does the cubic spline generate an interpolation function where the derivative exists for evey point in the domain of the interpolation function
Added by Howard R.
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In cubic spline interpolation, The third derivatives of the splines are continuous at the interior data points. The first derivatives of the splines are continuous at the interior data points. The first and the second derivatives of the splines are continuous at the interior data points. The second derivatives of the splines are continuous at the interior data points.
Sri K.
In the theory of cubic splines, we choose cubic polynomials whose values, derivatives, and second derivatives all agree at the points they meet. This ensures that the joins look smoother than for quadratic splines. Suppose we have the points A = [0, 1], B = [1, 3], C = [2, 2], and D = [3, 2] (you may have to move D in the GeoGebra app): These points are joined by the functions p(x) = 1 + 2x + x^2 - x^3 (blue) on the interval [0, 1], q(x) = 5x - 2x^2 (red) on the interval [1, 2], and r(x) = ax^3 + bx^2 + cx + d (black) on the interval [2, 3]. The cubic polynomials p and q meet at the point B. They share the same first derivative at B, namely p'(1) = q'(1) = 1. They also share the same second derivative at B, namely p''(1) = q''(1) = -4. Now we can try to find the cubic polynomial r(x) which passes through C and D and agrees with q(x) at C up to its second derivative. Can you find it?
Definition 3.10 Given a function f defined on [a, b] and a set of nodes a = x0 < x1 < ... < xn = b, a cubic spline interpolant S for f is a function that satisfies the following conditions: (a) S(x) is a cubic polynomial, denoted Sj(x), on the subinterval [xj, xj+1] for each j = 0, 1, ..., n - 1; (b) Sj(xj) = f(xj) and Sj(xj+1) = f(xj+1) for each j = 0, 1, ..., n - 1; (c) Sj+1(xj+1) = Sj(xj+1) for each j = 0, 1, ..., n - 2; (Implied by (b).) (d) S'j+1(xj+1) = S'j(xj+1) for each j = 0, 1, ..., n - 2; (e) S''j+1(xj+1) = S''j(xj+1) for each j = 0, 1, ..., n - 2;
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