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25 Dec 2021	Mathchan is launched into public

File: Saudi combined e.png ( 334.65 KB , 2033x1433 , 1663822754492.png )

Anonymous September 22, 2022 (Thu) 04:59:14 No. 308

Reply

Peak oil.

It's mathematically impossible for you to not die within a year.

I posted with the alias mustang on pol, you can see in archive.

File: r_triangle_1.jpg ( 4.38 KB , 242x208 , 1655398539385.jpg )

Anonymous June 16, 2022 (Thu) 16:55:45 No. 189

Reply

Trigonometric functions are used to relate the angles of a right triangle to its sides.

\qquad \sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\quad\cos(\text{angle}) =\frac{\text{adjacent}}{\text{hypotenuse}}\quad\tan(\text{angle}) = \frac{\sin(\text{angle})}{\cos(\text{angle})} = \frac{\text{opposite}}{\text{adjacent}}\quad

With respect to any of the two angles

\alpha

or

\beta

(the third angle of a right triangle is, of course,

90^\text{o}

thereofre it's irrelevant) three sides can be identified with the following names: hypotenuse, opposite and adjacent. The hypotenuse is easy to identify and it's the longest i.e.

c

. The adjacent side is the one that's "touching" the angle; for angle

\alpha

that would be

b

, while for angle

\beta

that would be

a

. The opposite side is the one that's NOT "touching" the angle; for angle

\alpha

that would be

a

, while for angle

\beta

that would be

b

.

Therefore, the complete set of trigonometric functions for the triangle in the pic related is:

\qquad \sin(\alpha) = \frac{a}{c}\quad \cos(\alpha) = \frac{b}{c}\quad \tan(\alpha) = \frac{a}{b}

\qquad \sin(\beta) = \frac{b}{c}\quad \cos(\beta) = \frac{a}{c}\quad \tan(\beta) = \frac{b}{a}

Thus, just knowing the angle

\alpha

allows conversion of any side to any other side:

	$a$	$b$	$c$
$a=$	$a$	$b\tan(\alpha)$	$c\sin(\alpha)$
$b=$	$\frac{a}{\tan(\alpha)}$	$b$	$c\cos(\alpha)$
$c=$	$\frac{a}{\sin(\alpha)}$	$\frac{b}{\cos(\alpha)}$	$c$

And similarly, just knowing the angle

\beta

also allows conversion of any side to any other side:

	$a$	$b$	$c$
$a=$	$a$	$\frac{b}{\tan(\beta)}$	$c\cos(\beta)$
$b=$	$a\tan(\beta)$	$b$	$c\sin(\beta)$
$c=$	$\frac{a}{\cos(\beta)}$	$\frac{b}{\sin(\beta)}$	$c$

Intuition behind this is straightforward: For angles of a triangle, sine and cosine are bounded functions between zero and one.

(0 \leq \sin(x)\leq 1,\; 0\leq \cos(x) \leq1)

. Multiplying something with these functions should produce a smaller or equal value, while dividing something with these functions should produce a larger or equal value. Since hypotenuse is the longest, multiplying it with sine or cosine will "reduce" it to a leg. Similarly, dividing a leg by sine or cosine will "grow" it into a hypotenuse. A leg can be converted to another leg by "growing" it into a hypotenuse first, then "reduciPost too long. Click here to view the full text.

7 posts omitted. Click here to view.

>>

Anonymous June 16, 2022 (Thu) 17:01:28 No. 197

Taylor series of trigonometric functions

Sine and cosine have the following Maclaurin series (which are Taylor series around

x_0 = 0)

:

\qquad \sin(x) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n + 1)!}x^{2n + 1}

\qquad \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}

This is the same as writing:

\qquad \sin(x) = x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} -\frac{x^{11}}{11!} +\dots

\qquad \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10}+\dots

By summing the Maclaurin series of

\cos(x)

and

i\sin(x)

it's possible to prove Euler's formula

e^{ix} = \cos(x) + i\sin(x)

\qquad \cos(x) + i\sin(x) = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots = e^{ix}

Maclaurin series can be used to approximate trigonometric functions to an arbitrary degree
For small angles (i.e.

