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0

$$ \int_\Omega f =\int_{\Re^n} f \chi_\Omega$$ where $\chi_\Omega$ is the charcteristic function of the domain $\Omega$. Then the inverse can be given as a convolution of the function with the characteristic function.


1

The $\int_\pi^{2\pi}$ part is substituted via $s = 2\pi - t$, yielding $f(x_0 - t) = f(x_0 - (2\pi - s)) = f(x_0 + s - 2\pi) = f(x_0 + s)$. Finally the variable of integration is renamed to $t$. $$\begin{align*} \ldots & = \frac1{2\pi} \int_0^\pi f(x_0 - t) D_n(t)\ \mathrm dt + \frac1{2\pi} \int_\pi^{2\pi} f(x_0 - t) D_n(t) \ \mathrm dt\\ & ...


0

Your exponentials are very well-behaved with respect to $y$, so Cauchy--Schwarz seems like a good thing to do : $$ |\varphi(x, n, t)|\leq \Bigg(\frac1n\sum_{y=1}^n\Bigg|\sum_{k=1}^n {\rm e}^{2\pi ik(x-y)/n + 2i \sin(2k\pi/n) t}\ \Bigg|^2\Bigg)^{1/2} $$ Then you can expand the squares and compute the sum over $y$.


1

Simplifying a bit the expression we have: $$ G(\nu) = \frac{e^{-2\pi \nu i t_a}}{\Delta t}\int_{-\Delta t/2}^{\Delta t/2}f(-t)e^{-2\pi\nu i t}\,dt$$ and since $\operatorname{sinc}(x)$ is an even function we have: $$ G(\nu) = \frac{2 e^{-2\pi \nu i t_a}}{\Delta t}\int_{0}^{\Delta t/2}f(-t)\cos(2\pi\nu t)\,dt$$ and since in a right neighbourhood of zero both ...


1

Assuming that $f(x)$ is supported on $[-1,1]$ we have that $\widehat{f}(s)$ is an entire function of exponential type $1$ by Paley-Wiener theorem. If $\widehat{f}$ is not allowed to have any complex root out of $\pm t_0$, then $$g(s)=\frac{\widehat{f}(s)}{s^2-t_0^2}$$ is an entire function of exponential type $1$ with no roots, hence $g(s)=\exp(as+b)$ by ...


0

Consider $c_1,c_2,c_3$ and linear combination of the above vectors $c_1i+2c_2+3c_3=0$ $2c_1+ic_1-ic-2-c_3=0$ $3c_1+4c_2-ic_2+2c_3=0$ Equating like terms we get $c_1=0,2c_2+3c_3=0$ ;$2c_1-c_3=0,c_1-c_2=0$ from there we get $c_1=c_2=c_3=0$


0

Here is an attemt to proove that $u(k,x)=1$ is the only possible solution. The proof uses Fourier integration basics only. We first write $u(k,x)=1+w(k,x)$ and show that $w(k,x)=0$. By using the dirac delta function $f(x)=\delta(x-x_0)$ as test function, the equation becomes $$0=\frac{1}{2\pi}\int dk e^{-ikx'}\int dx e^{ikx}w(k,x)\delta(x-x_0) \\ = ...


0

(We replace $\theta$ throughout by $x$.) Let's look at the full Fourier series of $f(x) = x(\pi-x)$ on $[-\pi,\pi]$ first, where $f$ is defined outside of this interval by periodicity, i.e., $f$ is of period $2\pi$ outside $[-\pi,\pi]$. In this case $\displaystyle a_0 = \frac{1}{\pi}\,\int_{-\pi}^{\pi}\,x(\pi-x)\, dx = -\frac{2\pi^2}{3},$ where the first ...


2

$$a_n=\frac{2}{\pi/5}\int_0^{\pi/5}\sin^2(5t)\cos\left(\frac{2\pi nt}{\pi/5}\right)dt=a_n=\frac{10}{\pi}\int_0^{\pi/5}\sin^2(5t)\cos\left(10 nt\right)dt=...$$ $$b_n=\frac{2}{\pi/5}\int_0^{\pi/5}\sin^2(5t)\sin\left(\frac{2\pi nt}{\pi/5}\right)dt=a_n=\frac{10}{\pi}\int_0^{\pi/5}\sin^2(5t)\sin\left(10 nt\right)dt=...$$


0

I think the problem is that those two was are actually not different. I try to give an example of what I mean: If $f$ is a $T_0$-periodic function and you integrate $$b_k = \int_0^{2T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt = \int_0^{T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt+\int_{T_0}^{2T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt$$ $$= ...


