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/sci/ - Science & Math

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9319219 No.9319219 [Reply] [Original] [archived.moe]

>be some Indian who shits on the streets
>teach yourself math for shits and giggles
>ignore the rules, do a bunch of operations on infinity
>get pic related
>be hailed as a math genius

Is this real life?

>> No.9319225

ramanujan was neither the first nor last to come up with that sum.
there is also literally nothing wrong with analytical continuation. It is, as far as anyone knows, a perfectly consistent mathematical tool.

>> No.9319234

Nobody actually thinks that formula is true.
Zeta(s) is only defined by the sum for re(s)>1.
Zeta(-1) is obtained by analytic continuation.
Ur just mad because an Indian from 100 years ago with limited resources was way smarter than you will ever be.

>> No.9319247

>there is also literally nothing wrong with analytical continuation. It is, as far as anyone knows, a perfectly consistent mathematical tool.

Except it comes up with incorrect conclusions, such as the one provided in the OP.

>> No.9319258
File: 254 KB, 1300x3900, math trench.jpg [View same] [iqdb] [saucenao] [google] [report]

you forgot:
>die of dysentery

>> No.9319261
File: 54 KB, 197x259, leface.png [View same] [iqdb] [saucenao] [google] [report]


>> No.9319268

incorrect how? because you don't like it? because there is some real world example that contradicts it? because what? none of that matters, stop being autistic. Math is nothing more than playing around with axioms. If those axioms don't lead to contradictions, then it's a perfectly valid mathematical model. Sometimes those models resemble a real world situation closely enough and they become useful, but that has no bearing on whether it's valid.

>> No.9319284
File: 4 KB, 270x186, download.png [View same] [iqdb] [saucenao] [google] [report]

OP take complex analysis.

>> No.9319289

Just look at the geometric series 1+x+x^2.... It obviously only makes sense for x within the unit circle and converges to 1/(1-x). But 1/(1-x) is well defined for x outside of the unit circle as well. You can get sums like 1+2+4+8...= -1 which also seem absurd. Then you learn about doing subtraction via two's complement and things seem less spooky.

>> No.9319321

There's nothing incorrect about it.

>> No.9319330
File: 55 KB, 617x347, 1509035736738.png [View same] [iqdb] [saucenao] [google] [report]

>performing arithmatic with infinity
>performing arithmatic with infinity and expecting an answer in finite time
>performing truncated assumptions with a lazy implementation of infinity and then pretending you did arithmatic at all

>> No.9319336

Could quantum mechanics be a continuation of an incorrect axiom that has no bearing on real life and our observations are simply limited by hard?

>> No.9319338

That's a dumb way to attempt to prove it, implying it's standard math. this is like proving 1=.999... by dividing 1 by 3. you can't group an infinite series arbitrarily and get constant results. You also can't perform arithmetic operations on them and maintain a consistent system. You HAVE to start by modifying your model to one where infinite summations are well defined.
tl;dr: fuck off numberphile

>> No.9319340

>performing arithmatic with infinity
Literally nothing wrong with that except for the part where you're a brainlet who doesn't know how to spell the word "arithmetic."

>> No.9319344

>That's a dumb way to attempt to prove it
That's the way Ramanujan did it, by relating it back to that alternating series so he could use its already known 1/4 value. You're pretty retarded if you think you're smarter than he was desu.

>> No.9319346

okay enjoy doing fake math to the benefit of no one, retard.

>> No.9319355


>> No.9319370

and you're pretty retarded if you think an argument with nothing but an appeal to authority backing it is any argument at all, even assuming what you said was correct.
Even if I followed the retarded logic that because he was smart everything he said is infallible, you're still wrong. Ramanujan clearly stated "under my theory" (referring to Ramanujan summation), doing exactly what I had just said you have to do. Even assuming you're right, you're still wrong.

>> No.9319381

It wasn't an argument for the sum working, it was an argument for why you're pretty retarded for calling Ramanujan's summation dumb. I wouldn't need to point out his intelligence in the first place if you didn't make the retarded decision to call his derivation approach dumb.

>> No.9319390
File: 4 KB, 156x232, mathBad.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9319401

I didn't call Ramanujan summation dumb, I called your counter-argument to >>9319268 dumb. Again, what you showed is only correct when you work in a different system. If you present it as proof for any standard system were infinite sums are not well defined, you're retarded. Ramanujan clearly knew that
>But as a fact I did not give him any proof but made some assertions as the following under my new theory. I told him that the sum...=-1/12 under my new theory
>I wouldn't need to point out his intelligence in the first place
You don't need to call it out at all, no matter what I do, because an appeal to authority isn't an argument. Yet here is your retarded ass doing it again. It doesn't matter how smart he is, he can still be wrong. He wasn't in this case, so you think in that tiny little brain of yours that your argument makes sense, but what if we were talking about his "number of primes under x" formula? You're the worst kind of retard. The kind that memorizes random things (or just googles them) in order to appear intelligent yet has no idea about what's actually going on.

>> No.9319403

The Ramanujan summation doesn't make the mistake that picture depicts.

>> No.9319408

what mistake?
it's my theory so fuck off

>> No.9319413

>I called your counter-argument to >>9319268 dumb.
What? First of all that isn't my post, second of all why are you talking about a counter-argument to it when it was an argument against one of your posts, and third you didn't even reply to that post when you wrote "that's a dumb way to attempt to prove it," you replied to one of my posts, and the way you were calling "dumb" was in fact Ramanujan's way of deriving that value.

>> No.9319418


-1/12 is the y intersection of 1 + 2 +... when transformed into a smooth curve. the transformation allows divergent series to be assigned values other than infinity.

all of this is from fucking wikipedia. just read wikipedia you lazy fucks.

>> No.9319422

That's not the only way to get to that -1/12 value.

>> No.9319437

woops, I meant counter-argument to >>9319247
I clicked on the wrong reply
>and the way you were calling "dumb" was in fact Ramanujan's way of deriving that value
No because Ramanujan wasn't trying to prove to anyone that the sum is -1/12. Again,
>But as a fact I did not give him any proof but made some assertions as the following under my new theory.
He wasn't proving anything. He was clearly talking about a different system and once you accept his axioms that sum trivially follows. So "proving" to someone that the sum is -1/12 by using >>9319321 is dumb since the issue is clearly not that they don't know arithmetic, but that they are working under a different system.
And for the millionth fucking time, none of that even matters because
>durr that's how Ramanujan did it
is not an argument.

>> No.9319443


that is the way you're supposed to get it, not that numberphile bullshit

>> No.9319454

you mean a complex curve, smooth curves aren't rigid like that

>> No.9319459

I'm really sure that's not how ramanujan did it. this is just retarded

>> No.9319461

Well the mathematical models of quantum mechanics produces consistent predictions of what will happen in different situations. Pretty low chance that the models will be wrong but make the correct predictions.

>> No.9319463

>haha wow check out this AMAZING math trick and the squiggly lines and then oh man cut all this shit out and paste this stuff in and use the little magic pixie dust WOW look at this amazing result
I hate numberphile so much

>> No.9319464

street shitters invented your number system, R haplo rapebaby.

>> No.9319468

doesn't look complex to me


>> No.9319473

smooth curves aren't rigid like that. it has much more properties than just "a smooth curve", which could give literally any value.

>> No.9319486


>> No.9319488

Numberphile's explanation of this was completely retarded. It didn't explain shit, and actually implied that was the correct solution to the sum of all natural numbers. I guess to serve as clickbait? By far their worst video that I've seen.

>> No.9319514

better tell Wikipedia then that they have been gaslighted by the Illuminati

>> No.9319664

Wrong. That is absolutely retarded and a completely unacceptable proof.
The first line alone would get you laughed out of any University, since you obviously do not understand even the basics of analysis.

>> No.9319850
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>> No.9319866

somtimes brady spontaneously becomes a brainlet

>> No.9319899
File: 59 KB, 645x729, 1507205929387.png [View same] [iqdb] [saucenao] [google] [report]

It is really fucking weird that >>9319321 just happens to get the same answer. This can't be just a coincidence. This is me by the way >>9319850 . No shit you can't manipulate infinite series like that. But how does making that mistake produce the same answer? That is weird as fuck, actually. Really fucking weird. Why is not interested in THAT?

>> No.9319962

The [math]-\frac1{12}[/math] meme is dead, anon.

>> No.9320109

Because if you use invalid operations, you can get anything you want.

>> No.9320376

Now write that Argument without those ... more formally.

