - \sin(6\pi x) + 60\pi x \right]
\end{eqnarray}
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\fermi\pars{x} &= \sum_{k = n + 1}^{\infty}a_{k}x^{k}
=
\sum_{\ell = n + 1}^{\infty}b_{\ell}\pars{x - 1}^{\ell}
=
\sum_{\ell = n + 1}^{\infty}b_{\ell}\pars{-1}^{\ell}
\sum_{k = 0}^{\infty}{\ell \choose k}\pars{-1}^{k}x^{k}
\\[3mm]&=
\sum_{k = 0}^{\infty}x^{k}\pars{-1}^{k}
\sum_{\ell = n + 1}^{\infty}{\ell \choose k}b_{\ell}\pars{-1}^{\ell}
=
\sum_{k = 0}^{\infty}x^{k}\pars{-1}^{k}
\sum_{\ell\ =\ \max\braces{n + 1, k}}^{\infty}{\ell \choose k}b_{\ell}\pars{-1}^{\ell}
\end{align}

$$
a_{k}
=
\pars{-1}^{k}
\sum_{\ell\ =\ \max\braces{n + 1, k}}^{\infty}{\ell \choose k}b_{\ell}\pars{-1}^{\ell} 
$$

At first sight, it seems the only way is "guessing" !!!.
A: Instead of trig functions, how about monomials?  If you want $k$ derivatives to vanish, take $$\begin {cases} f(x)=2^kx^{k+1} &0\le x \le \frac 12\\1-2^k(1-x)^{k+1} & \frac 12 \lt x \le 1 \end {cases}$$  It is continuous, but the derivatives are not.  It looks like a step function as you suggested, more so as $k$ increases.  The first derivative is even continuous.
The classic example of a function with all derivatives zero at $x=0$ is $\exp(\frac {-1}x)$  As you take derivatives you multiply the function by various polynomials in $\frac 1x$, but the exponential goes to zero faster as you approach zero.  Once you have that, you can do many things with it.  The easiest way to get both ends is $$f(x)=\begin {cases} 0& x=0 \\\exp\left(\frac {-1}x\right) & 0 \lt x \le\frac 14\\\exp (-4)+(2-4\exp(-4))(x-\frac 14)& \frac 14 \lt x \lt \frac 34\\1- \exp\left(\frac {-1}{1-x}\right)& \frac 34 \le x \lt 1 \\1& x=1\end{cases}$$  This is continuous, but the derivatives are not (at $\frac 14, \frac 34$).  With some more work, you can make all the derivatives continuous as well.
