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/math/ - Mathematics

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25 Dec 2021Mathchan is launched into public
Catalog Archive

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

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  No. 232
Reply
What are some good books to learn symplectic geometry with?
>>
  No. 233
Have you had a look at Abraham and Marsden? Alternatively, Berndt?
>>
  No. 252 >>256
could symplectic geometry replace a classical mechanics course?
>>
  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 )

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  No. 241
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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 )

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  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 )

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  No. 134
Reply
>>138 >>174
does anyone have good resources to learn the notation? i could figure out basic things like
((λx.x)(λa.a))((λ x . x) (λ a . a))

but got filtered by
(((λf.(λx.(fx)))(λa.a))(λb.b))(((λ f . (λ x . (f x))) (λ a . a)) (λ b . b))
>>
  No. 138 >>139
>>134
What's your background
>>
  No. 139 >>141
>>138
mathematics, i managed to figure out the notation by fucking around with emacs lisp
>>
  No. 141

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

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

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  No. 159
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>>162
how do i start learning advanced math, ahead of my classes? any good textbooks/sources? what is your advice?
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  No. 162

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

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>>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?
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  No. 166

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

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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!
>>
  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 )

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  No. 145
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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
>>
  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.
>>
  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.
>>
  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.
>>
  No. 158
i love primes bro

File: image.png ( 736.33 KB , 800x647 , 1643053076107.png )

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  No. 113
Reply
>>115 >>114
What the heck is a Euclid distance? Why is it different from the Minsk distance and Cosine distance?
If i have a matrix as so, i'm expected to make all three distance matrices from it

X=[02203151] X = \begin{bmatrix} 0 & 2 \\ 2 & 0 \\ 3 & 1 \\ 5 & 1 \end{bmatrix}


Also this probably has something to do with statistics idk
>>
  No. 114 >>116
>>113
Cosine similarity is
∑k=1nxkyk∑k=1nxk2⋅∑k=1nyk2\frac{\sum_{k=1}^n x_ky_k}{\sqrt{\sum_{k=1}^n x_k^2}\cdot\sqrt{\sum_{k=1}^n y_k^2}}


Minkowski distance is
(∑k=1n∣xk−yk∣p)1p\left(\sum_{k=1}^n|x_k - y_k|^p\right)^\frac{1}{p}


Euclidean distance is a speacial case of Minkowski distance when
p=2p=2
i.e.
∑k=1n(xk−yk)2\sqrt{\sum_{k=1}^n (x_k - y_k)^2}



Your matrix is
XX
means:
[x1y1x2y2x3y3x4y4]\begin{bmatrix} x_1 & y_1\\x_2 & y_2\\x_3 & y_3\\x_4 & y_4\\\end{bmatrix}
>>
  No. 115

File: image.png ( 891.54 KB , 707x1000 , 1643060754913.png )

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>>113
I'm a dumbass, it's actually minkowski distance

Anyways, for the euclidian distance, apparently you do:
(x1−y1)2+(x2−y2)2\sqrt{ (x_1-y_1)^2+(x_2-y_2)^2}

or for those playing along in calc, 
=SQRT((B$3-B3)^2+(C$3-C3)^2)
and apply to the appropriate cells.

[0.02.83.25.12.80.01.43.23.21.40.02.05.13.22.00.0] \begin{bmatrix} 0.0 & 2.8&3.2&5.1\\ 2.8&0.0&1.4&3.2\\ 3.2&1.4&0.0&2.0\\ 5.1&3.2&2.0&0.0 \end{bmatrix}


Since i was working in calc, the table containing the data spans over the 
B3:C6
range

Then, for the minkowski distance, there's something called the P value, and if it's equal to 1, then it becomes the manhattan distance, so in this example the equation would be
∣x1−y1∣ + ∣x2−y2∣|x1-y1|\ +\ |x2-y2|

and the corresponding calc formula for a particular cell (i'm not saying which one hehe) 
=(($B$3-$B3)^1+($C$3-$C3)^1)^(1/RIGHT($E$14,1))


[00240024−2−202−4−4−20]\begin{bmatrix} 0&0&2&4\\0&0&2&4\\-2&-2&0&2\\-4&-4&-2&0 \end{bmatrix}



And for the cosine similarity or distance idk, just do a vector dot product
x⋅y∣x∣⋅∣y∣\frac{x \cdot y}{|x|\cdot|y|}

This time it was a bit more difficult in calc, and the formula for the left upper cell is 
=SUMPRODUCT(B$3:C$3,B3:C3)/(SQRT(SUMSQ(B$3:C$3))*SQRT(SUMSQ(B3:C3)))


[1.000.000.320.200.001.000.950.980.320.951.000.990.200.980.991.00]\begin{bmatrix} 1.00&0.00&0.32&0.20\\0.00&1.00&0.95&0.98\\0.32&0.95&1.00&0.99\\0.20&0.98&0.99&1.00\\ \end{bmatrix}



Cool.
>>
  No. 116
>>114
Thanks for the info. I probably made some mistake here but it's probably good

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