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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 ...


2

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


2

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

Some comments on the first one: It's as far as i can see not an good idea to switch to polar coordinates. Regularisation of the expression after performing one of the integrations seems horrible An indirect way is much faster here: First: It's known that $G(\vec{x})=-2\pi\log(|\vec{x}|)$ (1) is the Greensfunction of the $2$-D Poisson equation. $$ ...


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\\ & ...


1

We have: $$ u' = -2\pi x u, $$ so transforming both sides we have: $$ 2\pi i\xi\, \widehat{u}(\xi) = -i \widehat{u}'(\xi).$$


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 $\sum_n n^2 a_n^2 \leq \frac{1}{\pi} \int_{-\pi}^{\pi}|f'|^2$.


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

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 ...


1

$$ \int_\Omega f =\int_{\Re^n} f \chi_\Omega$$ 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

Denote by $\mathcal{F}$ the Fourier transform in $\mathbb{R}^n$. Given $u\in L^1(\Omega)$ define its $\Omega$-Fourier transform as $$ \mathcal{F}_{\Omega}u(q)=\int_\Omega u(x)\,e^{-iqx}\,dx. $$ Let $\tilde u\colon\mathbb{R}^n\to\mathbb{R}$ be $$ \tilde u(x)=\begin{cases}u(x) & \text{if }x\in\Omega,\\0 & \text{if }x\not\in\Omega.\end{cases} $$ Then $$ ...


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

$$f(t)=a_0+ \sum_{n=1}^{\infty}a_n \cos(n x)+\sum_{n=1}^{\infty}b_n \sin(n x)$$ 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.



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