text:
[ 3 / biz / cgl / ck / diy / fa / g / ic / jp / lit / sci / tg / vr ] [ index / top / reports / report a bug ] [ 4plebs / archived.moe / rbt ]

2017/01/28: An issue regarding the front page of /jp/ has been fixed. Also, thanks to all who contacted us about sponsorship.

# /sci/ - Science & Math

[ Toggle deleted replies ]
File: 15 KB, 479x282, natural number.jpg [View same] [iqdb] [saucenao] [google] [report]

>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?

 >> Anonymous Fri Nov 24 21:55:35 2017 No.9319225 >>9319219ramanujan 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.
 >> Anonymous Fri Nov 24 22:03:12 2017 No.9319234 >>9319219Nobody 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.
 >> Anonymous Fri Nov 24 22:13:19 2017 No.9319247 >>9319225>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.
 >> Anonymous Fri Nov 24 22:23:20 2017 No.9319258 File: 254 KB, 1300x3900, math trench.jpg [View same] [iqdb] [saucenao] [google] [report] >>9319219you forgot:>die of dysentery
 >> Anonymous Fri Nov 24 22:24:25 2017 No.9319261 File: 54 KB, 197x259, leface.png [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Fri Nov 24 22:32:57 2017 No.9319268 >>9319247incorrect 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.
 >> Anonymous Fri Nov 24 22:41:51 2017 No.9319284 File: 4 KB, 270x186, download.png [View same] [iqdb] [saucenao] [google] [report] >>9319247OP take complex analysis.
 >> Anonymous Fri Nov 24 22:45:10 2017 No.9319289 >>9319247Just 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.
 >> Anonymous Fri Nov 24 23:10:56 2017 No.9319321 >>9319247There's nothing incorrect about it.s=(1+2+3+4+...)4s=(4+8+12+16+...)s-4s=(1+2+3+4+...)-(4+8+12+16+...)-3s=1-2+3-4+...4*-3s=4*(1-2+3-4+...)-12s=(1-2+3-4+...)+(1-2+3-4+...)+(1-2+3-4+...)+(1-2+3-4+...)-12s=(1-2+3-4+...)+1+(-2+3-4+5+...)+1+(-2+3-4+5+...)+(1-2)+(3-4+5-6+...)-12s=(1-2+3-4+...)+1+(-2+3-4+5...)+1+(-2+3-4+5+...)+-1+(3-4+5-6+...)-12s=1+(1-2+3-4+...)+(-2+3-4+5+...)+(-2+3-4+5+...)+(3-4+5-6+...)-12s=1+((1-2-2+3)+(-2+3+3-4)+(3-4-4+5)+(-4+5+5-6)+...)-12s=1+(0+0+0+0+...)-12s=1s=-1/12
 >> Anonymous Fri Nov 24 23:15:55 2017 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
 >> Anonymous Fri Nov 24 23:19:02 2017 No.9319336 >>9319268Could quantum mechanics be a continuation of an incorrect axiom that has no bearing on real life and our observations are simply limited by hard?
 >> Anonymous Fri Nov 24 23:19:28 2017 No.9319338 >>9319321That'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
 >> Anonymous Fri Nov 24 23:20:05 2017 No.9319340 >>9319330>performing arithmatic with infinityLiterally nothing wrong with that except for the part where you're a brainlet who doesn't know how to spell the word "arithmetic."
 >> Anonymous Fri Nov 24 23:23:14 2017 No.9319344 >>9319338>That's a dumb way to attempt to prove itThat'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.
 >> Anonymous Fri Nov 24 23:23:43 2017 No.9319346 >>9319340okay enjoy doing fake math to the benefit of no one, retard.
 >> Anonymous Fri Nov 24 23:31:04 2017 No.9319355 >>9319346>DURR NO BENEFITShttp://lmgtfy.com/?q=signal+processing+ramanujan+sum
 >> Anonymous Fri Nov 24 23:37:58 2017 No.9319370 >>9319344and 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.Retard.
 >> Anonymous Fri Nov 24 23:44:29 2017 No.9319381 >>9319370It 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.
 >> Anonymous Fri Nov 24 23:54:56 2017 No.9319390 File: 4 KB, 156x232, mathBad.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Sat Nov 25 00:03:29 2017 No.9319401 >>9319381I 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 placeYou 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.
