Continuous Version of Roots of Unity Filter
by Altheman, Dec 27, 2009, 1:52 AM
Geometrically, we can see that the roots of unity filter is the discrete version of the Cauchy Integral Formula.
Suppose we have a function,
, that takes in complex numbers and outputs complex numbers.
Suppose we are given an analytic function,, on some neighborhood about 
,
(That basically says that the function is well behaved in the sense of a complex derivative. Another way of thinking about it is that is very much like a polynomial. )
The roots of unity filter says that the average of on a regular polygon centered about is 
.
If we were to have a continuous analogue to the above, we would want a weighted sum of about a circle.
Consider a small circle![\[ z=z_0+r e^{i\theta}\implies dz=ri e^{i\theta}d\theta=i (z-z_0)d\theta\]](//latex.artofproblemsolving.com/4/c/9/4c94fdf4977dd366041ec843f38b75559dd91690.png)
One might guess that 
would be an appropriate weight if we are using a circle. Indeed, if we rearrange the previous line, we get a perfect match to the Cauchy Integral Formula: ![\[ \frac{d\theta}{2\pi}= \frac{1}{2\pi i}\cdot \frac{dz}{z-z_0}\]](//latex.artofproblemsolving.com/2/d/0/2d097df1786f08b355c768340b4583f9d1014e6c.png)
Maybe when I get time, I'll expand this article. There is a lot to be thought about here
--RUF extracts terms (x^n)^k k=0,1,... what about the linear functional that extracts terms that are nk+1, etc? how does this correspond to continuous version (ex. multply by something?)
-- weighted sums of things are important.... ex. fourier theory. there is also a lot to be said about the correspondence of discrete and continuous operators. ex. again, fourier transform
--what do higher order cauchy integral formulas correspond to geometrically
--what do the weights look like for a more general path that surrounds?
Suppose we have a function,
, that takes in complex numbers and outputs complex numbers.Suppose we are given an analytic function,
, on some neighborhood about 
,(That basically says that the function is well behaved in the sense of a complex derivative. Another way of thinking about it is that
is very much like a polynomial. )The roots of unity filter says that the average of
on a regular polygon centered about is 
.If we were to have a continuous analogue to the above, we would want a weighted sum of
about a circle.Consider a small circle
![\[ z=z_0+r e^{i\theta}\implies dz=ri e^{i\theta}d\theta=i (z-z_0)d\theta\]](http://latex.artofproblemsolving.com/4/c/9/4c94fdf4977dd366041ec843f38b75559dd91690.png)
One might guess that 
would be an appropriate weight if we are using a circle. Indeed, if we rearrange the previous line, we get a perfect match to the Cauchy Integral Formula: ![\[ \frac{d\theta}{2\pi}= \frac{1}{2\pi i}\cdot \frac{dz}{z-z_0}\]](http://latex.artofproblemsolving.com/2/d/0/2d097df1786f08b355c768340b4583f9d1014e6c.png)
Maybe when I get time, I'll expand this article. There is a lot to be thought about here
--RUF extracts terms (x^n)^k k=0,1,... what about the linear functional that extracts terms that are nk+1, etc? how does this correspond to continuous version (ex. multply by something?)
-- weighted sums of things are important.... ex. fourier theory. there is also a lot to be said about the correspondence of discrete and continuous operators. ex. again, fourier transform
--what do higher order cauchy integral formulas correspond to geometrically
--what do the weights look like for a more general path that surrounds
?
Your post was very interesting. I'd enjoy seeing more. Especially an example or two... 
June 2011
March 2011