= Solution
A <natural cubic spline interpolant> $s$ satisfies $s(x_i)=y_i$, is twice continuously differentiable, and is a <polynomial> of degree at most three on each interval $[x_i,x_{i+1}]$. Its natural <boundary conditions> are
$$
\boxed{s''(x_1)=s''(x_n)=0.}
$$
If defined outside the data interval, it continues linearly on both exterior intervals. Thus function values and the first two <derivatives> match at the knots, but the third <derivative> may jump.
For an explicit construction, let $h_i=x_{i+1}-x_i>0$ and $M_i=s''(x_i)$. Set $M_1=M_n=0$ and solve
$$
h_{i-1}M_{i-1}+2(h_{i-1}+h_i)M_i+h_iM_{i+1}
=6\left\{\frac{y_{i+1}-y_i}{h_i}-\frac{y_i-y_{i-1}}{h_{i-1}}\right\},\quad2\le i\le n-1.
$$
On $[x_i,x_{i+1}]$, put
$$
\begin{aligned}
s(x)={}&\frac{M_i(x_{i+1}-x)^3+M_{i+1}(x-x_i)^3}{6h_i}\\
&+\left(y_i-\frac{M_ih_i^2}{6}\right)\frac{x_{i+1}-x}{h_i}
+\left(y_{i+1}-\frac{M_{i+1}h_i^2}{6}\right)\frac{x-x_i}{h_i}.
\end{aligned}
$$
This formula interpolates the values and matches second <derivatives>; the tridiagonal equations enforce first-derivative matching. The homogeneous system has only the zero solution: in a row with maximal nonzero $|M_i|$, its diagonal coefficient strictly exceeds the sum of its neighbour coefficients, making cancellation impossible. Thus the natural interpolant exists uniquely.
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