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Math Help - Splines

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    131

    Splines

    Hello,

    i try to solve this problem:

    Let a=x_0 < x_1 <...<x_n =b be a partition of [a,b]. and let V={v \in C^0[a,b] : v_{|[a,b]} \in C^4[x_i-1,x_i] with v(x_i)=y_i for some y_i \in \mathbb{R}. Show that every solution of argmin_{v \in V} \int |v'(x)|^2 dx must be a spline of degree 1.

    Do you have a idea, how i can solve this?
    It is obvious, that there is a 1 Spline in V (the only one).
    Now we have to show that any other element in V is greater than this 1-spline. But how can i do this?

    Regards
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  2. #2
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    May 2011
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    The only spline of degree one that interpolates the points (x_i,y_i) is in fact the piecewise linear interpolant. Let's assume that the data points are taken from some function f such that y_i = f(x_i) and denote the piecewise linear interpolant by g = If. Now, let h \in V be some other function that interpolates the data points, and define e = h - g. Then,

    \int_a^b (h')^2 dx = \int_a^b (e'+g')^2 dx = \int_a^b (e')^2 dx + 2\int_a^b e'g' dx + \int_a^b (g')^2 dx

    If you can show that the second term on the right, i.e. \int_a^b e'h' dx is zero, then you may conclude that
    \int_a^b (g')^2 dx \leq \int_a^b (h')^2 dx.

    In other words, now you must show that e' and the degree zero spline g' are orthogonal. This is quite easy to show:
    \int_a^b e'g' dx = \sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} e'g' dx
    First, we have divided the interval [a,b] into parts [x_i,x_{i+1}], then we use integration by parts
     \sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} e'g' dx = \sum_{i=1}^{n-1} \left( eg' \mid_{x_i}^{x_{i+1}} - \int_{x_i}^{x_{i+1}} eg'' dx \right)
    Since g' is a degree zero spline, its derivative is obviously zero, so the integral on the right is zero. Only the sum remains, and now we use the fact that g=Ih=If, so we may write e=h-Ih. We find that
    \sum_{i=1}^{n-1} e(x)g'(x) \mid_{x_i}^{x_{i+1}} = \sum_{i=1}^{n-1} (h(x_{i+1}) - g(x_{i+1}))g'(x_{i+1}) - (h(x_{i}) - g(x_{i}))g'(x_{i}) = 0
    This last part is obviously zero because both h and g satisfy the same interpolation conditions at the points x_i.

    I based this on a similar proof in a compendium that was written for a course in splines which I'm taking. Here the proof was about the cubic spline interpolant which minimizes the integral of the second derivative squared. If you're curious, send me a PM and I'll see if I can find a link.
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