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$\int_a^b \left |X(x)-\sum c_i X_i(x)\right |^2dx =\int_a^b \left [X^2(x)+\sum_{i=1}^n c_i^2 X_i^2-2\sum_{i=1}^n c_iX(x)X_i(x)+2\sum_{i<j}c_ic_jX_iX_j \right ] dx$ because $\begin{array}\\ |X(x)-\sum c_i X_i(x) |^2 &=(X(x)-\sum c_i X_i(x))^2\\ &=X^2(x)-2X(x)\sum c_i X_i(x)+(\sum c_i X_i(x))^2\\ &=X^2(x)-2\sum c_i X(x)X_i(x)+\sum_i c_i ... 0 I'm basically going to paraphrase from Mallat's wavelet tour text where you can find this proof. You can fill in the gaps, make it more rigorous, change assumptions etc... I'm going to assume that the window function,$g(t)$, is real, symmetric, and has unit norm, i.e.$g(t) \in \mathbb{R}$,$g(t) = g(-t)$, and$||g||_2 = 1$. Using Parseval's identity on ... 1 without any other specification than that cited we assume that we are working over a period with$-\pi \le \omega_0t \le \pi$so$T_0 = \frac{2\pi}{\omega_0}$so we are looking for an expansion: $$f(t) = \sum_{n=-\infty}^\infty c_n e^{ni\omega_0 t}$$ inspection generally means "just by looking at" - usually when we can directly guess the correct ... 1 Use the real form of Fourier series. The first thing that we examine is the integral$\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f(t)g(nt)dt} $. If we can determine this integral, then we can work out the limit. Both f and g are continuous and therefore square integrable. Note that since they are both periodic of period 2$\pi$, the convergence properties of the ... 1 Re 2: It is not true that$f \in C^\infty$(for general values of$a$). Note that we are looking at the$2\pi$-periodic extension of the given$f$. For most values of$a$,$f(\pi) \neq f(-\pi)$which implies that the periodic extension of$f$is not continuous at odd multiples of$\pi$. However,$f$is smooth enough to guarantee that the Fourier series of ... 1 I haven't contributed anything, just spelling it all out. We have the assumption that$a=T/4$which implies$T/2=2a$. So using the formular for the$0$'th coefficient, we get: $$c_0=\frac{1}{T}\int_{0}^{T}s(t)\;\mathrm{d}t=\frac{1}{T}\int_{-T/2}^{T/2}s(t)\;\mathrm{d}t.$$ If we define$\Pi_k(t)$to be$1$, if$|t|<k/2$and$0$, if$|t|\geq k/2$, then ... 1 Firstly, the function is zero on both intervals$[-T/2, -a]$and$[a, T/2]$; Also, Integral kernel of Fourier series, the$e^{inω}$, is periodic for each$n∈ \mathbb N$that means integral can be taken on$[-T/2, T/2]$; Since$C_0$is a DC component which means$n$= 0, whole above make the integration of$C_0$stand on$[-a, a]$. 0 This result proves the conjecture is true. Consider a stationary complex random function$\zeta(t)$. Stationarity requires that the autocorrelation function depend on the difference$t –t’$only and that the mean value must be a constant (not a member of a distribution function nor depend on$t$). Assume the power spectral density (hereafter called the ... 0 While I accepted the answer above, this is how my lecturer (and later my friend) explained it to me (the exam is tomorrow). We first define $$g_n(x):=f(x) - S_n f(x)$$ just to remind ourselves that we need to be careful of cancellations. Then uniform convergence of$S_nf$to$f$is equivalent to showing$g_n→ 0$uniformly; since we know (part 2) that$g_n(x) ...

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The discrete-time Fourier transform (DTFT) is defined as $X(\theta)=\sum_n f[n]e^{-i\theta n}$ and it is the equivalent Fourier transform for discrete time series. The resulting $X(\theta)$ is in continuous time and is $2\pi$-periodic. In your case, $f[n]=1$ and you're asking for $X(\theta)|_{\theta=-\omega T_0}$. The DTFT of $1$ should be a delta ...

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The function $f(\omega) = \frac{T_0}{2\pi}\sum_{k=-\infty}^\infty e^{i k \omega T_0}$ is a Fourier series of $\sum_{k=-\infty}^\infty \delta\left(\omega - \frac{2\pi k}{T_0}\right)$. This function is most commonly known under the names the Dirac Comb and the Shah Function.

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A clear connection between Fourier series and the Fourier transform can be formulated in the framework of distribution theory. It turns out that the maximal domain of the Fourier transform is the space of so-called tempered distributions. These are linear functions from a space $S$, called the Schwartz class, into $\mathbb{R}$, with some continuity ...

