# Polynomials, Kernels and Derivatives

1. Sep 10, 2010

### psholtz

Is there a simple way to show that when we differentiate the following expression (call this equation 1):

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

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

$$Y'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt$$

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

$$Y(x) = \int_0^x (x-t)f(t)dt = x\int_0^x f(t)dt - \int_0^x tf(t)dt$$

$$Y'(x) = \int_0^x f(t)dt + xf(x) - xf(x) = \int_0^x f(t)dt$$

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. Sep 10, 2010

### psholtz

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:

$$(x-t)^n = \left(\array{c}n \\ k\endarray\right)x^{n-k}(-t)^k$$

so the function Y can be rewritten as the sum:

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

Take the kth term in this sum:

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

Differentiating w.r.t. x:

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

Consider the second term. It can be written as:

$$\frac{x^nf(x)}{n!}\left(\array{c}n \\ k\endarray\right)\left(-1\right)^k$$

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

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

Since we can write:

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

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):

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

and specifically, the coefficient on this term:

$$\frac{n-k}{n!}\left(\array{c}n\\k\endarray\right) = \frac{n-k}{n!} \cdot \frac{n!}{k!(n-k)!}$$

$$= \frac{1}{n\cdot (n-1)!} \cdot \frac{n \cdot (n-1)!}{k!((n-1)-k)!}$$

$$= \frac{1}{(n-1)!} \cdot \frac{(n-1)!}{k! ((n-1)-k)!}$$

$$= \frac{1}{(n-1)!} \cdot \left(\array{c}n-1\\k\endarray\right)$$

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

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

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

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

which was to be proved..

3. Sep 10, 2010

### LCKurtz

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:

$$\frac d {dx}\int_0^x f(x,t)\, dt = f(x,x) + \int_0^x f_x(x,t)\, dt$$