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1

When $\alpha \neq 0$ you'll need to deform the contour over the new saddle point located at $z = i\operatorname{arctanh} \alpha$. At this saddle point we have $$ \cos z + i\alpha \sin z = \sqrt{1-\alpha^2} - \frac{1}{2}\sqrt{1-\alpha^2} (z-i\operatorname{arctanh} \alpha)^2 + O(z-i\operatorname{arctanh} \alpha)^3, $$ so the Laplace method yields $$ ...


1

The support of $\psi$ must appear somewhere, otherwise $x\to 1$ is not a distribution, for example. Part A. We bound things: $$ \left|\int_{|x|>a} \frac{1}{x}\psi dx\right|\leq \sup_{|x|>a} \frac{1}{|x|} |\int_{|x|\geq a} \psi dx| \leq \frac{1}{a}\|\psi\|_{\infty}|\textrm{supp} \psi|. $$ For the second part, $$ ...


1

What happens depends on what variable(s) you are applying the Fourier transform to; if we suppose we are making the Fourier transform with regards to $x \rightsquigarrow \xi$, then if we are using the one-dimensional convention that $$ \hat{f}(\xi) = \int^{\infty}_{-\infty}f(x)e^{-i\xi x} dx$$ so in higher dimensions we generalise to $$ \hat{f}(\xi,y,t) ...


1

I think that as long as everything is well defined (for example, being in the Schwartz space), then the result is what you think it should be. For example, $$F\left(\frac{\partial^2}{\partial x \partial y}\right) u(x,y,t) = (ix)(iy) \hat{u}(x,y,t)$$ (assuming you are doing the transform on $x$ and $y$). This is stated in Stein's book on Fourier Analysis ...


2

The Fast Fourier Transform is a particularly efficient way of computing a DFT and its inverse by factorization into sparse matrices. The wiki page does a good job of covering it. To answer your last question, let's talk about time and frequency. You are right in saying that the Fourier transform separates certain functions (the question of which functions ...


2

Fourier Transform is a function. Fast Fourier Transform is an algorithm. It is similar to the relationship between division and long division. Division is a function, long division is a way to compute the function.


1

Let $a>0$. Put $$T_a= \int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ and for $\varepsilon\in ]0,a[$: $$T_a(\varepsilon)=\int_{|x| \geq a} \frac{\psi(x)}{|x|} dx + \int_{\varepsilon<|x|<a} \frac{\psi(x)-\psi(0)}{|x|}dx$$ We have that $T_a(\varepsilon)\to T_a$ if $\varepsilon\to 0$. Now: ...


1

The first computation is correct, but very inefficient. No integration or changes of variable are needed. The distribution $k(1-e^{x/k})\delta_{1/k}$ means: we multiply a test function by $k(1-e^{x/k})$ and then plug in $x=1/k$. This is exactly the same as plugging in $x=1/k$ and multiplying by $k(1-e^{1/k^2})$. The result will not exceed ...


1

Your attempt went astray at this step: $$ \left|\sum_{n \epsilon N} \psi(x) -\psi(0)\right| =\sum_{n \epsilon N} |\psi(x) -\psi(0)| \tag{wrong} $$ The triangle inequality gives $\le$, but this does not help you demonstrate your claim (since you wanted to show the sum on the left is large). Instead, consider a partial sum: $$\sum_{n=1}^M (\psi(x) -\psi(0)) ...


1

For $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$ and $y\in\mathbb{R}^{d}$, let us write $\left(M_{y}f\right)\left(x\right):=e^{2\pi i\left\langle x,y\right\rangle }\cdot f\left(x\right)$. It is then easy to see $\left\Vert M_{y}f\right\Vert _{1}=\left\Vert f\right\Vert _{1}$ for $f\in L^{1}(\mathbb{R}^{d})$ [in fact, this remains valid for every ...


0

Thanks to @mathematican For any ψ with ψ′=1 on [0,1] the sum isn't even finite.


1

Your notes are messed up; maybe you dozed off while the lecturer changed the assumptions on $f$? :) As others said, $f\in L^2$ implies that the Fourier series converges to $f$ a.e., by Carleson's theorem. The way your statement is set up is also weird. Setting $$\displaystyle f(a)=\frac{1}{2}\frac{f(a^+)-f(a^-)}{2}\tag1$$ at the jump points of a piecewise ...


