Re: [math-fun] Egyptian Fractions (was Re: Messages in pi (cont'd))
Yes, any divergent series whose terms approach 0. Say we're choosing the sign of each term. Then all possible sign choices amount to the countable product K (cartesian power) of {-,+}, which has the size of 2^aleph_0 = the continuum. The product K naturally has the topology of the Cantor set. So for each point of the Cantor set K — i.e., for each choice of signs S : Z+ —> {-,+} — this S gives rise to the real number Sum_{n in Z+} S(n)/n. So assuming the probability of convergence is 1 (which it is), we get a map K —> R from the Cantor set onto the real numbers. This should by rights be continuous, so it's a funny map — a continuous surjection from the totally disconnected Cantor set onto the real numbers. But basic theorems in topology rule this out (the image must be compact). Therefore, mathematics is inconsistent. Another related thing is to look at the distribution of values obtained when choosing signs randomly. Or for more fun, use points on the unit circle S^1 in the complex plane, chosen independently at random from S^1, as the coefficient for each term of the harmonic series. What does the distribution of sums in the plane look like? —Dan Cris Moore wrote: ----- This applies to any divergent series, right? And the fun thing is that any C has a series with any initial subsequence. I suppose this means any C has countably many sequences, although it’s not clear that taking a finite initial subsequence and then turning greedy covers all of them…
On Dec 10, 2018, at 3:43 PM, Dan Asimov <dasimov@earthlink.net> wrote:
There is the greedy-harmonic-series-with-signs method: ...
I’ve often considered using a ratio of 2 finite sums of finite products of finite powers of primes (positive integer powers only) to express and calculate with non transcendentals exactly…..and transcendentals to a given finite accuracy. Was just never into it enough to get past the idea stage, fractals are more interesting ;)
On Mon, Dec 10, 2018 at 8:21 PM Dan Asimov <dasimov@earthlink.net> wrote:
Yes, any divergent series whose terms approach 0. Say we're choosing the sign of each term. Then all possible sign choices amount to the countable product K (cartesian power) of {-,+}, which has the size of 2^aleph_0 = the continuum. The product K naturally has the topology of the Cantor set. So for each point of the Cantor set K — i.e., for each choice of signs
S : Z+ —> {-,+}
— this S gives rise to the real number
Sum_{n in Z+} S(n)/n.
So assuming the probability of convergence is 1 (which it is),
An interesting claim; can you prove it? Can you prove it?
we get a map
K —> R
from the Cantor set onto the real numbers.
More generally, the map K gives a measure on the reals, by pushing forward the standard measure on K. I would conjecture that under this measure, every individual point in R has measure 0. More strongly, I suspect that this measure is absolutely continuous (that is, the measure of any set with Lebesgue measure 0 is 0 under this measure. If this is true, what does its Radon Nikodym derivative look like?
This should by rights be continuous, so it's a funny map — a continuous surjection from the totally disconnected Cantor set onto the real numbers. But basic theorems in topology rule this out (the image must be compact). Therefore, mathematics is inconsistent.
Taking this as a find-the-fallacy puzzle, here's the solution in rot13: Gur enatr bs X vf abg gur erny ahzoref, ohg gur rkgraqrq erny ahzoref, gung vf, gur erny ahzoref jvgu gjb nqqvgvbany inyhrf, cbfvgvir naq artngvir vasvavgl, nqqrq. X zncf qviretrag frdhraprf gb vvasvavgl be artngvir vasvavgl. Gur rkgraqrq erny ahzoref ner pbzcnpg.
Another related thing is to look at the distribution of values obtained when choosing signs randomly. Or for more fun, use points on the unit circle S^1 in the complex plane, chosen independently at random from S^1, as the coefficient for each term of the harmonic series.
What does the distribution of sums in the plane look like?
—Dan
Cris Moore wrote: ----- This applies to any divergent series, right? And the fun thing is that any C has a series with any initial subsequence. I suppose this means any C has countably many sequences, although it’s not clear that taking a finite initial subsequence and then turning greedy covers all of them…
On Dec 10, 2018, at 3:43 PM, Dan Asimov <dasimov@earthlink.net> wrote:
There is the greedy-harmonic-series-with-signs method: ...
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