\alpha\approx 0

), good approximations (often used in physics) are:

\qquad \sin(\alpha)\approx x

\qquad \cos(\alpha)\approx 1

Which are just Maclaurin series with only 1 term.

>>

Anonymous June 16, 2022 (Thu) 17:02:01 No. 198 >>211

Computation of trigonometric functions

Maclaurin series lend an easy way to numerically compute.
We need a helper function

ifac(n)

that computes

n!,\; n\in\mathbb{N}

and a helper function

fpow(x, n)

that computes

x^n,\;x\in\mathbb{R}, n\in\mathbb{N}

:

int ifac(int n) {

    return (n > 0) ? (n * ifac(n - 1)) : 1;

}

float fpow(float x, int n) {

    return (n < 0) ? fpow(1/x, -n)

        : (n == 0) ? 1

        : (n == 1) ? x

        : (n > 1)  ? x * fpow(x, n - 1)

        : 1;

}

Now

\sin(x)

can be computed with

fsin(x, prec)

:

float fsin(float x, int termCount) {

    int n, sign = 1;

    float numer  = x,

          denom  = 1,

          result = sign * (numer / denom);



    for (n = 1; n < termCount; n++)  {

        sign *= -1;

        numer = fpow(x, 2*n+1);

        denom = ifac(2*n+1);

        result += sign * (numer / denom);

    }



    return result;

}

While

\cos(x)

can be computed with

fcos(x, prec)

:

float fcos(float x, int termCount) {

    int n, sign = 1;

    float numer = 1,

          denom = 1,

          result = sign * (numer / denom);

    for (n = 1; n < termCount; n++)  {

        sign *= -1;

        numer = fpow(x, 2*n);

        denom = ifac(2*n);

        result += sign * (numer / denom);

    }

    return result;

}

While Maclaurin series for

\sin(x)

and

\cos(x)

converge on entire

\mathbb{R}

, they converge the fastest when

x\in(-\pi,\pi]

so its good to bound

x

to that interval considering trigonometric functions are periodic. In fact, only 5-15 are iterations are usually good enough in double-precision floating point precision.

>>

Anonymous June 23, 2022 (Thu) 18:40:06 No. 211

File: 8ce0ff601f99fdbf275b63c9b3b1944e.jpg ( 172.93 KB , 1040x1136 , 1656009606752.jpg )

>>198
You touched on what nobody actually mentions, how these functions are computed. Excellent exposition.

>>

Anonymous September 04, 2022 (Sun) 09:48:21 No. 289

File: 1648530726809.png ( 25.95 KB , 644x800 , 1662284901464.png )

gemmy though, thanks for the full info

>>

Anonymous September 04, 2022 (Sun) 21:57:36 No. 292

great thread

File: kotberet.png ( 154.62 KB , 393x403 , 1642718192578.png )

The probability of kot id on /bant/ Anonymous January 20, 2022 (Thu) 22:36:33 No. 106

Reply

>>129

As well as "Jew" or "gay" id on /pol/. I mean it's a three-letter word where second letter can be written as a number (k0t) and first and last letter aren't the same.

Now let's talk about IDs. They are 8-character sequences consisting of base64 characters (numbers, lowercase, uppercase letters, / and +). So, there can be

10 + 26 + 26 + 2 = 64

possible characters for one position in id and thus

K_0 = 64^8 = 2^{48} = 281\ 474\ 976\ 710\ 656

possible IDs (BIG number :o).

Now back, to KOT.
You can write letters k, o and t in uppercase or lowercase and you can also write o as zero. That means you have

2\times3\times2 = 12

combinations of base64 characters that can make up the word "k0T".

