New answers tagged

1

You've calculated the variable shift incorrectly: if $f(x)=\cos(ax)$, then $f(x-b) = \cos(ax-ab)$. So in fact $$ \mathcal{F}( \cos(a(x-\pi/(2a))) ) = e^{-i(\pi/2)(\omega/a)} \sqrt{\frac{\pi}{2}} (\delta(\omega-a) + \delta(\omega+a)). $$ But by the defining property of $\delta$, the exponential on the first term is the same as $e^{-i\pi/2}$, that on the ...


0

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Denote the dot product of vectors $x,y,s$ by $x\cdot s,\;y\cdot s$. Then $f\in L^2(\mathbb{R}^3)$ can be written in terms the Fourier transform and its inverse: $$ f = \frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^3} \left(\int_{\mathbb{R}^3} f(y)e^{-is\cdot y}dy\right)e^{is\cdot x}ds $$ This is a "continuous" eigenfunction expansion in terms ...


0

I would think boundedness comes from the mean value theorem, where $|\phi(x) - \phi(-x)| \leq 2 x \sup_y |\partial_x \phi(y)|$.


1

You can use the following basic properties of characters: $$ \forall a \in G,\ \sum_{\chi \in \hat{G}} \chi(a) = \begin{cases} 0 & \mbox{if } a \not = 1 \\ |G| & \mbox{otherwise} \end{cases}\ \mbox{ and }\ \bar{\chi} = \frac{1}{\chi} $$ Try to solve the problems before reading the solutions. You're wrong. Being an isometry for $\mathcal{F}$ means $$ ...


1

$$f(a,b)=\int_{-1}^1 \left(-a-b x+x^2\right)^2 \, dx=2 a^2-\frac{4 a}{3}+\frac{2 b^2}{3}+\frac{2}{5}$$ $$\frac{\partial f }{\partial a}=4 a-\frac{4}{3};\;\frac{\partial f }{\partial b}=(4 b)/3$$ Derivatives are both zero at $\left(\frac{1}{3},0\right)$ It is a minimum because Hessian matrix of second derivatives is constant and positive.


0

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0

Here's mine. I'm answering late, I wish this should get attention. We'll use the expansion of $\tanh^{-1}$: $$\frac{1}{2}\log\frac{1+y}{1-y}=\sum_{n\geq0}\frac{y^{2n+1}}{2n+1},\quad|y|<1$$ We start with this inequality: $$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\frac{1}{1+2xy+y^2}dx\,dy$$ The LHS of this equality gives: $$\int_{-...


0

Simple attempt: $$let\ \ f(x)=\left\{\begin{matrix} & 1-x^{2}\ \ \ ,\ |x|< 1& \\ & 0\ \ \ \ \ \ \ \ \ \ , \ \ |x|> 1 & \end{matrix}\right \\$$ by fouier transform of f(x) we have : $$F(a)=\int_{0}^{\infty }f(t).cos(at)dt=\int_{0}^{1}\ (1-t^{2}).cos(at)\ dt=\int_{0}^{1}cos(at)dt-\int_{0}^{1}t^{2}cos(at)dt$$ $$\therefore F(a)=\...


0

Before some comments adding some background to @S.H.W.'s very efficient approach, it may be worthwhile to give an (iconic!) argument that is easier to justify: Since cosine is a linear combination of complex exponentials, it suffices to evaluate $\int_{-\infty}^\infty { e^{itx}\over 1+x^4 }dx$ for real $t$. There are two cases, depending on the sign of $t$. ...


2

Let $$\mathcal{F}{f}(s) = \int_{-\infty}^{+\infty}f(x)e^{-2\pi isx }dx$$We have $$\mathcal{F}(fg) = \mathcal{F}{f}*\mathcal{F}g$$Where $*$ denotes convolution. Also we have $$\mathcal{F}\cos 2\pi a t = \frac{1}{2}(\delta(s-a)+\delta(s+a))$$ Here $a = 1$ and the result is $$\mathcal{F}g(s) = (\frac{1}{2}(\delta(s-1)+\delta(s+1)))*\frac{1}{1 + s^{4}} = \frac{1}...


