Going from Fourier sum to Fourier integral - confusion on intermediate step

Question

How does it solve the $A_0$ term? Can one prove this with an example? I don't know how to prove it to myself with an example

In my lecture notes, my lecturer is trying to justify $C_n$ as weighting the sine and cosine terms in the Fourier integral differently despite how it looks explicitly in the Fourier infinite sum, that is, in:

$$\sum_{n=-\infty}^{\infty} C_n \ e^{ik_nx} \ $$

I have two main confusions with this:

1) How does this rewriting with $a_n$ and $b_n$ resolve the $A_0$ term?

2) I don't see how the substitution of $A_n$ and $B_n$ for $a_n$ and $b_n$ is valid, even given the changed summation ranges. I also don't know how I'd prove it with an example. How can one do this?

Answer 1

The "trick" is in the sum from $-\infty$ to $\infty$ where $|n|$ is counted once for $|n|=0$ , and twice for $|n| \ge 1$.

Answer 2

To clarify I'll derive the $f(x)$ expression directly. $C_ne^{ik_nx}=C_n(cos(k_nx)+isin(k_nx))$. Therefore $a_n=C_n$ and $b_n=iC_n$, for all n.
For $n \gt 0$ use the fact that $k_{-n}=-k_n$ and cos is even while sin is odd. Therefore for $n \gt 0, A_n=a_n+a_{-n}$ while $B_n=b_n-b_{-n}$.
$A_0$ has to be treated as a special case. $k_0=-k_{-0}$ forces $k_0=0$, so there is no sin term and the cos term $=1$, while $A_0=a_0+a_{-0}$ means $a_0$ is counted twice, so that $\frac{A_0}{2}$ is what is needed for the final form $f(x)=\frac{A_0}{2}+\sum _{n=1}^{\infty}A_ncos(k_nx)+B_nsin(k_nx)$.

I can see that the lecture notes are somewhat confusing, since they work backwards from $A_n$ and $B_n$ rather than forward.