Example 18 Let the quadratic function $f(x)=a x^{2}+b x+c(a \neq 0)$ have values whose absolute values do not exceed 1 on the interval $[0,1]$, find the maximum value of $|a|+|b|+|c|$.
Solve: From

we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(0)=c \\
f(1)=a+b+c \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c
\end{array}\right. \\
\left\{\begin{array}{l}
a=2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1) \\
b=-3 f(0)+4 f\left(\frac{1}{2}\right)-f(1) \\
c=f(0)
\end{array}\right. \\
\begin{array}{l}
f=f(0)
\end{array} \\
b=
\end{array}
$$

Therefore,
$$
\begin{aligned}
|a|+|b|+|c|= & \left|2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1)\right|+\mid-3 f(0)+ \\
& \left.4 f\left(\frac{1}{2}\right)-f(1)|+| f(0)|\leqslant 2| f(0)|+4|-f\left(\frac{1}{2}\right) \right\rvert\,+ \\
& 2|f(1)|+3|f(0)|+4\left|-f\left(\frac{1}{2}\right)\right|+|f(1)|+|f(0)|= \\
& 6|f(0)|+8\left|-f\left(\frac{1}{2}\right)\right|+3|f(1)| \leqslant
\end{aligned}
$$
$$17$$

Equality holds if and only if $\left\{\begin{array}{l}f(0)=1 \\ f\left(\frac{1}{2}\right)=-1 \\ f(1)=1\end{array}\right.$, i.e., $\left\{\begin{array}{l}a=8 \\ b=-8 \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}f(0)=-1 \\ f\left(\frac{1}{2}\right)=1 \\ f(1)=-1\end{array}\right.$, i.e., $\left\{\begin{array}{l}a=-8 \\ b=8 \\ c=-1\end{array}\right.$. In this case, the quadratic function is $f(x)=8 x^{2}-8 x+1$ or $f(x)=-8 x^{2}+8 x-1$. If we let $x=\cos ^{2} x$, then at this time, $|5(x)|=|\cos 4 \alpha| \leqslant 1$.

Note: Example 18 can be generalized: Suppose the cubic function $f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$ on the interval $[0,1]$, the absolute value of its values is no greater than 1, find the maximum value of $|a|+|b|+|c|+|d|$.

Solution: The maximum value can be obtained as $99$. Specifically, $a=\frac{1}{3}\left[-16 f(0)+32 f\left(\frac{1}{4}\right)-32 f\left(\frac{3}{4}\right)+16 f(1)\right], b=$ $\frac{1}{3}\left[32 f(0)-56 f\left(\frac{1}{4}\right)+40 f\left(\frac{3}{4}\right)-16 f(1)\right], c=\frac{1}{3}\left[-19 f(0)+24 f\left(\frac{1}{4}\right)-\right.$ $\left.8 f\left(\frac{3}{4}\right)+3 f(1)\right], d=f(0) ; f(0)=-1, f\left(\frac{1}{4}\right)=1, f\left(\frac{3}{4}\right)=-1, f(1)=1$; or $f(0)=1, f\left(\frac{1}{4}\right)=-1, f\left(\frac{3}{4}\right)=1, f(1)=-1$, the cubic function is $f(x)=32 x^{3}-$ $48 x^{2}+18 x-1$ or $f(x)=-32 x^{3}+48 x^{2}-18 x+1$.

More generally, for an $n$-degree function $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, on the interval $[0,1]$, if $|f(x)| \leqslant 1$, then when
$$
\begin{array}{c}
f\left(\cos ^{2} \frac{n \pi}{2 n}\right)=-1, f\left(\cos ^{2} \frac{(n-1) \pi}{2 n}\right)=1 \\
f\left(\cos ^{2} \frac{(n-2) \pi}{2 n}\right)=-1, \cdots, f\left(\cos ^{2} 0\right)=(-1)^{n+1}
\end{array}
$$

or when
$$
\begin{array}{c}
f\left(\cos ^{2} \frac{n \pi}{2 n}\right)=1, f\left(\cos ^{2} \frac{(n-1) \pi}{2 n}\right)=-1 \\
f\left(\cos ^{2} \frac{(n-2) \pi}{2 n}\right)=1, \cdots f\left(\cos ^{2} 0\right)=(-1)^{n}
\end{array}
$$

$\sum_{i=1}^{n}\left|a_{i}\right|$ has the maximum value.