2

By integration by parts, we have: $$ n\, a_n = \int_{-\pi}^{\pi} f(t)\, n\cos(nt)\, dt = -\int_{-\pi}^{\pi} f'(t)\sin(n t)\,dt \tag{1}$$ and since $f'\in C^0([-\pi,\pi])$, the RHS of $(1)$ converges to zero by the Riemann-Lebesgue lemma. $f'\in C^0([-\pi,\pi])$ implies $f'\in L^2([-\pi,\pi])$, hence $$ \sum_{n\geq 1} n^2 a_n^2 <+\infty $$ follows from ...


1

Yes, the first question is answered by the celebrated Riemann-Lebesgue Lemma. Since the Fourier coefficients of $f'$ are $na_n$, by applying the Bessel inequality to $f'$ you get $\sum_n n^2 a_n^2 \leq \frac{1}{\pi} \int_{-\pi}^{\pi}|f'|^2$.


0

Translation of a signal is reflected by a (complex) rotation in the Fourier domain as the well-known property states ($\mathcal{F}(s(t-\tau))=e^{-j\omega\tau}\mathcal{F}(s(t))$). Therefore, the statement that the Fourier transform is not able to tell the difference due to time shifts is incorrect since, if the latter was the case, the translated signals ...


0

It's implicit that the period is $P$ as you can gather from the partial sums of the FS (read line 2 under "Definition"). The important thing here is that the Fourier coefficients are always defined by integrating over a symmetric interval of length $P$, the period of $f$. Here you are looking for a so-called "half range expansion", so you need to extend ...


1

$$f(t)=a_0+ \sum_{n=1}^{\infty}a_n \cos(n x)+\sum_{n=1}^{\infty}b_n \sin(n x)$$ Now you can compare what you have in the comment with the formula to read off coefficients $a_2,b_2,a_6,b_6$. All the other coefficients seemed to be zero.


1

For $(2)$ use the identities $$ \cos(A+B) = \cos(A)cos(B) - \sin(A)\sin(B) $$ $$ \sin(A-B) = \sin(A)cos(B) - \cos(A)\sin(B). $$ Added: Based on what you put in the comment down you can write $f(t)$ as $$ f(t) = - \frac{\sqrt2}{2}\sin(2t)+\frac{\sqrt2}{2}\cos(2t) -\cos(6t) $$ and Fourier series is $$ f(t) = a_0 + a_1\cos(t) ...


1

The Fourier sine coefficients should be given by a formula of the form $b_n = \frac{2}{\pi}\,\int_0^{\pi} \, \theta(\pi - \theta)\, \sin (n\theta) \, d\theta = \frac{8}{\pi\,n^3}$ for $n$ odd (and $0$ for $n$ even). The stated formula now follows. The factor of $2$ comes from the fact that we have an odd extension of $f$ to $[-\pi,\pi]$ so that the ...


1

If your goal to find the Fourier transform of the function $f(x) = e^{-\pi ax^{2} + 2\pi ibx} $ then here is an easier approach $$ F(k) = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + 2\pi ibx}e^{-ikx}dx = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + (2\pi b-k)ix}dx. $$ To evaluate the last integral complete the square and then use Gaussian integrals.


1

Hint to solve the integral: $$ \exp \left(ax^2 +bx\right) = \exp \left(\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2 -\frac{b^2}{4a} \right),$$ for given constants $a$ and $b$.


0

The steps missing: $$\begin{align}f(x^{\prime})&=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x) \,\,\mathrm{d} x \right) \,\,\mathrm dk\tag{0}\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dx\,dk\tag{1}\\ ...


0

It looks like you want $\sin(mp)=0$ when $m$ is a multiple of 2 and 3 respectively. One way of achieving this is to set $p=\pi/2$ and $p=\pi/3$.


2

It is a consequence of Poisson summation formula. You just have to prove that, if $$ f(x) = \exp\left(-\pi a x^2+2\pi i b x\right), $$ then its Fourier transform is: $$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi a}}\exp\left(-\frac{(2b\pi+s)^2}{4a\pi}\right).$$


3

I assume you refer the continuous-time Fourier transform (CTFT) $$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$ and to the discrete Fourier transform (DFT) $$X[k]=\sum_{n=0}^{N-1}x[n]e^{-i2\pi nk/N}\tag{2}$$ where $x[n]$ and $X[k]$ are sequences, whereas $x(t)$ and $X(\omega)$ are functions. The DFT can be computed efficiently by the ...


0

In order to find its Fourier series, a periodic function with period $R$ should be thought of as a function defined on a circle of circumference $R$, call it $S^1_R$. The Fourier series of the function is then its representation in the basis of $L^2(S^1_R)$ given by orthonormal eigenfunctions of the Laplace operator. If two functions have incommensurate ...