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I think I know what's bothering you! Yes, if I have understood your point of view, a Fourier series and a Fourier transform are two completely different things. Short answer: the Fourier transform is a continuous version of the coefficients in a Fourier series - and just as you have to sum the coefficients alongside with the sines and cosines to get back a ...

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Yes. Given a continuous function f(t) that is periodic with period T, and the Fourier transform of that function F(w), then there will necessarily always be some time t where F(0) == f(t). (That time t will be different for different functions f(t).) F(0) is also called the DC component, also called the average value of f(t). This can be proven using the ...

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I'm assuming you are working with Riemann integrable functions. Any such function $f$ is uniformly bounded on $[0,2\pi]$ by some constant $M$. You can choose the approximating function $g$ to also be bounded by $M$, and that gives you a reduction: $$\int_{0}^{2\pi}|f-g|^{2}dx \le M\int_{0}^{2\pi}|f-g|dx.$$ Now you only have to approximate in the ...

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If $\phi\in C^\infty (\mathbb {R}),$ and $\phi = 0$ on $[-1,1],$ then $\phi (x)/x^2 \in C^\infty (\mathbb {R}).$ Proof: Being $C^\infty$ is a local property. This function is certainly in $C^\infty (\mathbb {R}\setminus \{0\})$ and it equals the $C^\infty$ function $0$ near $0.$ So now you have to check that the various growth conditions defining the ...

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You could use the heat kernel. For example, the following holds for all $f\in L^{1}(\mathbb{R})$: $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{-s^2t}e^{isx}ds = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(y)e^{-(x-y)^{2}/4t}dy.$$ Under your conditions, $\hat{f} \ge 0$ and $f$ is continuous at $0$. Setting $x=0$ gives $$... 1 The function is defined on the interval (- \pi, 0) \cup (0,\pi), which means that the length of the interval (which happens to be the period) is \boxed{L = 2\pi}. We know that:$$\displaystyle a_n= \dfrac 2L \cdot \int_{-\pi}^{\pi} f(x)\cdot \cos\left(\dfrac{2n\pi x} L\right) \,dx.$$However,$$\displaystyle \int_{-\pi}^0 -1\cdot \cos (nx)\,dx = ...

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Not in general. A set of conditions is given in this paper: Hayes, Monson H., Jae S. Lim, and Alan V. Oppenheim. "Signal reconstruction from phase or magnitude." Acoustics, Speech and Signal Processing, IEEE Transactions on 28.6 (1980): 672-680.

