/sci/ - Science & Math » Thread #12192077

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Super math

Anonymous Sun 04 Oct 11:20:16 2020 No.12192077 View Reply Original Report

Quoted By: >>12192088 >>12192093 >>12192098 >>12192108 >>12192116 >>12192211 >>12192214

Do any of you have any experience with super mathematics?

It's described as transcendent functions that have been integrated or producted infinitely.

Anonymous

Anonymous Sun 04 Oct 2020 11:24:01 No.12192088 Report

Quoted By:

>>12192077

Anonymous

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Anonymous Sun 04 Oct 2020 11:25:06 No.12192093 Report

Quoted By: >>12194935

>>12192077
Forgot obligatory pict

Anonymous

Anonymous Sun 04 Oct 2020 11:28:43 No.12192098 Report

Quoted By:

>>12192077
These are just a few steps I used using super calculus to create a novel integral with a trailing 5.

Anonymous

Anonymous Sun 04 Oct 2020 11:32:09 No.12192108 Report

Quoted By: >>12192112 >>12192113

>>12192077
Yes try $^ \inftyx$ . It's equal to y=x^y. You can plot it

Anonymous

Anonymous Sun 04 Oct 2020 11:33:22 No.12192112 Report

Quoted By:

>>12192108
* $^\infty x$

Anonymous

Anonymous Sun 04 Oct 2020 11:33:45 No.12192113 Report

Quoted By: >>12192120

>>12192108
https://fractional-calculus.com/super_integral.pdf
Here are some basics on the super integral. It's great lead into probability space.

Anonymous

Anonymous Sun 04 Oct 2020 11:36:26 No.12192116 Report

Quoted By: >>12192123 >>12192414

>>12192077
Ive done horseshit math before. Does that count.

Anonymous

Anonymous Sun 04 Oct 2020 11:39:32 No.12192120 Report

Quoted By:

>>12192113
There's a smoothing method that uses a super integral. the super integral from $x-\delta$ to $x+\delta$ of f(x) smooths a function.

Anonymous

Anonymous Sun 04 Oct 2020 11:40:15 No.12192123 Report

Quoted By: >>12192132 >>12192134

>>12192116
I'll teach you some super math (algebra) that'll out pace a computer. You'll no longer know horseshit math.
How many numbers are in 2 to the power of 100,000,000,000,000,000?

Anonymous

Anonymous Sun 04 Oct 2020 11:43:03 No.12192132 Report

Quoted By: >>12192136

>>12192123
100,000,000,000,000,000 ln(2) / ln(10)

Anonymous

Anonymous Sun 04 Oct 2020 11:43:19 No.12192134 Report

Quoted By:

>>12192123

The answer is 693147180559945309 numbers.

Anonymous

Anonymous Sun 04 Oct 2020 11:44:33 No.12192136 Report

Quoted By: >>12192142

>>12192132
That's close?

Anonymous

Anonymous Sun 04 Oct 2020 11:47:34 No.12192141 Report

Quoted By: >>12192145 >>12192405 >>12192541

Alright here's a tougher question. When does 2^x equal e^2

Keep in mind the like x->inf for (2+1)^(1/x)/2^x is the differential that equals e^1/2

Anonymous

Anonymous Sun 04 Oct 2020 11:48:01 No.12192142 Report

Quoted By:

>>12192136
floor(100,000,000,000,000,000 ln(2) / ln(10)) + 1

Anonymous

Anonymous Sun 04 Oct 2020 11:48:35 No.12192145 Report

Quoted By:

>>12192141
Lim x->inf (2+1/x)^(1/x)/2^x -.-

Anonymous

Anonymous Sun 04 Oct 2020 11:51:00 No.12192157 Report

Quoted By:

https://www.wolframalpha.com/input/?i=Lim+x-%3Einf+%282%2B1%2Fx%29%5Ex%2Flog%28e%5E2%29%5Ex

A little algebra and you can see what 2^x equals at infinity

Anonymous

Anonymous Sun 04 Oct 2020 11:52:57 No.12192160 Report

Quoted By:

Remember the rule that logs cancel out at the limit of infinity

Anonymous

Anonymous Sun 04 Oct 2020 12:00:48 No.12192173 Report

Quoted By: >>12192176

The reason it's important is for predicting the value of 3^x and 4^x... Etc at their infinity limits. If you multiply these limits together you can see geometric gamma space. Gamma(x)^x.

