Fourier series representation of piecewise function

$ $ {Expand} \; f(x)= \begin{cases} 2{A\over L}x & 0\leq x\leq {L\over 2} \ \ 2{A\over L}\left(L-x\right) & {L\over 2}\leq x\leq L \end{cases} $ $

I have determined $ A_0$ (but omitted) to be $ A_0=A$ . For $ \ A_n \ $ and $ \ B_n$ , I’m rather confused where to start. For $ A_n$ : $ $ A_n={2\over L}\left[2{A\over L}\int_{0}^{L\over 2}x\cos\left({2n\pi x}\over L\right)dx+2{A\over L}\left(L\int_{L\over 2}^L\cos\left({2n\pi x}\over L\right)dx-\int_{L\over 2}^Lx\cos\left({2n\pi x}\over L\right)dx\right)\right]$ $ And a similar case to $ B_n$ , I’m not quite sure if I’m on the right path.

Fourier series Parseval equality with partial sums

Let $ f \in \mathcal{R}_\left[-\pi,\pi\right]$ be function with period $ 2\pi$ . We denote n-th partial Fourier series sum of function $ f$ with $ S_n(x)$ . Prove that: $ $ \int_{-\pi}^{\pi}\left(f(x)-S_n(x) \right)^2dx = \pi\sum_{k=n+1}^{\infty}(a_k^2+b_k^2) $ $

I know that Parseval equality(is it called identity?) is true: $ $ \frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = \frac{a_0^2}{2} + \sum_{k=1}^{\infty}(a_k^2+b_k^2) $ $

I tried the following (without much loss of generality I have assumed that $ a_0 = 0$ ) to obtain Parseval equality back: $ $ \int_{-\pi}^{\pi}\left(f(x)-S_n(x) \right)^2dx = \pi\sum_{k=n+1}^{\infty}(a_k^2+b_k^2) $ $ $ $ \frac{1}{\pi}\left(\int_{-\pi}^{\pi}f(x)^2dx -2 \int_{-\pi}^{\pi}f(x)S_n(x)dx + \int_{-\pi}^{\pi}S_n(x)^2dx \right)= \sum_{k=n+1}^{\infty}(a_k^2+b_k^2) $ $ Now, we have: $ $ \int_{-\pi}^{\pi}f(x)S_n(x)dx = \int_{-\pi}^{\pi}f(x)\left[\frac{a_0}{2} + \sum_{k=1}^n\left(a_kcos(kx) + b_kcos(kx)\right)\right]dx = $ $ $ $ = \sum_{k=1}^n\int_{-\pi}^{\pi}f(x)a_kcos(kx)dx+ \sum_{k=1}^n\int_{-\pi}^{\pi}f(x)b_kcos(kx)dx = $ $ $ $ = \sum_{k=1}^na_k\int_{-\pi}^{\pi}f(x)cos(kx)dx+ \sum_{k=1}^nb_k\int_{-\pi}^{\pi}f(x)cos(kx)dx = $ $ $ $ = \sum_{k=1}^na_ka_k+ \sum_{k=1}^nb_kb_k = $ $ $ $ = \sum_{k=1}^na_k^2+ \sum_{k=1}^nb_{k}^2 = \sum_{k=1}^na_k^2+b_k^2 $ $

So the main equation is now: $ $ \frac{1}{\pi}\left(\int_{-\pi}^{\pi}f(x)^2dx -2\sum_{k=1}^na_k^2+b_k^2 + \int_{-\pi}^{\pi}S_n(x)^2dx \right)= \sum_{k=n+1}^{\infty}(a_k^2+b_k^2) $ $

I feel like I got closer, because I have the $ a_k^2 + b_k^2$ sums from 1 to n and from n+1 to infinity. Is that correct line of reasoning? How can I proceed from here?

$L^2$ functions with compactly supported fourier transforms form a Hilbert space

Given a fixed compact subset of $ \mathbb{R}$ , I want to show that square integrable functions on the real line whose fourier transforms are supported in the given compact set form a Hilbert space in the $ L^2$ norm. It was easy to show that the functions form a subspace. However I am stuck at the closedness of the subspace. Could anyone please help me?