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Here's my attempt: I'm not sure I get those answers either - are you sure they are right? The formula for $a_{n}, b_{n}$ are: $$a_{n} = \frac{1}{L} \int\limits_{-L}^L f(x)cos(\frac{n\pi x}{L})dx$$ $$b_{n} = \frac{1}{L} \int\limits_{-L}^L f(x)sin(\frac{n\pi x}{L})dx$$ Since L is half the period, L = $\pi$ in this example. $$f(x) = \begin{cases} ... 0 To avoid any confusion, let us state that a (pure) sine has the form A\sin(\omega t), a (pure) cosine has the form A\cos(\omega t), a sinusoid has an arbitrary phase and one of the equivalent forms A\sin(\omega t+\phi) or A\cos(\omega t+\psi) - where \phi and \psi differ by a quarter turn. So the sine and cosine are special cases of the ... 0 Start with convergence criterion: A necessary and sufficient condition for the Fourier series T(x) of f to converge pointwise to c(x) on E is that there exists a fixed \delta such that  0 < \delta < \pi and \int_0^\delta {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0}  pointwise on E. Here {g_{c(x)}}(u) = \frac{1}{2}\left( ... 1 Since the integrand is even, your integral is 2\int_0^\infty\dots dt. For this MathWorld (38) gives a closed form in terms of Eulerian numbers$$\int_0^\infty \left(\frac{\sin x}{x}\right)^{2n} \mathrm{d} x = \frac{\pi}{2(2n-1)!} \genfrac{<}{>}{0}{}{2n-1}{n-1}$$0 You have the relationship$$ \mathcal{F}[f](0) = \int_{-\infty}^{\infty} f(x) e^{-i0x} \, dx = \int_{-\infty}^{\infty} f(x) \, dx $$and, for the inverse transform, the Fourier Inversion Theorem gives$$ f(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{F}[f](k) \, dk, $$putting the 2\pi all on the inverse transform (the result is different if you ... 1 The displacement u(x,t) of the string is described by the wave equation for t \ge 0 and 0 \le x \le 1:$$ u_{tt} = c^{2}u_{xx},\\ u(x,0)=f(x),\;\;u_{t}(x,0)=0,\\ u(0,t)=u(1,t)=0. $$The constant c > 0 is a physical constant associated with the properties of the string. You start by looking for separated solutions T(t)X(x) with ... 1 Certainly convergence of the Riemann sums does not follow from integrability and continuity of g. Consider this function g: Spikes of decreasing width. Area of the spike centered at m\pi is 1/2^m. We have: \bullet  g is continuous \bullet  \int_{-\infty}^{+\infty} |g(\xi)|\;d\xi = 1 \bullet  g(m\pi) = 1 for m=1,2,3,\dots; ... 4 Since \hat{f}\in\ell^1\implies f\in L^\infty and \hat{f}\in\ell^2\implies f\in L^2, Riesz-Thorin interpolation guarantees that If \hat{f}\in\ell^q where \frac1p+\frac1q=1 and 1\le q\le2, then f\in L^p Therefore, since the \hat{f} you give above is in \ell^q for all q\gt1, we have that f\in L^p(\mathbb{T}) for all p\lt\infty. ... 0 Adding to the answer before that the Fourier transform T : L^2 \rightarrow L^2 is indeed surjective, the answer is: the Fourier transform is bijective iff p=2 (considering it on all L^p, p \in [1,2]). The most important igredient to show this is Plancherel's Theorem (unitarity of T) which is first used to extend T continuously from L^1 to L^2 ... 5 Knowing this series is convergent, you can find its sum using the formula$$ \sum_{k=1}^{\infty} \frac{z^k}{k} = -\log{(1-z)}. $$Then the sum is$$ -\frac{1}{2}\left( \log{(1-e^{2\pi i \theta})}+\log{(1-e^{-2\pi i \theta})} \right), $$and the limit of this as \theta \to 0 does not exist, since \log{z} diverges as z \to 0. 2 Write the Fourier transform as \mathcal{F}: \mathcal{S} \to \mathcal{S}. Observe that \mathcal{F}^{-1} \neq \mathcal{F} (assuming we choose the unitary definition), but that$$ \mathcal{F}^{-1} f(x) = \mathcal{F} f(-x) $$Therefore we conclude that$$ \mathcal{F}\mathcal{F}\mathcal{F}\mathcal{F} = \mathrm{Id}.This tells you that the eigenvalues must ... 1 Here's a proof of Morlet's inversion formula. We define \begin{align*} C_\psi &:= \int_0^\infty \dfrac{\overline{\hat{\psi}(u)}}{u}\, \mathrm{d}u \end{align*} and assume that this quantity is finite. Assume that x,\psi \in L^2(\mathbb{R}). I make this assumption so that Parseval's identity holds. Then, using Parseval's identity and basic properties of ... 2 The reason is as follows: by the Fourier inversion theorem, if f,\widehat{f}\in L^1, then we know that f agrees almost everywhere with a continuous function f_0. Moreover, f_0 = (f^{\vee})^{\wedge} so, since f^{\vee} is in L^1 (being that it is related to \widehat{f} via reflection), f_0 is bounded by the Riemann-Lebesgue Lemma. Hence f is ... 1 You say you get to -\int_{-l}^l e^{ikx} \phi''(x) \, \mathrm{d} x $$but then apply the Riemann-Lebesgue lemma. 1 In view of this comment clarifying the question, we can give a counter example. Consider$$ f = \exp( - |x|) $$which is in any L^p. Its Fourier transform is well known to be (up to a constant depending on the dimension)$$ (1 + |\xi|^2)^{-(n+1)/2} $$This function is smooth and decays much faster (with all derivatives) than what you supposed. (In ... 2 First, let me say that you can show this pretty easily be writing the convolution as a matrix vector product, with the matrix being a circulant one defined by one of the vectors. The result falls out due to the DFT diagnolizing circulant matrices. Anyway, you can also show this directly substituting the discrete convolution formula, and playing with ... -1 This is only a hint. Try the following problem first: Define g(t) = e^{1/t} on (0,1] and g(0) = 0. Can you show that g(t) \in C^\infty([0,1])? What estimates can you make on the sup norm of the derivatives of g. Now the problem at hand. To show f is C^\infty on an interval like [0, \epsilon] with \epsilon < 2\pi, write f(t) = ... 1 You've got to know a bit about generalized functions for this type of integral to make sense. The generalized function 1(x) can be defined by the sequence 1_n(x)=\exp(-x^2/n^2). The identification with 1 comes from the limit,$$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} f(x) 1_n(x) \mathrm{d}x = \int_{-\infty}^{\infty} f(x) \mathrm{d}x,$$... 3 A correct way to do this with the formalism of the distributions: Let T \in \mathcal S' , then \hat{T},the Fourier transform of T is defined by$$\forall \varphi \in \mathcal S, \langle \hat{T}, \varphi \rangle = \langle T, \hat{\varphi} \rangle $$So the Fourier transform of \delta_{x_0} is given by$$\forall \varphi \in \mathcal S, \langle ...

