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## Main Question or Discussion Point

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.