I hope that this is the appropriate forum to ask something about Fourier series.
My question is a little intuitive.Say I expand a function in Fourier series with
n=-∞ to n=∞.The graph of the function is available.

Now suppose,I cut off some terms for which |n|>N and expand the function.It will not be a Fourier series any more.But I am not worried about that.All I want to know whether this process is capable to preserve the essential feature of the graph.If terms like that (|n|>N) contribute very small to the actual series,then what I am telling is possible with a good approximation.Please let me know...

I encountered this problem in deriving the Parsevals formula in a Quantum Mechanics book where they have folowed the procedure in "approximation in the mean".

I hope that this is the appropriate forum to ask something about Fourier series.
My question is a little intuitive.Say I expand a function in Fourier series with
n=-∞ to n=∞.The graph of the function is available.

Now suppose,I cut off some terms for which |n|>N and expand the function.It will not be a Fourier series any more.But I am not worried about that.All I want to know whether this process is capable to preserve the essential feature of the graph.If terms like that (|n|>N) contribute very small to the actual series,then what I am telling is possible with a good approximation.Please let me know...

Let me be sure I understand what you mean. Suppose x_n is the first function, with n \in (-\infty, \infty). The discrete-time Fourier transform of x_n is then denoted X(j\omega) . Now you form a second time-domain function y_n which equals x_n for |n|<N and 0 otherwise. Call the transform of this function Y(j\omega). To see how Y(j\omega) relates to X(j\omega), notice that we can write y_n as the product of x_n and a window function defined as w_n = 1 if |n|<N and 0 otherwise. Then, the windowing theorem tells us that Y(j\omega) is given by the convolution of X(j\omega) and W(j\omega):

Y(j\omega) = \frac{1}{2\pi}\int_{-\pi}^{\pi} X(j\omega) W(j(\omega - \theta)) \, d\theta

Also, note that the W(j\omega) = \frac{\sin(N\omega)}{\sin(\omega /2)}, which is a sync function. So, then, Y(j\omega) looks like the convolution of X(j\omega) with a sync function, which will "smear out" the transform in a non-recoverable way. I.e., you can't take the windowed transform and get back to the original, non-time-limited function, cause you can't undo the smearing. As N gets larger, the sync function looks more and more like an impulse, and Y(j\omega) gets closer and closer to X(j\omega). So, whether this preserves the essential features of x_n depends on what you consider essential. You're certainly not going to get anything but zeros in the region that was windowed out, but if this is insignificant, its possible that Y(j\omega) would look very similar to X(j\omega).

neelakash---

I think the only problem is around discontinuities in the graph. So, supposing you have a step function or a square wave or something, you'll get trouble around the breaks in the graph if you truncate the series. The Fourier series will oscillate wildly at these points.

Other than that, I think it's just an engineering question. How accurate does the representation really need to be? For most nice (read: continuous) functions, ten terms is way more than you'll ever need. If you don't believe me, just take a program like mathematica and graph the first few terms and see how close you get.

I think the only problem is around discontinuities in the graph. So, supposing you have a step function or a square wave or something, you'll get trouble around the breaks in the graph if you truncate the series. The Fourier series will oscillate wildly at these points.

I had the impression he was talking about a discrete function, in which case the continuity issue doesn't come up (I assumed this because the time index was n instead of t, which is the typical pedagogy for these things). But the fact that he keeps calling it a Fourier Series, instead of Transform, would imply that we're talking about a periodic function, which would change the story... time for neelakash to clarify!!

Ahh fair enough. I misread the original post.

It is very much advaned mathematics for me.I do not know convolution and wndow theorems etc.I have the knowledge of Fourier Transform only.Hoowever,I think I can appreciate this.
you can't take the windowed transform and get back to the original, non-time-limited function, cause you can't undo the smearing. As N gets larger, the sync function Y(jw) looks more and more like an impulse, and gets closer and closer to X(jw). So, whether this preserves the essential features of x_n depends on what you consider essential. You're certainly not going to get anything but zeros in the region that was windowed out, but if this is insignificant, its possible that Y would look very similar to X.

