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Yes integration is a 1/f Fourier multiplier. But if you want to do (1/f)^alpha then it's not so straightforward in the time domain.
by dls2016 5y ago
Yes integration is a 1/f Fourier multiplier. But if you want to do (1/f)^alpha then it's not so straightforward in the time domain.
- abnry 5y agoIn the theoretical PDEs world, non-integer alpha represents a fractional derivative.
- dls2016 5y agoWord. I did my time in the Sobolev spaces.
- wyager 5y ago> Yes integration is a 1/f Fourier multiplier. Can you explain this? I don’t see the connection. I can see how the zero-frequency value would be equal to the integral (well, the average). Edit: figured it out. Derivative operator multiplies each basis function by its index. D exp(inx) = inexp(inx). Apply the inverse operation (divide by index) to get the integral.