# Tag Info

1

The way I view the use of the FT using periodic BCs is that the problem is really about expressing the solution in terms of Fourier series with the "transform" being the Fourier coefficients. The periodic BC's imply that $$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(t) e^{i n \pi x}$$ where $$f(x) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi x}$$ Thus, ...

0

Indeed, as you were told, there is no proof of the formula you're asking, this is a definition. However, here is the way why the expression on the left is used in so many contexts. Consider a sequence of function $$\delta_\epsilon(x)=\begin{cases}\frac{1}{2\epsilon},&-\epsilon<x<\epsilon,\\ 0,&\mbox{otherwise}. \end{cases}$$ It is ...

2

We are computing here the Fourier transform of a Gaussian. It is a basic fact of Fourier theory that the result is again a Gaussian, appearing as a function of the "frequency variable". This can be proven using residue calculus, or applying Green's theorem to a suitable setup in ${\mathbb R}^2$. In this way one obtains, e.g., $$\int_{-\infty}^\infty ... 1 Call this integral I\left( f\right) so I\left( 0\right)=\int_{\mathbb{R}}\exp\left(-\pi x^2\right)=1 and$$\frac{\partial I}{\partial f}=-i\pi\sqrt{2}\int_\mathbb{R}x \exp\left(-\pi x^2-i\pi\sqrt{2}fx\right)dx.$$Rewrite I,\,\frac{\partial I}{\partial f} using x = y - \frac{if}{\sqrt{2}} to prove that \frac{\partial I}{\partial f} = -\pi fI. (You'll ... 1 Using your definitions of the \text{sinc}-function (\text{sinc}(t)=\sin(\pi t)/(\pi t)) and of the \text{rect}-function (a rectangle of height 1 in the range -\frac12<t<\frac12, and zero otherwise), you have the Fourier correspondence$$T\;\text{sinc}(Tt)\longleftrightarrow\text{rect}\left(\frac{t}{T}\right)\tag{1}$$The given function can ... 1 Yes, there is such a tempered distribution. In fact, a unique such tempered distribution. Recall that the Fourier transform is an automorphism of S(\mathbb{R}), and the inclusion \iota \colon D(\mathbb{R}) \hookrightarrow S(\mathbb{R}) has dense image. Therefore \hat{D}(\mathbb{R}) = \mathscr{F}(D(\mathbb{R})) is a dense subspace of S(\mathbb{R}). ... 1 No, you need to actually evaluate the convolution of the inverse transforms, which looks like$$\begin{align}\sqrt{\frac{\pi}{2}} \int_{-1}^1 dk' \, e^{-(k-k')^2/2} &= \sqrt{ \pi} \int_{(k-1)/\sqrt{2}}^{(k+1)/\sqrt{2}} du \, e^{-u^2} \\ &= \frac{\pi}{2} \left [\operatorname{erf}{\left (\frac{k+1}{\sqrt{2}}\right )} - \operatorname{erf}{\left ...

0

Hint 2: use a test function $e^{i(\omega t-k x)}$ and see what the differential operators do in this function. Then the (whole) Fourier solution is the (function represented by the) sum over all these (fourier basis) functions. In other words if the required solution is expanded in a 2-D fourier space of the above basis functions: $$f(x,t) = \sum_k ... 0 Because you are assuming that f'\in L^1(\mathbb{R}), then the following limits exist$$ \lim_{x\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\int_{0}^{x}f'(t)dt+f(0). $$The above limits must be 0 because f \in L^1(\mathbb{R}); otherwise, for example, if \lim_{x\rightarrow\infty}f(x)=L \ne 0, then |f(x)| \ge L/2 for large enough x, ... 1 The function f(t)=\sin(6\pi t)/t is square integrable on \mathbb{R}. So the Fourier transform is$$ \hat{f}(s)= \lim_{R\rightarrow \infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{\sin(6\pi t)}{t}e^{-ist}dt. $$Because \sin(6\pi t)/t is even and e^{-ist}=\cos(st)-i\sin(st) has odd imaginary part, then the integral for the Fourier transform ... 0 I understood the FFT and IFFT parts of my question. I am still not sure about very first part part: defining nonlinear forces in the time domain. If it is a harmonic motion and is nonlinear restoring forces are a hysteresis, and then I can directly convert this force-deformation over one period of the applied force (T=2*pi/w). Then I can obtain the fourier ... 1 You can use Parseval's Theorem: If f(x) and g(x) have respective FT's$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x} G(k) = \int_{-\infty}^{\infty} dx \, g(x) e^{i k x} $$Then$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) In your case, f(x) = \frac{\sin{x}}{x} and g(x) = ... 1 \begin{align} \int_{-\infty}^{\infty}\frac{\sin x}{x^3+x}dx & =\Im\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x^3+x}dx \\ & = \Im\left(\frac{1}{2i} \int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}\left[\frac{1}{x-i}-\frac{1}{x+i}\right]dx\right) \end{align} The integrand on the right decays when x=z is in the upper half plane. So you can close the ... 0 \begin{align} u(t,x) & = \mathcal{F}^{-1}(e^{-t|\xi|^2}\mathcal{F}(f)) \\ & = \frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n}e^{i\xi\cdot x}e^{-t|\xi|^2}\left(\int_{\mathbb{R}^{n}}f(y)e^{-iy\cdot\xi}dy\right)d\xi \\ & = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}f(y)\left(\int_{\mathbb{R}^{n}}e^{i\xi\cdot(x-y)}e^{-t|\xi|^2}d\xi\right)dy \\ & = ... 1 we know that f'(x) = -2 a x f(x) \implies \ln f(x) = -a x^2 + cste \implies f(x) = C e^{-a x^2}. now if g(t) = e^{-a t^2} then g'(t) = -2 a t g(t) and if a > 0, |g(t)| and |g'(t)| fastly decreases when t \to \infty so :i \omega \int_{-\infty}^\infty g(t) e^{- i \omega t }dt = \int_{-\infty}^\infty g'(t) e^{- i \omega t }dt = ...