[math]\sum\limits_{n=1}^{\infty} n = \lim_{m\to\infty} {\left(\sum\limits_{n=1}^{m} n\right)} = \infty[/math]

>> No.9320652

Adding/Substracting/Multiplying/Dividing infinite sums is only defined if the limit is a real number (no, infinity is not a number)

Try again, sweetie

>> No.9320665

>Nobody actually thinks that formula is true.
It's not more or less true than a statement like

e^3 > 4

Those are mathematical statements that depend on a context.

>Zeta(s) is only defined by the sum for re(s)>1.
This is technically false, as the analytical continuation is part of the zeta function definition. The infinite sum of reals is only defined for those numbers, however.

Finally, 1+2+3+...=-1/12 isn't just true in the analytical continuation context, there are several theories that lead there (like Ramanujan summation, not apriori tied to analytic continuation at all)

>> No.9320837
File: 46 KB, 600x630, 1509575677092.jpg [View same] [iqdb] [saucenao] [google] [report]

Let us define the decimal of 1/3 as small s

(A) [s×1] 0.333r3 × 3 = 0.999r9
(B) [s×2] 0.666r6 × 3 = 1.999r8
(C) [s×3] 0.999r9 × 3 = 2.999r7
(D) [s×4] 1.333r2 × 3 = 3.999r6
(E) [s×5] 1.666r5 × 3 = 4.999r5
(F) [s×6] 1.999r8 × 3 = 5.999r4
(G) [s×7] 2.333r1 × 3 = 6.999r3
(H) [s×8] 2.666r4 × 3 = 7.999r2
( I ) [s×9] 2.999r7 × 3 = 8.999r1

Upon truncating the repeating pattern in attempts to define the infinite sequence in finite time, we develop rounding errors within the decimal beyond (A) .
It can be agreed that (A) is closer to 1 than (B) is closer to 2, and how both are closer to their assumed ceiling than (C) is to 3. As we continue all the way to s×9, the rounding error now gives 8.999r1, which is 8 orders of infinity further from 9 than (A) is from 1.

The truncation process assumes an upper limit of infinity. The action is made even more futile when each increment of s requires an even greater order of accuracy, thus truncating the infinite sequence of 1/3 as s to satisfy s×3=1 will immediately cause rounding errors at s×4 that further increase in inaccuracy at every s×n beyond 4. Instead, s would need to be redefined at every step of the way to remove the rounding errors, yet s is always defined as the decimal of 1/3 as 0.333r3

the awful assumtion that 0.999r9 = 1 is not the worst issue here, rather instead that 0.333r3 must equal 0.333rN to eradicate rounding errors by truncation, which is the essential equivalent of saying 0.3333 = 0.333N, or 0.3 = 0.N, or 3 = N.
If 0.999r9 = 1, then 3 is equal to 1, 2, 3, 4, 5, 6, 7, 8, or 9; and if 3 can be equal to any number, then any number can be equal to any other number.
If 1 = 3
and 5 = 3
then 1 = 5

To get 0.999r9 = 1, means to literally make any number equal to any other number, which is a false statement, therefore 0.999r9 cannot equal 1 as claiming otherwisr is also a false statement.

>> No.9320841

But 0.999r9 = 1 is true by definition.

>> No.9320843

Someone needs to take some analysis. It is true, however, that every sequence converges in an ultrafilter, so perhaps you have that [math] \sum^n_{k=1} k \to \frac{-1}{12}[/math] in that sense.

>> No.9320847

I don't think you understand how numbers work.

>> No.9320854
File: 336 KB, 629x468, 4bOGtMo.png [View same] [iqdb] [saucenao] [google] [report]

Prove it wrong

>> No.9320860

Read about Casimir effect, it is an experimentally proved physical effect whose theory uses directly this result.

>> No.9320862

see >>9320841

>> No.9320882

>how does making that mistake produce the same answer?
Because it's not a mistake.
>Because if you use invalid operations, you can get anything you want.
Ramanujan didn't know in advance to try to get -1/12, he expressed surprise in that result and said something along the lines of how people would try to get him committed to a psych ward for ending up with that value. So no, that's not an explanation. I also don't know of any actual mathematician making a formal argument that the Ramanujan summation was "invalid." At best you'll find complaints that it isn't "rigorous," which isn't the same thing. You can introduce this missing "rigor" by explicitly limiting how the terms can be manipulated, but that's not the same thing as saying Ramanujan went against what these sorts of explicit limitations would have allowed. It works because he didn't go outside of those limitations even though he also didn't explicitly establish those limitations either.

>> No.9320891

1/3 * 3 = 1, by definition

1/3 cannot be written as a decimal, however, as it would take an infinite amount of time, an eternity, to write out the repeating 3's.

This is where shit retard math comes into play thinking you can truncate that infinite sequence as something like 0.333 repeating.
0.333 REPEATING * 3 = 1, In theory, however even if you had four orders of eternity to spend writing out 0.333 infinitely three times, adding them together, your final result would still be 0.999 infinitely.

In truth, the act of calling upon the REPEATING assumption adds any minute amount to make the statement 0.333r×3=0.999r9=1 true, but each individual call to the REPEATING assumption INCRREMENTS the accuracy dependancy by an order of infinity. If 0.333r3 has an infinite amount of 3's, 0.666r6 would have squared the amount of 6's and 0.999r9 wpuld have cubed the amount of 9'd, which is then going into the territory of literally trying to perform arithmetic on infinity rather than simply with loose concepts of infinity. What is infinity squared, or infinity cubed?

>> No.9320899
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This is the dumbest post I've ever read on /sci/.
If you were just pretending to be retarded I gotta congratulate you on mastery of your craft.

>> No.9320901

If you reject standard mathematical axioms (like AC or the Axiom of infinity) then you can get any result want.

There is no point in your proof, it is about as worthwhile as defining 1=2.

>> No.9320905
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>> No.9320912

Thats the fucking point retard. Truncating an infinitely repeating sequence immediately betrays accuracy, and any number can equal any other number, which as you understand is retarded, yet you would still gladly defend 0.333r×3 = which uses the same exact faulty logic.

You believe 0.999r9 = 1 yet think that is somehow different than saying 2=1

By logic, 0.999r9 does not actually equal 1.

>> No.9320914

There is no proof.
You reject the standard axiomatic of ZFC, after that anything goes.

You could prove anything that way just by defining it to be true.

>> No.9320918

0.9999... = 0.3333... * 3 = 1/3 * 3 = 1

There's no "in theory" - in the common definition of "..." the above is true, if you don't like it you can make up your own field of math where it's not

>> No.9320919

Inside ZFC and the standard definition of the real numbers, 0.999r9 = 1 and 2=/=1.
These immediately follow from the definition of the reals.

If you reject ZFC anything you do is practically worthless, unless you build it up from an alternative framework, you are essentially just defining things to be true.

>> No.9320926

>prove it wrong
Whatever axioms you're using, they're not the one's mathematicians typically use. You can base your number system off of whatever it is you're doing there if you want.

>> No.9320929

>If 0.999r9 = 1, then 3 is equal to 1, 2, 3, 4, 5, 6, 7, 8, or 9

Why is that true?

>> No.9320930

How fucking dumb are you. Do you just not understand what "truncating" means?
Look it up faggot.

You cannot ACTUALLY write the decimal of 1/3. You CAN write a TRUNCATED ASSUMPTION as something like (0.333...) or (0.333r3), whatever way you want to WRITE it. Howeverx this truncated assumption is not actually equal to the decimal of 1/3.

do you not understand the concept of infinity either? you could write a decimal one hundred 3's and still be inaccurate, one thousand 3's and still be inaccurate, one million, one billion, one trillion 3's and still be inaccurate.
By truncating in calling upon a REPEATING sequence, you may as well simply remove the REPEATING sequence.

Speaking in terms of decimal, 3/10 = 0.3
0.3×3 = 0.9
claiming 0.999... = 1 is literally the same claim as 0.9 = 1, or 9/10 = 10/10

you cannot perform arithmetic on infinite sequences, regardless of ANY AND ALL mental gymnastics claiming otherwise.

>> No.9320932

You can give up here, this doesn't mean anything.

>> No.9320934

Because 0.333r3 must equal to 0.333rN to eradicate rounding errors from truncating the infinite sequence as shown in the (A)~( I ) results.

>> No.9320937

Your mistake, if you're being serious, is thinking that there's any rounding going on at all.

>> No.9320942

>You CAN write a TRUNCATED ASSUMPTION as something like (0.333...) or (0.333r3), whatever way you want to WRITE it. Howeverx this truncated assumption is not actually equal to the decimal of 1/3.
>you could write a decimal one hundred 3's and still be inaccurate, one thousand 3's and still be inaccurate, one million, one billion, one trillion 3's and still be inaccurate.
You're confusing symbols of numbers with numbers themselves.
It's completely irrelevant that you can't personally scribble an infinite number of "3" symbols on a piece of paper. That was never a requirement for representing a number with an infinitely repeating decimal part.