 >> Anonymous Sat Nov 25 00:04:08 2017 No.9319403 >>9319390The Ramanujan summation doesn't make the mistake that picture depicts.
 >> Anonymous Sat Nov 25 00:05:41 2017 No.9319408 >>9319403what mistake?it's my theory so fuck off
 >> Anonymous Sat Nov 25 00:08:03 2017 No.9319413 >>9319401>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.
 >> Anonymous Sat Nov 25 00:11:27 2017 No.9319418 >>9319219-1/12 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.
 >> Anonymous Sat Nov 25 00:13:57 2017 No.9319422 >>9319418That's not the only way to get to that -1/12 value.
 >> Anonymous Sat Nov 25 00:31:48 2017 No.9319437 >>9319413woops, I meant counter-argument to >>9319247I clicked on the wrong reply>and the way you were calling "dumb" was in fact Ramanujan's way of deriving that valueNo 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 itis not an argument.
 >> Anonymous Sat Nov 25 00:37:14 2017 No.9319443 >>9319422that is the way you're supposed to get it, not that numberphile bullshit
 >> Anonymous Sat Nov 25 00:54:35 2017 No.9319454 >>9319418you mean a complex curve, smooth curves aren't rigid like that
 >> Anonymous Sat Nov 25 01:02:05 2017 No.9319459 >>9319321I'm really sure that's not how ramanujan did it. this is just retarded
 >> Anonymous Sat Nov 25 01:06:40 2017 No.9319461 >>9319336Well 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.
 >> Anonymous Sat Nov 25 01:11:39 2017 No.9319463 >>9319443>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 resultI hate numberphile so much
 >> Anonymous Sat Nov 25 01:12:15 2017 No.9319464 street shitters invented your number system, R haplo rapebaby.
 >> Anonymous Sat Nov 25 01:18:22 2017 No.9319468 >>9319454doesn't look complex to mehttps://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#/media/File:Sum1234Summary.svg
 >> Anonymous Sat Nov 25 01:22:06 2017 No.9319473 >>9319468smooth curves aren't rigid like that. it has much more properties than just "a smooth curve", which could give literally any value.
 >> Anonymous Sat Nov 25 01:31:30 2017 No.9319486 >>9319330This
 >> Anonymous Sat Nov 25 01:32:45 2017 No.9319488 >>9319463Numberphile'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.
 >> Anonymous Sat Nov 25 01:50:23 2017 No.9319514 >>9319473better tell Wikipedia then that they have been gaslighted by the Illuminati
 >> Anonymous Sat Nov 25 03:59:24 2017 No.9319664 >>9319321Wrong. 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.
 >> Anonymous Sat Nov 25 06:33:56 2017 No.9319850 File: 50 KB, 645x729, 1507450280553.png [View same] [iqdb] [saucenao] [google] [report] >>9319321>s-4s=(1+2+3+4+...)-(4+8+12+16+...)>-3s=1-2+3-4+...
 >> Anonymous Sat Nov 25 06:44:15 2017 No.9319866 >>9319488somtimes brady spontaneously becomes a brainlet
 >> Anonymous Sat Nov 25 07:21:16 2017 No.9319899 File: 59 KB, 645x729, 1507205929387.png [View same] [iqdb] [saucenao] [google] [report] >>9319850>>9319866>>9319488>>9319418It 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?
 >> Anonymous Sat Nov 25 08:16:47 2017 No.9319962 The $-\frac1{12}$ meme is dead, anon.
 >> Anonymous Sat Nov 25 10:27:46 2017 No.9320109 >>9319899Because if you use invalid operations, you can get anything you want.
 >> Anonymous Sat Nov 25 12:33:54 2017 No.9320376 >>9319321Now write that Argument without those ... more formally.>>9319219$\sum\limits_{n=1}^{\infty} n = \lim_{m\to\infty} {\left(\sum\limits_{n=1}^{m} n\right)} = \infty$
 >> Anonymous Sat Nov 25 14:12:56 2017 No.9320652 >>9319321Adding/Substracting/Multiplying/Dividing infinite sums is only defined if the limit is a real number (no, infinity is not a number)Try again, sweetie
 >> Anonymous Sat Nov 25 14:18:23 2017 No.9320665 >>9319234>Nobody actually thinks that formula is true.It's not more or less true than a statement likee^3 > 4Those 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)
 >> Anonymous Sat Nov 25 15:10:03 2017 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 = 8.999r1Upon 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 = 3and 5 = 3then 1 = 5To 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.