0

The Cesaro means of the Fourier series converges to $f$ everywhere that $f$ is continuous, which is almost everywhere for a Riemann integrable function. So, if the Fourier series converges at a point of continuity of $f$, then it must converge to $f$. So under your assumptions, that would mean that $f$ equals the Fourier series at least at every point of ...


1

Here is an attempt at a constructive answer. The claim is that a pure sinusoid is the only periodic waveform that has the property $\forall A_1,\tau_1 \exists A_2,\tau_2$ such that $x(t) + A_1x(t+\tau_1) = A_2 x(t+\tau_2)\;\forall t$ where $x(t)$ is $T$-periodic. I am going to limit myselfto "well-behaved" functions which can be uniquely expressed ...


1

$\require{cancel}$ \begin{eqnarray} X(f) &=& \int_{-\infty}^{+\infty}{\rm d}t~ \frac{1}{T}e^{-(t-T)/T}u(t- T) e^{-2\pi i f t} \\ &=& \frac{1}{T} \int_{T}^{+\infty}{\rm d}t~ e^{T/T} e^{-(1 + 2\pi i f T)t/T} \\ &=& \frac{e}{\cancel{T}}\frac{\cancel{T}}{1 + 2\pi i f T} e^{-(1 + 2\pi i f T)} \\ &=& \frac{\cancel{e}}{1 + 2\pi i f T}...


2

This is badly false but of course you need to pick examples where the integrals actually converge. A simple example is to take $f, g$ to be step functions ("square waves") with disjoint supports, say $f$ the indicator function of $[0, 1]$ and $g$ the indicator function of $[2, 3]$; then $\int fg = 0$ but $(\int f)(\int g) = 1$. More conceptually, ...


0

Somewhere you need convergence of the Fourier series of $x(t)$ to $x$ in some sense. If you use a normalized inner product on $[-T_0/2,T_0/2]$ given by $$ \langle f,g\rangle = \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}f(t)\overline{g(t)}dt, $$ then the Fourier series for $f$ becomes $\sum_{n=-\infty}^{\infty}\langle f,e^{ik\omega t}\rangle e^{ik\...


1

This is similar to looking at a symmetric or hermitian matrix $A$, and expanding a vector in the eigenvectors of this matrix $A$. The operator $$ Lf=-f'',\;\;\; 0 \le x < \infty,\;\; f(0)=0, $$ is a self-adjoint operator on the domain consisted of all twice-absolutely continuous functions $f\in L^2[0,\infty)$ for which $f''\in L^...


2

I'll use $z=x+iy$ instead of $\lambda$. One can show that $|F(z)|$ is uniformly bounded on circles of radius $n + 1/2$, $n = 1, 2, 3, \ldots$, and $\lim_{n \to \infty } F((n+\frac 12)y) = 0$. (1) implies that $F$ is constant (using the maximum modulus principle and Liouville's theorem). (2) then implies that $F$ is identically zero. Without loss of ...


1

This seems to be exactly what Jeff Rauch does in his notes “Fourier Analysis from Complex Analysis”, pag.7: http://www.math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf The only difference is that Rauch uses $\sin \pi \lambda$ in place of your $e^{-i2\pi \lambda}-1$. To obtain the key uniform boundedness, he uses the “Cauchy inequalities”, but to be honest I ...


1

I consulted with my senior in university and got the answer. Let $ j = n $ for simplicity and define $$ G(x_n) = \exp(2\pi ibx_n - \pi r(x_n - a)^2). $$ Then, $ u \in L^2(\mathbb{R}^n) $ satisfies the condition if and only if $$ u(x) = v(x_1, \dots, x_{n - 1}) G(x_n) \quad (\text{for a.e. $ x \in \mathbb{R}^n $}) \tag{$ * $} $$ for some $ v \in L^2(\mathbb{R}...


0

Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. For $f(x)=e^{ic\log x}$ with $c=2\pi ka$, we have $f'(x)=icf(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{ic\...


