Timeline for What's the relationship between the roots of a function and that of a filtered Fourier series representation?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2016 at 12:15 | comment | added | Rajesh D | I felt this is needless to say, as OP knows that Fourier series converges pointwise at all points except jumps,in this case. | |
| Feb 24, 2016 at 12:09 | comment | added | Rajesh D | @user1952009 : original function $M$ is binary and hence has no zero set but only sign change points. Thats why all I have said is that the number of zeros of $f(M)$ would be equal to the sign changes, as long as cutoff is high enough. I agree the location of zeros of $f(M)$ do not exactly coincide with the jump points but convergence as the cutoff goes to $\infty$. (I could not comment right below your comment as I don't have enough rep now. | |
| Feb 20, 2016 at 3:39 | history | edited | Rajesh D | CC BY-SA 3.0 |
added 16 characters in body
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| Feb 20, 2016 at 3:14 | history | answered | Rajesh D | CC BY-SA 3.0 |