[math-fun] Name of this transform?
What's the name of this transform? It basically treats the even part of the function as "real", the odd part as "imaginary", and rotates it by 90 degrees: T(f)(z) = z even(f)(z) - odd(f)(z)/z = z(f(z) + f(-z))/2 - (f(z) - f(-z))/2z It's somewhat related to the Hilbert transform, but that splits the function into positive and negative frequencies rather than even and odd part. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
I don't know, but: Why is multiplication or division by z interpreted as rotation? Or what am I missing? I thought your text meant to describe this: (Tf)(z) = i (f(z) + f(-z)) / 2 - (f(z) - f(-z)) / 2 --Dan
On Dec 18, 2014, at 10:46 AM, Mike Stay <metaweta@gmail.com> wrote:
What's the name of this transform? It basically treats the even part of the function as "real", the odd part as "imaginary", and rotates it by 90 degrees:
T(f)(z) = z even(f)(z) - odd(f)(z)/z = z(f(z) + f(-z))/2 - (f(z) - f(-z))/2z
It's somewhat related to the Hilbert transform, but that splits the function into positive and negative frequencies rather than even and odd part.
On Thu, Dec 18, 2014 at 12:25 PM, Daniel Asimov <asimov@msri.org> wrote:
I don't know, but: Why is multiplication or division by z interpreted as rotation? Or what am I missing?
I thought your text meant to describe this:
(Tf)(z) = i (f(z) + f(-z)) / 2 - (f(z) - f(-z)) / 2
No. Assume f is continuously differentiable so it has a Taylor series. This transform swaps neighboring coefficients in the Taylor series and negates one of them. Split f into even and odd parts fe and fo. Think of f as a 2-d vector |fe| |fo| Then rotate via the matrix |0 -1| |1 0| which is like i in that applying the matrix twice is minus the identity. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Multiplication and division by z is for changing even to odd and vice-versa. On Thu, Dec 18, 2014 at 12:41 PM, Mike Stay <metaweta@gmail.com> wrote:
On Thu, Dec 18, 2014 at 12:25 PM, Daniel Asimov <asimov@msri.org> wrote:
I don't know, but: Why is multiplication or division by z interpreted as rotation? Or what am I missing?
I thought your text meant to describe this:
(Tf)(z) = i (f(z) + f(-z)) / 2 - (f(z) - f(-z)) / 2
No. Assume f is continuously differentiable so it has a Taylor series. This transform swaps neighboring coefficients in the Taylor series and negates one of them.
Split f into even and odd parts fe and fo. Think of f as a 2-d vector |fe| |fo|
Then rotate via the matrix |0 -1| |1 0| which is like i in that applying the matrix twice is minus the identity.
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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