Wikipedia:Reference desk/Archives/Mathematics/2012 October 30

= October 30 =

Relationship between $$IJ$$ and $$I \cap J$$
Let $$I$$ and $$J$$ be ideals of a ring. Is there a relationship between $$IJ$$ and $$I \cap J$$? Other than the obvious one $$IJ \subset I \cap J$$?--AnalysisAlgebra (talk) 01:09, 30 October 2012 (UTC)
 * If the ring is commutative, they have the same radical: $$\sqrt{IJ} = \sqrt{I \cap J}$$. This also means that if P is a prime ideal, then P contains IJ if and only if it contains I∩J.  Rckrone (talk) 05:07, 30 October 2012 (UTC)

Canonical map
What is the canonical map from a ring $$R$$ to $$R[x]/(ax-1)$$ where $$a \in R$$?--AnalysisAlgebra (talk) 05:26, 30 October 2012 (UTC)

(p.s. and showing that its kernel is $$\{b\in R|\exists n\in\mathbb{N} : a^nb=0\}$$ would be nice too...)--AnalysisAlgebra (talk) 05:29, 30 October 2012 (UTC)
 * There is a natural inclusion of R in R[x]. R[x] is the ring of polynomials in x with coefficients in R.  The elements of R can also be considered polynomials (with degree 0).  Then there is a natural surjection from R[x] to R[x]/(ax-1) mapping a polynomial to its equivalence class mod (ax-1).  The composition of these two is the canonical map from R to R[x]/(ax-1). Rckrone (talk) 05:54, 30 October 2012 (UTC)

Generalization of the Laplacian and partial derivative
Let $$u(x,y)$$ be the solution of the steady-state heat equation $$\frac{\partial^2u}{\partial y^2}+\sum_{j=1}^d \frac{\partial^2u}{\partial x_j^2}=0$$ with $$u(x,0) = f(x)$$

A generalization of repeatedly applying the Laplacian is $$(-\Delta)^a f(x) = \int_{\mathbb{R}^d} (2\pi|\xi|)^{2a}\widehat{f}(\xi)e^{2\pi i\xi \cdot x}d\xi$$ and it is quite easy to show that this coincides with the regular Laplacian when a is a positive integer, if f is in the Schwartz space ($$\widehat{f}$$ is the Fourier transform of $$f$$). Show that $$(-\Delta)^{k/2}f(x) = (-1)^k\lim_{y \rightarrow 0} \frac{\partial^k u}{\partial y^k}(x,y)$$ for positive integer $$k$$.Widener (talk) 11:07, 30 October 2012 (UTC)


 * Take the Fourier transform of u wrt to the x variables and solve the steady state heat equation: $$\hat{u}=e^{-2\pi y|\xi|}\hat{f}$$ (this has the correct initial conditions in y=0 and boundary conditions at infinity). Apply an inverse Fourier transform and again differentiating enough times under the integral gives $$(-\Delta)^{k/2}f(x) = (-1)^k\lim_{y \rightarrow 0} \frac{\partial^k u}{\partial y^k}(x,y)$$.  Sławomir Biały  (talk) 00:36, 31 October 2012 (UTC)

Heisenberg Principle Implies Theorem
Show that the Heisenberg uncertainty principle implies $$\int_{-\infty}^\infty (-\frac{d^2f}{dx^2}(x)+x^2f(x))\overline{f(x)}dx \ge \int_{-\infty}^\infty f(x)\overline{f(x)}dx$$ for every $$f$$ in the Schwartz space.