 >> Anonymous Sat Nov 25 15:12:15 2017 No.9320841 >>9320837But 0.999r9 = 1 is true by definition.
 >> Anonymous Sat Nov 25 15:12:51 2017 No.9320843 >>9319321Someone needs to take some analysis. It is true, however, that every sequence converges in an ultrafilter, so perhaps you have that $\sum^n_{k=1} k \to \frac{-1}{12}$ in that sense.
 >> Anonymous Sat Nov 25 15:15:37 2017 No.9320847 >>9320837I don't think you understand how numbers work.
 >> Anonymous Sat Nov 25 15:19:33 2017 No.9320854 File: 336 KB, 629x468, 4bOGtMo.png [View same] [iqdb] [saucenao] [google] [report] >>9320847Prove it wrong
 >> Anonymous Sat Nov 25 15:21:09 2017 No.9320860 >>9319247Read about Casimir effect, it is an experimentally proved physical effect whose theory uses directly this result.
 >> Anonymous Sat Nov 25 15:21:30 2017 No.9320862
 >> Anonymous Sat Nov 25 15:28:05 2017 No.9320882 >>9319899>how does making that mistake produce the same answer?Because it's not a mistake.>>9320109>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.
 >> Anonymous Sat Nov 25 15:31:31 2017 No.9320891 >>93208621/3 * 3 = 1, by definition1/3 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?
 >> Anonymous Sat Nov 25 15:33:31 2017 No.9320899 File: 37 KB, 569x506, wj.jpg [View same] [iqdb] [saucenao] [google] [report] >>9320891This 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.
 >> Anonymous Sat Nov 25 15:35:00 2017 No.9320901 >>9320891If 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.
 >> Anonymous Sat Nov 25 15:35:36 2017 No.9320905 File: 66 KB, 554x400, 1473433322140.png [View same] [iqdb] [saucenao] [google] [report] >>9320899PROVE IT WRONG MONGOLOIDSAYING "UR DUMB" IS NOT AN ARGUMENT
 >> Anonymous Sat Nov 25 15:38:25 2017 No.9320912 >>9320901Thats 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=1By logic, 0.999r9 does not actually equal 1.
 >> Anonymous Sat Nov 25 15:38:42 2017 No.9320914 >>9320905>PROVE IT WRONGThere 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.
 >> Anonymous Sat Nov 25 15:40:18 2017 No.9320918 >>93209050.9999... = 0.3333... * 3 = 1/3 * 3 = 1There'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
 >> Anonymous Sat Nov 25 15:40:25 2017 No.9320919 >>9320912Inside 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.
 >> Anonymous Sat Nov 25 15:44:35 2017 No.9320926 >>9320854>prove it wrongWhatever 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.
 >> Anonymous Sat Nov 25 15:45:38 2017 No.9320929 >>9320854>If 0.999r9 = 1, then 3 is equal to 1, 2, 3, 4, 5, 6, 7, 8, or 9Why is that true?
 >> Anonymous Sat Nov 25 15:47:19 2017 No.9320930 >>9320918How 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 = 0.30.3×3 = 0.9claiming 0.999... = 1 is literally the same claim as 0.9 = 1, or 9/10 = 10/10you cannot perform arithmetic on infinite sequences, regardless of ANY AND ALL mental gymnastics claiming otherwise.
 >> Anonymous Sat Nov 25 15:48:26 2017 No.9320932 >>9320854>1.999r8You can give up here, this doesn't mean anything.
 >> Anonymous Sat Nov 25 15:48:50 2017 No.9320934 >>9320929Because 0.333r3 must equal to 0.333rN to eradicate rounding errors from truncating the infinite sequence as shown in the (A)~( I ) results.
 >> Anonymous Sat Nov 25 15:50:22 2017 No.9320937 >>9320934Your mistake, if you're being serious, is thinking that there's any rounding going on at all.
 >> Anonymous Sat Nov 25 15:52:05 2017 No.9320942 >>9320930>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.
