What's the Paradox in Combining Sine and Cosine Series for Fourier Expansion?

In summary, the conversation discusses the use of sine and cosine functions as eigenfunctions of the hermition operator and their role in expanding functions in L2. The paradox arises when considering the coefficients of the sine and cosine series, as it seems that they should be the same but in fact they are defined differently. It is important to note that both sine and cosine series are needed to span the entire space of L2 functions.
  • #1
ShayanJ
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We know that because [itex] \sin{nx} [/itex] and [itex] \cos{nx} [/itex] are degenerate eigenfunctions of a hermition operator(the SHO equation),and eachof them form a complete set so we for every [itex] f(x) [/itex] ,we have:

[itex]
f(x)=\frac{a_0}{2}+\Sigma_1^{\infty} a_n \cos{nx}
[/itex]
and
[itex]
f(x)=\Sigma_1^{\infty} b_n \sin{nx}
[/itex]
But here,because for every [itex] m [/itex],[itex] \sin{mx} [/itex] and [itex] \cos{mx} [/itex] are orthogonal,we also can have:
[itex]
f(x)=\frac{a_0}{2}+\Sigma_1^{\infty} a_n \cos{nx} + \Sigma_1^{\infty} b_n \sin{nx}
[/itex]
And its easy to understand that the [itex] a_n [/itex]s and [itex] b_n [/itex]s are the same.
So it seems we reach to a paradox!
What's wrong?
thanks
 
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  • #2
It's clear that for example the sum of sin(nx) is an odd function, so you can't expand a non-odd function with the sin(nx) only. Your statement in the first lime should be incorrect.

Sorry, I can't help solve the paradox.
 
  • #3
A cosine series is a even function. A sine series is odd function.
If the 2nd function was called g(x) instead, then the third function is (f+g)(x).

And no, its not "easy to understand" what you mean by the coefficients being the same. They are defined differently. Why should they be the same?
 
  • #4
The point is,[itex] \sin{nx} [/itex] are eigenfunctions of the hermition equation below:
[itex]
\frac{d^2 y}{dx^2}+n^2 y=0
[/itex]
So they should form a complete set and this means that every function can be expanded as a linear combination of them.The same is true for [itex] \cos{nx} [/itex]

And about the coefficients
We have:
[itex]
f(x)=\Sigma_1^{\infty} b_n \sin{nx}
[/itex]
if we multiply by [itex] \sin{mx} [/itex] and integrate from [itex] 0 [/itex] to [itex] 2 \pi [/itex] we get:
[itex]
\int_0^{2\pi} f(x) \sin{mx} dx = \Sigma_1^{\infty} \int_0^{2 \pi} b_n \sin{mx} \sin{nx} dx=b_m \int_0^{2 \pi} \sin^2{mx} dx=\pi b_m [/itex]
And so:
[itex]
b_n=\frac{1}{\pi} \int_0^{2 \pi} f(x) \sin{nx} dx
[/itex]
if you do the same for the Fourier series consisted of both sine and cosine terms,you get the same formula and Its the same for cosine.
 
  • #5
Both sinnx and cosnx are the eigenfunctions of the hermitian equation. Doesn't this mean that sinnx and cosnx complete each other ? You expanded your function by half of a complete set.
 
  • #6
Shyan said:
So they should form a complete set and this means that every function can be expanded as a linear combination of them. The same is true for [itex] \cos{nx} [/itex]

What I highlighted is plural, not singular they. You need both sine and cosine to span L2. You can't just discard half your answer.

And about the coefficients
We have:
[itex]
f(x)=\Sigma_1^{\infty} b_n \sin{nx}
[/itex]
if we multiply by [itex] \sin{mx} [/itex] and integrate from [itex] 0 [/itex] to [itex] 2 \pi [/itex] we get:
[itex]
\int_0^{2\pi} f(x) \sin{mx} dx = \Sigma_1^{\infty} \int_0^{2 \pi} b_n \sin{mx} \sin{nx} dx=b_m \int_0^{2 \pi} \sin^2{mx} dx=\pi b_m [/itex]
And so:
[itex]
b_n=\frac{1}{\pi} \int_0^{2 \pi} f(x) \sin{nx} dx
[/itex]
if you do the same for the Fourier series consisted of both sine and cosine terms,you get the same formula and Its the same for cosine.

So? And how is this relevant to your claim that there is a paradox?

Edit: Reading your post again, you haven't explained what your paradox actually is! Are you claiming that [itex]a_n = b_n[/itex]? Because that is simply wrong.
 
Last edited:
  • #7
The sin(nx) are eigenfunctions of that equation. This doesn't mean they are spanning.
In fact, just sinx is an eigenfunction, but just sinx is not enough. They point is that we need all the eigenfunctions, so we need all the sin(nx) and all the cos(nx).

The whole point of these decompositions is that the eigenfunctions form a basis. If you remove even a single one, then the set is no longer spanning, and you don't have a basis anymore.
 
  • #8
But for every [itex] m [/itex] , [itex] \sin{mx} [/itex] and [itex] \cos{mx} [/itex] are degenerate eigenfunctions.Doesn't that make a difference?
I mean making each of them spanning alone?
 
  • #9
Each of them spans its spanning space not the entire space. You seem to be under the impression that any function in L2 can be written using a sine series or using a cosine series. That is not true. As pwsnafu said, cosine series span the space of all even functions in L2 and sine series span the space of all odd functions in L2. To get all functions in L2, you need both sine and cosine.
 

FAQ: What's the Paradox in Combining Sine and Cosine Series for Fourier Expansion?

What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine waves with different amplitudes, frequencies, and phases. It is named after French mathematician Joseph Fourier, who developed the concept in the early 19th century.

How is a Fourier series calculated?

To calculate a Fourier series, the function must be decomposed into its individual sine and cosine components using trigonometric identities. These components are then weighted and summed to create the overall Fourier series representation of the function.

What is the purpose of a Fourier series?

A Fourier series is used to approximate a periodic function with a finite number of sine and cosine waves. This can be useful in analyzing and understanding complex periodic phenomena, such as sound waves, electrical signals, and mechanical vibrations.

Can any function be represented by a Fourier series?

No, not all functions can be represented by a Fourier series. The function must be periodic and have a finite number of discontinuities in order for a Fourier series to accurately represent it.

How is a Fourier series used in other fields?

Fourier series have applications in various fields such as physics, engineering, signal processing, and image processing. They are used to analyze and manipulate periodic data, and can also be used to solve differential equations.

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