0

First of all, the claim of the theorem should be that $$\widehat{f'}(t)=2\pi i t \widehat{f}(t).$$ And yes, I don't know why your textbook would use Sobolev inequality as your idea is correct and $f' \in L^1$ is enough (as you need the Fourier transform of the derivative to exist). The proof, which is the one you describe, goes like this: Fourier ...

1

$$\left|\frac{e^{-ixh}-1}{h}\right|= \left|\left.\frac{1}{h}e^{-iyh}\right|_{y=0}^{y=x}\right|=\left|\int_{0}^{x}-ie^{-iyh}dy\right| \le |x|.$$

0

It looks like I need to proceed a little bit about what I understand from this problem: I understand the use of fft to calculate the coeff of displacement in fourier series from time domain to freq domain, which results in sines and cosines function approximation. And using the IFFT to calculate the response in time domain makes sense too, but what happens ...

2

$\forall a\in \mathbb{R}^{+}$, \begin{align*} \int_{-a}^{a} \left[ \color{red}{\int_{-\infty}^{\infty} e^{-ikx} dx} \right] dk &= \int_{-\infty}^{\infty} \left[ \int_{-a}^{a} e^{-ikx} dk \right] dx \\ &= \int_{-\infty}^{\infty} \left[ -\frac{e^{-ikx}}{ix} \right]_{k=-a}^{a} dx \\ &= \int_{-\infty}^{\infty} \frac{2\sin ax}{x} dx \\ ...

0

You can understand it using the formalism of generalized functions. The generalized function $\mathrm{1}$ is defined by the sequence of functions $1_n(x)=\exp(-x^2/n^2)$. In this formalism the Fourier transform of $\mathrm1$ will be another generalized function defined by the sequence of functions, $$\int_{-\infty}^\infty \mathrm{1}_n(x)e^{-ikx} \ ... 1 We should clarify that this formula is not a calculus formula. Instead it is a distribution formula. The right hand side means$$ 2\pi h\delta(p-p')(f)=2\pi h \delta_{p,p'}f(p) $$In other words, you can think the right hand side as a 'function' that operates on another function and outputs a value. If p\not=p' then this 'function' vanishes, otherwise it ... 0 Let$$ \phi(t)=\begin{cases}e^t & \text{if }t\le0,\\0 & \text{if }t>0.\end{cases} $$Then$$ \int_{-\infty}^{t} e^{\tau-t} u(\tau)\,d\tau=\int_{-\infty}^{\infty} \phi(\tau-t)\, u(\tau)\,d\tau=\phi\ast u(t). $$Since the Fourier transform of a convolution is the product of the Fourier transforms, taking the Fourier transform on the equation we get ... 1 This answer simply adds some details to Ron Gordon's answer. Use symmetry to get rid of the absolute values.$$ \begin{align} \int_{-\infty}^\infty\left|x\right|^{-1/2}e^{-\left|x\right|}e^{ixy}\,\mathrm{d}x &=2\operatorname{Re}\left(\int_0^\infty x^{-1/2}e^{-x}e^{ixy}\,\mathrm{d}x\right)\\ &=2\operatorname{Re}\left(\int_0^\infty ...

