- #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.
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.