2

You might want to take a look at the answers to this question. In less technical language: The delta “function” has the defining property that $$\int_{-\infty}^\infty \delta(x)f(x)\,dt=f(0)$$ for any continuous function $f$. Substituting in $x=t-t_0$ with $f(x)=e^{-j\omega x}$ immediately yields the desired result. Your rewrite of the integral “according ...


0

I'm not entirely sure what your work is supposed to convey. One thing to note is that you should never associate delta distributions to integration in this way. It is not a function. Do not treat it as such. Note that if $f\in C_c^{\infty}(\mathbb{R})$, then $\langle u,f\rangle = \sum_{k=1}^{\infty} \frac{1}{k^2}f\left(\frac{1}{k}\right)$. You know that $f$ ...


0

isn't the following a counter-example to the claim that the Fourier transform is continuous: Fourier transform of sin(x)/x is the box function?


0

With $f(t)=\sin(2t)e^{-|t|}$ one simple option to compute the Fourier transform is to use the Fourier transform $G(\omega)$ of $g(t)=e^{-|t|}$ (I leave this part up to you), and then use the modulation property of the Fourier transform: $$F(\omega)=\frac{1}{2i}\left[G(\omega-2)-G(\omega+2)\right]$$ (because $\sin(2t)=\frac{1}{2i}(e^{2it}-e^{-2it})$)


0

We want to evaluate $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}\sin(2t)e^{-|t|}\,dt.$$ The simplest approach - I think - is to introduce the complex exponential. Recognize that $\sin(x) = \text{im}(e^{ix})$, where $\text{im}$ means to take the imaginary part. So the equivalent problem is to evaluate ...


1

What is going on is - after splitting the integral into two by distributing the factor $\frac{i}{z} - \frac{i}{z-i}$ - a shift of contour. Let us suppose that $f$ is holomorphic (or more generally meromorphic) on $\{ z : u < \operatorname{Im} z < v\}$ and satisfies some appropriate growth restrictions as $\lvert\operatorname{Re} z\rvert \to \infty$. ...


1

The Fourier transform ${\cal S}(\mathbb{R})\to {\cal S}(\mathbb{R})$ is a linear isometry (Fourier inversion). If $C\subseteq W$ where $C$ is compact and $W$ is open, then we can find a smooth function $g$ such that $g\equiv 1$ on $C$ and $g\equiv 0$ outside $W$ (why?). Let us take $W=\{x\in\mathbb{R}: d(x,C)<1\}$ for concreteness and observe now that $g$ ...


1

You already know that $\int_\mathbb{R}|\widehat{f}(\omega) |(1+|\omega|^p) d\omega<+\infty$ implies $f$ is $p$ times continuously differentiable. Now suppose $\hat f$ has compact support. Under mild assumptions on $f$ (namely, $f\in L^1+L^2$) we know that $\hat f$ is locally integrable. The integral $\int_\mathbb{R}|\widehat{f}(\omega) |(1+|\omega|^p) ...


2

The Fourier transform is a unitary operator on your space. This means that its transpose is its inverse, $\mathcal F ^* = \mathcal F^{-1}$. The typical thing to do is to replace $T$ with $\mathcal F T \mathcal F^*$. Observe that with this convention, you have $$ (\mathcal F T \mathcal F^*)\hat f=(\mathcal F T \mathcal F^*)\mathcal F f=\mathcal F ...


5

To compute the limit, take a test function $\varphi$ and integrate: $$\int_{-\frac{1}{2k}}^{\frac{1}{2k}} k(1+\cos (2\pi kx))\varphi(x)\,dx = k\int_{-\frac{1}{2k}}^{\frac{1}{2k}}\varphi(x)\,dx + \int_{-1/2}^{1/2}\cos (2\pi y)\varphi\left(\frac{y}{k}\right)\,dy.$$ The first integral tends to $\varphi(0)$, and the second to $$\varphi(0)\int_{-1/2}^{1/2} ...


3

You have supposed that $a>0$. You have to take cases for $a$. In the case of $a<0$ the absolute value will appear. $a<0:$ $$\int_{-\infty}^{+\infty}x(a \cdot t) \cdot e^{-j \omega t}dt$$ $$y=a \cdot t \Rightarrow t=\frac{y}{a}$$ $$dy=adt$$ $$\text{ When } t \rightarrow -\infty \Rightarrow y \rightarrow +\infty$$ $$\text{ When } t \rightarrow ...