>> No.9320944

>You cannot ACTUALLY write the decimal of 1/3.
Yeah, I can, it's 0.3333...
More formally it's [eqn]\lim_{m\to \infty} \sum_{n=1}^n \frac 3 {10^n} = \sum_{n=1}^\infty \frac 3 {10^n}[/eqn]

>you cannot perform arithmetic on infinite sequences, regardless of ANY AND ALL mental gymnastics claiming otherwise.
Wrong. If you had any experience with calculus you would know

>> No.9320945

It can't be true. Because if S=1+2+3+...
So S=1+(1+1)+(1+1+1)+...
Ramanujan was a bastard

>> No.9320947

0.333r3×2 = 0.666r6

0.666r6×3 = 1.999r8
6×3 = 18 <- see that 8?

holy fuck i hope you arent attending a university much less stolen a degree.

>> No.9320953

You can't do that. This is why zeta regularization limits you to working with term variables, so you don't do shit like that.

>> No.9320965

underrated, saved

>> No.9320967

The bigger problem here is to get that 8, you have to work from multiplying with 3 from the end of the infinite sequence, which there is no end to an infinite sequence, so you can't ever even actually get 1.999r8; we can assume there is an 8 at the end because 6×3, but we cannot assume where the end is, and becomes the equivalent of turning any number like 0.382726102827633828627 into 0.3827261028r7

we cut out information by assuming a limit of infinity, which leads to the wrong answer outright. The retardation stems from instead of realizing the answer is wrong, pretending the answer is right in lieu, giving us 0.999=1

>> No.9320992

Infinite repeating sequences should be seen as inficators that the math is faulty. Pi doesn't repeat and look how much that has enabled the world.

I dont think anyone here can give a good reason for 0.999=1 existing.

>> No.9321010

Infinite repeating sequences follow directly from the definition or limits, and limits are used to define numbers like pi and e or concepts like derivatives which are used pretty much everywhere

>> No.9321018

Get 6.5 then. Go!

>> No.9321040

No, I am pretty sure it is. You can't manipulate infinite series. In fact, if you just translate those series into summation notation you can get a contradiction: you can't combine the lists since the 4n will always be longer. It looks cool, but my understanding is that it isn't a valid operation, and even just using variables instead of numbers with "..." will show that.

>> No.9321042

>4n will always be longer
offset, rather
longer was the wrong word.

>> No.9321088
File: 41 KB, 300x400, Emma-Stone-normalize.jpg [View same] [iqdb] [saucenao] [google] [report]

The -1/12 also exists in classical analysis and already Euler had the most general result on it.

Consider the difference between integral and sum,

[math] \int _m^n f(x)\, {\rm d}x=\sum _{i=m}^n f(i)-\frac 1 2 \left( f(m)+f(n) \right) -\frac 1{12}\left( f'(n)-f'(m)\right) + \frac 1{720}\left( f'''(n)-f'''(m)\right) + \cdots [/math]

where the coefficients are in simple way computed from Bernoulli numbers. A child of this would be e.g.

[math] \frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1) [/math]

but using the above I can generate infinitely many rigid identities that will all have that magic factor.
It's only Ramanujan who studies the same things and make the identification. A classical analogon would be

[math] \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1)) [/math]

You can type that into wolfram alpha for confirmation. The limit z to infinite gives an expression for 1+2+3+..., however, classically there's the log.
>this kills the sum

>> No.9321154

>shits and giggles
found the urban redneck

>> No.9321159

i'll do you one better

[math]\displaystyle\ln\left(1+x\right)=x-\frac {x^2}
{2} + \frac {x^3} {3} - \frac {x^4} {4} ...[/math]
[math]\displaystyle \ln\left(2\right) = 1-\frac {1}
{2} + \frac {1} {3} - \frac {1} {4} ...[/math]
consider the series: [math]\displaystyle S = 1-\frac {1} {2} - \frac {1} {4} + \frac {1} {3} - \frac {1} {6} - \frac {1} {8} +
\frac {1} {5} - \frac {1} {10} - \frac {1} {12} + ... [/math]
it's obvious that S and ln(2) have the same terms, just reordered in a new pattern.
for clarification, here's it in series form
[math]\displaystyle \ln(2) = \sum \{1, -2^{-1}, 3^{-1}, -4^{-1}, 5^{-1}, -6^{-1}, 7^{-1}, -8^{-1}, 9^{-1}, -10^{-1}... \}[/math]
[math]\displaystyle S = \sum \{1, -2^{-1}, -4^{-1}, 3^{-1}, -6^{-1}, -8^{-1}, 5^{-1}, -10^{-1}, -12^{-1}... \}[/math]
it might seem like they aren't even equivalent, but note that instead of 1/odd, -1/even, 1/odd, -1/even, we have 1/odd, -1/even, -1/even, 1/odd. all we did was shifted the evens over by one, which explains why the number 1/7 isn't even in the first 9 terms of S (it's the 10th term, actually)
this pattern can be represented by: [math]\displaystyle
\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)} -
\frac {1} {4x} [/math]
x - x/2 = x, and similarly, [math]\displaystyle
\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)}=\frac {1} {2\left(2x-1\right)}[/math]
so we can rewrite S as the following: [math]\displaystyle \frac {1} {2} - \frac {1} {4} + \frac {1} {6} - \frac {1} {8} + \frac {1} {10} -
or [math]\displaystyle \frac {1} {2} \left( 1- \frac {1} {2} + \frac {1} {3} - ... \right) [/math] = [math]\displaystyle \frac {1} {2} \ln \left( 2 \right)

even though ln(2) and S have the same terms, just in a different order.

>> No.9321176
File: 393 KB, 342x342, 1501701688178.gif [View same] [iqdb] [saucenao] [google] [report]

1/3 = 0.333[3
2/3 = 0.666[6
3/3 = 0.999[9
3/3 = 1
1 = 0.999[9

this works when referencing the thirds up to the whole. Referencing thirds beyond the whole is different.

4/3 = 1.333[2

By motive of proving 0.999[9 = 1, adding 0.333[3 to 0.999[9 results in 1.333[2

1.333[2 - 0.999[9 should give us 0.333[3, which it does, and 0.999[9 × 2 = 1.999[8, but what is 1.999[9 then?

One note of providing acceptance for 0.999[9 = 1 is how close it is to 1, but 1.999[8 = 2 is less accurate. It would be more reasonable to say 1.999[9 = 2,
but 1.999[9 is not the result of 0.333[3 × 6. So if there is a number greater than the infinite sequence that is more accurate, this creates a problem of how to interpret the infinite sequence.

The infinite sequence is more than just infinitely repeating. It could be any number and will be any number by nature of calculating each successive digit. Even when it repeats a single digit, the potential for the next digit to be any other number remains.

by nature of truncating an infinite amount of decimal 3's from the fraction 1/3 as something like 0.333[3, we invoke this special property of infinity that increments an imaginary value to satisfy not only 0.333[3 × 3 = 0.999[9 = 1, but also 0.333[3 × 9 = 2.999[ㅌ = 3.
Disregarding the special property would make ㅌ = 8, yet defining the end limit as 8 would leave room for the larger number ㅌ = 9, meaning 2.999[8 is not actually as equal or more equal to 3 than 2.999[9 would be. This is where the special property of infinity can be utilized to fill in the gap between 2.999[8 and 3, giving us 2.999[9 towards 3 instead. Realistically, it would fill in the gap from the time of its first invocation at 0.333[3 where every increment of 1/3 also increments the value of the property defining the gap, thus 0.333[3 × 3 is actually understood as (0.333[3 × 3) + (ㅌ × 3)

0.333[3 × 3 = 0.999[9
(0.333[3 × 3) + (ㅌ × 3) = 1
1 - (ㅌ × 3) = 0.999[9
0.999[9 =/= 1
ㅌ × 3 =/= 0

>> No.9321184

>1 + 0.33333... = 1.333332...
Really convinced me there

>> No.9321193

Oh but you dont have any problem when using limits right fucking faggot?

>> No.9321201

Ok guys i found out something very marvelous:
the sum of all positive integers is the same thing as the sum of all positive integers below infinity, and as we know the sum of all positive integers below n is
(n-1)/2 * n
and we also know that the sum of all positive integers is -1/12
(infinity-1)/2 * infinity = -1/12
1/2*infinity^2 -1/2*infinity + 1/12 = 0
infinity^2 - infinity + 1/6 = 0
infinity = +1/2 +/- sqrt((1/2)^2 - 1/6)
= either 1/2+1/sqrt(12) or 1/2-1/sqrt(12) if i calculated correctly.

