Q about Fourier Coefficient Derivation

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SUMMARY

The discussion centers on the derivation of Fourier coefficients, specifically the calculation of cn through the multiplication of the equation ∑ cn * ejnx by e-jmx. This process is essential because it involves integrating the resulting product function, which leads to the orthonormality condition expressed as ∫_{0}^L exp((2π/L)(m-n)u) du ∝ δ_{mn}. The participants emphasize the importance of understanding the orthonormal system formed by the elements {e^{jnx}} within the context of Hilbert spaces.

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Enigma322
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For the calculation of cn u have to multiply the equation

∑ cn * ejnx
by
e-jmx

what is the reason for this? in my textbook it says nothing about it and on some sites it just said "without justification"
i guess what I am asking is why does this do what we want?
ps: how do u properly make equations here?
 
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Enigma322 said:
For the calculation of cn u have to multiply the equation

∑ cn * ejnx
by
e-jmx

what is the reason for this? in my textbook it says nothing about it and on some sites it just said "without justification"
i guess what I am asking is why does this do what we want?
ps: how do u properly make equations here?
It's not just multiplication, but you also have to integrate the resulting product function. It's done this way because
$$
\int_{0}^L \exp\left(\frac{2\pi}{L}(m-n)u\right) du \propto \delta_{mn}
$$
 
Yes, the elements ##\{e^{jnx}\}## form an orthonormal system on your Hilbert space ...
 

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