1

False. Example : $\hat{f}$ : the rect function / $f$ : sinc


0

If $T$ is the period of your input signal $y(t)$, and $\omega_0=2\pi/T$ then the complex Fourier series of $y(t)$ is $$y(t)=\sum_{n=-\infty}^{\infty}c_ne^{jn\omega_0t}$$ where $c_n$ are the Fourier coefficients of $y(t)$. For each complex exponential $e^{jn\omega_0t}$ the corresponding output of an LTI system with frequency response $H(\omega)$ is ...


0

Hint: use the time-shifting property ("translation" here) of the Fourier transform.


1

Please check page 7.7.3 of ERDELYI_Higher Transcendental Functions_Volume 2. The integral closest to the one that you need is:


0

Hint: $$\hat{f}(k_x,k_y)=\int_{-\infty}^{\infty}\, dx \int_{-\infty}^{\infty}\, dyf(x,y)\ e^{-ixk_x-iyk_y}$$ $$\hat{f}(k,\alpha)=\int_{0}^{\infty}\, rdr\int_{0}^{2\pi}\, d\phi f(r)\ e^{-ir k \cos(\phi-\alpha)} $$


0

Use Parseval's theorem to complete the proof : Fourier transform of $f-f^*$ is zero so the function must be zero.


1

Since $p\ge 1$, we have $-2p+1\le -1$. Therefore, the integral $\int_0^1 r^{-2p+1}\,dr$ is ... not what you wrote. Divergence of integral in Part B follows from the divergence of integral in Part A. Frankly, these are strange questions. Re-check for typos.


0

This is a problem I also faced as a student. This happens everywhere in mathematics and perhaps in other subjects too. It is better to use different names when you have a different meaning or a different formula. It is better to use one standard than every one use their own formula and use the same name. Matrix canonical forms also different ...


1

As long as you don't consider the inverse transformation it is unimportant whether you write $e^{i\lambda t}$ or $e^{-i\lambda t}$ in the definition of $F$. As soon as you look at $F$ and $F^{-1}$ at the same time you have to put a minus sign either in the definition of $F$ or in the resulting $F^{-1}$. This is so because the Fourier transform is sort of a ...


3

The information you are given says that $\lvert\lvert f\rvert\rvert_{L^{\infty}(\mathbb{R})}=1$ and $\lvert\lvert f\rvert\rvert_{L^{2}(\mathbb{R})}=\sqrt{\pi}$. Let $\frac{1}{p_{t}}=\frac{1-t}{2}+\frac{t}{\infty}=\frac{1-t}{2}$. We use the log convexity of $L^{p}$ norms (sometimes referred to as an interpolation result) to conclude: $$\lvert\lvert ...


2

The DTFT of the sequence $x_n$ is $$X(\omega)=\sum_{n=-\infty}^{\infty}x_ne^{-in\omega}$$ Its complex conjugate is (given that $x_n$ is real-valued) $$X^*(\omega)=\sum_{n=-\infty}^{\infty}x_ne^{in\omega}=X(-\omega)$$ So for real-valued $x_n$ you have $$X^*(\omega)=X(-\omega)\tag{1}$$ and if you take the magnitude on both sides of (1) you get ...


0

What about this?: \begin{align} \\ &=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-(x_1^2\lambda_1+\cdots+x_n^2\lambda_n)}e^{-i<t,Sx>} \, \mathrm{d}x \\&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! \prod_{j=1}^{n}e^{-\lambda_jx_j^2}\prod_{j=1}^{n}e^{-i((S^{T}t)_jx_j)} \, \mathrm{d}x ...


0

Can I do $\mathbb{R}^{2}$ and you do $\mathbb{R}^{n}$? Since $\mathbf{A}$ is symmetric and positive definite, there exists an orthonormal basis for $\mathbb{R}^{2}$ such that $\mathbf{A} = \begin{pmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{pmatrix}$. Rotate the coordinate system so the positive $x$-axis points in the same direction as the ...


5

We have $$ e^{-i\frac{4}{3}\pi n} - e^{-i\frac{2}{3}\pi n}=e^{i\frac{2}{3}\pi n} - e^{-i\frac{2}{3}\pi n}=2i\sin\left(\frac{2\pi n}3\right)=\left\{\begin{array}\\0&\text{if}&n\equiv0\mod3\\ \pm i\sqrt3&\text{if}&n\equiv\pm1\mod3 \end{array}\right. $$


0

Use the sum formula $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ and compute the Fourier coefficients on just the part with the variables.