Like 1/(1-x) expressing infinity polynomials we have identities for gamma space that take on algebraic expression.

The consequences of this are having a useful meaning to e^(1/gamma(x)).

Anonymous

Anonymous Sun 04 Oct 2020 12:03:31 No.12192176 Report

Quoted By: >>12192177

>>12192173
For the limit of x->inf for 2^x*3^x*4^x...n^x we get the lim->x for gamma(x)^x

Here's another hint: the limx->inf for (gamma(x)+1/x)^x/(gamma (x)^x is e^(1/gamma(x))

Anonymous

Anonymous Sun 04 Oct 2020 12:04:59 No.12192177 Report

Quoted By: >>12192188

>>12192176
You'll see an identity that's a large product of fractions of e that equals e to the power of a fraction of gamma.

E.g. e^(1/gamma(x))

Anonymous

Anonymous Sun 04 Oct 2020 12:08:40 No.12192188 Report

Quoted By:

>>12192177
Like I said before it's much like how 1/1-x can represent a huge series of polynomials.

e^(1/gamma(x)) represents a large series of products taken to their limit.

Anonymous

Anonymous Sun 04 Oct 2020 12:16:48 No.12192198 Report

Quoted By: >>12192201

Once you realize the size of the limit of x as it goes towards infinity for gamma(x)^x you can see the difference in size for the limit as x^x.

Protip: the limit of x^x is a step function at infinity. It changes from a limit similar to 10^x to infinity to a large number that resembles e^2. It's next step goes to e^8. It's next step to e^81.

Anonymous

Anonymous Sun 04 Oct 2020 12:18:45 No.12192201 Report

Quoted By:

>>12192198
That is x^x is a step function for every differential near any whole number. This is why it cannot be integrated naturally.

Anonymous

Anonymous Sun 04 Oct 2020 12:25:04 No.12192211 Report

Quoted By:

>>12192077
You should really be asking yourself what the hell does (e-1)^x^2 have to do with x^x as it's limit goes to infinity. Why the perfect repeating number?

Anonymous

Anonymous Sun 04 Oct 2020 12:27:05 No.12192214 Report

Quoted By:

>>12192077
If you're wondering why 50.5^x is there, it's because it's the sum from 1 to 100.

Anonymous

Anonymous Sun 04 Oct 2020 12:29:17 No.12192221 Report

Quoted By:

Is super math working out??

Anonymous

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Anonymous Sun 04 Oct 2020 12:31:51 No.12192226 Report

Quoted By:

Maybe we need some examples here

Anonymous

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Anonymous Sun 04 Oct 2020 12:32:52 No.12192227 Report

Quoted By:

Anonymous

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Anonymous Sun 04 Oct 2020 12:34:33 No.12192234 Report

Quoted By:

In case you didn't get the identity

Anonymous

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Anonymous Sun 04 Oct 2020 12:35:40 No.12192237 Report

Quoted By:

In case you didn't get the identity...

Anonymous

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Anonymous Sun 04 Oct 2020 12:38:19 No.12192241 Report

Quoted By:

So predictable

Anonymous

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Anonymous Sun 04 Oct 2020 12:40:58 No.12192242 Report

Quoted By:

Remember that it's simplifying the fraction

Anonymous

Anonymous Sun 04 Oct 2020 12:43:40 No.12192244 Report

Quoted By: >>12192245

Eventually the computer breaks and it cannot jump to the next conclusion that this equals e^(1/gamma(x))

Anonymous

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Anonymous Sun 04 Oct 2020 12:44:46 No.12192245 Report

Quoted By: >>12192258

>>12192244
Notice that the super computing cloud has failed us.

