## A question about Fourier series

We know that because $\sin{nx}$ and $\cos{nx}$ are degenerate eigenfunctions of a hermition operator(the SHO equation),and eachof them form a complete set so we for every $f(x)$ ,we have:

$f(x)=\frac{a_0}{2}+\Sigma_1^{\infty} a_n \cos{nx}$
and
$f(x)=\Sigma_1^{\infty} b_n \sin{nx}$
But here,because for every $m$,$\sin{mx}$ and $\cos{mx}$ are orthogonal,we also can have:
$f(x)=\frac{a_0}{2}+\Sigma_1^{\infty} a_n \cos{nx} + \Sigma_1^{\infty} b_n \sin{nx}$
And its easy to understand that the $a_n$s and $b_n$s are the same.
So it seems we reach to a paradox!
What's wrong?
thanks
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 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.
 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?

## A question about Fourier series

The point is,$\sin{nx}$ are eigenfunctions of the hermition equation below:
$\frac{d^2 y}{dx^2}+n^2 y=0$
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 $\cos{nx}$

We have:
$f(x)=\Sigma_1^{\infty} b_n \sin{nx}$
if we multiply by $\sin{mx}$ and integrate from $0$ to $2 \pi$ we get:
$\int_0^{2\pi} f(x) \sin{mx} dx = \Sigma_1^{\infty} \int_0^{2 \pi} b_n \sin{mx} \sin{nx} dx=b_m \int_0^{2 \pi} \sin^2{mx} dx=\pi b_m$
And so:
$b_n=\frac{1}{\pi} \int_0^{2 \pi} f(x) \sin{nx} dx$
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.
 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.

 Quote by Shyan 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 $\cos{nx}$
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: $f(x)=\Sigma_1^{\infty} b_n \sin{nx}$ if we multiply by $\sin{mx}$ and integrate from $0$ to $2 \pi$ we get: $\int_0^{2\pi} f(x) \sin{mx} dx = \Sigma_1^{\infty} \int_0^{2 \pi} b_n \sin{mx} \sin{nx} dx=b_m \int_0^{2 \pi} \sin^2{mx} dx=\pi b_m$ And so: $b_n=\frac{1}{\pi} \int_0^{2 \pi} f(x) \sin{nx} dx$ 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 $a_n = b_n$? Because that is simply wrong.
 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.
 But for every $m$ , $\sin{mx}$ and $\cos{mx}$ are degenerate eigenfunctions.Doesn't that make a difference? I mean making each of them spanning alone?
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.