Suppose that first three letters of your ID is one of 12 possible K0Ts. We have 5 letters of ID left and they can be anything from base64 alphabet.
So, there are

K_1 = 12 \times 64^5 = 12 \times 2^{30} = 12\ 884\ 901\ 888

So, it's also many, many ids. More than all cats, all humans, but less than all chickens in the world. So, maybe kot ID isn't really that rare.

Also there are 6 positions on which KoT id can happen:
KOTxxxxx
xKOTxxxx
xxKOTxxx
xxxKOTxx
xxxxKOTx
xxxxxKOT

We can assume that there are the same number of ids with KoT on n-th position where

n = 1 ... 6

.

Remember, that KOT id can happen twice, so we must exclude duplicates:
KOTKOTxx
KOTxKOTx
KOTxxKOT
xKOTKOTx
xKOTxKOT
xxKOTKOT

Each of these duplicate positions exists in

K_2 = 12 \times 12 \times 64^2 = 144 \times 2^{12} = 589824

configurations.

Because one duplicate can belong to two sets of IDs with at least one kot id (KOTxxKOT can belong to KOTxxxxx and xxxxxKOT), our equation for number of all KOT ids must be:

6 \times K_1 - 6 \times K_2

Let's calculate it:

6 \times K_1 - 6 \times K_2 =

6 \times 12 \times 2^{30} - 6 \times 144 \times 2^{12} =

72 \times 2^{30} - 864 \times 2^{12} =

77\ 305\ 872\ 384

For every human on this planet there are approximately 10 KOT ids. :o

Now we can calculate probablility of having koT id:

\frac{77305872384}{K_0} = \frac{77305872384}{281474976710656} \approx 0.00027464563 \approx \frac{1}{3641}

Thus, k0t id happens approximately one time for every 3641 ids.
Now you can calculate daily probability, because I don't know if ids on /bant/ are boardwise or just one for every poster and thread. x)

2 posts omitted. Click here to view.

>>

Anonymous January 29, 2022 (Sat) 10:17:18 No. 126 >>127

>>110
can you calculate the probability of "jak" id on /qa/ though

>>

Anonymous January 29, 2022 (Sat) 21:34:57 No. 127

>>126

Assuming that you can write second 'a' of 'jak' as 4 it's the same probability.

Thoughever, you can write it also as 'jaq' or 'jac', so there are 6 possibilities of writing the last letter there are:

2 \times 3 \times 6 = 36

Possible variation of previous constants for JaK:

K_0 = the\ same

K_1 = 36 \times 64^5 = 36 \times 2^{30} = 38\ 654\ 705\ 664

K_2 = 36 \times 36 \times 64^2 = 1296 \times 2^{12} = 5\ 308\ 416

The final equation for the number of jAc IDs:

6 \times K_1 - 6 \times K_2 =

6 \times 36 \times 2^{30} - 6 \times 1296 \times 2^{12} =

216 \times 2^{30} - 7776 \times 2^{12} =

231\ 896\ 383\ 488

And the probability:

\frac{231896383488}{K_0} = \frac{231896383488}{281474976710656} \approx 0.000823861453 \approx \frac{1}{1214}

So it's just pretty much 3 times bigger than "KoT". :o

/qa/ is dead tohough :(

>>

Anonymous January 30, 2022 (Sun) 10:11:07 No. 129 >>131

>>106
Let's eneralize this to id of length

n

(for 4chan:

n=8

) with set of

q

different possible characers (for base64:

q=64

) and substring of interest of length

\ell\ (\ell\leq n)

(for k0t:

\ell=3

) with

p

different variations of the substring (for k0t:

p = 12

)

The number of all possible IDs is:

\qquad K_0 = q^n

If

k0t

appears in at least one place, we must consume

\ell = 3

characters to assemble them into

k0t

, then we return the

k0t

into the bunch (so we now have

n -\ell + 1

items in the bunch) and then we arrange the

k0t

in

n - \ell + 1\choose 1

ways. We have

p

variations of k0t, plus

n-\ell

other characters which can each be

q

different vays (so

q^{n-\ell}

) (even tho I said "plus" we use

\times

bcoz of thing called multiplication principle but w/e)