1

Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. For $f(x)=e^{2\pi ib\log x}$, we have $f'(x)=2\pi ib f(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{2\pi ib\log ...


1

Using the fact that $-|t-x| = x-t$ if $x≤t$ and $t-x$ when $x≥t$, you get $$ \begin{align*} 2\,f(-t) &= e^{-t}\int_{-\infty}^t e^{x-x^2} \mathrm d x + e^t \int_t^{\infty} e^{-x-x^2} \mathrm d x \\ &= e^{-t}\int_{-\infty}^t e^{-(x-1/2)^2+1/4} \mathrm d x + e^t \int_t^{\infty} e^{-(x+1/2)^2+1/4} \mathrm d x \\ &= e^{1/4-t}\int_{-\infty}^{t-1/2} e^{-...


0

Let $\{ P_n \}_{n=0}^{\infty}$ denote the Legendre polynomials, normalized so that $\|P_n\|_{L^2[-1,1]}=1$ for all $n$. Let $f \in L^2[-1,1]$ and let $\epsilon > 0$ be given. Then there is a continuous function $g \in C[-1,1]$ such that $\|f-g\|_{L^2} < \epsilon/2$. The Weierstrass approximation theorem gives the existence of a polynomial $p$ such that ...


1

$$\begin{align} \int_0^{\pi/2}\frac{|\sin(2nx)|}{x}dx &= \int_0^{n\pi}\frac{|\sin(x)|}{x}dx \tag{i} \\ &= \sum_{k=1}^n \int_{(k-1)\pi}^{k\pi} \frac{|\sin (x)|}{x}dx \tag{ii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{|\sin (x + (k-1)\pi)|}{x + (k-1)\pi}dx \tag{iii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{\sin (x)}{x + (k-1)\pi}dx \tag{iv} \end{...


0

In order to use separation of variables, we can propose a solution of the form $$ \omega(t,\eta)=𝑔(𝑡)ℎ(\eta)+\frac{𝐶𝑡}{A} $$ where $g,h$ are two unknown functions, that we want to find out. Then we have: $$ \frac{\partial\omega}{\partial t}=𝑔'(𝑡)ℎ(\eta)+\frac{𝐶}{A} \quad and \quad \frac{\partial^2\omega}{\partial \eta^2}=𝑔(𝑡)ℎ''(\eta). $$ Finally ...


1

Using the Fourier transform $$ \hat{f}(t,\xi) = \int_{-\infty}^{\infty} f(t,\eta) \, e^{-i\xi\eta} \, d\eta. $$ we get $$ A\partial_t\hat{f}(t,\xi) = -B\xi^2\,\hat{f}(t,\xi) + C\,2\pi\,\delta(\xi). $$ The homogeneous equation, $$ A\partial_t\hat{f}(t,\xi) = -B\xi^2\,\hat{f}(t,\xi) $$ has solutions $$ \hat{f}_h(t,\xi) = \hat{R}(\xi)\,e^{-B\xi^2t/A}, $$ where $...


1

Requested Form of the Integral Let $T$ be an orthogonal linear transformation where $T(\xi)=|\xi|(1,0,0,\dots,0)$. $$ \begin{align} \int_{\mathbb{R}^d}\frac{\left|\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y &=\int_{\mathbb{R}^d}\frac{\left|\,e^{i|\xi|y_1}+e^{-i|\xi|y_1}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y\...


1

Define $F(\xi) = \int_{\mathbb{R}^d} \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}dy$. Show that $F(\alpha \xi) = |\alpha|^{\alpha} F(\xi)$ for some $\alpha \in \mathbb{R}$ (independent of $\xi \in \mathbb{R}^{d} \setminus \{0\}$), and then show that $F(O(\xi)) = F(\xi)$ for each orthogonal transformation $O$. ...


0

This is getting here a little late (after the acceptance of a solution), but I thought I would expand on @uniquesolution's answer. To me, it seems that it is the continuity of $\sigma_N f$ that is the issue. For that I will make use of the following result: Proposition: Let $g \in L^1([-\pi,\pi])$ and $h \in C^0([-\pi,\pi])$, then $f \ast g \in C^0([-\pi,\...