0

What you call frequency shift is actually partly undetermined because the phase is neglected. Exchanging the order of the two operations merely makes this more evident: the arbitrary phase factor chosen in one procedure differs from the arbitrary phase factor in the other procedure by a phase $\omega t_0.$ Note that this applies to any function $f(t),$ not ...

4

It is easier to do the direct computation. Forgive me for using my own notation for the time being. Let $$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x}$$ where $$f(x) = (2 \pi |x|)^{-1/2} \, e^{-|x|}$$ Then $$F(k) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} dx \, x^{-1/2} \, e^{-x} \cos{k x} = \sqrt{\frac{8}{\pi}} \int_0^{\infty} dx \, e^{-x^2} ... 1 Even though Matt has solved the original problem properly, I feel obligated to give a proper solution to the fixed version. This is just for completeness: Step 1: By scaling, we may suppose that t=1. This is a calculation just like the one done by Matt above. Step 2: By translating f if necessary, we may restrict ourselves to the evaluation of |T_1 ... 2 So we're in dimension n=3 according to the problem statement. I agree with you that we can get the inequality without the 1/t factor basically by Cauchy-Schwarz. But am I being silly or does the stated inequality not hold by scaling? Suppose the inequality were true. Let f_{t}=f(\cdot/t). Then ... 1 The Fourier transform of f(x) doesn't exist in the usual sense, but since f can be viewed as a tempered distribution, we can interpret the Fourier transform in that setting. (I'm using the normalization \hat f(\omega) = \int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt. If you're using something else, the answer is a little different.) First of all, ... 0 I could derivate the formula for radix-4 NTT following this. In radix-4, each butterfly's element is given by$$Y_{p,q} = \sum_{l=0}^3\left[W_N^{l\cdot q}X\left(l,q\right)\right]W_4^{l\cdot p}.$$This way, let P be the prime that generates the finite field used by NTT and W_4 and WInv_4 respectively it's twiddle factors for forward and inverse NTT. ... 0 Note that the real part X_R[k] satisfies$$X_R[k]=X_R[N-k]\tag{1}$$where N is the DFT length. The imaginary part satisfies$$X_I[k]=-X_I[N-k]\tag{2}$$Combining (1) and (2) gives the following condition for the complex DFT X[k]=X_R[k]+iX_I[k]:$$X[k]=X^*[N-k]\tag{3}$$where * denotes complex conjugation. Equation (3) means that the ... 2 Homomorphism property A big advantage of the kernel e^{i\omega t} over \operatorname{cas}(\omega t) is that the former is a homomorphism of the group (\mathbb{R},+) into the multiplicative group of unimodular complex numbers: e^{i\omega (t+s)} = e^{i\omega t}e^{i\omega s}. This identity leads to Shift formula: the transform of f(t-\tau) is ... 0 I have dug up some book titles that i found useful, so i am writing my own answer. Namely: -Mathematical methods for physics and engineering by Riley, Hobson, Bence. -Many books or chapters from books of the author Vladimir Arnold. -Strang's Calculus -Hildebrand's Methods of Applied Mathematics -Churchill's Fourier Series and Boundary Value Problems The ... -1 Fourier transform does not exist for every signal application.So by introducing the region of convergence in Fourier transform which is known as Laplace Transform one may have indirectly the Fourier transform of signal. 3 The ''function'' as it is has no right to be called a distribution, since it has non-summable singularities in the form 1/y at x=\pm1: therefore I'm going to assume that your teacher understood it as Cauchy principal value:$$ u(x) = PV \frac{x}{x^2-1}. $$According to your book, a function f\in L^1_{\text{loc}}(\mathbb R) is a tempered distribution in ... 5 To get you started:$$| f_n(x) - f(x)| =\left| (1/\pi) \int_{-\infty}^{\infty} f(x-t) P_{n}(t) \; dt - f(x)\right| = (1/\pi)\left| \int_{-\infty}^{\infty}\left[f(x-t)- f(x)\right] P_n(t) \; dt \right|$$Now because f is continuous on \mathbb{T} and 2\pi-periodic, we can essentially deduce its properties by considering its restriction f_r to ... 