Central motion excludes the center

Central motion is the motion of a particle under the influence of a central force. I have known for some year that the particle cannot not pass through the origin. Mimicking a proof in my multivariable calculus book, I'll formalize the fact.

Theorem
If \(r:(a, b)\to\mathbb{R}^3\) is twice differentiable, \(r(t)\times r''(t) = \vec{0}\) for all \(t\) and \(r(t_0)\times r'(t_0) \not=\vec{0}\) for some \(t_0\), then \(r(t)\) is never \(\vec{0}\).

Proof
\[\frac{d}{dt}(r(t)\times r'(t)) = r'(t)\times r'(t) + r(t)\times r''(t) = \vec{0}\]Thus \(r(t)\times r'(t)=c\) for some \(c\). Because \(r(t_0)\times r'(t_0) \not=\vec{0}\), \(c\not=\vec{0}\). If \(r(t_1)\) were \(\vec{0}\) for some \(t_1\) then \(c\) would be \(\vec{0}\) so there is no such \(t_1\).

I wonder how to generalize this to higher dimensions.

Continuity at infinity

Usually \(f(x)\to L\text{ as }x\to \infty\) is defined in an epsilon N fashion. With a projective point of view, infinity becomes a point among the usual numbers, and the definitions of convergence at finite and infinite points can be unified. What strikes me is that this means that if f(x) converges as x goes to infinity, then f can be continuously extended at infinity.

Cauchy–Schwarz

Finally I understand the Cauchy–Schwarz inequality! No induction needed.

Theorem
\(a, b\in \mathrm{Rat}^n\)
\(n\in\mathrm{Nat}\)
                     
\((a\bullet b)^2 \leq \mathop{\rm Q}(a)\mathop{\rm Q}(b)\)

Proof
Recall that by definition
\[a\bullet b=\sum_{i=1}^n a_ib_i\]\[\mathop{\rm Q}(a)=\sum_{i=1}^n a_i^2\]\[\mathop{\rm Q}(b)=\sum_{i=1}^n b_i^2.\]
The claim, reformulated as
\[\left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right),\]
is equivalent to (by distributing terms)
\[\left(\sum_{i=1}^n a_i^2b_i^2\right)+\sum_{1\leq i<j\leq n}2a_ib_ia_jb_j \leq \left(\sum_{i=1}^n a_i^2b_i^2\right)+\sum_{1\leq i<j\leq n}a_i^2b_j^2+a_j^2b_i^2\]
and (by collecting terms)
\[0 \leq \sum_{1\leq i<j\leq n}a_i^2b_j^2+a_j^2b_i^2-2a_ib_ia_jb_j = \sum_{1\leq i<j\leq n}(a_ib_j-a_jb_i)^2.\]
The last statement is true because squares are positive and sums of positive numbers are positive.

A spiral

Let \(\mathbb{O}\) be the countable set of all infinite rational number sequences which are ultimately but not identically zero, and let \(H\) contain those elements in \(\mathbb{O}\) whose last nonzero element is positive. With respect to \(\mathbb{O}\), define \(H^\complement\). Then \(f(x) = -x\) is a bijection from \(H\) to \(H^\complement\).

Also, \(H\) looks like a spiral. To see that, require the last nonzero element to be the first, then the second, and then the third one, and watch this sequence of subsets twist into a new dimension at each step.

That's really something! A spiral whose reflection in the origin is its complement.

Elegant matrix inversion

I had an exam today. One question inspired me. Let \(\mathbf{E}_n\hspace{-3pt}: n\times n\) and \(\mathbf{e}_{ij}=1\). Then the inverse of \(\mathbf{A}_n = \mathbf{E}_n - \mathbf{I}_n\) is \(\mathbf{A}_n^{-1} = \frac1{n-1}\mathbf{E}_n-\mathbf{I}_n\) if \(2\leq n\). There are two interesting things about this: The inverse can be expressed using matrix operations, and the matrix operations can be used to verify that \(\mathbf{A}^{-1}\) is actually an inverse.

To verify the result, note first that \(\mathbf{E}_n^2=n\mathbf{E}_n\). This identity can be seen as the workhorse of the method. Check that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}\).
\[\begin{align}\mathbf{A}_n^\phantom{-1}\hspace{-4pt}\mathbf{A}_n^{-1} &= (\mathbf{E}_n - \mathbf{I}_n)\left(\frac1{n-1}\mathbf{E}_n-\mathbf{I}_n\right) \\&= \frac n{n-1}\mathbf{E}_n-\mathbf{E}_n-\frac1{n-1}\mathbf{E}_n+\mathbf{I}_n = \mathbf{I}_n\end{align}\]

I'll have to think about this some more sometime. Can other simple matrices be inverted and expressed using matrix operations? What are some useful workhorses? Why has no one told me about this before?

First four bars

Today I wrote the first piece of music that sounded interesting and good. It's only four bars.

An open question

Is there a nonnegative function \(f\) such that \(\displaystyle\int\limits_0^\infty \cfrac{f(x)}{\int\limits_x^\infty f(y)\; dy} dx\) exists?