 >> Anonymous Sat Nov 25 15:52:27 2017 No.9320944 >>9320930>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
 >> French_doesn't_suxx Sat Nov 25 15:52:43 2017 No.9320945 It can't be true. Because if S=1+2+3+...So S=1+(1+1)+(1+1+1)+...S=1+1+1+...Ramanujan was a bastard
 >> Anonymous Sat Nov 25 15:53:02 2017 No.9320947 >>93209320.333r3×2 = 0.666r60.666r6×3 = 1.999r86×3 = 18 <- see that 8?holy fuck i hope you arent attending a university much less stolen a degree.
 >> Anonymous Sat Nov 25 15:56:57 2017 No.9320953 >>9320945You can't do that. This is why zeta regularization limits you to working with term variables, so you don't do shit like that.
 >> Anonymous Sat Nov 25 16:02:59 2017 No.9320965 >>9319261underrated, saved
 >> Anonymous Sat Nov 25 16:03:32 2017 No.9320967 >>9320947The 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 0.3827261028r7we 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
 >> Anonymous Sat Nov 25 16:13:33 2017 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.
 >> Anonymous Sat Nov 25 16:20:55 2017 No.9321010 >>9320992Infinite 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
 >> Anonymous Sat Nov 25 16:24:36 2017 No.9321018 >>9320109Get 6.5 then. Go!
 >> Anonymous Sat Nov 25 16:37:10 2017 No.9321040 >>9320882No, 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.
 >> Anonymous Sat Nov 25 16:38:12 2017 No.9321042 >>9321040>4n will always be longeroffset, ratherlonger was the wrong word.
 >> Anonymous Sat Nov 25 17:01:22 2017 No.9321088 File: 41 KB, 300x400, Emma-Stone-normalize.jpg [View same] [iqdb] [saucenao] [google] [report] >>9319899The -1/12 also exists in classical analysis and already Euler had the most general result on it. Consider the difference between integral and sum,$\int _m^n f(x)\, {\rm d}x=\sum _{i=m}^n f(i)-\frac 1 2 \left( f(m)+f(n) \right) -\frac 1{12}\left( f'(n)-f'(m)\right) + \frac 1{720}\left( f'''(n)-f'''(m)\right) + \cdots$where the coefficients are in simple way computed from Bernoulli numbers. A child of this would be e.g.$\frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$ 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$\sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1))$ 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
 >> Anonymous Sat Nov 25 17:26:58 2017 No.9321154 >>9319219>shits and gigglesfound the urban redneck
 >> Anonymous Sat Nov 25 17:29:06 2017 No.9321159 >>9319321i'll do you one better$\displaystyle\ln\left(1+x\right)=x-\frac {x^2} {2} + \frac {x^3} {3} - \frac {x^4} {4} ...$$\displaystyle \ln\left(2\right) = 1-\frac {1} {2} + \frac {1} {3} - \frac {1} {4} ...$consider the series: $\displaystyle S = 1-\frac {1} {2} - \frac {1} {4} + \frac {1} {3} - \frac {1} {6} - \frac {1} {8} + \frac {1} {5} - \frac {1} {10} - \frac {1} {12} + ...$it's obvious that S and ln(2) have the same terms, just reordered in a new pattern.for clarification, here's it in series form$\displaystyle \ln(2) = \sum \{1, -2^{-1}, 3^{-1}, -4^{-1}, 5^{-1}, -6^{-1}, 7^{-1}, -8^{-1}, 9^{-1}, -10^{-1}... \}$$\displaystyle S = \sum \{1, -2^{-1}, -4^{-1}, 3^{-1}, -6^{-1}, -8^{-1}, 5^{-1}, -10^{-1}, -12^{-1}... \}$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: $\displaystyle\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)} - \frac {1} {4x}$x - x/2 = x, and similarly, $\displaystyle\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)}=\frac {1} {2\left(2x-1\right)}$so we can rewrite S as the following: $\displaystyle \frac {1} {2} - \frac {1} {4} + \frac {1} {6} - \frac {1} {8} + \frac {1} {10} - ...$or $\displaystyle \frac {1} {2} \left( 1- \frac {1} {2} + \frac {1} {3} - ... \right)$ = $\displaystyle \frac {1} {2} \ln \left( 2 \right)$even though ln(2) and S have the same terms, just in a different order.
