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Math Help - Porperty of B-spline

  1. #1
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    Porperty of B-spline

    Hello,

    Assuming that a B-spline N_{m} is defined for some m \in \mathbb{N} then the B-spline N_{m+1} is defined by:

    N_{m+1}(x)=\int_{0}^{1}N_{m}(x-t))dt

    How can I, through induction, show that:

    \int_{-\infty}^{\infty}N_{m}(x)dx=1

    Suggestions would be greatly appreciated.
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  2. #2
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    How is N_1(x) defined?
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  3. #3
    Ase
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    Hi

    Working on the same problem. So far I have shown that the property is true for m = 1. My line of thought is then to show that the property is true for m+1. I am however struggling with a double integral.

    I do not remember exactly how N_m is characterized. That makes it a bit hard to integrate. (I know that there is an explicit formula for the B-splines, but good luck integrating that.) I was thinking that there must be a better way to do the integration.
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  4. #4
    Ase
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    Hi HallsofIvy

    Did'nt see your post there.

    N_1(x) is the characteristic function from [0,1].
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  5. #5
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    B-spline

    Hello,

    So we are considering the statement:

    S_{m}: \hspace{0,5cm}\int_{-\infty}^{\infty}N_{m}(x)dx=1

    As ase mentioned the statement S_{1} is true for m=1. This establishes the basis for the induction.

    Now we assume that S_{m} is true for a fixed value of m. Using the induction hypothesis we want to derive S_{m+1}. We get:

    \int_{-\infty}^{\infty}N_{m+1}(x)dx=\int_{-\infty}^{\infty}\int_{0}^{1}N_{m}(x-t)dt

    From here I'm not sure. This is where Ase got stuck (I presume). Suggestions are appreciated greatly. Thanks.
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  6. #6
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    Quote Originally Posted by surjective View Post
    Hello,

    Assuming that a B-spline N_{m} is defined for some m \in \mathbb{N} then the B-spline N_{m+1} is defined by:

    N_{m+1}(x)=\int_{0}^{1}N_{m}(x-t))dt

    How can I, through induction, show that:

    \int_{-\infty}^{\infty}N_{m}(x)dx=1

    Suggestions would be greatly appreciated.
    Are you familiar with the convolution of functions? because if so this problem is easy since N_m = f\ast N_{m-1} where f=1_{[0,1]} and it's well known that since both f and N_{m-1} are in L^1(\mathbb{R} ) then \int_{\mathbb{R} } N_m = \left( \int_{\mathbb{R} } f \right) \left( \int_{\mathbb{R} } N_{m-1} \right) = 1
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  7. #7
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    Property revisited

    Thanks for the reply. Not sure if I get the convultion-idea. I know the definition of convultion and that by definition:

    N_{m}=N_{1}*N_{1}*N_{1}*\cdots *N_{1}

    Could you maybe outline your comment one more time.

    Furthermore, I was wondering if the intended could be shown via induction? Is that possible?
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  8. #8
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    Quote Originally Posted by surjective View Post
    Furthermore, I was wondering if the intended could be shown via induction? Is that possible?
    Notice that the idea for a proof I gave uses an inductive argument.

    Now, just remember that for f,g\in L^1(\mathbb{R} ) we define f\ast g(x)= \int_{\mathbb{R} } f(x-t)g(t)dt = \int_{ \mathbb{R} } g(x-t)f(t)dt = g\ast f (x) where the middle equality is by the change of variable theorem. With this definition in hand, and letting f be as I defined it in the previous post we get:

    f\ast N_{m-1}(x) = \int_{ \mathbb{R} } N_{m-1}(x-t)f(t)dt = \int_{0}^{1} N_{m-1}(x-t)dt = N_m(x).

    Now, you're asked to prove that \int_{ \mathbb{R} } f\ast N_{m-1} =1 , but just note:

    \int_{\mathbb{R} } f\ast N_{m-1} = \int_ {\mathbb{R} } \left(\int_{\mathbb{R} } f(x-t)N_{m-1}(t)dt\right) dx = \int_{\mathbb{R} } \left( \int_{\mathbb{R} } f(x-t)N_{m-1}(t)dx\right) dt = \left( \int_{\mathbb{R} } f(y)dy \right) \left( \int_{\mathbb{R} } N_{m-1}(t)dt \right) =1 <br />

    Where we use Fubini (the fact that the theorem applies is left to you) for the second equality and our induction hypothesis for the last.
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  9. #9
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    Property

    Thanks for the response.

    It seems that you are well versed in this subject. I have posted another questions elsewhere about the centered B-spline. Would it be possible for you to have a look at it?

    Appreciate your time.
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