Sunday, February 15, 2015
Sunday, February 1, 2015
Theorem classification
I claim (as an experiment) that there are three kinds of theorems "If A then B". Let \(|A|\) be the number of arrangements of objects that fit the description, i.e. interpret \(A\) as a set of tuples.
- \(|A| = \aleph_0\) – these give exact answers
- \(|A| = c\) – these give approximate results, usually to arbitrary precision
- \(|A| > \aleph_0\text{ but }|A|\not=c\) – these are even deeper set-theoretical fantasies
If \(|A|\in\mathbb{N}\text{ or }|A| = 0\) then the theorem doesn't deserve to be called a theorem because the use of the implication can be avoided by enumerating all statements \((a_i\in A,b_i\in B)_{i=1}^n\).
A subtle issue is that my use of sets is not justified. For example, according to present set theory, there is no set of all sets, so my classification cannot handle statements like "All sets have a power set". I count all such theorems into category 3 because they are set-theoretic.
A subtle issue is that my use of sets is not justified. For example, according to present set theory, there is no set of all sets, so my classification cannot handle statements like "All sets have a power set". I count all such theorems into category 3 because they are set-theoretic.
How many functions are there?
Claim
There are more real functions than there are real numbers.
Explanation
\(\{f:\mathbb{R}\to\mathbb{R}\} \geq \{f:\mathbb{R}\to\{0, 1\}\} = \mathcal{P}(\mathbb{R}) > \mathbb{R}\)
Claim
There are just as many continuous functions as there are real numbers.
Explanation
\(\{f:\mathbb{R}\to\mathbb{R}, f\text{ continuous}\}\\\qquad\begin{cases}\geq\{f:\{0\}\to\mathbb{R}\} = \mathbb{R}\\\leq\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}=\mathbb{R}^{\mathbb{N}\cdot\mathbb{N}}=2^{\mathbb{N}\cdot\mathbb{N}\cdot\mathbb{N}}=2^\mathbb{N}=\mathbb{R}\end{cases}\)
The first step on the second line is justified by the fact that for every continuous function, there is a sequence of parallelogram functions converging pointwise to it. By parallelogram function, I mean a function whose value is specified at evenly distributed points and interpolated linearly in between. For example, a parallelogram function approximating \(\sin(x)\) is \[f(x) = \sin(\lfloor x\rfloor) + (x-\lfloor x\rfloor)\sin(\lfloor x\rfloor+1).\]
To narrow the number of parallelogram functions down to \(\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}\), require that the value at \(0\) be specified and that the distance between specified points for the \(n\)th function in the sequence be \(1/n\).
There are more real functions than there are real numbers.
Explanation
\(\{f:\mathbb{R}\to\mathbb{R}\} \geq \{f:\mathbb{R}\to\{0, 1\}\} = \mathcal{P}(\mathbb{R}) > \mathbb{R}\)
Claim
There are just as many continuous functions as there are real numbers.
Explanation
\(\{f:\mathbb{R}\to\mathbb{R}, f\text{ continuous}\}\\\qquad\begin{cases}\geq\{f:\{0\}\to\mathbb{R}\} = \mathbb{R}\\\leq\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}=\mathbb{R}^{\mathbb{N}\cdot\mathbb{N}}=2^{\mathbb{N}\cdot\mathbb{N}\cdot\mathbb{N}}=2^\mathbb{N}=\mathbb{R}\end{cases}\)
The first step on the second line is justified by the fact that for every continuous function, there is a sequence of parallelogram functions converging pointwise to it. By parallelogram function, I mean a function whose value is specified at evenly distributed points and interpolated linearly in between. For example, a parallelogram function approximating \(\sin(x)\) is \[f(x) = \sin(\lfloor x\rfloor) + (x-\lfloor x\rfloor)\sin(\lfloor x\rfloor+1).\]
To narrow the number of parallelogram functions down to \(\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}\), require that the value at \(0\) be specified and that the distance between specified points for the \(n\)th function in the sequence be \(1/n\).
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.
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.
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