How to Expand cx(x-l) into a Fourier Series?

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The discussion focuses on expanding the function $$cx(x-l)$$ into a Fourier series, specifically into the form $$\sum_{n=-\infty}^{\infty}a_{n}e^{-\alpha_{n} x}$$ where $$\alpha$$ is a constant. Participants noted a potential typo in the original summand, questioning the dependence of the exponential on the index $$n$$. The correct transformation was confirmed to resemble a Fourier series transform, with an example provided: $$cx(x-1) = c\left(\frac{\pi^2}3 + (-2-i)e^{-ix} - (2-i)e^{ix} + ...\right)$$.

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I have a function $$cx(x-l)$$ where c is constant

I want to expansion this function $$cx(x-l)$$ to $$\sum_{n=-\infty}^{\infty}a_{n}e^{-\alpha x}$$

how can i do it? you have a idea
 
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Is there perhaps a typo in the summand? The exponential does not seem to depend on $n$ and it is not clear to me what $\alpha$ is.
 
Janssens said:
Is there perhaps a typo in the summand? The exponential does not seem to depend on $n$ and it is not clear to me what $\alpha$ is.


$$cx(x-l)$$ to $$\sum_{n=-\infty}^{\infty}a_{n}e^{-\alpha_{n} x}$$ $\alpha$ is constant any constant
 
Another said:
$$cx(x-l)$$ to $$\sum_{n=-\infty}^{\infty}a_{n}e^{-\alpha_{n} x}$$ $\alpha$ is constant any constant

That looks like a Fourier series transform.

We have:
$$cx(x-1) = c\left(\frac{\pi^2}3 + (-2-i)e^{-ix} - (2-i)e^{ix} + ...\right)$$
 

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