 Solution

The printed product omits the node with index zero. The correct cardinal polynomial isFor example, with two nodes zero and one, values zero and one, and both derivative data zero, the printed index range gives $L_1=1$ and its displayed interpolant is the constant one, which fails the value condition at zero. Thus the source formula cannot be proved literally.

With the corrected product, $L_k(x_j)={\delta }_{kj}$. For $j\ne k$, $L_k^2$ has a double zero at $x_j$, so both basis functions and their derivatives vanish there. At $x_k$, put $\ell =L_k^{\prime }(x_k)$. The value basis $H_k=L_k^2[1-2(x-x_k)\ell ]$ satisfies $H_k(x_k)=1$ and $H_k^{\prime }(x_k)=2\ell -2\ell =0$. The derivative basis $K_k=L_k^2(x-x_k)$ satisfies $K_k(x_k)=0$ and $K_k^{\prime }(x_k)=1$. Both have degree at most $2n+1$. Consequentlyhas every prescribed value and derivative, proving membership in the interpolation class after the required index correction. This corrected Hermite interpolant is used in the following parts.