= Solution
{c}
Truncate the <Chebyshev series> at the full <degree of precision of a quadrature rule>:
$$
s_{2n+1}(x)=\sum_{j=0}^{2n+1}c_jT_j(x),\qquad R_n=f-s_{2n+1}.
$$
Since $|T_j(x)|\leq1$ on $[-1,1]$, the <geometric series> estimate gives
$$
\|R_n\|_\infty\leq\sum_{j=2n+2}^\infty e^{-Kj}
=\frac{e^{-K(2n+2)}}{1-e^{-K}}.
$$
The same bound proves <uniform convergence> of the original <Chebyshev series>, so its sum and the tail estimate are justified. <Gauss-Legendre quadrature> is exact on $s_{2n+1}$. Its positive <quadrature weights> sum to $2$, hence
$$
\begin{aligned}
|I-\widehat I_n|
&=\left|\int_{-1}^1R_n(x)\,dx-\sum_{i=0}^nw_iR_n(x_i)\right|\\
&\leq2\|R_n\|_\infty+\left(\sum_{i=0}^nw_i\right)\|R_n\|_\infty\\
&\leq\frac{4e^{-2K}}{1-e^{-K}}e^{-2Kn}.
\end{aligned}
$$
Therefore one explicit choice proving <exponential convergence of Gaussian quadrature> is
$$
\boxed{C=\frac{4e^{-2K}}{1-e^{-K}},\qquad L=2K.}
$$
Using exactness through <polynomial degree> $2n+1$ doubles the exponent compared with truncating at <polynomial degree> $n$.
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