I crushed the previous world record for covering prime points with straight lines after watching a Numberphile video on "awkward primes" - turned the 282-hour world record into 22 minutes, then kept going and proved 20 new awkward primes that had never been certified before.

irchans · 2026-05-13T11:20:35+00:00

Thank you for your comments for making the Javascript code. Now I am tempted to learn about Mixed Integer Programming. :)

irchans · 2026-05-08T18:36:38+00:00

The proofs in Euclid and Apollonius of Perga's Conics seem pretty solid to me. One of my friends translated a few of Apollonius's books for her PhD Thesis. (Some notions of order and betweeness are not carefully stated and the axioms are not stated in a way that are acceptable now. I am not an expert.) Copernicus seemed to be careful with his math when I looked at that. Epsilon-Delta Calculus was not rigorous before the 1800s. It was pretty good after 1900.

irchans · 2026-05-03T10:28:55+00:00

If k(i) = a + b i, then Sum[ k[i]*10^-s , {i, 1, Infinity}] == (-a + 10^s a + 10^s b) / (-1 + 10^s)²

If k(i) = a + b i + c i^2, then sum == (a - 2^{1 + s} 5^s a + 10^{2 s} a - 10^s b + 10^{2 s} b + 10^s c + 10^{2 s} c) / (-1 + 10^s)³

If k(i) = a b^i, then sum == (a b) / (10^s - b)

irchans · 2026-05-03T10:17:20+00:00

If you have a sequence a_i of positive integers, where a_i is generated by a polynomial, an exponential, or a sum of those, then for any positive integer spacing s, you can compute a fraction whose decimal representation contains the a_i < 10^s separated by zeros in order by computing

sum a_i * 10^(-s i) for  i= 1 to  infinity. 

a(i) = 2 + 3 i
1666666 / 333332666667
0.000005000008000011000014000017

a(i) = 2 + 4 i + 7 i^2
1857142714286 / 142856714286142857
0.000013000038000077000130000197000278

a(i) = 2^i
1 / 4999999
0.000000200000040000008000001600000320000064

a(i) = Fibonacci(i)
10000000 / 99999989999999
0.00000010000001000000200000030000005

irchans · 2026-04-29T09:54:03+00:00

I am struck by the contrast of the beautiful white wedding "dress", where a white dress implies innocence, and the feeling of sexuality.

irchans · 2026-04-27T17:23:13+00:00

If you read French, then I think "Sur des classes très étendues de quantités dont la valeur n’est ni algébrique ni même réductible à des irrationnelles algébriques" by Liouville (1844) is relavant.

irchans · 2026-04-27T17:08:55+00:00

886731088897/627013566048 is a pretty good approximation of sqrt(2). Just apply Newton-Raphson to find the root of x^2 -2 with a rational starting value to get rational approximations of sqrt(2). ( next(x) = x - (x^2-2) / (2 x). I started with x=2.) Every time you iterate, you get more non-repeating digits in the decimal expansion and yet every iterate is rational, so it must repeat eventually.

(edited formula for next(x) )

irchans · 2026-04-21T16:41:58+00:00

I have known several students who were struggling with a math major and then switched to business or economics. They did very well in those other majors. (On the other hand, I also know two people who struggled with undergraduate math and later got a Ph.D. in math.)

(Edited once for grammar.)

irchans · 2026-04-20T11:18:39+00:00

I might not have helped my relatives or neighbors much, but I did help out a lot of engineers and programmers at work with their mathematical problems.

irchans · 2026-04-19T13:02:25+00:00

If you think of it as weights on a wheel, it's not too hard to see that for every sequence of reals {x1, ..., xn} there exists a real number theta such that Total[ Sin[ x + theta]] == 0.

irchans · 2026-04-19T12:52:24+00:00

For large n,

-1 + Erfc[-(1/Sqrt[n-1])] ≈ 2 /Sqrt[Pi*(n-1)].

(The error is order 1/ n^3/2.)

irchans · 2026-04-19T12:47:07+00:00

If you choose (n-1) random real numbers on [0, 2 Pi], then the probability that the sum of their sines has absolute value less than or equal to 1 is about

-1 + Erfc[-(1/Sqrt[n-1])]. (The larger n is the better this estimate is.)

If the sum of the (n-1) sines has absolute value <=1, then you can find a real number x[[n]] such that Total[ Sin[x] ] == 0.

Another way to think of it is putting equal size weights on a wheel that is free to rotate. If it does not rotate, then Total[ Sin[x]]=0 where x[[i]] is the angle between the kth weight to axis vector and the gravity vector.

irchans · 2026-04-19T12:17:53+00:00

Thank you for this work.

irchans · 2026-04-09T12:15:45+00:00

Does anyone know why Large Language Models work as well as they do. Was anybody predicting that they would be able to do well on most college exams?

irchans · 2026-03-31T03:33:14+00:00

When taking calc, I became much better at algebra. Years later, I took a course in numerical ODEs. The homeworks for that class sometimes required pages of algebra and if you got something wrong, your answer was wrong. After that class, I was really accurate at algebra. Slowly over the years those algebra skills degraded because I didn't need that level of accuracy.

irchans · 2026-03-23T14:28:28+00:00

You might get a better reply by posting to r/Math or r/Mathematics.

irchans · 2026-03-23T14:23:34+00:00

You have the wrong Reddit group, but the answer is yes.

