Find the natural cubic spline interpolant to $f(x) = e^{x^2}$ at the nodes $\{x_i\}^2_{i=0} = \{−1, 0, 1\}$. Calculate the value of the interpolant at $x = 0.5$. What is the error at this point? S"(a) =0, S"(b) =0 : a=-1 , b=1.
Ideas so far: $S''(a) =0$, $S''(b) =0$ : $a=-1$ , $b=1$. Therefore we can say $S'(a) = f'(a)$ and $S'(b) = f'(b)$.
Then we are evaluating at $x= 0.5$. First we look at the cubic spline from $x=-1$, $x=0$ now we construct the polynomial $S_1(x) = 0.5+b_1(x-(-1))+c_1(x-(-1))^2+d_1(x-(-1)^3$. Now we take $x=0, x=1$ and construct the polynomial $S_2(x) = 0.25+b_2(x-0)+c_2(x-0)^2+d_2(x-0)^3$. The six coefficients $b_1,c_1,d_1,b_2,c_2,d_2$ are the unknowns that we need to determine. From the interpolation conditions, we get $S(0)$ and $S(1)$
The conditions for the natural spline are that the cubic polynomials $S_1(x)$, $S_2(x)$ over $[-1,0]$ resp. $[0,1]$ satisfy \begin{align} S_1(-1)&=f(-1)\\ S_1''(-1)&=0\\ S_1(0)&=S_2(0)=f(0)\\ S_1'(0)&=S_2'(0)\\ S_1''(0)&=S_2''(0)\\ S_2(1)&=f(1)\\ S_2''(1)&=0 \end{align} These are $8$ equations for the $8$ coefficients of the $2$ cubic polynomials.