function with zero first to n'th derivative at end points

Question

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

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \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" !!!.

Answer 3

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}

Answer 4

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.