>> No.9321202
File: 207 KB, 692x960, 1510723646424.jpg [View same] [iqdb] [saucenao] [google] [report]

>.9999 = 1 defacto!!!!
really niggers my noggin.

>> No.9321220

Theres no numbers between 0.99... and 1

But every real number haves at least one number between them if they are different, therefore they must be equal

>> No.9321228

0.991 is closer to 1 than 0.99
0.99999991 is closer to 1 than

>> No.9321261

I don't get this either. How can you combine the summations like this? If you set a limit of n, anywhere, even close to infinity, they sums don't match up this way. Also, if we are doing n to infinity, wouldn't both of those sums be equal to infinity? Maybe I am just dumb as fuck.

>> No.9321273
File: 6 KB, 250x250, wholesale-vib5928-triple-hooks-fishing-lure-vivid-fish-shaped-baits-steel-balls-within53a9295af3d14_1.jpg [View same] [iqdb] [saucenao] [google] [report]



-3s = -3-6-9-12-...
did you just shit your brains for a second or what.

>> No.9321492

No, you're just missing the point of introducing 4s and subtracting it in the first place which is that Ramanujan wanted to assign a value to the 1+2+3+4... series by relating it to the alternating 1-2+3-4... series that he already knew had the value of 1/4 assigned to it. And to get from 1+2+3+4... to 1-2+3-4... he subtracted 4 from the second term (2-4=-2), 8 from the fourth term (4-8=-4), 12 from the sixth term (6-12=-6), etc. It wouldn't be useful to subtract in the places you subtracted from, that wouldn't relate the series to something familiar he could work with.

>> No.9321503

>If you had any experience with calculus you would know
ironic. If you had any experience with calculus, you would know it has nothing to do with aritmetic on infinite series.

>> No.9321505

I'm not him. But I got that part. It is rather obvious. I don't get how you can use that notation and ignore that for N=4, you still have -12 and -16 terms. For N=9^99999999999 you still have dangling terms too. It doesn't seem like that operation is valid for summations both for infinite sums and for any other N. It looks like a slight of hand. Also, if we get to assume that infinity makes this all go away, why can we assume the preceding term is infinity minus infinity? I was always under the impression that you can't manipulate infinite series with this way because of the aforementioned issues.

>> No.9321510

I could be a complete idiot though. I don't know, but it certainly doesn't make notational sense.

>> No.9321543

( r × n ) + ( ㅌ × n ) = j

Where r is an infinite repeating decimal, n is the multiplicator, and ㅌ is a single instance of "tending towards infinity" to define accuracy. But how could we further define ㅌ?
Working with 1/3 to 3/3=1, this would be
( 0.333[3 × 3 ) + ( ㅌ × 3 ) = 1
( 0.999[9 ) + ( ㅌ ) = 1
1 - ㅌ = 0.999[9
In this sense, ㅌ would only be an inverse of [math]\infty[/math]. If [math]\infty[/math] defines some garantuan unthinkable uncountable thing, then ㅌ defines an abstract remainder to be applied to end an infinite sequence, where [math]\infty[/math] + ㅌ = ㅐ, letting ㅐ be equal to this abstract concept of [math]\infty[/math] + ㅌ.
The problem with [math]\infty[/math] is that it describes an abstract concept, therefore you need abstract functions to properly work with it and deconstruct it back down to something that is actually usuable in real number arithmetic, though i'm gonna guess the benfit of all this is as meaningless as the original claim of 0.999[9 = 1, but at least better since it doesn't defy logic by saying x=y and at least attempts to identify aspects that were not previously addressed.

>> No.9321545


>> No.9321630

You cant perform finite arithmetic on infinity.

i dont know why this is such a hard concept for many people in the thread to have understood. They've spent too much time being lied to by professors and having their noses buried in books, and not enough time actually attempting a useful calculation using an infinite repeating decimal.

Performing calculations in finite time and further engineering computers to perform calculations leads to problems related to computing speed, defined as time, as well as memory registers for storing and performing math upon digits.

It's disinformation to truncate and concact an infinite decimal, regardless of how large or small the new implied repetition decimal is. 0.999[...] is as innacurate as 0.99999999999[...] or 0.9[...]
You can make the statement that 0.333[...] = 1/3, but 0.333[...] × 3 does not equal 3/3 or 1. Instead, it equals 0.999[...], and requires the intentional inclusion or addition of another abstract value to rival the concept of
[math]\infty[/math] in order to truely make it equal to 1.

>> No.9321672

0.333... × 2.999... = 0.999...
0.333... × 3 = 0.999...

one of these statements is very wrong.
the second statement can be boiled down to 3×3=9, and we see repeating 9's in the answer, so that checks out.
The first statement is boiled down to 3×9=27 where we would expect 7 at the end of the sequence, yet the answer claims the sequence is a string of 9's.
This alone disproves 0.999...=1, as 2.999... does not equal 3, and the first statement's answer of 0.999... is actually incorrect, instead being 0.999...7

if we use ㅌ as an abstravt concept to define some kind of inverse of [math]\infty[/math], then apply it back to those statements, we get this:
0.333... × 2.999... = 0.999...7+(2ㅌ) = 1
0.333... × 3 = 0.999...+ㅌ = 1
where, because [math]\infty[/math] is abstract, all we must do is merely pretend ㅌ has an abstract undefined value that would top off the infinite sequence of 9's, or factor in some ability to fill in the gap between 0.999...7 and 1 in a different way than between 0.999...9 and 1.
Each invocation of an infinite sequence in the problem would equate to an incremental value of ㅌ being applied to sum, and this is all if you just want 0.999...9 to equal 1, which it otherwise would only be equal to 0.999...

>> No.9321709


>> No.9321720
File: 13 KB, 210x240, skeeter-valentine-doug-8.55.jpg [View same] [iqdb] [saucenao] [google] [report]

Lets take your example and apply ㅌ to it
9.999...=10x +10ㅌ
10x-x +(10ㅌ-ㅌ) =9.999...-0.999...

Where your original problem makes a fault is here at

x = 0.999[9
10x = 9.999[9
But 10x-x, 9x, does not equal 9.0
9x = 8.999[1

Your problem derives from seeing the decimal as seperate from the whole, so you see 0.9, subtract that from 9.9, and get 9.

if it were seperate from the whole, you would consider x+x = 1.999[9, but that is only true if one and only one of those x values is equal to the whole number 1, otherwise it is

if one x is presupposed to equal 1 while the other x is presupposed to equal 0.999[9, that just doesnt work. If x=1, then the answer is 2, not 1.999[9. if x=0.999[9, then the answer is 1.999[8

>> No.9321744

>requires the intentional inclusion or addition of another abstract value to rival the concept of ∞
>if we use ㅌ as an abstravt concept to define some kind of inverse of ∞
A symbol for infinitesimals already exists, it's ε (epsilon).
You really shouldn't try self-teaching mathematics like that, you're coming to a bunch of retarded and faulty conclusions that would have been avoided if you studied this stuff formally.

>> No.9321793

Fuck off, nigger.

All i'm saying is you cant have infinites without some kind of equally abstract and seperated, intentionally included concept in the problem that would essentially nullify the infinite sequence.

The faulty retarded conclusion is saying 0.999...=1, when the truth is however you'd like to personally write your infinitesimal, the actual statement is 0.999...+ε = 1
0.999...+ㅌ = 1

excluding the acknowledgement of the infinitesimal is excluding information from the problem and pretending it all makes sense in the end. THAT is self-taught math. 1+1=3 and thats just okiedokie cause cleetus brought over a jug of moonshine and we're just gonna get drunk until our sisters are purrty.

>> No.9321926

>game theory harder than differential equations

What a load of shit

>> No.9321932

Infinite decimal expansions have rigorously defined meanings, they are equivalent to infinite summations.
[math]0.\overline{9} \equiv \sum_{k=1}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k[\math]
[math]0.\overline{9} = \sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k - 9[\math]
[math]0.\overline{9} = 9 \left( \frac{1}{1-\frac{1}{10}} \right) - 9[\math]
[math]0.\overline{9} = 1[\math]

>> No.9321936

It definitely is. There are way more people who can do DE decently than Game Theory. Game Theory only seems simple, but coming up with an interesting new finding is much harder than coming up with a 2nd degree DE to explain some phenomena. There are way more people who can do the later. Fact.

