Error estimation for splines in polynomial interpolation

Question

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At the bottom part, why is it 1/8 and not 1/2 ? With n=1. Even if n was =2, we would still have 1/6.. what am I missing?

Answer 1

Note that the interpolating spline is linear in this case. Let $s_i(x)$ be the $i$th piece of the spline, and let $f_i$ be the restriction of $f$ to this interval.

From the (Lagrange) error formula for interpolation, we know $$ f_i(x) - s_i(x) = \frac{f''(\xi)}{2} (x - x_{i+1})(x-x_i) $$ so that \begin{align*}|f_i(x) - s_i(x)| &\le \left\vert \frac{ (x - (x_{i+1}))(x- x_i)}{2} \right\vert \max \limits_{\xi \in [x_i,x_{i+1}]} |f''(\xi)| \\ & = \left\vert \frac{ (x - (x_i + h_i))(x- x_i)}{2} \right\vert \max \limits_{\xi \in [x_i,x_{i+1}]} |f''(\xi)| \end{align*}

Now, use elementary calculus to verify that $g(x)= (x - (x_i + h_i))(x- x_i)$ has a local extrema at the point $x_i + \frac{h_i}{2}$ with corresponding output $-h_i^2/4$.