Tag Info

New answers tagged

0

Following Fedja's hint, let $\phi$ be a nonnegative test function supported in $(1,2)$ such that $\phi=1$ on $(5/4,7/4)$. For $b>0$, let $\phi_{b}(x):=\phi(bx)$. Observe that $$\int_{\mathbb{R}}\phi_{b}(x)e^{1/x}dx=b^{-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx,\qquad\forall b>0$$ Suppose that there is a distribution $u\in\mathcal{D}'(\mathbb{R})$ of order ...


1

APPROACH 1: We can show that $J_{\nu}(kr)$ and $J_{\nu}(sr)$ are orthogonal by appealing to the governing ODE $$\frac{d}{dr}\left(r\frac{dJ_{\nu}(kr)}{dr}\right)+\left(k^2r-\frac{\nu}{r}\right)J_{\nu}(kr)=0 \tag 1$$ $$\frac{d}{dr}\left(r\frac{dJ_{\nu}(sr)}{dr}\right)+\left(s^2r-\frac{\nu}{r}\right)J_{\nu}(sr)=0 \tag 2$$ Multiplying $(1)$ by ...


0

So what you do is to realize that you can replace $(k_x,k_y,k_z)\cdot(x,y,z)$ by $\kappa r \cos(\theta)$, where $\kappa = \sqrt{k_x^2 + k_y^2 + k_z^2}$ and $r = \sqrt{x^2+y^2+z^2}$ and $\theta$ is the angle between $(k_x,k_y,k_z)$ and $(x,y,z)$. Also, you have that the Fourier transform of the constant function is the Dirac delta function, so we can ...


2

Ok, you said you knew how to prove it. Others might not. And you may like this better than what you have (it's the second thing I always think of when this comes up, and I like it a lot better than the first thing I think of...) Say $u$ is a distribution and $u'=0$. By definition $u(\phi')=0$ for any test function $\phi$. Hence $u(\phi)=0$ for any test ...


2

No, for instance if $S(x) = x$ then $\hat S = i\delta'$ (up to a multiplicative constant depending on how you define the fourier transform). More generaly if $S$ is a polynomial $S(x) = \sum a_k x^k $ then $\hat S = \sum a_k i^k\delta^{(k)} $


0

Let $g(x) = \operatorname{sgn}(x)e^{-|x|}$ and let $\delta$ be Dirac's delta "function". The problem is to find $(g*f)'(x)$. $$ (g*f)'(x) = \underbrace{(\delta'*(g*f))(x) = ((\delta'*f)*g)(x)} = (f'*g)(x). \tag 1 $$ The thing $\underbrace{\text{in the middle}}$ has no meaning outside the context of some theory in which we can speak of $\delta'$. But the ...


2

HINT: Split the integral as $$\int_{-\infty}^{\infty} \text{sgn} (x-y)e^{-|x-y|}f(y) \, dy = \int_{-\infty}^x e^{-(x-y)}f(y)\,dy-\int_x^{\infty} e^{(x-y)}f(y) \, dy$$ and use Leibnitz's Rule for differentiating under an integral. SPOILER ALERT: SCROLL OVER SHADED AREA TO SEE ANSWER


-4

Well the definition of the support is defined as the part of the domain where the distribution is non-zero. There are a few definitions of the Dirac delta but all agree that $x\neq 0 \implies \delta(x) = 0$.


1

Notation: For a fixed function $f$ defined on $\mathbb{R}^{n}$ and $y\in\mathbb{R}^{n}$, we write $\tilde{f}(x)=f(-x)$ and $(\tau^{y}f)(x)=f(x-y)$. I am assuming that you are familiar with the following result. Theorem. For $\phi\in\mathcal{S}(\mathbb{R}^{n})$, the tempered distribution $u\ast\phi$ coincides with a $C^{\infty}(\mathbb{R}^{n})$ ...


1

Tempered distributions have Fourier transforms. If $P(D)\,F=0$, take the Fourier transform to get $P(\xi)\,\hat F(\xi)=0$ for all $\xi\in\mathbb{R}$. Since $P(\xi)\ne0$ if $\xi\ne0$ the support of $\hat F$ is $\{0\}$. Can you take it fromhere?



Top 50 recent answers are included