>> No.9321937

[math]0.\overline{9} \equiv \sum_{k=1}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k[/math]
[math]0.\overline{9} = \sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k - 9[/math]
[math]0.\overline{9} = 9 \left( \frac{1}{1-\frac{1}{10}} \right) - 9[/math]
[math]0.\overline{9} = 1[/math]

>> No.9321948

[math] \displaystyle
\frac{1}{3} = \left (\frac{3}{10} + \frac{1}{30} \right )
&= 0.3 + \frac{1}{30} \\
= 0.3 + \left ( \frac{3}{100} + \frac{1}{300} \right )
&= 0.33 + \frac{1}{300}\\
= 0.33 + \left ( \frac{3}{1000} + \frac{1}{3000} \right )
&= 0.333 + \frac{1}{3000} \\
= 0.333 +\left ( \frac{3}{10000} + \frac{1}{30000} \right )
&= 0. \underset{n}{ \underbrace{3333}} + \frac{1}{3 \underset{n}{ \underbrace{0000}}} \\
\\ \displaystyle
\Rightarrow 0.\overline{3} = \frac{1}{3}

>> No.9321954

But in terms of understanding the subject in general, diff eq is way harder. Of course it's going to be hard to come up with a way to describe something with game theory than with diff eq, but that's just cause they're completely different tools. Actually learning game theory in a class is much easier

>> No.9323334

The issue i see here is performing with fractions that can avoid usage of writing the repeating sequence as a decimal.
In the top example, multiplying (3/10) by 3 then multiplying (1/30) by 3 would result in (9/10 + 1/10), which obviously equals 1. The fractions written as decimals are 0.90 + 0.10 = 1.0
if you alter the order and method of multiplying to (3/10 + 1/30) multiplied once by 3, you get 0.999[9 instead - which then leads to thinking the same exact method of multiplying by 3 was used and therefore 0.999[9 = 1

Your method even works beyond n/3 where n+=4, for example 6 gives us 1.999[9 where i'd otherwise shown 6 to work via
0.333[3 × 6 = 1.999[8 + ㅌ

It occurred to me last night that I was the one truly concatenating the infinite sequence. When writing the decimal as 0.333[3, i use the numbers left of the bracket to show pattern, the bracket to define repetition, and the number to the right of the bracket to define an 'end' to the sequence. I thought this method to work as it would be abled to be used to disprove 0.999[9 approaching infinity as equating a whole integer, to not conflict with other fields of mathematics. Could you imagine how many graphing solutions wouldn't even need to exist if a y approaching infinity would allow an x to equal 0 or go negative? In those situations, the y is taken defacto as ∞ and the x is never written as if it'd reached 0, thereby disproving by use of infinitesimals that 0.000[1 != 0, so why would 0.999[9 = 1?


>> No.9323361


that infinity is uncountable and an infinite sequence unwritable means it could never be utilized in arithmetic without making assumptions about an upper limit and truncating or capturing a portion of that infinity to substitute and represent infinity. Simply writing 0.333[...] to represent 1/3 denies some information on how to perform arithmetic on the infinite sequence, or rather it has not done enough to prepare the infinite value for use in finite, real arithmetic. 0.333[...] substitutes the eternity it would take to write out as many 3's as possible, but it then betrays arithmetic by offering no contained number with which to perform useful math that could give a real answer. If we know 3×4 = 12, 0.3×4 = 1.2, and 0.03×4 = 0.12, in each circumstance we can see the right-most digit is 2, therefore we expect the same of the final digit of an infinite sequence in 0.333[3 × 4 = 1.333[2

Performing this concatination of the infinite repetition is necessary to actually use the infinite decimal in arithmetic, else 0.333[...] × 4 should be considered undefined. For every 3 we write, we expect to write a 2 at the end, but since there is no end, we write no 2 and betray useful arithmetic, becoming the equivalent of writing something like 123,456,789 × 4 and pretending the answer suffices after only multiplying the first 5 digits 123,45 × 4 ; as a smaller scale example.

>> No.9323411

I thought you were trolling yesterday, but this is getting desperate.

>> No.9323463

This assumption of truncation exists in the real world, such as using PI. If we only decided to use PI once it had been fully calculated or never use it because it will never be fully calculated, we'd have made no progress in any field and yet for that same reason we write PI as π instead of trying to write it's infinite sequence. We know 2πr can be used to define the circumfrence of a circle, but defining it so exactly becomes unecessary. We can roll a 12 inch wheel and measure it to complete a full rotation at 3'feet and some centimeters and thats good enough because it will roll over again and again, while π never completes it's roll in needlessly defining even smaller and more accurate decimal places that for the purpose of our wheel's single rotation, 1rot = 3.1415[nnn[' feet of distance, are not useful.
so if that can be true, why can't 1 = 0.999[9?
because it's infinite repetition is written as repeating 9's instead of using some symbol like ㅌ, like pi's infinite decimal uses π?
Using π in computer math will usually have the number stored up to 15 decimal places as 3.141592653589793, which suffices for most practical math.
If we applied the same method to storing ㅛ as 0.333333333333333, 4ㅛ = 1.3333333333333332, and for the purpose of finite arithmetic this would be the equivalent of 0.333[3×4 = 1.333[2

But in a situation where we'd want our result to be a little more human readable and only accurate up to something like 4 decimal places, math.PI woud return 3.1415 and 4ㅛ would return 1.3333, in each case omitting any higher degree of accuracy so even 9ㅛ = 8.9999999999999997 or 8.999[7 would become 8.9999 but still not 9.

The real methodology here is accepting the infinite decimal sequence of PI does not round up to 7/22 or even 3.14, therefore 3ㅛ should not round up to 1 and only truncated to 0.99 at best.

>> No.9323475
File: 66 KB, 600x471, 1511085854684.jpg [View same] [iqdb] [saucenao] [google] [report]

Nah, you're just being a shitter stuck thinking 1/3 can be written as a decimal without taking liberties, while failing to define what the limits of those liberties are.

>> No.9323523

You need to get your head around irrational numbers, buddy.

>> No.9323625

How did you get from this:
to this:
Shouldn't it be -3s = 1-4+2-8+3-12+... ?

>> No.9323660

See >>9321505
I don't understand it either. Where do the terms go? Plug in any value for N and you N/2 terms that just disapear. You may be missing what 2-4 = -2 and 4-8 = -8, but that doesn't explain my question. And no, I don't think infinity solves this at all.

>> No.9323688

In that case he's just simplifying (+2-4), (+4-8), and so on. No idea what the implications of that are for an infinite series, but that's the idea.

>> No.9323900
File: 13 KB, 657x527, 1494486599769.png [View same] [iqdb] [saucenao] [google] [report]

If 0.99999999 approaching 1 actually ever made it to 1, the world would be a much better place than it actually is.

That is all the proof I need.

>> No.9323921 [DELETED] 

>>9321948 isn't uncountable
every line = 1
line #1 = 1
line #10 = 1
line #982376487236823498 = 1

at infinity, it still is 1

>> No.9323929 [DELETED] 

>>9321948 (You) isn't uncountable
every line is exactly 1/3
line #1 = 1/3
line #10 = 1/3
line #982376487236823498 = 1/3

at infinity, it still is 1/3

>> No.9323938

>>9321948 isn't uncountable
every line is exactly 1/3
line #1 = 1/3
line #10 = 1/3
line #982376487236823498 = 1/3

at infinity, it still is 1/3

>> No.9323945


>> No.9323950
File: 7 KB, 645x773, Thatface20110725-22047-wlaopv.png [View same] [iqdb] [saucenao] [google] [report]

0.999... approaching 1 is like an eternal road trip to disney world where you never actually get there and you also have to take a pee but your dad just keeps telling you to hold it. Then one day you make it to the parking lot and it's gigantic so you spend 7 weeks looking for a parking space and finally you find one but it's 10,000 miles walking distance from the park entrance and like, yea, you're "at" disney world in a sense, even if you're actually not in the park having fun riding rollercoasters, therefore this definition of being at disney world when you're going to spend the next 3 years walking across the parking lot doesn't exactly mean too much and every time you pass by a parking attendant valet greeter who says "welcome to disney world!" you'll probably just want to literally murder them.

>> No.9323961
File: 34 KB, 601x508, e7db9e4e33592de521b23e955084488e.jpg [View same] [iqdb] [saucenao] [google] [report]

0.999... approaching 1 is like living while thinking there is an after life.

>> No.9323982

wtf the limit of that sum is infinite, how does this make any sense?

>> No.9324720

This equation doesn't work for any number that isn't a single digit repeating decimal, therefore the method and proof is simply just incorrect.
1/n = ( ((floor(10/N))/10) + (1/(10×n)) ) is false.
1/7 ≠ (1/10 + 1/70), therefore 1/3 ≠ (3/10 + 1/30)

woulda been nice and made 4/3 = 1.333[3, but nope.
4/3 looks more like 1.333[2 now

>> No.9324725

It makes sense in a different system than the one you’re used to.