\qquad K_1 = {n - \ell + 1 \choose 1}\cdot p \cdot q^{n - \ell}

If there are two

k0t

s somewhere, we must consume

2\ell = 6

characters to assemble 2

k0t

s (both with

p=12

variations so

p^2

), then we return the k0t in the bunch (we now have

n -2\ell + 2

items) and we arrange the two kots in

n-2\ell + 2)\choose 2

ways. We have two kots which can each be

p

different ways (so

p^2

) and

n - 2\ell

unconsoomed characters which can each be

q

different ways (

q^{n - 2\ell}

)

\qquad K_2 = {n - 2\ell + 2)\choose 2}\cdot p^2\cdot q^{n - 2\ell}

Therefore if we have

k

k0t

s, the fomula is the following: (and I have a truly marevolous way to prove this formula by induction but I am also truly a lazy fuck.)

\qquad K_k = {n - k(\ell - 1)\choose k}\cdot p^k\cdot q^{n-k\ell}

Now if we want to count all the

k0t

s without duplicates, we must use inclusion-exclusion principle:

\qquad K_{\Sigma} = \sum_{k = 1}^{n - k(\ell - 1) > 0}(-1)^{k - 1}\cdot K_k

Then the probability of

k0t

appearing is

\qquad P = \frac{K_\Sigma}{K_0}

So if I plug in the numbers for

k0t

on /pol/ slash /biz/:

\begin{aligned} \qquad n &= 8\\q &= 64\\\ell&=3\\p&=12 \end{aligned}

We get the following result

\qquad P = \frac{K_\Sigma}{K_0} = \frac{1}{q^n}\sum_{k = 1}^{n - k(\ell - 1) > 0}(-1)^{k - 1}\cdot K_k = \frac{1}{64^8}\cdot\left({8 - 3 + 1\choose 1}\cdot 12\cdot 64^5 - {8 - 6 + 2\choose 2}\cdot 12^2\cdot 64^2\right) = \frac{12\cdot 64^2}{64^8}\cdot(6\cdot 64^3 - 6\cdot 12) = \frac{77305872384}{281474976710656} \approx 0.02746\%

Which matches OP's numbers!

>>

Anonymous January 30, 2022 (Sun) 18:50:14 No. 131

>>129
Nice, I treat the fact that our results are the same as a miracle because I always get something wrong during calculations. x)

Also I wonder if determining number for arbitrary word and

n

(or just equation for

K_k, k \in 1,...,n

) can have an elegant equation or generic brute-force algorithm is the best way.

For example the word SUS is kinda sus, because SUSUSUSx is a valid id with three occurences of the searched word.

We can assume that there are problems when

k

-long suffix of the word is

k

-long prefix like when 'S' is 1-long suffix and prefix of 'SuS' and I see hope there, but words like 'SSUSSUSS' have the same words of length 1,2 and 5 as prefixes and suffixes of respective lengths.

>>

Anonymous September 02, 2022 (Fri) 12:29:27 No. 281

test

File: 1638294292313.png ( 363.02 KB , 1195x1240 , 1640551533576.png )

Anonymous December 26, 2021 (Sun) 20:45:34 No. 73

Reply

>>269 >>271

Supose you have two baskets with one ball each. Probability that one basket has k ball in 0-th time is

p_{0,k}

therefore

p_{0,1} = 1

and for k != 1

p_{0,k} = 0

.

Now (You) choose random ball from any basket (every ball have the same probability of being chosen, not dependent on which basket is it placed in). You put new ball in a basket where this choosen ball was. :0 You repeat this t times.

