- 2,802
- 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