Polynomial interpolation I need to find the polynomial of degree $3$ with respect to these conditions:
$$\begin{cases} p(0) = 1\\
 p(1) = -1\\   
 p'(0) = 1\\   
 p''(0) = 0
\end{cases}$$
How do I deal with the condition on the second derivative?  
 A: More generally speaking, consider the cubic polynomial as $$p(x)=a x^3+bx^2+cx+d$$ for which the first and second derivatives are given by $$p'(x)=3a x^2+2bx+c$$ $$p''(x)=6ax+2b $$and now apply the conditions in the order they are given in the post. So,$$p(0)=d=1$$ $$p(1)=a+b+c+d=-1$$ $$p'(0)=c=1$$ $$p''(0)=2b=0$$ So, you have four simple equations to solve for $a,b,c,d$ from which $b=0$, $c=1$, $d=1$, $a=-3$.
Just to make the problem more general, suppose that instead you were given the conditions $p(0)=\alpha$, $p(1)=\beta$, $p'(0)=\gamma$, $p''(0)=\delta$, the same procedure would lead to $a=-\alpha +\beta -\gamma -\frac{\delta }{2}$, $b=\frac{\delta }{2}$, $c=\gamma$, $d=\alpha$.
A: The first, third and fourth constraints give:
$$ p(x) = 1+x+ax^3 $$
(just consider the RHS as a Taylor series in $x=0$), hence by plugging in the third constraint we get:
$$ p(x) = 1+x-3x^3.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$$
\,{\rm p}\pars{x}
=\overbrace{\,{\rm p}\pars{0}}^{\ds{=}\ \dsc{1}}\ +\ 
\overbrace{\,{\rm p}'\pars{0}}^{\ds{=}\ \dsc{1}}\ x +\ \ \half\,\
\overbrace{\,{\rm p}''\pars{0}}^{\ds{=}\ \dsc{0}}\ x^{2}\ +\
{1 \over 6}\,\,{\rm p}'''\pars{0}x^{3}
$$

$$
\,{\rm p}\pars{x}
=1 + x +  {1 \over 6}\,\,{\rm p}'''\pars{0}x^{3}
$$

$$
\,{\rm p}\pars{1}=-1
=1 + 1 +  {1 \over 6}\,\,{\rm p}'''\pars{0}1^{3}
=2 + {1 \over 6}\,\,{\rm p}'''\pars{0}\quad\imp\quad\,{\rm p}'''\pars{0}=-18
$$

$$\color{#66f}{\large\,{\rm p}\pars{x}}=\color{#66f}{\large 1 + x -  3x^{3}}
$$