I have a version of the Heisenberg uncertainty principle which states that $$\left(\int_{-\infty}^\infty x^2|f(x)|^2dx\right)\left(\int_{-\infty}^\infty \left| \frac{df}{dx}\right|^2dx\right) \ge 1/4$$. I can get the integral on the left hand side of the inequality to $$\int_{-\infty}^\infty \frac{df}{dx}(x)\overline{\frac{df}{dx}(x)}+x^2f(x)\overline{f(x)}dx$$ but this is a sum instead of a product, and I don't see how you can get the inequality anyway. Widener (talk) 14:57, 30 October 2012 (UTC)


 * Your version of Heisenberg is wrong. It should be
 * $$\left(\int_{-\infty}^\infty x^2|f(x)|^2dx\right)\left(\int_{-\infty}^\infty \left| \frac{df}{dx}\right|^2dx\right) \ge \frac{1}{4}\left(\int_{-\infty}^\infty |f(x)|^2\,dx\right)^2.$$
 * So, this is an inequality of the form $$AB\ge C^2/4$$ and you want to show an inequality of the form $$A+B\ge C$$. You can use the AM-GM inequality.   Sławomir Biały  (talk) 00:40, 31 October 2012 (UTC)

Properties of the generalized Laplacian
How do you verify smoothness of a vector valued function? In particular, I want to show that the function $$g(x) = \int_{\mathbb{R}^d} (2\pi|\xi|)^{2a}\widehat{f}(\xi)e^{2\pi i\xi \cdot x}d\xi$$, where $$f$$ is in the Schwartz class, and real $$a>0$$, is smooth.

Also, unlike the standard Laplacian, the function $$\int_{\mathbb{R}^d} (2\pi|\xi|)^{2a}\widehat{f}(\xi)e^{2\pi i\xi \cdot x}d\xi$$ is not necessarily in the Schwartz class if $$a$$ is not an integer. How do you prove this?Widener (talk) 18:56, 30 October 2012 (UTC)
 * Differentiation under the integral sign proves smoothness. Non-Schwartzness follows since the frequency domain representation, $$(2\pi|\xi|)^{2a}\widehat{f}(\xi)$$, is non-smooth.   Sławomir Biały  (talk) 22:21, 30 October 2012 (UTC)
 * Okay, but remember, this is a vector valued function. What exactly do you mean by "differentiation" in this context?
 * That is to say, x is a vector. Widener (talk) 23:23, 30 October 2012 (UTC)
 * Partial derivative Sławomir Biały  (talk) 23:40, 30 October 2012 (UTC)
 * Okay. $$\frac{\partial}{\partial x_i}(-\Delta)^af(x) = \int_{\mathbb{R}^d} 2\pi i\xi_i(2\pi|\xi|)^{2a}\widehat{f}(\xi)e^{2\pi i\xi \cdot x}d\xi$$. Hmm. The partial derivatives commute essentially because multiplication is commutative. Does this prove that these partial derivatives are $$C^1$$? I know the converse is true at least. But do you even know that the integral still converges?Widener (talk) 23:47, 30 October 2012 (UTC)
 * Integral converges because $$\hat{f}$$ is rapidly decreasing.  Sławomir Biały  (talk) 00:25, 31 October 2012 (UTC)
 * Yes, of course. Is my logic about the partial derivatives being commutative therefore $$C^1$$ correct? I know that partial derivatives that are $$C^1$$ commute, but I don't know if the converse is true. And does that indeed imply that the whole function is smooth? Widener (talk) 00:32, 31 October 2012 (UTC)
 * BY the way, how do you justify interchanging the partial derivative and the Integral?Widener (talk) 00:54, 31 October 2012 (UTC)
 * That's an argument that requires using dominated convergence theorem. Sławomir Biały  (talk) 01:28, 31 October 2012 (UTC)
 * Also, could you explain the non-Schwartzness in more elementary terms?
 * Thanks Widener (talk) 23:21, 30 October 2012 (UTC)
 * A function is Schwartz iff it's smooth and its Fourier transform is smooth.  Sławomir Biały  (talk)|
 * So, for instance, choosing $$a = 1/4$$, $$(2\pi|\xi|)^{1/2}\widehat{f}(\xi)$$ is not differentiable at $$\xi = 0$$? Widener (talk) 23:54, 30 October 2012 (UTC)