Anonymous

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Anonymous Sun 04 Oct 2020 12:46:20 No.12192248 Report

Quoted By:

I broke wolfram

Anonymous

Anonymous Sun 04 Oct 2020 12:52:54 No.12192258 Report

Quoted By: >>12192269

>>12192245
So we can see the limit of x as it goes to infinity for (gamma(x) +1/x)^x is equal to gamma(x)^x(e^(1/gamma (x)))

Anonymous

Anonymous Sun 04 Oct 2020 12:57:07 No.12192269 Report

Quoted By:

>>12192258
Simple algebra shows us the first step in mapping geometric gamma.

(Gamma(x)+1/x)^x/e^(1/gamma(x))=gamma(x)^x

Anonymous

Anonymous Sun 04 Oct 2020 12:58:15 No.12192273 Report

Quoted By:

Anyone following this??

Anonymous

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Anonymous Sun 04 Oct 2020 13:02:05 No.12192283 Report

Quoted By: >>12194841 >>12194967

Kryptonian math lol

Anonymous

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Anonymous Sun 04 Oct 2020 13:03:05 No.12192289 Report

Quoted By: >>12192297 >>12192307 >>12194841 >>12194967

Hehehe

Anonymous

Anonymous Sun 04 Oct 2020 13:06:18 No.12192297 Report

Quoted By:

>>12192289
The best part is if you ask wolfram if those two equations equal each other it just checks out

Anonymous

Anonymous Sun 04 Oct 2020 13:09:12 No.12192307 Report

Quoted By:

>>12192289
I had to solve n=np to figure this out guise. Common.

Anonymous

Anonymous Sun 04 Oct 2020 13:10:19 No.12192312 Report

Quoted By: >>12192319 >>12192354

I'm sure you weren't thinking about it but this directly related to solving the Riemann zeta hypothesis.

Anonymous

Anonymous Sun 04 Oct 2020 13:11:30 No.12192319 Report

Quoted By:

>>12192312
I'm not posting more until someone responds

Anonymous

Anonymous Sun 04 Oct 2020 13:15:44 No.12192325 Report

Quoted By:

We have touched algebraic expressions that touch on geometric probability. But we can go so much higher.

Anonymous

Anonymous Sun 04 Oct 2020 13:17:01 No.12192331 Report

Quoted By:

We need to cover various infinity product before we get to super geometric space.

Anonymous

Anonymous Sun 04 Oct 2020 13:21:03 No.12192339 Report

Quoted By: >>12192340 >>12192343

First I'd like you to consider that 2*2 is 2^2.

Then that (2*2*2) is 2^3.

Then I'd like you to consider (2*2)(2*2*2) is 2^(2+3).

Then I'd like you to consider the infinite product of 2 is 2^1*2^2*2^3*2^4....+2^n

This equals 2^(1+2+3+4...n)
This will translate into 2^((n^2+n)/2)

Anonymous

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Anonymous Sun 04 Oct 2020 13:22:49 No.12192340 Report

Quoted By:

>>12192339
So we look deep and try to figure what is .... Super gamma?

Anonymous

Anonymous Sun 04 Oct 2020 13:24:52 No.12192343 Report

Quoted By:

>>12192339
For those that don't remember... Which sum turns into an integral if it's (n^2+n)/2

Anonymous

Anonymous Sun 04 Oct 2020 13:28:15 No.12192354 Report

Quoted By: >>12192359

>>12192312
So we have a series of fractions 1 + 1/2^n + 1/3^n... 1/^n in which all numbers to their power of infinity recycle at e^n.