6

This is not true. Simply take $f = \chi_{[-1,1]}$ (the indicator function of the interval $[-1,1]$). Then $f \in L^p$ for all $p \in (0,\infty]$, but if $\widehat{f} \in L^1$ was true, then Fourier inversion would imply that $$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$ would be (almost everywhere equal to) a continuous function. This is clearly not the ...


1

Your professor is wrong and Umberto P. is right. If we take $$ f(x)= K_0(|x|) $$ where $K_0$ is a modified Bessel function of the second kind, we have $f\geq 0$ and: $$ \int_{-\infty}^{+\infty}K_0(|x|)\,dx = \pi,\qquad \int_{0}^{+\infty}K_0(|x|)^2\,dx = \frac{\pi^2}{2},$$ so $f\in L^1\cap L^2(\mathbb{R})$, but: $$ \widehat{f}(s) = ...


0

A Fourier series means the amplitude of the different harmonics, who are an integer multiple of a base frequency. It is easy to see, that a this base frequency simply doesn't exist in your case. Although a Fourier transform of a such function of course exist, which is trivially $$F(s)=\delta(t-\omega_1)+\delta(t-\omega_2)$$


1

The Fourier series of a non-periodic function is really the Fourier series of its periodic extension. For example, there is a Fourier series of $f(x)=x$ on $[0,\pi]$, which is actually the Fourier series of the sawtooth wave that is formed by periodically extending $f(x)=x$. The Fourier series for a non-periodic function will not converge at every point but ...


2

Here's my best guess: The statement in your homework solution is not giving you the Fourier transform of the dirac delta function. What it is saying is that we can define the dirac delta function as the inverse Fourier transform of $f(\omega) = 1$. That is, we can define $$ \delta(x) \equiv \frac 1{2 \pi} \int_{-\infty}^\infty d \omega \, 1 \cdot e^{i ...


-1

It is the sifting property of Dirac delta function $$\int_{-\infty}^{\infty} \delta(x) f(x) dx = f(0)$$ with suitable conditions on $f$.


1

Do you still require some input? If so, I can look at it and get back asap. Hi. First let's note that the assumption implies that $\overline{c_{-n}}=c_n$ for all $n$ (start with $n=0$ and move up by differentiating, setting $t=0$ etc.) This then implies that $w(z)=\overline{w(1/\bar{z}})$. Thus if $z$ is a root, then $\frac{1}{\bar{z}}$ is also a root and ...


2

In general the inverse Fourier transform is not well defined. Observe that for generic choices of $z_k$ there exists $\omega$ such that the denominator of your expression vanishes. This implies that $F(\omega)$ is not even a tempered distribution, so you can't really even define its (inverse) Fourier transform. Supposing the inverse Fourier transform is ...


1

$$g(x+2\pi)=f(k(x+2\pi))=f(kx+2k\pi)=f(kx)=g(x)$$


0

To include this as the answer that most helped me, since it was just as a comment to my original post. The most straightforward solution seemed to be by using the property of the Discrete Time Fourier Transform for when you are multiplying by n in the time domain, that corresponds to a derivation in the Fourier/frequency domain. Therefore, the solution ...


1

You haven't given a definition for the discrete Fourier transform but I presume it is something like $$X(\omega) = \sum_{n=-\infty}^{\infty}x(n) e^{-i\omega n}$$ which in this case would be $$X(\omega) = \sum_{n=0}^{\infty}n^2 a^n e^{-i\omega n} = \sum_{n=0}^{\infty}n^2 z^n$$ where $z = ae^{-i\omega}$. Now suppose we set $$f(z) = \sum_{n=0}^{\infty} z^n$$ ...


0

(I would usually post this as a comment, but apparently I need an account for that.) Suppose that $f \in \mathscr{S}'(\mathbb{R}) $ is a solution to the equation $$ f'-f = \delta_0 + 1. $$ Since you are following Gerd Grubb's textbook, have a look at the technique described on page 108, starting around equation (5.41). We can set the differential operator ...


3

Let $f(x) = \text{sinc}(x)$. We can rewrite $$ f(x) = \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} = \frac{1}{2\pi}\frac{e^{i \pi x}-e^{-i \pi x}}{i x} = \frac{1}{2 \pi}\int \limits_{- \pi}^{\pi}e^{i \omega x} \,d \omega = \mathcal{F}^{-1}(1_{[-\pi, \pi]}).$$


0

The $N^{th}$ root of unity is a complex number of the form $e^{j2\pi/N}$. A sinusoidal wave is a function like so: $$ e^{j\omega t} = \cos(\omega t) + j \sin(\omega t), $$ where $j=\sqrt{-1}$. The formula you are looking at uses the roots of unity for convenience/preference/cleanliness, though I don't like it and find it confusing. When you raise a root of ...