In math language it is probably:

p_{t,k} = p_{t-1,k-1}\frac{k-1}{t+1} + p_{t-1,k}\frac{t-k}{t+1}

Now

The most bone-chilling, slow-burn, atmosphere-oozing thing I discovered about it:

p_{t,k} = \frac{1}{t+1}

HOLY SCIENCE!!!!!!!!1

Probability of having k balls after t time is always the same!!!!!

5 posts omitted. Click here to view.

>>

Anonymous July 23, 2022 (Sat) 16:58:29 No. 254 >>255

>Probability of having k balls after t time is always the same!!!!!
I agree. And this probability is 1/2. It either happens or it doesn't.

>>

Anonymous July 24, 2022 (Sun) 10:48:48 No. 255

>>254
and now consider the conditional probability of it happening or not.

>>

Anonymous August 27, 2022 (Sat) 16:51:36 No. 269 >>270

>>73
op is a faggot (test)

>>

Anonymous August 27, 2022 (Sat) 16:52:08 No. 270

>>269
Your containment board >>>/test/

>>

Anonymous August 28, 2022 (Sun) 13:11:53 No. 271

>>73
Your post was kind of difficult to read so I will try to summarise the problem and solution in my own words. I also think you made a typo, so my reccurance relation looks very slightly different to yours, but the solution is the same.

Problem

You have two boxes, box #1 and box #2. You also have a bag with an unlimited number of balls. At time 0, box #1 has one ball and box #2 has one ball.

In every iteration:
- Choose a ball at random from the set of balls that are in the boxes.
- Take a new ball from the bag and put it into the box where the chosen ball was.

So at time

t

, there are

t+2

balls in both boxes. Each box has between

1

and

t+1

balls (inclusive).

Let

p_{t,k}

denote the probability that box #1 has exactly k balls at time t.

So by the initial conditions:
-

p_{0,k}=1

if

k=1

-

p_{0,k}=0

otherwise

Derive a closed form expression for

p_{t,k}

.

Solution

We know that at time

t-1

there were

(t-1)+2=t+1

balls in both boxes.

At time

t

, how can box #1 come to contain

k

balls? There are two cases:
- at time

t-1

it had

k-1

balls and was chosen
- at time

t-1

it had

k

balls and was not chosen

This leads to the following reccurence relation

p_{t,k} = p_{t-1,k-1}\frac{k-1}{t+1} + p_{t-1,k}\frac{t+1-k}{t+1}

which has the solution by induction on

t

p_{t,k} = \frac{1}{t+1}

where

1<=k<=t+1

.

Base case for

t=1

is trivial.

p_{t,k}

= p_{t-1,k-1}\frac{k-1}{t+1} + p_{t-1,k}\frac{t+1-k}{t+1}

= \frac{1}{(t-1)+1}\frac{k-1}{t+1} + \frac{1}{(t-1)+1}\frac{t+1-k}{t+1}

by inductive assumption and because

t>=1

so no problems with dividing by 0

= \frac{k-1}{t(t+1)} + \frac{t+1-k}{t(t+1)}

= \frac{k-1+t+1-k}{t(t+1)}

= \frac{t}{t(t+1)}

= \frac{1}{t+1}

which is what we wanted.

Now for some values of

k

, namely

1

and

t+1

this is somewhat an obvious result because in these cases, we have to look at the probability of choosing one box repeatedly which is just

\frac{t!}{(t+1)!}=\frac{1}{t+1}

, but for the other cases it is not so obvious. Drawing a tree diagram helps verify the result, I did it up to

t=3

.

Cool problem overall anon :)

File: Oyu.jpg ( 52.67 KB , 960x720 , 1657222704337.jpg )

Anonymous July 07, 2022 (Thu) 19:38:24 No. 232

Reply

What are some good books to learn symplectic geometry with?

>>

Anonymous July 10, 2022 (Sun) 14:54:27 No. 233

Have you had a look at Abraham and Marsden? Alternatively, Berndt?

>>

Anonymous July 23, 2022 (Sat) 08:50:24 No. 252 >>256

could symplectic geometry replace a classical mechanics course?

