Brunton39

2022-06-29

Let $f\left(x\right)$ be a polynomial function with non-negative coefficients such that $f\left(1\right)={f}^{\prime }\left(1\right)={f}^{″}\left(1\right)={f}^{‴}\left(1\right)=1$. Find the minimum value of $f\left(0\right)$

scoseBexgofvc

Expert

It is perhaps simpler to use Taylor's formula with fixed degree three and a Lagrange remainder:
$f\left(x\right)=1+\left(x-1\right)+\frac{1}{2!}\left(x-1{\right)}^{2}+\frac{1}{3!}\left(x-1{\right)}^{3}+\frac{{f}^{\left(4\right)}\left(\xi \right)}{4!}\left(x-1{\right)}^{4}.$
where $\xi$ is between 1 and x. For $x=0$ is $\xi \ge 0$ and the last term is non-negative, since ${f}^{\left(4\right)}$ has non-negative coefficients as well. This gives
$f\left(0\right)\ge 1-1+\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\phantom{\rule{thinmathspace}{0ex}}.$
The bound is sharp, equality holds for the function
$f\left(x\right)=1+\left(x-1\right)+\frac{1}{2!}\left(x-1{\right)}^{2}+\frac{1}{3!}\left(x-1{\right)}^{3}=\frac{1}{3}+\frac{1}{2}x+\frac{1}{6}{x}^{3}\phantom{\rule{thinmathspace}{0ex}}.$

Do you have a similar question?