- #1
- 2,810
- 605
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
[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