I am talking of a function a(k) which is the momentum representation of a wave function.This may be a Gaussian.Then,its F.T. which is the wave function itslf is also a Gaussian.I want to know if this Gaussian will significantly change if I cut short the infinite limit integral in F.T. to a large N limit summation.Ftrom what quadraphonics says,I think it is possibnle.

I'm just a lowly electrical engineer, but it sounds as if, for say - simulation or approximation methods - truncating the infinity of terms of the Fourier components of a periodic waveform (the way we do in real-world simulation and analysis) to something manageable, possibly 10 terms, as Ben suggested, would actually be a good idea.

I am talking of a function a(k) which is the momentum representation of a wave function.This may be a Gaussian.Then,its F.T. which is the wave function itslf is also a Gaussian..

Okay, we're talking about functions of continuous domains then? In that case, there may be issues related to continuity.


I want to know if this Gaussian will significantly change if I cut short the infinite limit integral in F.T. to a large N limit summation.Ftrom what quadraphonics says,I think it is possibnle.

What do you mean by "large N limit summation?" Supposing we're just talking about cutting the limits of the integral to some finite values, this is equivalent to multiplying by a window function, and then taking the regular Fourier Transform. So, again, the windowing theorem will tell you what happens, which is to say that the wave function you get is equal to the convolution of the original wave function with the transform of the window function (which will be a sync function for a rectangular window like I've described). Whether this change is significant or not depends on what you mean by "significant." Certainly, it will no longer be exactly Gaussian. However, it may be that some properties (say, the mean value or some other statistics) will not change significantly. Without knowing how to measure the significance of the disturbance, it's difficult to say anything more.

I don't think that the ringing due to truncation that Ben mentioned is going to be a factor here. That arises when you start with a discontinuous function, take its Fourier transform, truncate the transform, and then transform back. What you get is ringing in the region where the discontinuity originally was. Moreover, this rining doesn't decrease in amplitude as you include more and more of the transform, but rather shrinks in its extent in time. This isn't an issue here, though, since both the original function and its transform are continuous.

AAAAhhhh...
I appreciate your help...But not clear yet.Let me clarify:
Note that I am talking of a wave function,not only Gaussian.Let me write two examples:wave function representation in momentum space:

a(k)=(1/√ε) in |k|<ε/2 and
a(k)=0 in |k|>ε/2

or,

a(k)=(√σ√π)exp[(-σ²k²)/2]

So that by taking the Fourier Transform,we might get the corresponding Ψ(x,t).It will be an integral from -∞ to ∞.Alternatively we may express Ψ as a Fourier series expansion (Complex) from n=-∞ up to n=∞.

While performing approximation in the mean here,we try to approximate Ψ by the polynomial: (Sⁿ)=∑(dⁿ)exp[i2nπx/L] where the summation is from n=-N to n=N.

[dⁿ is the co-efficients,not a power series]

The approximation will be best if ∫[|Ψ-(Sⁿ)|²] dx is minimum
The calculation is performed by cutting the Ψ series from n=±∞ to n=±N.

This method gives a nice derivation of Parseval's formula.

What I am worried: what is the justification of cutting short the infinite series or the integral into a finite series?

AAAAhhhh...
I appreciate your help...But not clear yet.Let me clarify:
Note that I am talking of a wave function,not only Gaussian.Let me write two examples:wave function representation in momentum space:

a(k)=(1/√ε) in |k|<ε/2 and
a(k)=0 in |k|>ε/2

or,

a(k)=(√σ√π)exp[(-σ²k²)/2]


Okay.


So that by taking the Fourier Transform,we might get the corresponding Ψ(x,t).It will be an integral from -∞ to ∞.

