Find the minimum value of the constant term.  
Let $f(x)$ be a polynomial function with non-negative coefficients such that $f(1)=f’(1)=f’’(1)=f’’’(1)=1$. Find the minimum value of $f(0)$.
 
By Taylor’s formula, we can obtain
 
$$f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\cdots+\frac{f^{(n)}(1)}{n!}(x-1)^n.$$
 
Hence $$f(0)=\frac{1}{2}-\frac{1}{6}+\sum_{k=4}^n\frac{f^{(k)}(1)}{k!}(-1)^k.$$
 
This will help?
 A: It is perhaps simpler to use Taylor's formula with fixed degree three and a Lagrange remainder:
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\frac{f^{(4)}(\xi)}{4!}(x-1)^4.
$$
where $\xi$ is between $1$ and $x$. For $x=0$ is $\xi \ge 0$ and the last term is non-negative, since $f^{(4)}$ has non-negative coefficients as well. This gives
$$
 f(0) \ge 1 - 1 + \frac 12 - \frac 16 = \frac 13 \, .
$$
The bound is sharp, equality holds for the function
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3 = \frac 13 + \frac 12 x + \frac 16 x^3 \, .
$$
A: Motivation: First, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4$.
Second, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4 + 4a_5$.
Thus, we think we can eliminate $a_1, a_2, a_3$
from $f(1) = f'(1) = f''(1) = f'''(1) = 1$.
The operation $f(1) - f'(1) + \frac12 f''(1) - \frac16 f'''(1)$
can do this.
A solution without Taylor's theorem:
Let $f(x) = \sum_{k=0}^n a_k x^k$.
We have
\begin{align*}
 f'(x) &= \sum_{k=1}^n k a_k x^{k - 1}, \\
 f''(x) &= \sum_{k=2}^n k(k - 1) a_k x^{k - 2}, \\
 f'''(x) &= \sum_{k=3}^n k(k - 1)(k - 2) a_k x^{k - 3}.
\end{align*}
Then we have
\begin{align*}
 &f(x) - f'(x) + \frac12 f''(x) - \frac16 f'''(x)\\
 =\,\, & a_0 + a_1(x - 1) + a_2(x - 1)^2 + a_3(x - 1)^3\\
 &\quad + \sum_{k=4}^n a_k \left(x^k - k x^{k - 1} + \frac{k(k - 1)}{2}x^{k - 2} - \frac{k(k - 1)(k - 2)}{6}x^{k - 3}\right).
\end{align*}
Thus, we have
$$f(1) - f'(1) + \frac12 f''(1) - \frac16 f'''(1)
= a_0 - \sum_{k=4}^n \frac{(k - 1)(k - 2)(k - 3)}{6} a_k $$
which results in
$$a_0 = \frac13 + \sum_{k=4}^n \frac{(k - 1)(k - 2)(k - 3)}{6} a_k.$$
Thus, $f(0) = a_0 \ge \frac13$
with equality if $a_0 = \frac13$ and
$a_4 = a_5 = \cdots = a_n = 0$.
Using $f(x) = \frac13 + a_1 x + a_2 x^2 + a_3x^3$
and $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we have $a_1 = \frac12, a_2 = 0, a_3 = \frac16$.
Thus, $f(0) = \frac13$ is attained when $f(x) = \frac{1}{3} + \frac12 x + \frac16 x^3$.