0

$|| f ||_p^p = \int_{\{(x,y): x^2 + y^2 \leq 1\}} \frac {1}{|x + iy|^{2p}} dxdy = \int_0^{2\pi} \int_0^1 \frac {1}{r^{2p}} r dr d\theta$ I think you can do the rest.


0

The technique is basically the same as in Part A, but you need to be careful about using Fubini's theorem. Fix some set $U = [\epsilon, 1] \times[-1, 1]$ for $\epsilon > 0$. Then $$C_1 \int_{[-1, 1]} |x_2|^{-p/3}\,dx_2 \leq \int_U |f|^p\, \leq C_2 \int_{[-1, 1]} |x_2|^{-p/3}\, dx_2,$$ where $C_1 = \min_U |x|^p > 0$ and $C_2 =\max_U |x|^p > 0$. That ...


0

The Fourier transform of an integrable function is bounded, so having $x^2$ in the denominator is out of question. I used Wolfram Alpha to remind myself that the transform of $\exp(-|x|)$ is the Cauchy distribution: $$\mathcal F\left(e^{-|x|}\right) = \frac{2}{\pi}\frac{1}{\omega^2+1}$$ By the scaling property, $$\mathcal F\left(e^{-a|x|}\right) = ...


1

In order to make sense of sines with complex arguments, you need to decompose them into complex exponentials. After that notice you're working with two series, each a Taylor expansion of the natural logarithm function (written appropriately). Proceed from there.


1

One way to think about it is in terms of what you are discarding. The majority of the power in the image's spectrum is in the low frequency bins, right? So when you ignore the magnitude of the frequency samples by setting them all to, say, unity, you are attenuating those samples whose magnitude was higher than unity and amplifying those whose magnitudes ...


0

There's a few ways to solve it, but here's one: Change of variables: Work with asset prices $Y=\exp(y)$ and exercise prices $K=\exp(k)$ rather than log prices $y$ and log exercise prices $k$. Restrict range of integration: Option prices are bounded. Hint: $(Y-K)^+$ Apply the expression: $$ \int_u^\infty x^{-v} (x-u)^{w-1} dx = u^{w-v}\beta(v-w, w)$$ ...


0

The JPEG format is based on the Discrete Cosine Transform, a real form of the Fourier transform. You can also use SVD to compress images. See this Wolfram demonstrations for instance.


0

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


1

The integral does not converge in any usual sense. You want the Fourier transform of $\ln |x|$. The Laplacian of $\ln |x|$ is $2\pi \delta_0$, a multiple of the delta function at the origin. The Fourier transform of this $2\pi \delta_0$ is the constant function $2\pi$. On the other hand, taking the Laplacian amounts to multiplying the Fourier transform by ...


3

Pointwise convergence of Fourier coefficients is too weak for such conclusion. Let's take $f\equiv 0$ and $f_m = e^{m(1+ix)}$. Then the convergence of coefficients holds, but $|f_m|\to\infty$ pointwise. If you also assume that the coefficients are uniformly bounded in $\ell^2$ norm, then weak convergence in $L^2$ holds. That is, $\int f_m g\to \int fg$ for ...


1

Hint: If using the Fourier transform $\hat{f}(\xi) = \int\limits_{-\infty}^{\infty}f(x)e^{-2\pi i x\xi}dx$, then $x^nf(x) $ has the Fourier transform $\left(\frac{i}{2\pi}\right)^n\frac{d^n\hat{f}(\xi)}{d\xi}$, and $f(x) = e^{-\alpha x^2}$ has the Fourier transform $\hat{f}(\xi) = \sqrt{\frac{\pi}{\alpha}}e^{-(\pi\xi)^2/\alpha}$. What happens if you try to ...


1

The proof is correct but so wordy... How about this: There is a standard theorem in real analysis: if $\sum f_n'$ converges uniformly on some interval $I$ and $\sum f_n$ converges at some point of $I$, then $\sum f_n$ converges uniformly on $I$, its sum $f$ is differentiable, and $f'=\sum f_n'$. The hypotheses in 1 apply to $f_n(x)=a_n\cos nx+b_n\sin ...



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