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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 $\{b\in R|\mathrm{\exists }n\in \mathbb{N}:a^{n}b=0\}$ 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 $\frac{{\partial }^{2}u}{\partial y^2}+\sum _{j=1}^d\frac{{\partial }^{2}u}{\partial x_j^2}=0$ with $u(x,0)=f(x)$
A generalization of repeatedly applying the Laplacian is $(-\mathrm{\Delta }{)}^{a}f(x)=\underset{{ℝ}^d}{\int }(2\pi |\xi |{)}^{2a}\hat{f}(\xi )e^{2\pi i\xi \cdot x}d\xi$ 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 ($\hat{f}$ is the Fourier transform of $f$).
Show that $(-\mathrm{\Delta }{)}^{k/2}f(x)=(-1{)}^k\underset{y\to 0}{\lim }\frac{{\partial }^{k}u}{\partial y^k}(x,y)$ 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 $(-\mathrm{\Delta }{)}^{k/2}f(x)=(-1{)}^k\underset{y\to 0}{\lim }\frac{{\partial }^{k}u}{\partial y^k}(x,y)$. Sławomir Biały (talk) 00:36, 31 October 2012 (UTC)

Heisenberg Principle Implies Theorem

Show that the Heisenberg uncertainty principle implies $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(-\frac{d^{2}f}{d{x}^2}(x)+x^{2}f(x))\overline{f(x)}dx\ge \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\overline{f(x)}dx$ for every $f$ in the Schwartz space.
I have a version of the Heisenberg uncertainty principle which states that $(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2|f(x){|}^{2}dx)\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|\frac{df}{dx}\right|}^{2}dx\right)\ge 1/4$. I can get the integral on the left hand side of the inequality to $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{df}{dx}(x)\overline{\frac{df}{dx}(x)}+x^{2}f(x)\overline{f(x)}dx$ 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

$(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2|f(x){|}^{2}dx)\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|\frac{df}{dx}\right|}^{2}dx\right)\ge \frac{1}{4}{(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|f(x){|}^{2}dx)}^2.$

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 $g(x)=\underset{{ℝ}^d}{\int }(2\pi |\xi |{)}^{2a}\hat{f}(\xi )e^{2\pi i\xi \cdot x}d\xi$, where $f$ is in the Schwartz class, and real $a>0$, is smooth.
Also, unlike the standard Laplacian, the function $\underset{{ℝ}^d}{\int }(2\pi |\xi |{)}^{2a}\hat{f}(\xi )e^{2\pi i\xi \cdot x}d\xi$ 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, $(2\pi |\xi |{)}^{2a}\hat{f}(\xi )$, 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. $\frac{\partial }{\partial x_i}(-\mathrm{\Delta }{)}^{a}f(x)=\underset{{ℝ}^d}{\int }2\pi i{\xi }_i(2\pi |\xi |{)}^{2a}\hat{f}(\xi )e^{2\pi i\xi \cdot x}d\xi$. 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 $a=1/4$, $(2\pi |\xi |{)}^{1/2}\hat{f}(\xi )$ is not differentiable at $\xi =0$? Widener (talk) 23:54, 30 October 2012 (UTC)