1

Answering my own question here, so it may be false. Define, $$S_N(f)=\sum_{n=0}^{N}\langle f, \mathfrak{L_n}\rangle\mathfrak{L}_n(\theta)$$ Where $f$ is assumed to be Riemann integralable on $[-1,1]$ and $\mathfrak{L}_n(\theta)$ is defined in the question. Now consider the following inner product for $m \leq N$, $$\langle f-S_N(f),\mathfrak{L}_m \rangle=\...


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$\hat{f}(z)$ is $\frac{1}{2^n}$ times the sum of $2^n$ iid Bernoulli random variables taking on the values $1, -1$ with probability $\frac{1}{2}$ each, for every $z$ (the signs $(-1)^{x \cdot z}$ don't affect this!). So it has a binomial distribution: we have $$\mathbb{P} \left( \hat{f}(z) = \frac{2^n - 2k}{2^n} \right) = \frac{1}{2^{2^n}} {2^n \choose k}.$$ ...


0

It is actually simple: To the function $f(x)$ defined on $[0,L)$, attach the function $g(x)=f(-x)$ defined on $[-L,0)$. The resulting composite function $h(x)$ is even ($h(x) = h(-x)$) on the interval $[-L,L)$, therefore its Fourier series has cosines only. This captures the DCT of $f(x)$.


1

Looking at the integral it is obvious that this is a convolution of : $f(p)$ and $exp\left(\frac{-p^2}{2}\right)$ Now take fourier transform on both sides to get: $$\mathcal{F}(exp\left(\frac{-x^2}{4}\right))=\mathcal{F}(exp\left(\frac{-p^2}{2}\right))\mathcal{F}(f(p))$$ $$\mathcal{F}(exp\left(\frac{-x^2}{4}\right))=\int _{-\infty }^{\infty }e^{-0.25x^2-2\pi ...


0

This is related to the following, where $S_N^f(\theta)$ is the Fourier series truncated to the terms $1,\sin(n\theta),\cos(n\theta)$ for $n=1,2,3,\cdots,N$. $$ S_{N}^f(\theta)-L=\frac{1}{\pi}\int_0^{\pi}D_N(\theta')\left[\frac{f(\theta+\theta')+f(\theta-\theta')}{2}-L\right]d\theta. $$ The Dirichlet-Dini condition is formulated to make sure that the ...


1

Let $f$ be any function in $L^{1}$ which is not in $L^{2}$. Take $g(y)=f(-y)$ so that $g$ is also in $L^{1}$. Then then convolution does not exist at $x=0$.


4

Reduce the problem by looking at the equation solved by $$ v(x,t)=u(x,t)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i $$ This function $v$ satisifes $$ v_t = v_{xx} \\ v(x=0,t > 0) = u(x=0,t)-T_f=0 \\ v(x=L,t > 0) = u(x=L,t)-T_i=0 \\ v(x,0)= f(x)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i $$ With ...


3

That is not a function $\mathbb{R} \to \mathbb{R}$ since $\infty \not\in \mathbb{R}$. It is a function $\mathbb{R} \to \bar{\mathbb{R}} = \mathbb{R} \cup \{\infty\}$ though.


2

Let $f(t) = \frac{1}{(1+t^2)(4+t^2)}.$ Its Fourier transform is $$ \hat{f}(x) = \int_{-\infty}^{\infty} \frac{e^{-itx}}{(1+t^2)(4+t^2)}dt = \int_{-\infty}^{\infty} \frac{1}{3}\left( \frac{1}{1+t^2} - \frac{1}{4+t^2} \right) e^{-itx} dt \\ = \frac{1}{12} \int_{-\infty}^{\infty} \left(2 \frac{2\cdot 1}{1+t^2} - \frac{2\cdot 2}{4+t^2} \right) e^{-itx} dt = \...


1

Fourier considered the integral representation to be the limit of a discrete series as the period tended to $\infty$. The Fourier cosine version is $$ f(x) \sim \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(y)\cos(sy)dy\right) \cos(sx) ds $$ This form is correct. It can be derived from the exponential form after extending $f$ to ...


