Fourier Series from old Physics GRE

In summary, the conversation discusses a Fourier series problem from an old Physics GRE and the solution provided by a website. The problem is solved by eliminating choices and using reasoning to narrow down the correct answer. There is also a discussion on the incorrect solution provided by the website and the correct way to solve the problem.
  • #1
G01
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Homework Statement



Hi everyone. I have a Fourier series problem from an old Physics GRE that I would like to discuss.

Here is a link to the problem and a solution. The answer they get is correct, but their explanation is not:

http://grephysics.net/ans/9277/39

Homework Equations



Fourier Analysis, but we should be able to solve this without too much math.

The Attempt at a Solution



Since, you only have 1min. 15s for every problem on the test, usually these problems can be solved in that length of time.

So, I am trying to find a way to get to the correct answer by using reasoning and elimination instead of actually integrating to find the correct coefficients (time consuming). Here is where I am at:

You can eliminate all choices except for A and B since the terms in the series must be odd functions since the function we are expanding is odd.

Now, I can't seem to find a good way to narrow the answer down to B, which is the correct answer. Their explanation does not make sense, since if A is trivially 0 then so is B.

Does anyone here have any ideas? I have been staring at this for quite some time. Yet, I can't find a good way to eliminate A, except by brute calculation of the series. Any thoughts and help will be appreciated. Thank you.
 
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  • #2
Well, it is sort of obvious that the integral of the function times sin(2wt) vanishes, right? That rules out A. The 'clue' is clearly bogus. And it took me considerably longer the 1min 15s for that, so I guess I lose.
 
  • #3
Dick said:
Well, it is sort of obvious that the integral of the function times sin(2wt) vanishes, right? That rules out A. The 'clue' is clearly bogus. And it took me considerably longer the 1min 15s for that, so I guess I lose.

Yes. That makes sense Dick. Thanks for the help. I really hate these problems. Obviously this does not test your ability with Fourier series, but instead tests your ability to find a work-around alternative to a real solution in a short time. Well, if the ETS actually had tests which were good indicators of physics ability, life would just be too easy.:rolleyes:
 
  • #4
G01 said:
Yes. That makes sense Dick. Thanks for the help. I really hate these problems. Obviously this does not test your ability with Fourier series, but instead tests your ability to find a work-around alternative to a real solution in a short time. Well, if the ETS actually had tests which were good indicators of physics ability, life would just be too easy.:rolleyes:

It would have been much easier if I hadn't read the bogus solution first.
 
  • #5
Dick said:
It would have been much easier if I hadn't read the bogus solution first.

That website is hit and miss. They do have solutions to every released physics GRE question, but sometimes the solutions are not correct at all. That is one of the more blatant examples. Sometimes the website has mistakes which are very subtle and almost believable. You can see what happens to a "homework" website that does not have moderation like PF has.
 
  • #6
Ok, well I'd be guessing here. I did the "brute force method" to find Bn.
Bn=4/npi.

If I put this into Sum{Bn*sin(npi*t)}[1,2pi] I will get all zeros,ok we already know this so, that elimanates (A).

Look at (A) then look at (B). Try playing with the indices of (B), if i=0,...inf 1/(2i+1)

set m=1,...inf+1 1/(2m).

Geez, what I am TRYING to say is,we all know that our choice of Bn*sin(npi*t) 4ALL 'n' yields a zero. Wait a minute!

Just thought of something. Consider convergence, I mean |sin(x)|<=1,correct?
so if that be the case,then why not use these??


|sin(2n+1)|<=1 4ALL 'n' && so is |sin(n)|<=1.


look at the sums (A) & (B). Which converges faster??

1/(n) || 1/(2n+1)? For a 90 second problem,this might help.

*Always think in simplest terms*
 
  • #7
Oh boy,what a blunder!

(A) is ZERO

Thus 0<1/(2i+1), so (B) would be by default, the correct answer. Bn=4/(2i+1)*pi

So Sum{Bn*sin(2i+1)*wt}[0,inf] <=|sum{Bn}|[0,inf]
 
  • #8
But Jon, if (A) were zero,then that point is moot. Ok,ok,ok.

|1/n| < |1/(2n+1)|, I state my case.
 
  • #9
There's a discussion on that page that gives the easy way to spot that it's B:

[tex]V\left(t + \frac{\pi}{\omega}\right) = -V(t),[/tex]

hence, the answer cannot be A because it contains even multiples of [itex]n\pi/\omega[/itex], and so [itex]V(t + \frac{\pi}{\omega}) \neq -V(t),[/itex] for choice A.
 
  • #10
so mute what you are saying is (A) is NOT half wave symmetrical?
Therefore, by checking with the same rule,resolves the case for (B)?
Thanks
 
  • #11
Jon_ said:
so mute what you are saying is (A) is NOT half wave symmetrical?
Therefore, by checking with the same rule,resolves the case for (B)?
Thanks


Yes. Choice A contains terms [itex]\sin(2n\omega t)[/itex].

[tex]\sin\left(2n\omega \left(t + \frac{\pi}{\omega}\right)\right) =
\sin\left(2n\omega t + 2n\pi\right) = \sin(2n\omega t)[/tex]

Hence choice A does not have the [itex]V\left(t + \frac{\pi}{\omega}\right) = -V(t)[/itex] symmetry that the waveform shown in the problem has, whereas because choice B has only odd terms and so you get [itex]\sin\left(((2n+1)\omega t + (2n+1)\pi\right) = -\sin((2n+1)\omega t)[/itex] for every term in the sum, and so choice B does have the required symmetry.
 

What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function in terms of a sum of sine and cosine functions. It allows us to break down complex functions into simpler components and is widely used in physics and engineering.

How is a Fourier series used in physics?

Fourier series are commonly used in physics to analyze and describe periodic phenomena, such as sound waves and electromagnetic waves. They are also used in signal processing and image analysis to extract important information from complex signals.

What is the formula for a Fourier series?

The general formula for a Fourier series is given by: f(x) = a0/2 + Σ(an*cos(nx) + bn*sin(nx)), where a0 is the average value of the function, an and bn are the coefficients of the cosine and sine terms, and n is the frequency of the Fourier components.

What are the applications of Fourier series?

Fourier series have numerous applications in physics, engineering, and mathematics. Some common applications include analyzing vibrations and oscillations, solving differential equations, and decomposing signals and images into their frequency components.

What are the limitations of Fourier series?

Although Fourier series are a powerful tool, they have some limitations. They can only be used to represent periodic functions, and not all functions can be accurately represented by a finite number of terms in a Fourier series. They also have difficulty representing sharp changes or discontinuities in a function.

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