No problems here.


Alternatively we may express Ψ as a Fourier series expansion (Complex) from n=-∞ up to n=∞.

This is where you lose me. To express something as a Fourier series, that something needs to be periodic. Is the wave function periodic? If not, how do you apply a Fourier series to it? I'm particularly troubled here because the wave function derived from the FT of the momentum function should be infinite in extent, so you can't apply the usual tricks of periodic extension...


While performing approximation in the mean here,we try to approximate Ψ by the polynomial: (Sⁿ)=∑(dⁿ)exp[i2nπx/L] where the summation is from n=-N to n=N.

[dⁿ is the co-efficients,not a power series]

Okay; I take it this is a truncated version of the Fourier Series you were referring to? I'm still troubled by the periodicity of this expression, though...


The approximation will be best if ∫[|Ψ-(Sⁿ)|²] dx is minimum
The calculation is performed by cutting the Ψ series from n=±∞ to n=±N.

Okay.

What I am worried: what is the justification of cutting short the infinite series or the integral into a finite series?

Well, you can choose the N such that the probability that the momentum was greater than that is arbitrarily small; say one in a million. You're still going to be artificially reducing the uncertainty in the momentum, which will show up as added uncertainty in the position, but this can be very small for suitably large N. Said another way, the approximation error is a monotonically decreasing function of N, so you can always get an arbitrarily small error if you choose a sufficiently big N. A practical justification is that with finite N you can evaluate these expressions directly using a computer.

I had some wrong thing in my mind...Now that you have pointed them out I see them.However,I got my answer in a figure that illustrates the thing.I will explain this next morning.Good night.

I am not sure of your question either (and certainly have not made the effort required to fully follow quadraphonic's efforts to clarify it.) I just want to tell of a closely related and to me very interesting optical experient you can do (and I have done):

To some extent, all finite aperature optical systems "lose information" but it is not important for image quality unless the object is small (by human scales, but can still be relatively large numbe of wavelengths. For example of interesting object - a fine screen wire grid.

The light scattered by such a periodic grid is sort of like a 2D diffraction grading scattered light. I.e. the wires makes many parallel beams traveling in different directions spread over a small solid angle.*

Now if a lense intecepts most of these beams the small image formed at its focal length (on the other side from the object) will be a series of closely spaced discrete spots of light in a grid with the same rotational symetry as the original object. In a mathematical formulation these spots are the resolution of the objects image information into spacial fourier components -each spot corresponding to one of the difraction beams. (Of course, one must place a screen where these spots form to see them, otherwise the photons just keep going further away from the object. Also the room should be dark except for the light illuminating the object.)

Lets not place any screen there, but instead place a larger second lens still further away to reform an image of the original wire screen on a paper screen. (there will be again some loss of information because of the finate aperature of this second lens, but if it is large and captures most of the rays which passed thru the discrete points where the "spots" fromed by the first lense would have been, then the total loss of information is small and the image looks just like the original screen wire object.

Now for the surgical fun (which is closely related to your question about leaving out terms in the Fourier series, I think):

Take a fine wire probe with a tiny black ball (or opaque disk) tip and stick it into the location where the spots would form on a screen behind the first lense. What you are doing is selectively removing some of the various spatial Fourier components which when reassembled / summed by the second lense form a good image of the wire screen. Thus you no longer see on the screen an image of the screen wire. Instead you see and image of a non-existent but periodic object. - The one which, if it existed would have the spatial Fourier transform of all the components you did not block with the tiny black ball probe.

This is not a hard experiment to do. - If you like math and physics, do it. It is facinating to create these virtural objects by selectively subtracting components of the spatial Fourier transforms of the real object (the fine screen wire).
------------------
*All objects make their spatial Fourier transforms difraction patterns, not just highly periodic objects, but their transforms will a continuous intensity field, not discrete spots. Further more, when the characteristic dimentions of the object are larger (than the fine screen wire) the angular spread of their diffraction pattern (at leas 99+% of the intensity in it) will be very small. - I.e. each point of the object will be a extremely tight solid angle beam, not a spread out grid of intensity pattern produce by periodic intererence effects. Thus, you will only dim the entire image or block parts of it entirely depending on where you place the opaque stop.