0 I recommend checking out Gilbert Strang's book Introduction to Applied Math. Strang has great intuition and I think he explains things in a very clear, simple, and yet deep way. 0 The canonical definition of convolution of two arbitrary distributions with well associated supports is the following. Let S and T be two distributions, respectively, with supports A and B, and let \varphi is a test function with support K. You define$$ \langle S*T,\varphi\rangle=\langle S_x\otimes T_y,\alpha(x)\varphi(x+y)\rangle, $$where ... -1 To your question: Your definition of convolution is a map$$ *:(\mbox{test functions})\times (\mbox{distributions})\to \mbox{distributions}. $$So you're right, "\delta * g" does not make sense, as \delta is not a test function. 4 Subtracting 1 (whose Fourier transform is Dirac delta, up to a normalization constant) gives -1/(1+x^2), whose Fourier transform integral does converge, and can be evaluated by residues: a constant multiple of e^{-|x|}. 0 The expression \int_{-\infty}^{+\infty} \delta(x)f(x)dx has no meaning. The purpose of the right-hand side f(0) is precisely to give it a meaning, but by definition. In this case, indeed you require no properties on f other than it is a function with value at 0. However, you certainly will want \delta to interact with other objects. That is why it ... 0 First: This two equations$$ \begin{align} y_j &= \sum_{k=0}^{n-1}{x_k \cdot \omega^{jk}}\\ DFT[k] &= \sum_{n=0}^{N-1}{x[n] \cdot e^{-2 \pi i k \frac{n}{N}}} \end{align} are the same, if you choose \omega = e^{-\frac{2\pi i}{N}}. Of course you have to rename some varibles the k in the first equation is n in the second and so on. Second: ... 1 Let \alpha>n/2. Then, f can be split as the sum of two pieces: let B_a be the ball of radius a centred at the origin, then \begin{align} f&\equiv u+v\\ &= \chi_{B_a}(x)\frac{1}{|x|^{\alpha}}+\left(1-\chi_{B_a}(x)\right)\frac{1}{|x|^{\alpha}}; \end{align} u lies in L^1(\mathbb R^n), for 0<\alpha<n, and v lies in L^2(\mathbb ... 1 As discussed in the comments, we are only left to show that \Delta f_{\lambda}(x) = - \lambda^2 f_{\lambda}(x) The following proof will work if f \in C^2(\mathbb{R}^n) and it, together with all of its partial derivatives up to second order are in L^1, so that differentiation under the sign is justified. Also, we need \widehat{f} \in L^1, so that all ... 0 Let me sketch a solution by contradiction: We may assume t>0, as for the other case we simply invert the roles of the functions involved. STEP 1: We are going to use the inversion formula u(x,t) = \int_{\mathbb{R}} e^{8\pi^3 it \xi^3} \widehat{u_0}(\xi) e^{2\pi i x\xi} d\xi.$$Instead of integrating on \mathbb{R}, we may, by a contour ... 3 You're off on a small but important point. What Rudin is referring to as c_n is indeed$$ c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-int}~dt, $$and with this definition we have$$ f(x) \sim \sum_{n=-\infty}^\infty c_ne^{inx}. $$However, this definition of the Fourier coefficients depends on which particular inner product you use, and at this point Rudin ... 3 This is an improper integral near u=0. The numerator goes to 1 near u=0 but the denominator makes the integral divergent like \int_0^1 u^{-2}du . I suggest revisiting the two original functions whose Fourier transforms yielded your integral to see that they are in fact integrable and that their convolutions therefore is defined. (For example, are the ... 1 Just to put everything in the right place, since Hugo uses slightly different conventions. Here If \Delta u(x)=\delta(x) then u(x)=C |x|^{2-d} via Fourier transform? it is shown that if$$ f(x)={C_d}|x|^{1-d} $$then$$ \widehat f({\xi})=-\frac{1}{|\xi|^2}, $$which is what we want by the above calculation. Now, to fix the constant, one uses the ... 2 You can get something like that only in some kind of distributional context. As distribution, you can write, as one version$$ \sum_{n\in\Bbb Z}z^n=\lim_{q\to 1-0}\sum_{n\in\Bbb Z}q^{|n|}z^n =\lim_{q\to 1-0}\frac{1-q^2}{1-q(z+z^{-1})+q^2} $$which for |z|=1 and away from z=1 converges to zero while the integral over the unit circle always evaluates to ... 1 The first of your series converges for |z| < 1, the second for |z| > 1. In other words, there are no values of z for which it is meaningful to add the two series. 4 Let I(k) be the integral given by$$\begin{align} I(k)&=\int_0^\infty e^{ikx-1/2x} x^{-3/2}\,dx\\\\ \end{align}$$We assume that k>0 and leave the case for which k<0 to the reader. Let a=(1-i)\sqrt k. Then, we can write I(k) as$$I(k)=\int_0^\infty e^{-\frac a2 \left(ax+\frac1{ax}\right)}\,x^{-3/2}\,dx Enforcing the substitution ...

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