3

Suppose $x\ge0$. Let $$ I(x)=\int_0^\infty \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt.$$ Then \begin{eqnarray} I'(x)&=&\int_0^\infty \frac{t^2\cos(tx)}{(1+t^2)(4+t^2)}dt\\ &=&\int_0^\infty \frac{\cos(tx)}{4+t^2}dt-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt\\ &=&\frac\pi4e^{-2x}-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt \end{...


1

It is not clear from your question what you are trying to do. I think you have three options: Use a change of variable such as $\theta = 2 \tan^{-1} x$ which transforms $[0,\infty)$ to $[0,\pi)$. Then for a function $f:[0,\infty)$ you could obtain the cosine series for $f(\tan(\theta/2))=a_0/2 + \sum a_n \cos n \theta$. This results in a cumbersome ...


1

hint: With $f(x)=e^{-ax}$ the definition of fourier transform shows $$\hat{f}(w)=\sqrt{\dfrac{2}{\pi}}\dfrac{a}{a^2+w^2}$$ then $$f(x)=\int_{-\infty}^{\infty}\hat{f}(w)e^{2\pi ixw}dw=a\sqrt{\dfrac{2}{\pi}}\int_{0}^{\infty}\dfrac{2i \sin(2\pi xw)}{a^2+w^2}dw=e^{-ax}$$ so from $$\dfrac{1}{(1+w^2)(4+w^2)}=\dfrac13\dfrac{1}{1+w^2}-\dfrac13\dfrac{1}{4+w^2}$$ you ...


1

First $$I(x):=\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt$$ then applying a Laplace transform we get $$\mathscr{L}_{x\to s}\{I(x)\}=\int_0^{\infty} e^{-sx}\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dtdx$$ here by the $x\to s$ I just denote which vairables I change to which, to avoid confusion. Now since all integrals converge we change order of ...


4

1st Solution. Following @Claude Leibovici's suggestion, let us utilize complex-analytic technique. First, symmetrize the integral to write $$ I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}\,\mathrm{d}t = \frac{1}{2}\operatorname{Im}\biggl(\int_{-\infty}^{\infty}\frac{te^{itx}}{(1+t^2)(4+t^2)}\,\mathrm{d}t\biggr). $$ Now let $R > 2$...


3

Partial answer. $$I=\int_0^\infty \frac{t\sin{(tx)}}{(1+t^2)(4+t^2)}\,dt=\Im\left(\int_0^\infty \frac{t\,e^{itx}}{(1+t^2)(4+t^2)}\,dt \right)$$ $$\frac{t}{(1+t^2)(4+t^2)}=\frac{1}{6 (t+i)}-\frac{1}{6 (t-2 i)}-\frac{1}{6 (t+2 i)}+\frac{1}{6 (t-i)}$$ So, you basically face four integrals $$I_k=\int_0^\infty\frac {e^{itx}}{t+ki}$$ With simple change of ...


1

Perhaps item 1 should be written to imply that $f$ is equal, almost everywhere, to some continuous function $g$ (i.e., in the "equivalence class" (mod equality a.e.) of $f$, there is at least one continuous function.)


0

If the Fourier Series for $f$ converges uniformly to $g$, then $g$ is continuous and \begin{align} \hat{g}(n)&=\frac{1}{2\pi}\int_{0}^{2\pi}g(x)e^{-inx}dx \\ &=\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{k=-K}^{K}\hat{f}(k)e^{ikx}e^{-inx}dx \\ &= \lim_{N\rightarrow\infty}\sum_{k=-K}^{K}\hat{f}(k)\frac{1}{2\pi}\...


2

I find it suspicious that $a_n=0$ since the function $y-\dfrac{1}{2}$ is not odd. According to your calculations $y-\dfrac{1}{2}$ is a pure sinus wave. You should check the calculations of $a_n$ when $n=8$, the integral now involves $$\cos^2\left(\dfrac{16\pi t}{5}\right)$$ and is unlikely to vanish.


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