# Tag Info

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


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 ...

Top 50 recent answers are included