How is this possible? (simple fourier series)

In summary, the conversation discusses the equivalence of writing a well-behaved function as a sum of e^{2 \pi i n x} or as a sum of \sin{\pi n x}, and the implications of this in solving a particular partial differential equation. The response also clarifies that while the two methods are equivalent, they may result in different solutions due to the periodicity and oddness of the functions involved. Additionally, the conversation touches on the concept of the zero Fourier coefficient and its relationship to the arbitrary general solution.
  • #1
nonequilibrium
1,439
2
Hello,

Given a well-behaved function [tex]f:[0,1] \to \mathbb C[/tex] with f(0)=0, is it then totally equivalent to write it as a sum of [tex]e^{2 \pi i n x}[/tex] ([tex]n \in \mathbb Z[/tex]) or as a sum of [tex]\sin{\pi n x}[/tex] (n = 1,2,3,...) -- this last one by defining [tex]f:[-1,1][/tex] as an uneven function and then applying the first method?

The reason I ask is because when solving
[tex]-\frac{\partial^2 u(x)}{\partial x^2} = q(x)[/tex]
with the first method (to u(x), because q(x) is given), I gained no information about the zeroth Fourier coëfficient [tex]u_0[/tex] (because the relation between Fourier coëfficients is [tex]4 \pi n^2 u_n = q_n[/tex]),
while with the second method I had nothing of u left undetermined.

Can this be?

Thank you.
 
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  • #2
If u(x) is a solution of your example PDE, then u(x) + C is also a solution for any constant C.

So it makes sense that your Fourier solution method did not give you any information about the value of C (which is the zero Fourier coefficient of course). You need to use the boundary conditions to fix the value of C.

Your first statement is not quite true, because a function and the sum of its Fourier series can be different at a set of discrete points. So they are not "totally equivalent", but they are equivalent in the sense that you were probably thinking about, if you are an engineer or physicist (but not a pure mathematician).
 
  • #3
Hm, two questions regarding your response:

- Then shouldn't I also get the free parameter in the second method?

- Shouldn't I get two parameters in both methods? Because not only u(x) + C is a solution, but also u(x) + C + Dx.

NOTE: hm, you mean f(x) is not equivalent to its Fourier decomposition? I understand that, but I was asking if the exp(inx) decomposition is equivalent to the sine-decomposition.
 
  • #4
mr. vodka said:
Hm, two questions regarding your response:

- Then shouldn't I also get the free parameter in the second method?

- Shouldn't I get two parameters in both methods? Because not only u(x) + C is a solution, but also u(x) + C + Dx.

NOTE: hm, you mean f(x) is not equivalent to its Fourier decomposition? I understand that, but I was asking if the exp(inx) decomposition is equivalent to the sine-decomposition.

OK, I see what you were really asking now.

Yes the sine-series decomposition is equivalent to a full (sine and cosine) decomposition of an odd function over twice the interval, because the cosine terms are all 0 for an odd function.

But when you do this, you will only find solutions that are periodic odd functions. That is OK for a particular solution, because you are also extending q(x) to be a periodic odd function. But you lose the arbitrary general solution C + Dx, because the only values of C + Dx that satisfy the conditions you assumed are C = D = 0.
If C is non-zero, C is periodic, but not an odd function.
If D is non-zero, Dx is an odd function, but not periodic.

The reason you got the equation 0.u_0 = q_0 = 0, is because there is no "zero sine term" in a Fourier series. Sin 0x is identically zero, so the value of the coeffient is irrelevant.

If you have a general complex Fourier series, the e^(0iz) term is always equal to 1, but the coefficient of the "1" can be any complex number, so you don't lose information in that case.
 
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  • #5


Dear scientist,

The concept of a Fourier series is based on the idea that any well-behaved function can be represented as a sum of sinusoidal functions. This is known as the Fourier theorem. In the first method, you are using complex exponential functions to represent the function, while in the second method, you are using sine functions.

Both methods are valid and can be used to represent the function. However, the choice of which method to use may depend on the specific problem you are trying to solve. In the case of solving the equation -\frac{\partial^2 u(x)}{\partial x^2} = q(x), the first method may not provide information about the zeroth Fourier coefficient, but it may be more useful for solving the equation. On the other hand, the second method may provide information about the zeroth coefficient but may not be as useful for solving the equation.

In general, both methods can be used to represent a function, and it is up to the individual to choose which method is most suitable for their problem. It is possible to use both methods to represent the same function, as long as the chosen method is applied correctly. I hope this helps clarify the concept of Fourier series for you.
 

1. How does a simple Fourier series work?

A simple Fourier series is a mathematical tool used to represent a periodic function as a sum of sine and cosine functions with different amplitudes and frequencies. It is based on the idea that any periodic function can be broken down into simpler components, making it easier to analyze and manipulate. By using a finite number of terms in the series, a good approximation of the original function can be achieved.

2. What is the purpose of using a Fourier series?

A Fourier series is useful for many applications in science and engineering, such as signal processing, image compression, and solving differential equations. It allows us to represent complicated periodic functions in a simpler form, making it easier to analyze and manipulate them.

3. How is a Fourier series different from a Fourier transform?

While both Fourier series and Fourier transform are mathematical tools used to analyze and represent functions, they have different applications. A Fourier series is used for periodic functions, while a Fourier transform is used for non-periodic functions. The Fourier transform also provides a continuous representation of a function, while the Fourier series provides a discrete representation.

4. 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 with a finite period for a Fourier series to be applicable. Additionally, the function must also satisfy certain conditions, such as being continuous and having a finite number of discontinuities, for the Fourier series to converge.

5. What is the Fourier series formula?

The formula for a simple Fourier series is given by:
f(x) = a0 + ∑n=1(ancos(nx) + bnsin(nx))
where a0, an, and bn are the coefficients that determine the amplitude and frequency of the sine and cosine terms. These coefficients can be calculated using integrals or by solving a system of equations.

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