Fourier Series: To Factor or Not to Factor, That is the Question

  • Thread starter doive
  • Start date
  • Tags
    Fourier
In summary, there is an extra factor of 1/2 associated with a_0 in the Fourier series formula. This can be checked by setting f(x) = 1 and seeing that the resulting series has a 1/2 term. The other coefficients can also be checked by setting specific functions for f(x) and verifying that the coefficients match the formulas given.
  • #1
doive
15
0

Homework Statement



I have conflicting notes and even textbooks about where i should and shouldn't have factors of 2 when doing Fourier series. I just want to check once and for all I'm doing it right and not introducing erroneous factors of 2 or omitting any

Homework Equations


[tex]
\tilde{f}(x)= \frac{a_0}{2} + \sum^{\inf}_{n=1} [a_n cos(\frac{n \pi x}{L}) + b_n sin(\frac{n\pi x}{L})]
[/tex]

[tex]
a_n = \frac{1}{L} \int^{L}_{-L} cos \frac{n\pi x}{L} f(x) dx
[/tex]

[tex]
b_n = \frac{1}{L} \int^{L}_{-L} sin(\frac{n\pi x}{L}) f(x) dx
[/tex]

[tex]
a_0 = \frac{1}{2L} \int^{L}_{-L} f(x) dx
[/tex]

where the interval is [-L,L] and the period is 2L

The Attempt at a Solution



Is the above formulation correct (i'm particularly unsure about the 1/2L associated with a0)
 
Last edited:
Physics news on Phys.org
  • #2
Just plug in the series for f(x) into the integrals and see if the coefficients work out correctly.
 
  • #3
doive said:
Is the above formulation correct (i'm particularly unsure about the 1/2L associated with a0)

You have an extra factor of 1/2 associated with [itex]a_0[/itex]. To see that this is the case, set [itex]f(x) = 1[/itex]. Then according to your formulas, [itex]a_0 = 1[/itex] and the Fourier series is

[tex]\tilde{f}(x) = 1/2[/tex].

There should be a 1/2 either on the defining integral for [itex]a_0[/itex] or on the [itex]a_0[/itex] term in the series expansion, but not both.

You can check your other coefficients in a similar way by setting

[tex]f(x) = \cos(n\pi x/L)[/tex]

or

[tex]f(x) = \sin(n \pi x/L)[/tex]
 
  • #4
thank you guys!

I had tried checking it but couldn't be certain it wasn't my dodgy arithmetic!
 

FAQ: Fourier Series: To Factor or Not to Factor, That is the Question

What is Fourier factor frustration?

Fourier factor frustration is a phenomenon that occurs in materials with a periodic lattice structure, where the Fourier transform of the lattice potential has large peaks at non-zero values. This leads to difficulties in describing the material's properties using traditional methods.

How does Fourier factor frustration affect material properties?

The presence of Fourier factor frustration can result in complex and unconventional behavior in material properties, such as the emergence of unexpected phases or the suppression of long-range order.

What causes Fourier factor frustration?

Fourier factor frustration can arise from a variety of factors, including lattice defects, competing interactions, or geometric constraints. It is a common feature in many types of disordered or frustrated systems.

How is Fourier factor frustration studied?

Scientists use a variety of techniques, such as neutron scattering, X-ray diffraction, and computer simulations, to study the effects of Fourier factor frustration on material properties. These methods allow for the characterization of the material's lattice structure and the observation of any emergent phenomena.

Can Fourier factor frustration be harnessed for practical applications?

While it can lead to unconventional behavior, Fourier factor frustration is often seen as a challenge in materials research. However, some scientists are exploring ways to utilize this phenomenon for practical applications, such as in the design of new materials with unique properties or in the development of novel computing devices.

Back
Top