So that, in large numbers, we find 1 +1/e^2 + 1/e^3...+1/e^n

Anonymous

Anonymous Sun 04 Oct 2020 13:30:08 No.12192359 Report

Quoted By: >>12192362

>>12192354
I'm sorry I'm working on the limit to riemenn zeta. From there we can work backwards a bit

Anonymous

Anonymous Sun 04 Oct 2020 13:32:12 No.12192362 Report

Quoted By: >>12192364

>>12192359
From the first power which is the sum of all fractions we get the simple number (albeit very large) ln(10)

Anonymous

Anonymous Sun 04 Oct 2020 13:33:14 No.12192364 Report

Quoted By: >>12192368

>>12192362
It is an infinitely huge number but it's magnitude is not.

Anonymous

Anonymous Sun 04 Oct 2020 13:35:32 No.12192368 Report

Quoted By: >>12192382

>>12192364
What isn't described in modern mathematics is the use of an infinite sum or product to describe the power of an infinite number. So long as the algebra of an infinite sum and infinite product hold we can put precise values on hyper large numbers

Anonymous

Anonymous Sun 04 Oct 2020 13:41:43 No.12192382 Report

Quoted By:

>>12192368
By uhh using those equations to describe magnitudes.

You can thank superman later.

Anonymous

Anonymous Sun 04 Oct 2020 13:51:53 No.12192405 Report

Quoted By:

>>12192141
>graphs intersection
Pshhhh nothin personal kid

Anonymous

Anonymous Sun 04 Oct 2020 13:55:27 No.12192414 Report

Quoted By:

>>12192116
What's your definition of "not Horseshit" math

Anonymous

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Anonymous Sun 04 Oct 2020 15:00:44 No.12192541 Report

Quoted By: >>12193295 >>12194793 >>12194835 >>12194898

>>12192141
$2^x = e^2$
$x \ln 2 = 2$
$x = \frac{2}{\ln 2}$

Anonymous

Anonymous Sun 04 Oct 2020 19:33:46 No.12193295 Report

Quoted By: >>12193342

>>12192541
Great now integrate 2^x

Anonymous

Anonymous Sun 04 Oct 2020 19:49:39 No.12193342 Report

Quoted By: >>12194793 >>12194835

>>12193295
$\int 2^x dx = \frac{2^x}{\ln 2}$

Anonymous

Anonymous Mon 05 Oct 2020 05:45:26 No.12194793 Report

Quoted By:

>>12193342
>>12192541
These are related

Anonymous

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Anonymous Mon 05 Oct 2020 06:03:55 No.12194835 Report

Quoted By:

>>12192541
>>12193342
Common ppl does it take superman to point this out

Anonymous

Anonymous Mon 05 Oct 2020 06:05:57 No.12194841 Report

Quoted By:

>>12192283
>>12192289
This isn't just a random identity

Anonymous

Anonymous Mon 05 Oct 2020 06:17:57 No.12194864 Report

Quoted By:

How the fuck are you not interested in super math

Anonymous

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Anonymous Mon 05 Oct 2020 06:28:27 No.12194898 Report

Quoted By:

>>12192541
2^x equals e^2 at it's integral... Common now

Anonymous

Anonymous Mon 05 Oct 2020 06:37:22 No.12194925 Report

Quoted By:

Hey someone else contribute

Anonymous

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Anonymous Mon 05 Oct 2020 06:41:12 No.12194935 Report

Quoted By: >>12194964 >>12194973

>>12192093
Legit
But how does he know what that exact jar weighs? Including the lid. He'd be off by a bit.

Anonymous

Anonymous Mon 05 Oct 2020 06:52:31 No.12194964 Report

Quoted By:

>>12194935
:/

Anonymous

Anonymous Mon 05 Oct 2020 06:53:54 No.12194967 Report

Quoted By:

>>12192283
>>12192289
Common this alone is worth a million

Anonymous

Anonymous Mon 05 Oct 2020 06:56:01 No.12194973 Report

Quoted By: >>12194987

>>12194935
>But how does he know what that exact jar weighs? Including the lid.
That's what makes it super-mathematics.

Anonymous

Anonymous Mon 05 Oct 2020 07:01:39 No.12194987 Report

Quoted By:

>>12194973
:/

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