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

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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 $\{b\in R|\exists n\in \mathbb {N} :a^{n}b=0\}$ 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 $(-\Delta )^{a}f(x)=\int _{\mathbb {R} ^{d}}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$ 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 $\int _{-\infty }^{\infty }(-{\frac {d^{2}f}{dx^{2}}}(x)+x^{2}f(x)){\overline {f(x)}}dx\geq \int _{-\infty }^{\infty }f(x){\overline {f(x)}}dx$ for every $f$ in the Schwartz space.
I have a version of the Heisenberg uncertainty principle which states that $\left(\int _{-\infty }^{\infty }x^{2}|f(x)|^{2}dx\right)\left(\int _{-\infty }^{\infty }\left|{\frac {df}{dx}}\right|^{2}dx\right)\geq 1/4$. I can get the integral on the left hand side of the inequality to $\int _{-\infty }^{\infty }{\frac {df}{dx}}(x){\overline {{\frac {df}{dx}}(x)}}+x^{2}f(x){\overline {f(x)}}dx$ 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

$\left(\int _{-\infty }^{\infty }x^{2}|f(x)|^{2}dx\right)\left(\int _{-\infty }^{\infty }\left|{\frac {df}{dx}}\right|^{2}dx\right)\geq {\frac {1}{4}}\left(\int _{-\infty }^{\infty }|f(x)|^{2}\,dx\right)^{2}.$

So, this is an inequality of the form $AB\geq C^{2}/4$ and you want to show an inequality of the form $A+B\geq C$. 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. ${\frac {\partial }{\partial x_{i}}}(-\Delta )^{a}f(x)=\int _{\mathbb {R} ^{d}}2\pi i\xi _{i}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$. 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)