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Suppose $M$ is a piecewise constant function on an interval $T$ taking values $+1$ and $-1$, and that $M$ exhibits all the properties sufficient to ensure the existence of some converging Fourier series decomposition on $T$. Make no assumptions about the evenness or oddness of $M$, merely that all discontinuities of $M$ on $T$ occur where $M$ changes sign.

Write $F(M)$ for the presentation of $M$ as a converging Fourier series on $T$. Write $f(M)$ for a 'low-passs' filtered $F(M)$, i.e. $f(M)$ is $F(M)$ where all the terms in $F(M)$ having frequencies above some fixed predetermined value have been removed from the sum (so $f(M)$ is a partial sum).

What is the relationship between the zero sets of $F(M)$ and $f(M)$?

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  • $\begingroup$ I modified the formatting to use Latex for mathematical symbols. $\endgroup$
    – Paul Siegel
    Commented Jan 25, 2016 at 21:23
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    $\begingroup$ Isn't your $M$ a linear combination of "unit step functions"/"square waves"? Since Fourier coefficients are linear, your truncated Fourier series is a linear combination of truncated Fourier series squares waves. At every step you get some Gibbs phenomenon and these add up. Do you expect anything more than that? $\endgroup$
    – Dirk
    Commented Jan 26, 2016 at 10:31
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    $\begingroup$ @Justin Greenoug you have to learn more on Fourier series and filtering. the bandwith of your low-pass filter is proportional to how much your filter is concentrated in time, so when the filter is more concentrated than the minimum distance between two discontinuity points, there are as many zero in filtered $M$ as in $M$, and because the filter is symmetric, they are indeed at the same localization. that's exactly the Gibbs phenomenon Paul told about. $\endgroup$
    – reuns
    Commented Jan 28, 2016 at 1:48
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    $\begingroup$ @RajeshDachiraju "function" is a proper term in mathematics. "Binary wave" seems too informal. Anyone who takes the trouble to read the question will figure out the authors' meaning; I am very much not a fan of people changing other people's turn of phrase unless there is clear reason to do so $\endgroup$
    – Yemon Choi
    Commented Feb 20, 2016 at 17:03
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    $\begingroup$ @Rajesh : given what I answered above, I was thinking to a non-ideal low-pass filter that has compact support in time and is symmetric. but if the filter is the Dirichlet kernel, the $0$ won't be at the same exact location as in the original function, they will only converge to the original zeros ? $\endgroup$
    – reuns
    Commented Feb 20, 2016 at 23:32

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If the cut off for the low pass filtering (partial sum) is sufficiently high, the number of zeros of $f(M)$ is equal to the number of sign changes of the binary wave $M$. Although the Gibb's phenomenon comes into play, as long as the cut off frquency is high enough, the Gibb's oscillations do not affect the zero crossings of $f(M)$ as the wave $M$ is binary. Remember that the max over shoot of Gibbs overshoot is only approximately $9%$ percent of the jump amount.

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  • $\begingroup$ @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. $\endgroup$
    – Rajesh D
    Commented Feb 24, 2016 at 12:09
  • $\begingroup$ I felt this is needless to say, as OP knows that Fourier series converges pointwise at all points except jumps,in this case. $\endgroup$
    – Rajesh D
    Commented Feb 24, 2016 at 12:15

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