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Polynomials, Kernels and Derivatives

  1. Sep 10, 2010 #1
    Is there a simple way to show that when we differentiate the following expression (call this equation 1):

    [tex]Y(x) = \frac{1}{n!} \int_0^x (x-t)^n f(t)dt[/tex]

    that we will get the following expression (call this equation 2):

    [tex]Y'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

    It's simple enough to prove "by hand", for low-order polynomials. For instance, for n=1 we would have:

    [tex]Y(x) = \int_0^x (x-t)f(t)dt = x\int_0^x f(t)dt - \int_0^x tf(t)dt[/tex]

    [tex]Y'(x) = \int_0^x f(t)dt + xf(x) - xf(x) = \int_0^x f(t)dt[/tex]

    and it can be proved just as easily for n=2, 3, etc.. But is there a straightforward way to prove it for general n? By induction perhaps? I'm reading a text where the author just goes from Equation 1 to Equation 2, and keeps rolling, w/o giving much in the way of explanation.
  2. jcsd
  3. Sep 10, 2010 #2
    OK, I think I answered my own question.

    Do a binomial expansion on the polynomial term. The kth term in this expansion is given by:

    [tex](x-t)^n = \left(\array{c}n \\ k\endarray\right)x^{n-k}(-t)^k[/tex]

    so the function Y can be rewritten as the sum:

    [tex]Y(x) = \frac{1}{n!}\int_0^x \left[\sum_{k=0}^n \left(\array{c}n \\ k\endarray\right)x^{n-k}(-t)^k\right] \cdot f(t)dt[/tex]

    Take the kth term in this sum:

    [tex]Y_k(x) = \frac{(-1)^k}{n!}\left(\array{c}n \\ k \endarray\right)x^{n-k}\int_0^x t^kf(t)dt[/tex]

    Differentiating w.r.t. x:

    [tex]Y_k'(x) = \frac{(-1)^k}{n!}\left(\array{c}n \\k \endarray\right)(n-k)x^{(n-1)-k}\int_0^x t^kf(t)dt + \frac{(-1)^k}{n!}\left(\array{c}n \\ k\endarray\right)x^nf(x)[/tex]

    Consider the second term. It can be written as:

    [tex]\frac{x^nf(x)}{n!}\left(\array{c}n \\ k\endarray\right)\left(-1\right)^k[/tex]

    and consider the sum of these terms, from 0 to n:

    [tex]S_{n} = \frac{x^nf(x)}{n!}\sum_{k=0}^n \left(\array{c}n\\k\endarray\right)(-1)^k[/tex]

    Since we can write:

    [tex]\sum_{k=0}^n \left(\array{c}n\\k \endarray\right)(-1)^k
    = \sum_{k=0}^n \left(\array{c}n \\ k \endarray\right)(1)^{n-k}(-1)^k = (1-1)^n = 0

    Then clearly the "second term" in Y_k'(x) drops out when we sum the components up.

    Consider now the first term in Y_k'(x):

    [tex]\frac{n-k}{n!}\left(\array{c}n\\k\endarray\right)\int_0^x x^{(n-1)-k} (-t)^kf(t)dt[/tex]

    and specifically, the coefficient on this term:

    [tex]\frac{n-k}{n!}\left(\array{c}n\\k\endarray\right) = \frac{n-k}{n!} \cdot \frac{n!}{k!(n-k)!}[/tex]

    [tex]= \frac{1}{n\cdot (n-1)!} \cdot \frac{n \cdot (n-1)!}{k!((n-1)-k)!}[/tex]

    [tex]= \frac{1}{(n-1)!} \cdot \frac{(n-1)!}{k! ((n-1)-k)!}[/tex]

    [tex]= \frac{1}{(n-1)!} \cdot \left(\array{c}n-1\\k\endarray\right)[/tex]

    So that, ignoring the "second" term on Y_k'(x), which we already showed sums to 0, we can write Y_k'(x) as:

    [tex]Y_k'(x) = \frac{1}{(n-1)!}\int_0^x \left(\array{c}n-1 \\ k\endarray\right)x^{(n-1)-k}(-t)^kf(t)dt[/tex]

    so now summing over 0 to n-1, to get the full expression for Y'(x), we get:

    [tex]Y'(x) = \sum_{k=0}^{n-1} Y_k'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

    which was to be proved..
  4. Sep 10, 2010 #3


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    To differentiate

    Y(x) = \frac{1}{n!} \int_0^x (x-t)^n f(t)dt

    with respect to x, just use the appropriate version of Leibnitz's rule:

    [tex]\frac d {dx}\int_0^x f(x,t)\, dt = f(x,x) + \int_0^x f_x(x,t)\, dt[/tex]

    This gives your result immediately.
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