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Just in order to fix notations, let: $$\mathcal{F}(f)(\omega) = \widehat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)\, e^{-i\omega x}\,dx.\tag{1}$$ Assuming $f_1(x)=e^{-|x|}$ and $f_2(x) = \mathbb{1}_{[-1,1]}(x)$ we have: $$\widehat{f_1}(\omega)=\sqrt{\frac{2}{\pi}}\frac{1}{\omega^2+1},\qquad ... 3 (need to fix the end. This is the right idea, but the end conclusion appears wrong.)$$-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$$when |z|\leq 1 and z\neq 1. Then \sum_{n=1}^\infty \frac{\sin nx}{nx} is the imaginary part of:$$\frac{1}{x}\sum_{n=1}^\infty \frac{e^{inx}}{n}=-\frac{\log(1-e^{ix})}{x}$$When x=\frac{\pi}{4}, this is the ... 0 It is a standard result for the sum of a geometric series (where x may be real or as we need here, complex):$$ \sum_{k=0}^n x^{k}= \frac{1-x^{n+1}}{1-x} $$which you can see by multiplying out (1-x)\sum_{0}^nx^k which gives a telescoping sum equal to 1-x^{n+1}. Now if |x|<1 the sum converges as n\to \infty and the result is:$$ ...

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As for (2), look at Proposition 1.3 of: M. J. Marsden, F. B. Richards, and S. D. Riemenschneider, Cardinal spline interpolation operators on $\ell_p$ data, Indiana Univ. Math. J. 24(1975), 677-689; Erratum, ibid., 25(1976), 919. The result of this is that it is bounded on $\ell_p$ for every $1<p<\infty$. For (1), I can't give you a proof at the ...

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I guess $\beta>0$ to have $G_{\beta}\left(\omega\right)=e^{-\beta\omega^{2}}$ integrable. The fourier transform $\widehat{G_{\beta}}$ has the following property :$$-ix\widehat{G_{\beta}}\left(x\right)=\widehat{\frac{d}{d\omega}G_{\beta}}\left(x\right).$$ Computing the derivative and using the above property again yields to an ODE ...

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Plancherel's Theorem gives $$\int u\overline{v}dx = \int \hat{u}\overline{\hat{v}}ds.$$ So that gives you $$\int uv dx = \hat{u}\overline{\hat{\overline{v}}}ds.$$ And, $$\overline{\hat{\overline{v}}}(s)=\overline{\frac{1}{\sqrt{2\pi}}\int e^{-isx}\overline{v(x)}dx} =\frac{1}{\sqrt{2\pi}}\int e^{isx}v(x)dx,$$ ...

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You may just perform the change of variable $u:=\dfrac x a$, $dx=adu$, to get $$\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{a}f(\frac{x}{a}) e^{iwx} dx= \frac{1}{2\pi}\int_{-\infty}^\infty f(u) e^{iawu} du=F(aw).$$

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Hint: $$uv = \frac{1}{4} ((u+v)^2-(u-v)^2).$$ Apply Plancherel's theorem (twice) and conclude. Note: In your question, it should read $\int \hat{u} \bar{\hat{v}}$ instead of $\int \hat{u} \hat{v}$.

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Hopefully, third time's the charm. Goal Let $f\in L^{1}(\mathbb{R})$ be continuous function, such that $\text{supp}(\hat{f})\subset [-1,1]$ and $f(n)\geq 0$ for all $n\in\mathbb{Z}$. Then $$\int_{\mathbb{R}}f(y)dy\geq 0$$ Set $g(x):=\hat{f}(-x)=f^{\vee}(x)$. Clearly, $\text{supp}(g)\subset [-1,1]$, and since $g$ is continuous, $x^{n}g(x)\in ... 0 Hint :$f\ast g(x)=\int _\mathbb{R}f(x-y)g(y)dy=\int _{-1}^1 f(x-y)dy=\int _{-1}^xe^{y-x}dy+\int _x^1 e^{x-y}dy= \ldots $2 Both integrals can be computed through integration by parts, since: $$\int_{\mathbb{R}}\frac{\sin^2 x}{x^2}\,dx = \int_{\mathbb{R}}\frac{\sin(2x)}{x}\,dx = \pi$$ and: $$\int_{\mathbb{R}}\frac{\sin^4 x}{x^4}\,dx = \int_{\mathbb{R}}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{3x^3}\,dx=\int_{\mathbb{R}}\frac{\cos(2x)-\cos(4x)}{3x^2}\,dx=\frac{2\pi}{3}.$$ The general ... 2 Parseval in this case states that $$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$ where$F$and$G$are the respective FTs of$f$and$g$. (Integrability conditions such as absolute integrability over the real line must be satisfied for both pairs of functions.) When$f(x) = g(x) = \sin{x}/x\$, then ...

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