>> No.9324730

>1=.999... by dividing 1 by 3.
1= 3/3
1/3 + 1/3rd +1/3rd = 3/3rd
.333r+.333r+.333r= .999r
therefor .999r=1
does this upset you

>> No.9324734

if it works, it works
no matter what bs you say

>> No.9324738

the truth is 1 is rounding of .999r9

>> No.9324740 [DELETED] 

> 1/3 ≠ (3/10 + 1/30)

wew lad,
[math] \frac{3}{10} + \frac{1}{30} = \frac{9}{30} + \frac{1}{30} = \frac{10}{30}= \frac{1}{3} [/math]

you just went full retard

>> No.9324741

Thats not how proof works you retard. It doesn't actually work with single digit repetition decimals, it just masks the value.
2/3 = (6/10 + 2/30) = 0.666[7 even. Hell it literally only works for 1/n, but again it just masks the value. With 1/3 = 0.333[3, the method adds a {3} in front of the natural result returning 0.{3}333[3

>> No.9324743

>1/3 ≠ (3/10 + 1/30)

wew lad
\frac{3}{10} + \frac{1}{30} = \frac{9}{30} + \frac{1}{30} = \frac{10}{30}= \frac{1}{3}

you just went full retard

>> No.9324744

>it doesn't work

what line is faulty

>> No.9324747

You think a broken clock being right for two seconds a day suffices as an accurate clock. Plug 1/7 into that equation shitstick.

>> No.9324750

nah I'm good you admitted it's right

>> No.9324752

can someone take a look at >>9323660 and>>9321505 >>9319850
maybe >>9321088?
The problem with that manipulation is that for any given N terms in the summations there is always N/2 leftover terms from the second summation being left out (-12 and -16 form N=4, and so on). I don't see how anyone could think going to infinity would solve this as the value of the left out terms keeps increasing.

>> No.9324754
File: 33 KB, 600x700, Pool.jpg [View same] [iqdb] [saucenao] [google] [report]

Jesus, how delusional are you. I've called it wrong multiple times now.

How fucking sad is it to exist as you?

>> No.9324756


>> No.9324774

The method is false because it cannot solve as written. It presumably gives the right answer only on single digit repeating decimals, and only when it's o/n. Any number greater than 1 in place of o gives the wrong answer.

A proof has to be testable under different conditions to give the expected result, but most conditions fail this equation
1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
therefore any answer it does give should be discarded. You're just shifting the natural result of 1/n to the right a decimal place and injecting another number in front, and in a single digit repetition the number it injects is the same number so by first glance it looks like the same result, yet truthfully you've taken an infinite repetition and added one which results in infinity+1

>> No.9324788

>The method is false
well it's fucking easy to point out the faulty line then, isn't it

>> No.9324792
File: 142 KB, 617x347, 1509768568403.png [View same] [iqdb] [saucenao] [google] [report]


O K ?

you are injecting 3's into the answer then adding on an infinite repeating 3 decimal.
1/30 = 0.0333[3
1/300 = 0.00333[3
1/3000 = 0.000333[3
and so on and so forth. You are just substituting the 0's with 3's while not addressing the existence of the infinite, AND IF YOU TRIED 1/7 LIKE I TOLD YOU, YOU'D HAVE REALIZED YOU'RE JUST SHIFTING THE NATURAL ANSWER OVER AND INJECTING GARBAGE INTO THE FRONT OF IT.

we've reached a pinnacle of brainletism

>> No.9324802

omg we went over this already

>> No.9324810

does it hurt you that much that someone with only a fraction of resources compared to you accomplished something you couldn't in multiple lifetimes?

>> No.9324814
File: 73 KB, 334x319, 1508566033111-1.jpg [View same] [iqdb] [saucenao] [google] [report]

1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
1/3 = (3/10 + 1/30) = 0.{3}333[3
1/7 = (1/10 + 1/70) = 0.{1}142857[142857
When you plug in 1/7, you get
meanwhile, 1/7 = 0.142857...
Real 1/7 = 0.142857
your 1/7 = 0.114285

do you comprende

>> No.9324817

yeah your method is shit is seems

>> No.9324822
File: 7 KB, 420x420, b36.png [View same] [iqdb] [saucenao] [google] [report]

You should invest in bitcoin

>> No.9324823

>you are injecting
1 = 1/2 + 1/2

not "injecting" anything, just cutting it up

>> No.9324829

0.5 is not an infinite sequence. 1/3, 1/30, 1/300, 1/3000, etc, are all infinite sequences.

Nevermind the obvious flaw that the method does not work regardless of what numbers you plug in, shown by >>9324814

>> No.9324832
File: 10 KB, 604x658, NegativeZeta.gif [View same] [iqdb] [saucenao] [google] [report]

It's black magic.

>> No.9324843

you 10-based retard

>> No.9324844 [DELETED] 

Here's your fucking 7, snow-white
\frac{1}{7} = \left (\frac{142857}{1,000,000} + \frac{1}{7,000,000} \right )
&= 0.142857 + \frac{1}{7,000,000} \\
= 0.142857 + \left ( \frac{142857}{1,000,000,000,000} + \frac{1}{7,000,000,000,000} \right )
&= 0.142857142857 + \frac{1}{7,000,000,000,000}\\
= 0.142857142857 + \left ( \frac{142857}{1,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000} \right )
&= 0.142857142857142857 + \frac{1}{7,000,000,000,000,000,000} \\
= 0.142857142857142857142857 +\left ( \frac{142857}{1,000,000,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000,000,000} \right )
&= 0. \underset{n}{ \underbrace{142857142857142857142857}} + \frac{1}{7, \underset{n}{ \underbrace{000,000,000,000,000,000,000,000}}} \\
\\ \displaystyle
\Rightarrow 0.\overline{142857} = \frac{1}{7}

>> No.9324846
File: 60 KB, 1024x558, 1509533757591m.jpg [View same] [iqdb] [saucenao] [google] [report]

>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>>1/n = ( ((floor(10/N))/10) + (1/(10×n)) )
>"m-my math is sound, y-you're the retard"

>> No.9324850

you and your stupid floors

just point out which line in >>9321948 fails

>> No.9324853 [DELETED] 

here is the 1/7 case
\frac{1}{7} = \left (\frac{142857}{1,000,000} + \frac{1}{7,000,000} \right )
&= 0.142857 + \frac{1}{7,000,000} \\
= 0.142857 + \left ( \frac{142857}{1,000,000,000,000} + \frac{1}{7,000,000,000,000} \right )
&= 0.142857142857 + \frac{1}{7,000,000,000,000}\\
= 0.142857142857 + \left ( \frac{142857}{1,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000} \right )
&= 0.142857142857142857 + \frac{1}{7,000,000,000,000,000,000} \\
= 0.142857142857142857142857 +\left ( \frac{142857}{1,000,000,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000,000,000} \right )
&= 0. { \underbrace{142857}142857142857142857} + \frac{1}{7, {\underbrace{000,000},000,000,000,000,000,000}} \\
\\ \displaystyle
\Rightarrow 0.\overline{142857} = \frac{1}{7}

>> No.9324857

\frac{1}{7} = \left (\frac{142857}{1,000,000} + \frac{1}{7,000,000} \right )
&= 0.142857 + \frac{1}{7,000,000} \\
= 0.142857 + \left ( \frac{142857}{1,000,000,000,000} + \frac{1}{7,000,000,000,000} \right )
&= 0.142857142857 + \frac{1}{7,000,000,000,000}\\
= 0.142857142857 + \left ( \frac{142857}{1,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000} \right )
&= 0.142857142857142857 + \frac{1}{7,000,000,000,000,000,000} \\
= 0.142857142857142857 +\left ( \frac{142857}{1,000,000,000,000,000,000,000,000} + \frac{1}{7,000,000,000,000,000,000,000,000} \right )
&= 0. { \underbrace{142857}142857142857142857} + \frac{1}{7, {\underbrace{000,000},000,000,000,000,000,000}} \\
\\ \displaystyle
\Rightarrow 0.\overline{142857} = \frac{1}{7}

>> No.9324867

in base 3
one-third is "0.1"

So now what, is 10 magic because that's what jesus gave us?

>> No.9324912
File: 20 KB, 306x306, 79e0f76f22f429dc6a79ebbe8f3b484d.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9324917


>> No.9324951
File: 110 KB, 318x275, 1508479619038.png [View same] [iqdb] [saucenao] [google] [report]

No such thing.

>> No.9324954

>If you reject standard mathematical axioms (like AC or the Axiom of infinity) then you can get any result want.
Where can I read more about this deep theorem?

>> No.9324958

>never taken a real math course in my life the thread

>> No.9324960

>real math
In what way are "numbers" real math?

>> No.9324961

hares overtake tortoises erry day

>> No.9324967

Your IQ must be at least 30 to post in these threads.