>>

Anonymous July 25, 2022 (Mon) 04:36:36 No. 256

>>252
No. You would lack almost all physical understanding of what is happening. It would likely seem very poorly motivated too, if you didn't know where the idea of a symplectic manifold came from and why. I'm sure some people do study symplectic geometry without studying classical mechanics, but it seems pointless to me. And to actually replace a course on classical mechanics (the implication being that this is someone studying physics), I think would be a bad idea.

File: dbqOJ.gif ( 239.17 KB , 1838x1304 , 1658138309873.gif )

One Gaussian dominates another when I fit them in a natural way Anonymous July 18, 2022 (Mon) 09:58:29 No. 241

Reply

The picture below shows (in red) a sum of Gaussian kernels with different means, but all having variance 1, i.e.
f(x)=12π−−√∑i=1ne−12(x−μi)2 .
. The green curves show what I get if I select a point x0, and then "fit" a scaled Gaussian distribution g to f about x0, in the sense that I determine c and μ0 in the function
g(x):=c2π−−√e−12(x−μ0)2
so that g(x0)=f(x0) and g′(x0)=f′(x0). From numerical experimentation, it seems to always be the case that g(x)≤f(x) for all x. Does anyone know why this is so?

File: IMG_20220519_220921.jpg ( 3.11 MB , 4000x3000 , 1653047795771.jpg )

Anonymous May 20, 2022 (Fri) 11:56:36 No. 177

Reply

Are there mathemathics i can use to fix the silhouette of this image? like the back tip of the helmet?
to make it look "correct"?

File: image.png ( 126.74 KB , 2048x2048 , 1643699011807.png )

lambda calculus Anonymous February 01, 2022 (Tue) 07:03:32 No. 134

Reply

>>138 >>174

does anyone have good resources to learn the notation? i could figure out basic things like

((λ x . x) (λ a . a))

but got filtered by

(((λ f . (λ x . (f x))) (λ a . a)) (λ b . b))

>>

Anonymous February 08, 2022 (Tue) 03:34:55 No. 138 >>139

>>134
What's your background

>>

Anonymous February 08, 2022 (Tue) 04:14:19 No. 139 >>141

>>138
mathematics, i managed to figure out the notation by fucking around with emacs lisp

>>

Anonymous February 08, 2022 (Tue) 05:58:04 No. 141

File: da9ffae136e360d515303cbbb8af99e1.jpg ( 350.26 KB , 744x1052 , 1644299884658.jpg )

>>139
https://www.mscs.dal.ca/~selinger/papers/papers/lambdanotes.pdf

>>

Anonymous May 05, 2022 (Thu) 15:53:43 No. 174

>>134
Usually repeated abstraction is notated as just
lam f x. ...
instead of
lam f. lam x. ...
Lambda calculus isn't usually used in practice.
Not sure what you want to use it for.

File: 00000.jpg ( 81.33 KB , 289x293 , 1649020837918.jpg )

Anonymous April 03, 2022 (Sun) 21:20:37 No. 159

Reply

>>162

how do i start learning advanced math, ahead of my classes? any good textbooks/sources? what is your advice?

>>

Anonymous April 14, 2022 (Thu) 22:59:12 No. 162

File: orthocenter.png ( 329.63 KB , 1521x840 , 1649977152884.png )

>>159
Mathematics at an advanced level diverges into many different topics, each something you could study for years. What do you want to know more about?

>>

Anonymous April 25, 2022 (Mon) 11:36:55 No. 166

File: 1636870494174.jpg ( 202.36 KB , 2048x2048 , 1650886614985.jpg )

If you are just saying something as vague as "learning advanced math", then we can only tell you to google good introductory textbooks on various subjects. Maybe first open up Wikipedia and just check different genres there and see if something clicks and then find intro books on that!