Fourier coefficient of a periodic distribution

Let $ \tau>0$ , and let $ T\in \mathcal{D}'(\mathbb{R})$ be a $ \tau$ -periodic distribution (that is, $ \langle T, \varphi(\cdot+\tau)\rangle= \langle T,\varphi\rangle $ for all $ \varphi \in \mathcal{D}(\mathbb{R})$ ). Then $ $ T=\sum_{n\in \mathbb{Z}} c_n e^{i 2\pi t/\tau}, $ $ for some $ c_n\in \mathbb{C}$ , and where the equality means that the symmetric partial sums of the series on the right hand side converge in $ \mathcal{D}'(\mathbb{R})$ to $ T$ . What are the $ c_n$ s in terms of $ T$ ? One would think that they are given by $ c_n=\langle T, e^{-in2\pi /\tau}\rangle/\tau$ , but $ e^{-in2\pi/\tau}$ is not a test function in $ \mathcal{D}(\mathbb{R})$ .

On a simple discrete Fourier transform

Given an integer N, and two non-negative integer valued variables x,y which take values in $ {0,1,…,N-1}$ . Is it possible to obtain a close form for the following summation?

$ f(x,y)=\frac{1}{N}\sum_{z=0}^{N-1} e^{-i \frac{2\pi z}{N}( x + \frac{y}{N} )} $

Clearly when y=0, $ f(x,0)=\delta_{x,0}$ , but I am not sure how to evaluate the other limiting case $ f(0,y)$ , not even mention the general case. If I take $ N\rightarrow\infty$ , I think I should have roughly $ \delta(x+y/N)$ , but numerically it does not seem to be case.

Any help is greatly appreciated.

What is the meaning of the “constant term of Eisenstein series” in terms of Fourier analysis

Let $ G$ be a connected, reductive group over $ \mathbb Q$ , with parabolic subgroup $ P = MN$ . Let $ \pi$ be a cuspidal automorphic representation of $ M(\mathbb A)$ . For a smooth, right $ K$ -finite function $ \phi$ in the induced space $ \operatorname{Ind}_{P(\mathbb A)}^{G(\mathbb A)} \pi$ (realized in a suitable way as a function $ \phi: G(\mathbb Q) \backslash G(\mathbb A )\rightarrow \mathbb C$ ), we can associate the Eisenstein series

$ $ E(g,\phi) = \sum\limits_{\delta \in P(\mathbb Q) \backslash G(\mathbb Q)} \phi(\delta g)$ $ Assuming $ \pi$ is chosen so that this series converges absolutely, one can define the constant term of the Eisenstein series along a parabolic subgroup $ P’$ with unipotent radical $ N’$ :

$ $ E_{P’}(g,\phi) = \int\limits_{N'(\mathbb Q) \backslash N'(\mathbb A)}E(n’g,\phi)dn’ \tag{0}$ $

I see the constant term defined in this way without reference to Fourier analysis. Is it possible to always realize this object as the constant term of an honest Fourier expansion on some product of copies of $ \mathbb A/\mathbb Q$ ?

This can be done when $ G = \operatorname{GL}_2$ and $ P = P’$ the usual Borel. The unipotent radical identifies with the additive group $ \mathbb G_a$ , and for fixed $ g \in G(\mathbb A)$ the function $ \mathbb A/\mathbb Q \rightarrow \mathbb C, n \mapsto \phi(ng)$ has an absolutely convergent Fourier expansion

$ $ E(ng,\phi) = \sum\limits_{\alpha \in \mathbb Q} \int\limits_{\mathbb A/\mathbb Q} E(n’ng,\phi) \psi(-\alpha n’)dn’ \tag{1}$ $ where $ \psi$ is a fixed nontrivial additive character of $ \mathbb A/\mathbb Q$ . The constant term is

$ $ \int\limits_{\mathbb A/\mathbb Q} E(n’ng,\phi) dn’$ $ Setting $ n = 1$ in (1) gives us a series expansion for $ E(g,\phi)$ and (0) is the constant term of this series.

Fourier transform exponent

How would i complete the Fourier transform in the following image.

My understanding is that in this situation the 1 in the red circle in the image must be negative, in order for e^-infinity to approach 0, thus equalling 0.

This is not the case. As the 1 is positive. I have found online that the Fourier transform of this function is X(W) = (1)/(1+jW). although i cannot get this result due to the e^infinity.

Also what is the effect of e^-jw(infinity). this will be cosine and imaginary sin function with he same w but evaluated at time infinity which is undefined?