>> No.9324977
File: 38 KB, 645x729, 1509035922690.png [View same] [iqdb] [saucenao] [google] [report]

And whats 1/2 as a decimal in base-3?

>> No.9324979

I hate when people do this. Maybe you aren't one, but 99% of people who say shit like this don't understand shit and just memorize proofs or cheat their way to passing grades.

>> No.9324980

Especially when they claim that "analysis" or anything related to it is "real math".

>> No.9324984

>"nullify the infinite sequence"
Engineer spotted.

>> No.9324988


>> No.9324989
File: 82 KB, 645x729, 1506487478785[1].png [View same] [iqdb] [saucenao] [google] [report]

>epsilon is just like, a really really small number but not zero dude
please, please, brainlet, give us your definition of "epsilon".

>> No.9324996
File: 485 KB, 1050x1080, 1495467753515.png [View same] [iqdb] [saucenao] [google] [report]

No such thing.

>> No.9324999

I guess you've solved it.

>> No.9325001

Solved what exactly?

>> No.9325002

deci is 10, retard

>> No.9325003


>> No.9325006

But what I replied to has nothing to do with math.

>> No.9325009

Well you solved it, anon, you deserve it.

>> No.9325010

ever heard of Zeno?

>> No.9325013

Interesting. Could you point out the paper which proves that this so-called "paradox" is actually a paradox?

>> No.9325014
File: 973 KB, 312x213, x.gif [View same] [iqdb] [saucenao] [google] [report]

1/2 doesn't exist!!!

>> No.9325016

I don't see the "..." here, is this a typo?

>> No.9325018

it isn't
because infinities are everywhere

>> No.9325020

You got BTFO by reddit boi


>> No.9325023

>because "infinities" are everywhere
I see, so they aren't relevant to mathematics? That's what I thought.
It's fine if you use fairy-tales in "engineering", just don't bring your garbage elsewhere.

>> No.9325025

>I thought

>> No.9325027

That's pretty low of you.

>> No.9325029
File: 53 KB, 403x448, 1509935607777.png [View same] [iqdb] [saucenao] [google] [report]

0.1 is 0.333[3
0.2 is 0.666[6
0.1222[2 is 0.407[407
but i guess we can just round and approximate that 0.407[407 = 1/2 cause theres no other number higher than 0.407[...] and less than 0.5

this tard math seems mighty familiar

>> No.9325031

>higher than
Engineer spotted.

>> No.9325032

There is no rounding going on, stop this and go read a book.

>> No.9325033
File: 33 KB, 500x500, 1506140326977.jpg [View same] [iqdb] [saucenao] [google] [report]

What is 1/3 in base 12?

>> No.9325035


>> No.9325038
File: 89 KB, 363x475, 1504229489395.png [View same] [iqdb] [saucenao] [google] [report]


>> No.9325040

Your kind isn't welcome here.

>> No.9325054


0 1 2 3 4 5
0 1 2 10 11 12
0.5 ~> 0.12

>> No.9325063

[math]0.12_3 \times 2_3 > 1_3[/math]

>> No.9325075

top kek

>> No.9325082

your math is retard , u are 100% american

>> No.9325180
File: 13 KB, 389x500, x.png [View same] [iqdb] [saucenao] [google] [report]

>0.1222[2 is 0.407[407

christ, this idiot can't even count

>> No.9325796
File: 55 KB, 600x719, db782446bcb491d356c77e13bd4b98a0.jpg [View same] [iqdb] [saucenao] [google] [report]

[math]1=\sum_{k=1}^{\infty} k-\sum_{k=2}^{\infty} k=\infty-\infty=0[/math]

>> No.9325837

0.1000003= 0.333(...) = 1/3
0.111(...)3 = 0.500000= 1/2
0.2000003= 0.666(...) = 2/3
0.222(...)3 = 1.000000= 1/1

>> No.9325844
File: 7 KB, 211x239, stump wojak.png [View same] [iqdb] [saucenao] [google] [report]

>strokes theorem

>> No.9325921
File: 71 KB, 731x720, 1508571001225.jpg [View same] [iqdb] [saucenao] [google] [report]

No such thing.

>> No.9326537

Well you have to in order to perform any useful math on it.

If you don't set a cutoff limit, a computing calculating program will just freeze and never be able to finish writing out all the repeating 3's of 0.333...
I just made my own program with an internal cutoff limit of 80,000 digits and it took 57 seconds just to process all the 3's and display them on screen.
20,000 digits could be done in 3.8 seconds, so the computational speed issue is logorithmic.

The problem with setting a cutoff amount is that it presents inaccuracy. For example if the cutoff was set to 3 digits, i'd get 0.333000000000000, and even at the 20,000th or 80,000th digit, when you cut off there is still implied 0's there. I could add 3 instances of 80,000 digit 0.333... and only return 0.999... with 80,000 digits of accuracy, but never actually equal to 1.0 as it's 80,000 9's followed by a 0 at index 80,001

there would need to be some kind of expression to definite how to handle infinite repeating numbers, and sure enough these expressions actually exist in computer math explicitly because of the logic issue of defining 0.999...=1, yet outside of computer math these expressions are not used, which makes the 0.999...=1 concept fully abstract and incalculable. If a computer can't calculate it in finite time, why should any human think they could?

When it comes to math, nothing should be abstract. That would just defeat the purpose of math.

>> No.9326673

>When it comes to math, nothing should be abstract.
Stopped reading right here. Good thing I started from the bottom.
I'm sorry brain-dead engineers such as yourself aren't able to manipulate basic abstractions.

>> No.9326694


imagine if all you had to do to pass math tests was just muh abstraction and all the questions on the test were simply multiple choice plugged by a finalizing component to each question of "Do you have the answer?" and the choices were just yes and no.

Engineers are the only reason you can plug (1/3) × 3 into a computing device and get 1 as an answer.

>> No.9326698

Well, I'm in grad school right now at arguably the best institution in my country for my field (operator algebras).

>> No.9326699

abstraction doesn't mean
>hurr I did these mindless calculations in my head

>> No.9326710

Stopped reading right there. Take your subhuman engineer self somewhere else if you can't handle the most basic abstractions.

>> No.9326741

x1 = 0.99...9
x2 = 0.99...9 - 0.00...9
x2 = 0.99...90 \\ truncate the zero
x2 = 0.99...9 \\ same number of 9's as before
therefor 0.00...9 = 0
x1 = x2

>> No.9326746

That doesn't mean that characterization was inaccurate. Calling people idiots and flashing credentials should never mater in math. Either disprove what they say or point them to sources that will help them.

>> No.9326826

none of you fucking brainlets know anything about advanced topics in divergent series. Neck yourselves.

>> No.9327169

Yet it is only in your mind that you can assume 0.999... = 1.

Any attempt to write out the infinite 9's would go nowhere. It'd take you a lifetime to write a billion 9's and you'd die before ever getting to write " = ".

therefore, 0.999 != 1 for the purpose of arithmetic producing an answer in finite time.

Blame the decimal system, blame base-10, idc. Just don't shit on the practical realists who had to expand beyond your haughty abstraction complex to produce methods that could actually be computed. Again to mention, they've even worked with you so you can plug in 0.999... and get 1 back. They could have just as easily done what I have and said DOES NOT COMPUTE but no, they'll give you a 16-bit register, a 32-bit register, a 64-bit register, and they'll let you fill up to 8, 16, 32 decimal places of accuracy, they'll let you plug in 1/3 × 3 and they'll add 0.333333333333333 together 3 times and get 0.999999999999999 and then they'll say "oh he's doing the thing" and swap it out for 1.0 instead.

And this is the thanks we get.

>> No.9327226

0.1 = 1/10^1
0.01 = 1/10^2
0.001 = 1/10^3
0.000...1 = 1/10^inf = 0

>> No.9327232

>Any attempt to write

1 = 9/10 + 1/10 = 0.9 + 10/100
= 0.9 + 9/100 + 1/100 = 0.99 + 10/1000
= 0.99 + 9/1000 + 1/1000 = 0.999 + 10/10000
and so on, ad infinitum.
No slow creeping up towards one, each line already =1

>> No.9327255
File: 245 KB, 1063x1063, 1508010693769.png [View same] [iqdb] [saucenao] [google] [report]

>ad infinitum

>> No.9327274

oosh, that's a niche meme

>> No.9327361
File: 90 KB, 785x629, x.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9327511
File: 963 KB, 990x1146, 85a.png [View same] [iqdb] [saucenao] [google] [report]

You cannot do anything infinitely, and if you could it would be probably be a waste of your time.

>> No.9327518

>they'll let you plug in 1/3 × 3 and they'll add 0.333333333333333 together 3 times and get 0.999999999999999 and then they'll say "oh he's doing the thing" and swap it out for 1.0 instead.
No, that's bullshit. If it goes to 1 it's because of floating point limitations.