1

To supplement the answer already given, i will use the trick of adding a small real part to the exponent $i\omega t \rightarrow i\omega t-\delta|t|$ so that the integral is convergent for large $t$. We then have to calculate the following expression: $$ \lim_{\delta\rightarrow0}\int_0^{\infty}e^{i\omega t-\delta t}\log ...


0

No. For example take your measure space to be $[0,1]$ with Lebesgue measure. Take $f(x)=2x$, $p=1,q=2$. Then $||f||_p=1$. To normalize $f$ you can only multiply it by a constant- and that would disturb the $||.||_p$.


4

For suitable $f$, we have $$ \mathcal{F}(f')(\xi)=2\pi i\xi\mathcal{F}(f)(\xi) $$ Therefore, if one exists, it would be $$ \mathcal{F}(f'/f)(\xi)=2\pi i\xi\mathcal{F}(\log(f))(\xi) $$ which would lead to $$ \mathcal{F}(\log(f))(\xi)=\frac1{2\pi i\xi}\mathcal{F}(f'/f)(\xi) $$ if $f'/f$ has a Fourier Transform. To show a bit more care, we can make the ...


0

If $f(x)$ is odd and $g(x)$ is even (or visa versa) then $f(x)g(x)$ is odd. If $f(x)$ is even and $g(x)$ is even, then $f(x)g(x)$ is even. If $f(x)$ is odd and $g(x)$ is odd, then $f(x)g(x)$ is even. This particular example can be seen by using the formula: $$\sin(a)\cos(b) = \frac{1}{2}\left(\sin(a+b)-\sin(a-b)\right)$$ Then ...


3

The definite integral over a fundamental period annihilates all of the complex exponentials except for the constant term. Thus, to determine "how much of $e^{2\pi in t/T}$ is in $f(t)$" you have to do something to $f$ to turn the "$a_\omega e^{2\pi in t/T}$ term" of interest into a constant term; this is achieved by multiplying by the reciprocal of the ...


0

I'm gonna simplify the problem by assuming that $\ell=x_0=0$: notice that $$ \dfrac{f(x_0+t)+f(x_0-t)}{2}-\ell = \dfrac{(f(x_0+t)-\ell)+(f(x_0-t)-\ell)}{2}. $$ Hence Dini hypothesis is the same as saying that the function $f_1(t) = f(x_0+t)-\ell$ satisfies $$ \int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right|\dfrac{dt}{t}<\infty. $$ As it's proved later ...


0

I think this article might help you. Pointwise Convergence of Fourier Series, Charles Fefferman, Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 551-571: http://www.jstor.org/discover/10.2307/1970917?sid=21105651264483&uid=4&uid=2&uid=3737760


1

Let's work it out from first principles. Recall that $$\sin(a+b)=\cos(b)\sin(a)+\cos(a)\sin(b)$$ Now, we consider $\sin(\pi(\zeta-n))=\sin(\pi\zeta-\pi n)$, i.e. $a=\pi\zeta$ and $b=-\pi n$: \begin{align}\sin(\pi\zeta-\pi n)&=\cos(-\pi n)\sin(\pi\zeta)+\cos(\pi\zeta)\sin(-\pi n)\\ &=\cos(\pi n)\sin(\pi\zeta)-\cos(\pi\zeta)\sin(\pi n)\end{align} where ...


1

$\sin{\pi n} = 0$ when $n \in \mathbb{Z}$. Thus, because $\cos{\pi n} = (-1)^n$, we have $$\sin{\pi (\zeta - n)} = \sin{\pi \zeta} \cos{\pi n} - \sin{\pi n} \cos{\pi \zeta} = (-1)^n \sin{\pi \zeta} $$


3

Whenever this topic comes up, I find a lot of conjecture that reverses the order of the History of this subject. Some real History can be found in Dieudonne's "A History of Functional Analysis". I'll summarize part of what is found there. Fourier was looking at Heat Conduction. He was fascinated by Heat. And he was able to come up with a macroscopic theory ...


2

If you have an integral transform $Uf = \int_{-\infty}^{\infty} P(s,x)f(x)dx$ with an inverse integral transform $Vg = \int_{-\infty}^{\infty} Q(x,s)g(s)ds$, then you'll have something similar to the Fourier transform $\delta$-function behavior because $$ \begin{align} f(y) & =(VUf)(y) \\ & ...


1

You're seeing two peaks in the magnitude plot, just as you should. The reason you're not seeing two peaks in the angle plot is that the complex-argument function is ill-conditioned near zero. That is, a number that is "close to zero" doesn't necessarily have an angle that's close to zero. So, even though fa and fb are good approximations of the ...



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