# 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?

• If you're really stuck, start with a generic 3rd degree polynomial and see what conditions you get on the coefficients. You'll see there's lots of simplification. Commented Dec 12, 2014 at 23:47
• OK thans, I'll try that way Commented Dec 12, 2014 at 23:53

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 $a=-\alpha +\beta -\gamma -\frac{\delta }{2}$, $b=\frac{\delta }{2}$, $c=\gamma$, $d=\alpha$.

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: $$p(x) = 1+x-3x^3.$$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\,{\rm p}\pars{x} =\overbrace{\,{\rm p}\pars{0}}^{\ds{=}\ \dsc{1}}\ +\ \overbrace{\,{\rm p}'\pars{0}}^{\ds{=}\ \dsc{1}}\ x +\ \ \half\,\ \overbrace{\,{\rm p}''\pars{0}}^{\ds{=}\ \dsc{0}}\ x^{2}\ +\ {1 \over 6}\,\,{\rm p}'''\pars{0}x^{3}$$

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

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

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