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Thread: Splines

  1. #1
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    Oct 2010
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    Splines

    Hello,

    i try to solve this problem:

    Let $\displaystyle a=x_0 < x_1 <...<x_n =b$ be a partition of [a,b]. and let $\displaystyle V={v \in C^0[a,b] : v_{|[a,b]} \in C^4[x_i-1,x_i]$ with $\displaystyle v(x_i)=y_i $for some $\displaystyle y_i \in \mathbb{R}$. Show that every solution of $\displaystyle argmin_{v \in V} \int |v'(x)|^2 dx$ 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 $\displaystyle (x_i,y_i)$ is in fact the piecewise linear interpolant. Let's assume that the data points are taken from some function $\displaystyle f$ such that $\displaystyle y_i = f(x_i)$ and denote the piecewise linear interpolant by $\displaystyle g = If$. Now, let $\displaystyle h \in V$ be some other function that interpolates the data points, and define $\displaystyle e = h - g$. Then,

    $\displaystyle \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. $\displaystyle \int_a^b e'h' dx$ is zero, then you may conclude that
    $\displaystyle \int_a^b (g')^2 dx \leq \int_a^b (h')^2 dx$.

    In other words, now you must show that $\displaystyle e'$ and the degree zero spline $\displaystyle g'$ are orthogonal. This is quite easy to show:
    $\displaystyle \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 $\displaystyle [a,b]$ into parts $\displaystyle [x_i,x_{i+1}]$, then we use integration by parts
    $\displaystyle \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 $\displaystyle g'$ 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 $\displaystyle g=Ih=If$, so we may write $\displaystyle e=h-Ih$. We find that
    $\displaystyle \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 $\displaystyle h$ and $\displaystyle g$ satisfy the same interpolation conditions at the points $\displaystyle x_i$.

    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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