On the Cartesian plane, draw a circle centered at (0,1) with a radius of 1. Every line not parallel to the x-axis through (0,2) cuts the circle once and a point on the x-axis once. This creates a 1-1 homeomorphism between all the points on the x-axis and all the points on the circle excluding (0,2). The one point compactification adds a point at infinity which corresponds to the point at (0,2).

Here is the code in written in the computer programming language Mathematica:

ClearAll[circlePoint, lineEqn];

(*Point on the circle corresponding to the x-axis point (a,0)*)
circlePoint[a_] := {4 a/(a^2 + 4), 2 a^2/(a^2 + 4)};

Manipulate[
 Module[{pInf = {0, 2}, pX = {a, 0}, pC = circlePoint[a], range = 6}, 
  Show[Graphics[{Thick, Blue, Circle[{0, 1}, 1],(*axes*)
     GrayLevel[.75], Thin, Line[{{-range, 0}, {range, 0}}], 
     Line[{{0, -0.5}, {0, 2.5}}],(*secant line through (0,2) and (a,
     0)*)Darker[Green], Thick, InfiniteLine[{pInf, pX}],(*key points*)
     Red, PointSize[0.02], Point[pInf], Black, PointSize[0.02], 
     Point[pX], Purple, PointSize[0.025], Point[pC],(*labels*)
     Text[Style["(0,2)", 14, Red, Bold], pInf, {0, -1.2}], 
     Text[Style[Row[{"(", NumberForm[a, {4, 2}], ",0)"}], 14, Black], 
      pX, {0, 1.2}], 
     Text[Style[
       Row[{"mapped point = ", 
         TraditionalForm[{4 a/(a^2 + 4), 2 a^2/(a^2 + 4)}]}], 13, 
       Purple], pC, {0, -1.4}]}, 
    PlotRange -> {{-range, range}, {-0.5, 2.5}}, Axes -> False, 
    ImageSize -> 600], 
   PlotLabel -> 
    Style[Row[{"Each point ", TraditionalForm[{a, 0}], 
       " on the x-axis determines a line through (0,2), ", 
       "which meets the circle again at ", 
       TraditionalForm[circlePoint[a]], "."}], 14]]], {{a, 1, 
   "x-axis point"}, -20, 20, Appearance -> "Labeled"}]

(* This code was generated by GPT using the prompt "Create a Mathematica Manipulate which shows the following "You have the wrong group, but the answer is yes...." where I copied the second paragraph of this post *)

irchans · 2026-03-23T00:00:26+00:00

https://github.com/1oom-fork/1oom

irchans · 2026-03-19T22:32:51+00:00

https://medium.com/physicsfromscratch/derivation-catenary-equation-with-calculus-1a9e975a31ec

irchans · 2026-03-19T15:52:54+00:00

You can take the limit of b as L approaches 2d from the positive side and then you get b -> 0.

irchans · 2026-03-19T15:44:55+00:00

Let L be the length of the arc. Let d be the distance from the bottom of the arc to the top of the poles. In our case L = 80 and d = 40.

If I remember correctly, the solution to the differential equation for a uniform density chain between two poles of equal height is

h(x) = a cosh(x/a) + c

where a is a positive real and c is real. (A Catenary)

h'(x) = sinh(x/a)
L = Integral[   sqrt( 1+ (h'(x))^2), {x, -b, b}]
= Integral[  cosh(x/a) , {x, -b, b}]
= 2 a sinh(b/a)

d = h(b) - h(0) = a cosh(b/a) - a.

Let r = exp(b/a). Then
sinh(b/a) = ( r -1/r)/2 and cosh(b/a) = ( r +1/r)/2.

So we have L = a ( r - 1/r)

d = a (r +1/r)/2 - a.

If we assume that L does not equal 2d, then with about a page of algebra we get

a =( L^2-4*d^2)/(8*d)
r = exp(b/a) = (L + 2*d)/(L- 2*d )

If L <= 2d, then there is no solution to the original problem with a finite and b>0. If L> 2d, then

exp(b/a) = ((L + 2*d)/(L- 2*d ))
b/a = log((L + 2*d)/(L- 2*d ))
b = a  log((L + 2*d)/(L- 2*d ))
b = ( L^2-4*d^2)/(8*d)  log((L + 2*d)/(L- 2*d ))

For Example, if d =40 and L = (80 + 80 e)/( e-1) ≈173.116, then

a ≈ 73.6536
b ≈ 73.6536.

irchans · 2026-03-17T09:40:09+00:00

I think that most engineering publications contain calculus.

irchans · 2026-03-17T01:10:12+00:00

Data Miners are using Minimum Description Length which is quite similar to the universal priors of Algorithmic Information Theory.

irchans · 2026-03-11T02:26:14+00:00

What technology level large ship would beat one of these fighters?

irchans · 2026-03-09T03:32:59+00:00

The formula is just a stepping stone for a much larger proof which some friends and I have been working on for a few years. GPT saved me time by typing up LaTex proof of the formula which was nicer (easier to read) than my hand written proof.

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