Any explanation on this?  How can we rely on is or test relation?
 

Download equivalence_discrepancy.mw

@Rouben Rostamian  

You did not address my problem in totality.  You did not bother to open my worksheet.  Nonetheless, I tried your solution, but still NO results.  Check the link below.  I am getting errors & warnings about solutions may have been lost.

solving_transcendental.mw

@rlopez 

I modified the worksheet you provided to handle multiple functions other than the simple step function you provided as an example to find the overshoot due to the GIBBS effect.  I included a link below with a subset of functions I analyzed.  Numerically, as you or someone else stated the ratio peak value does seem to approach 1.08949 for all cases.  Having said that I have shown numerically the rate at which the series approaches the ratio 1.08949 is different for all of the functions.  In other words, for a specific k, say 1000, the ratio is different for each function.

I have an additional question which I thought I would post here since the topic so closely related, but I will post it as a separate question if you feel it merits that.  My question is based on the fact that as k-->infinity the location of xpeak-->0 which is the location of the jump discontinuity.  So how do I show in MAPLE analytically the value of the series as k-->infinity & location xpeak-->0?  When I tried this I got no answer because I think what I fed into MAPLE is not definable.

GIBBS_from_Lopez.mw

@Axel Vogt 

Is this because d(S2)/dx is symbolic instead of numeric?

@acer 

The sum command attempts symbolic summation, and the add command adds up a finite number of things. You're trying to do the latter, and so add is the suitable command for your task. Do you really want your addition of many sin and cos calls to be turned into LerchPhi or whatever other special function calls that sum command might come up with?! There's at least one thread from the past week (same coursework?) where someone asked how to stop that from happening.

That was actually from me.

@acer 

This appears to be a matter of experience with MAPLE to understand these limitations.  It would be great if there was some type of documentation in the HELP menu to identify these shortcomings.  Then again who would actually read & comprehend that.

Anyway, I actually need to proceed on this further with a more intricate signal.  I chose the function due to its familiaty.  So I might have further questions or difficulties.  I suppose I should make new posts as opposed to responding to this question?

@acer 

You used the add command opposed to my summation.  I took your worksheet & changed the add command to the summation & again got no results using fsolve.  Why is that & how do you know what fsolve can handle?

Another matter I noticed the Student Calculus Root command takes considerably more time than the RootFind command.  Why is that?

@Kitonum 

I tried using the commands in your example & apply it to finding the derivative of a series to find the roots.  I am not having success for the case k=100.  It worked for the case for k=10.  Judging from the plot of the series itself the peak value for 100 terms occurs around x~0.02.  I tried changing the interval to find the roots of the derivative, but I get nothing.

Are you able to find the roots & if so what am I doing wrong?

fsolve_not_working.mw

@_Maxim_    @vv    @rlopez

There are a number of convoluted responses to my question.  I am trying to make sense of them in context to my central question.  That question is:

Is the overshoot the same (1.089) near the discontinuity regardless of the shape of the function as the partial sum, Sn, approaches an infinite number of terms?  Let's start by saying the function is such that it is square integrable over the period of interest, and that quantity is finite.  So the function is bounded.  I surmise you are all trying to tell me YES that is TRUE.

My 2ndary question:

Are there any MAPLE scripts available to demonstrate the answer to the 1st question?  I surmise NO.

Let me add a 3rd question:

How is the overshoot actually defined?  Is it defined as the ratio of the max of Sn to the max of the function now matter the location of those peaks?  Or is it defined as the ratio of the max of Sn occuring at location x = Xpeak to the value of f(x) at Xpeak?

@_Maxim_ 

The proofs of the results related to the Gibbs phenomenon require just some basic calculus, they boil down to the fact that a certain sum is in fact a Riemann sum for one particular function. So the limiting behavior is always the same.

If you have links to any of these proofs could you forward them to me?  I have seen some handwaving for the rectangle function, but nothing for anything other than that.

@_Maxim_    @vv    @rlopez

There are a number of convoluted responses to my question.  I am trying to make sense of them in context to my central question.  That question is:

Is the overshoot the same (1.089) near the discontinuity regardless of the shape of the function as the partial sum, Sn, approaches an infinite number of terms?  Let's start by saying the function is such that it is square integrable over the period of interest, and that quantity is finite.  So the function is bounded.  I surmise you are all trying to tell me YES that is TRUE.

My 2ndary question:

Are there any MAPLE scripts available to demonstrate the answer to the 1st question?  I surmise NO.

Let me add a 3rd question:

How is the overshoot actually defined?  Is it defined as the ratio of the max of Sn to the max of the function now matter the location of those peaks?  Or is it defined as the ratio of the max of Sn occuring at location x = Xpeak to the value of f(x) at Xpeak?

@_Maxim_ 

You can see for k = 1 the overshoot is less than UNITY.  For all of the other k values from 11 & above it is greater than UNITY.  This is not the case for the rectangle function of Figure 1.

@_Maxim_ 

I'm not sure what you mean about the rectangle and 1-x being different.

If you look at the 2 plots for the Fourier series representation of the rectangle & 1/2 triangle function you will note the overshoot for the square wave is always greater than UNITY even for k = 1.  However, for the 1/2 triangle the overshoot is is less than UNITY for k values less than 10.  Then after that the overshoot increases above unity.  For the decaying exponential functiion, 3rd plot, the same behavior occurs.  That is what I deem as different.

So that is why I am questioning if the overshoot approaches 1.09 as k--> infinity for all discontinuities regardless of the waveform.

The infinite Fourier series either converges to the midpoint or diverges at the jump. In the latter case, it's still summable to the midpoint. So there is no Gibbs phenomenon for the infinite series.

My impression from the literature at least for the rectangle function is that the overshoot is 1.09 no matter how high the number of terms are in the series, whether 100, 1000, or infinite.  Are you suggesting the overshoot is UNITY for an infinite series?  If so could you demonstrate this or point me in a direction showing this?

@Mac Dude 

If x could never be negative for fundamental reasons then I'd argue both Fourier series are equally valid; your data cannot discriminate between the two.

If you need to known the spectral composition of the Fourier series then you need to tread carefully here. Your two series have a significantly different spectrum; if you do not have enough data to discriminate between the two you may not be able to determine the spectrum of the underlying distribution.

Hi MD,

Your 2 statements above seem to be contradictory.  I will elaborate on my situation & maybe you can give me a more informed opinion.

I have both pressure & strain collected over time so obviously I have no data for negative time.  However, I can take my data & use the "flip" command to extend my data into the (-) domain.  I understand the spectrum (Fourier coefficients vs k) for asymmetric vs symmetric are different, but the fundamental frequency of 2*pi*k/T is common to both.  If both spectra are inverted back into the time domain it appears the spectrum for the symmetric case actually does a better job representing the function in time that is actually asymmetric.

@Markiyan Hirnyk 

I copied & pasted from a master worksheet.  I did not mean to include the Matrix commands.  They are irrelevant.  I edited the worksheet & got rid of those commands & uploaded to the link.

If you simply examine the plots you clearly see the approximation of the 2 triangles for the same number of terms is far superior for the symmetric case.

My question is would I be making a mistake representing the asymmetric case with the symmetric approximation & simply make 0<=t<=1?  It appears the fit would be better for n=50 if I used the symmetric approximation.