[Math] function with zero first to n’th derivative at end points

boundary value problemcalculusfunctions

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,
enter image description here

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

Best Answer

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.

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