>>

Anonymous May 03, 2022 (Tue) 15:39:31 No. 172

A graphical approach to Algebra and Trigonometry basically gets you into Calculus

File: Tensegrity_simple_3_RL.png ( 10.89 KB , 384x256 , 1647295247452.png )

Prime factorization. Mathew White March 14, 2022 (Mon) 22:00:47 No. 145

Reply

The brain dump of a managerie of often contracdictory and fantastic characters, from a mossad member turned professor, to an indian student turned part time street criminal, a homeless poet in philly, an ex-skin head, a former chinese investment banker, the owner of a multimillion dollar startup, two military veterans, and a cast of other folks.

The result of a a simple discovery that a rich set of algebraic identities underlay the product of large primes.

For example, a series of variables related to n unknown value, d4, easily derived from known variables. d4a, d4u, d4z, d4H, etc.
Or c/d4, neither known by themselves, but the ratio of which is easily found, and the *product* of which, cd4, is the ratio of our product's factors, b/a.

Be warned the code is thick, mostly written by an indian guy, and translated by the rest of us over time. It is 198 pages of dense work, defining so many variables the original authors resorted to greek letters and elements from the periodic table for naming.

Good luck.

https://pastebin.com/Ad46Awcp

>>

Anonymous March 21, 2022 (Mon) 03:22:41 No. 154

>list of primes
>Miller test (deterministic Miller-Rabin)
>a bunch of tests and variables I have no reason to memorize, followed by a schizo string list including elements of the periodic table..
>...
I'd like to converse with you about your findings, but I have no paper, no lemmas and no theorems.

>>

Some observations Anonymous March 24, 2022 (Thu) 23:39:42 No. 155

"I'd like to converse with you about your findings, but I have no paper, no lemmas and no theorems. "

It appears there is no lemma or theorem and OP is working at best in 8th grade math.

I'm just a lowly front end developer but I'll take a crack at it.

All the key variables are on two apparent slopes, d4, and 'gold'.

I could see getting estimates on unknown factors by simply taking known products with common variables (d4a, d4u, d4z, d4H, etc), and then finding the smallest of the set, on the basis that would tightly constrain the upper and lower bound of the factors of those variables, but that assumes none of the unknown factors like u, d4, c, z, or H, are less than 1.

I also wrote some functions for converting variables between the gold line, the _gold line, and the d4 line.

def _gold_to_d4(var):
return (var**2)*Hc

def d4_to__gold(var):
return (var/Hc).sqrt()

This followed from the observation that

d4u/Hc/G
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')

(_goldc**2/G)
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')

meaning

d4u/Hc == some _gold**2

Shows d4 and _gold series are interconnected.

Taking any _gold, multiplying by the power of 2, and then multiplying by Hc should give
a d4 variable.

While dividing a d4 variable by Hc, and then rooting it, should give a _gold series variable

if d4 is inverted (less than 1), then (((_goldc)**2)*Hc)/d4 > d4u
To test this I ran the script a few times with various factors.

a
Decimal('84780491')
b
Decimal('7022437637')

>>> (((_goldc)**2)*Hc)/d4 > d4u
True

>>> a
Decimal('247017779')
>>> b
Decimal('8552169383')

(((_goldc)**2)*Hc)/d4 > d4u
True

>>> a
Decimal('39420301')
>>> b
Decimal('8490912585839')
(((_goldc)**2)*Hc)/d4 > d4u
False

As you can see with the above example, both c and d4 are normal, but c is less than d4,
so _cd4 is less than 1, and the outcome is of course false in this case.
Good to know.

And what this means is that when _cd4 is less than 1, Post too long. Click here to view the full text.

>>

Anonymous March 25, 2022 (Fri) 00:07:22 No. 156

Making sense of all of what looks like dead code, the elListI() function appears to list elements based on some mysterious "I" value.
The elements that have matching I values tend to share lots of identities. Thats all I could tell about it though.

>>

Anonymous April 01, 2022 (Fri) 09:59:56 No. 158

i love primes bro