I am sorry for the delay...
The mistake that I made was to say:
Alternatively we may express Ψ as a Fourier series expansion (Complex) from n=-∞ up to n=∞.

Actually what I said previous to this was to show you the things I was dealing with and unfortunately I was not careful enough.The book from where I ws reading starts like this:
they consider Ψ to be defined within an interval and considers another periodic function φ from n=-∞ up to n=∞ and that conforms to Ψ in Ψ's interval.Now they expand φ in Fourier Series and apply the approximation in the mean process.There they cut short the infinte sum to finte sum of very large number N (This was the site of my trouble).This leads to Parseval's theorem...
I apologize again to make trouble.

the approximation error is a monotonically decreasing function of N, so you can always get an arbitrarily small error if you choose a sufficiently big N
I see.
I referred to a book where they show the effect of combining more and more terms in the Fourier series.Higher terms increase accuracy.That is you will recognize a square wave more if you add first 10000000 terms than first 100 terms...
That's it.
Thank you all for sharing the discussion.
I may refer to Billy T's write up a bit later.

Actually what I said previous to this was to show you the things I was dealing with and unfortunately I was not careful enough.The book from where I ws reading starts like this:
they consider Ψ to be defined within an interval and considers another periodic function φ from n=-∞ up to n=∞ and that conforms to Ψ in Ψ's interval.Now they expand φ in Fourier Series and apply the approximation in the mean process.There they cut short the infinte sum to finte sum of very large number N (This was the site of my trouble).This leads to Parseval's theorem...

Yeah, I suspected it was something like that. This is what I'd call "periodic extension;" it's the standard way to apply Fourier Series to non-periodic functions. You simply repeat them to make them periodic, take the Fourier Series, and then remember to throw away all of the copies when you take the inverse. The only rub here is that the original wave function must be limited in time (or space or whatever) in order for the periodic extension to work out. So this process can't be applied to wave functions with infinite support (i.e., Gaussian). So, strictly speaking, this approach is only applicable to the "particle in a box" situation; i.e., an extremely deep potential well.

I referred to a book where they show the effect of combining more and more terms in the Fourier series.Higher terms increase accuracy.That is you will recognize a square wave more if you add first 10000000 terms than first 100 terms...

Yes, the higher-frequency terms represents features with short time-scales (like the edges of the square wave) while the low-frequency ones represent more "global" features. Also, not to burst your bubble, but it's very easy to prove Parseval directly. Start with one side (say, the time-domain one):

\int_{-\infty}^{\infty}|x(t)|^2\, dt = \int_{-\infty}^{\infty}x(t)x^*(t)\, dt

Now, write one of the x(t)'s as an inverse Fourier Transform:

\int_{-\infty}^{\infty} x(t) \left( \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} \, d\omega \right)^*\, dt = \frac{1}{2\pi}\int_{-\infty}^{\infty} x(t) \int_{-\infty}^{\infty} X^*(j\omega) e^{-j\omega t} \, d\omega \, dt

Now reverse the order of integration:

\frac{1}{2\pi}\int_{-\infty}^{\infty} X^*(j\omega) \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt\, d\omega

Now notice that the inner (dt) integral is the FT of x(t):

\frac{1}{2\pi}\int_{-\infty}^{\infty} X^*(j\omega) X(j\omega) \, d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty} |X(j\omega)|^2 \, d\omega

And we're done. The same basic approach works for the other versions of Parseval, which apply to Fourier Series, discrete time, etc. You can also prove the more general version of Parseval that applies to cross-correlations instead of just the energy of a single signal.