>> No.9327583

no need to
the process shows that increasing the decimals changes nothing

>> No.9327650
File: 24 KB, 800x450, IKIFEEL.jpg [View same] [iqdb] [saucenao] [google] [report]

> 0.99 = 0.9

>> No.9327698

how about you actually read >>9327232
protip: use your calculator

>> No.9327922

>use bad notation
>blame the player
the problem is that there are a bunch of theorems and assumptions made so that shit can even work. Using the summation sum and the arrow for "limit" is definetly misleading.

>> No.9328452

>Yet it is only in your mind that you can assume 0.999... = 1.
I couldn't care less about such uninteresting garbage. This is not even remotely related to math, only an engineer would think it is.

>> No.9328495

>All i'm saying is
>self-taught math
>just okiedokie
found the urban redneck retard

>> No.9328533
File: 9 KB, 211x239, wojak headlet.png [View same] [iqdb] [saucenao] [google] [report]

>you cannot perform arithmetic on infinite sequences

>> No.9328555 [DELETED] 
File: 42 KB, 479x720, UvGfe7Y.jpg [View same] [iqdb] [saucenao] [google] [report]

You can't.

Here's your fedora

>> No.9328581

>reddit filename
Subhuman trash is not welcome here.

>> No.9328655
File: 15 KB, 221x250, 1510711638411.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9328661 [DELETED] 
File: 42 KB, 645x729, tfw no brain.png [View same] [iqdb] [saucenao] [google] [report]

(a_n)_n + (b_n)_n \equiv (a_n + b_n)_n
here you go brainlet

>> No.9328664
File: 42 KB, 645x729, tfw no brain.png [View same] [iqdb] [saucenao] [google] [report]

[math](a_n)_n + (b_n)_n \equiv (a_n + b_n)_n[/math]
here you go brainlet

>> No.9328692
File: 82 KB, 680x680, f5753870a40ccef114a6cb88e7f48531.jpg [View same] [iqdb] [saucenao] [google] [report]

It's honestly a little depressing how there is an entire branch of math dedicated to truncating infinity while pretending they're still working with infinity. This is like the gender studies of maths.

>> No.9328792
File: 18 KB, 211x239, wojak spastic.gif [View same] [iqdb] [saucenao] [google] [report]

>WTF is limit

>> No.9328807

>truncating infinity
The fuck are you rambling about nigger, it's just elementary abstract algebra. You just define a group structure on the set of functions from the naturals to the reals, aka sequences, based on the group structure of the reals. Is universal quantification too magical for you?

>> No.9328811

>doing subtraction via two's complement
You mean 2-adic numbers.

>> No.9328870
File: 90 KB, 848x480, 1494816445491.png [View same] [iqdb] [saucenao] [google] [report]

>"truncating "
In general, the set of functions from any fixed set to a group has a group structure. Sequences are merely a special case of this.
No such thing.

>> No.9328991

I know what sums are, but this is so absurd that I just want to make sure.

It says 1+2+3+4+... is -1/12 when you never stop summing up, right?

How can a sum be negative if none of it's addends is negative and how can it's absolute be smaller than it's smallest addend? Explain for a brainlet please

>> No.9329000

it's not convergence of series of real numbers. it's a much more subtle notion, analytic continuation of a complex function. it turns out a power series uniquely defines an analytic function, and so that series on the left defines a function even outside the area where it converges

>> No.9329004
File: 157 KB, 1024x576, 1470910645307.jpg [View same] [iqdb] [saucenao] [google] [report]

>I know what sums are

>> No.9329540

fuck off back to >>>/pol/

>> No.9329549

I don't like left-wing boards, sorry.

>> No.9329615

If you could point to when I called anyone an idiot, that'd be enlightening.

>> No.9329620
File: 125 KB, 500x358, 1510270439396.png [View same] [iqdb] [saucenao] [google] [report]

Why was the post deleted

>> No.9329706

>No such thing.
how does the hare catch the tortoise then?

>> No.9329721

That's more of an engineering question, ask the subhumans.

>> No.9329755

>t. I don't know shit

>> No.9329796

How can anybody be this delusional?

>> No.9329798

>This is like the gender studies of maths.
Now that's an expression I don't hear very often.

>> No.9329861

just answer the fucking question

>> No.9329879

>When it comes to math, nothing should be abstract. That would just defeat the purpose of math.
please don't tell me engineers unironically believe this

>> No.9330732

learn to fucking read, faggot

>> No.9330755
File: 93 KB, 1194x884, ping pong anime screenshot.png [View same] [iqdb] [saucenao] [google] [report]


>> No.9330762

you're a retard

>> No.9330772

>even if you had four orders of eternity to spend writing
just one eternity would suffice, faggot

>> No.9330775

>literally trying to perform arithmetic on infinity
have you literally never heard of ordinal arithmetic you colossal waste of matter?

>> No.9330779

AC is independent of ZF, jackass
i can reject it all day long and have valid results

>> No.9330827

You can reject ZF all day long and have valid results.

>> No.9330859

not within ZF

>> No.9330868

>implying anyone mentioned doing something internally to ZF
Engineer spotted.

>> No.9331079

save us anon with the beacon of your cranium

>> No.9331149
File: 67 KB, 600x738, 600px-Omega-exp-omega-labeled.svg.png [View same] [iqdb] [saucenao] [google] [report]

This is unfuckingironically the most retarded thing I have seen for 2017.
Holy shit, I thought the new nintendo was bad but this takes the cake.

Math needs government regulation so retards can't get away with making up garbage like this.

>> No.9331154

And yet you'll never be rid of that fraction on the end no matter how far you go.

>> No.9331162

don't need to
the sum is already at 1

your job is to prove that it ever deviates from 1

>> No.9331165
File: 81 KB, 315x360, 1511761761718.jpg [View same] [iqdb] [saucenao] [google] [report]

.999... > 1.0

>> No.9331169

try again after you've taken your medication

>> No.9331170

What is that in the image?

>> No.9331177

>colossal waste of matter
underage edgelords like you have polluted this board

>> No.9331178
File: 18 KB, 384x384, 1487360360836.jpg [View same] [iqdb] [saucenao] [google] [report]

Its right though. Your shitmath has to assume an end. In finite arithmetic it works, caus there will be an end. In infinite arithmetic it will not work. There will always be one more 9 to calculate, so any truncating of the infinite repetition to make it finite would carry the implication that there was one more 9 you couldn't do arithmetic with, thus 1.000[...]9
And 1.000[...]9 > 1.0

I'm not sorry, i have nothing to apologize for. You should be disapppointed and angry with the tryhard brainlet professors who lie to you and say you can operate on infinite amounts.

>> No.9331183

>has to assume an end
nope, at infinity there are infinite 9s
and the last part is 1/inf = 0

and all the time, the sum = 1

>> No.9331191
File: 175 KB, 600x600, 58b.png [View same] [iqdb] [saucenao] [google] [report]

> 1 / ∞ = 0
Oh i see where you were confused. See your problem was that you were just retarded the whole time.
1 / ∞ = ω

0.999[...] + ω = 1.0

>> No.9331226

The point is that there is no drifting in the sum.
Of course we are talking about limits here.
That is obvious from the very beginning, what else would "0.999..." even mean.

>> No.9331228

>brainlets can't into the best field of mathematics
>set theory is appropriately reserved for those with the highest taste and intelligence
all is well in the world

>> No.9331238

1/3 = (3/10 + 1/30)... = 0.333r
3/3 = (9/10 + 1/10) = 1.0
+ (9/100 + 1/100) = 1.1 ... = 1.111r


>> No.9331241

3/10 + 1/30
= 9/30 + 1/30
= 10/30 = 1/3

congratz, you just went full retard

>> No.9331250
File: 102 KB, 300x256, 1486363850107.png [View same] [iqdb] [saucenao] [google] [report]

You wanna take another look at that post?

>> No.9331258


>> No.9331279

Serious question: what would anons that have taken it say are appropriate prerequisites for complex analysis? Not just to understand it, but to appreciate it?

Assuming of course, the typical undergraduate calculus courses are completed. What else should be taken prior? Or is it okay to take it right off like my college lets me do?

>> No.9331763

real analysis and a GOOD grasp of topology (point-set)

>> No.9331889

e is an infinite series
so are the trig functions

so when you do sin(2) you're doing arithmetic with infinity that is generally considered beneficial

>> No.9331895

but that's a sin
jesus told me so

>> No.9332239

No such thing.
>r*ddit image
Fuck off.

>> No.9332272

The question isn't math related so I don't give a fuck.

>> No.9332340

>t. don't know a fuck

>> No.9333412

He's awesome

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