 >> Anonymous Sat Nov 25 17:41:02 2017 No.9321176 File: 393 KB, 342x342, 1501701688178.gif [View same] [iqdb] [saucenao] [google] [report] 1/3 = 0.333[32/3 = 0.666[63/3 = 0.999[93/3 = 1Therefore1 = 0.999[9this works when referencing the thirds up to the whole. Referencing thirds beyond the whole is different.4/3 = 1.333[2By motive of proving 0.999[9 = 1, adding 0.333[3 to 0.999[9 results in 1.333[21.333[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) = 11 - (ㅌ × 3) = 0.999[90.999[9 =/= 1ㅌ × 3 =/= 0
 >> Anonymous Sat Nov 25 17:47:04 2017 No.9321184 >>9321176>1 + 0.33333... = 1.333332...Really convinced me there
 >> Anonymous Sat Nov 25 17:52:28 2017 No.9321193 >>9319219Oh but you dont have any problem when using limits right fucking faggot?
 >> Anonymous Sat Nov 25 17:55:34 2017 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 * nand we also know that the sum of all positive integers is -1/12so(infinity-1)/2 * infinity = -1/121/2*infinity^2 -1/2*infinity + 1/12 = 0infinity^2 - infinity + 1/6 = 0infinity = +1/2 +/- sqrt((1/2)^2 - 1/6)= either 1/2+1/sqrt(12) or 1/2-1/sqrt(12) if i calculated correctly.
 >> Anonymous Sat Nov 25 17:55:38 2017 No.9321202 File: 207 KB, 692x960, 1510723646424.jpg [View same] [iqdb] [saucenao] [google] [report] >>9321184>.9999 = 1 defacto!!!!really niggers my noggin.
 >> Anonymous Sat Nov 25 18:03:56 2017 No.9321220 >>9321202Theres no numbers between 0.99... and 1But every real number haves at least one number between them if they are different, therefore they must be equal
 >> Anonymous Sat Nov 25 18:08:04 2017 No.9321228 >>93212200.991 is closer to 1 than 0.990.99999991 is closer to 1 than0.9999999>haves
 >> Anonymous Sat Nov 25 18:25:11 2017 No.9321261 >>9319850I 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.
 >> Anonymous Sat Nov 25 18:31:37 2017 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] >>9319321s=(1+2+3+4+...)4s=(4+8+12+16+...)s-4s=(1+2+3+4+...)-(4+8+12+16+...)-3s=1-2+3-4+...>-3s=1-2+3-4-3s = -3-6-9-12-...did you just shit your brains for a second or what.
 >> Anonymous Sat Nov 25 20:13:40 2017 No.9321492 >>9321273No, 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.
 >> Anonymous Sat Nov 25 20:17:57 2017 No.9321503 >>9320944>If you had any experience with calculus you would knowironic. If you had any experience with calculus, you would know it has nothing to do with aritmetic on infinite series.
 >> Anonymous Sat Nov 25 20:18:20 2017 No.9321505 >>9321492I'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.
 >> Anonymous Sat Nov 25 20:19:27 2017 No.9321510 >>9321505>>9321492I could be a complete idiot though. I don't know, but it certainly doesn't make notational sense.
 >> Anonymous Sat Nov 25 20:33:05 2017 No.9321543 >>9321176( r × n ) + ( ㅌ × n ) = jWhere 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[9In this sense, ㅌ would only be an inverse of $\infty$. If $\infty$ defines some garantuan unthinkable uncountable thing, then ㅌ defines an abstract remainder to be applied to end an infinite sequence, where $\infty$ + ㅌ = ㅐ, letting ㅐ be equal to this abstract concept of $\infty$ + ㅌ.The problem with $\infty$ 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.
 >> Anonymous Sat Nov 25 20:33:59 2017 No.9321545 0.999...=x9.999...=10x 10x-x=9.999...-0.999...9x=9x=10.999...=1
 >> Anonymous Sat Nov 25 21:08:12 2017 No.9321630 >>9321545You 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 $\infty$ in order to truely make it equal to 1.
 >> Anonymous Sat Nov 25 21:26:24 2017 No.9321672 >>93216300.333... × 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...7if we use ㅌ as an abstravt concept to define some kind of inverse of $\infty$, then apply it back to those statements, we get this:0.333... × 2.999... = 0.999...7+(2ㅌ) = 10.333... × 3 = 0.999...+ㅌ = 1where, because $\infty$ 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...
 >> Anonymous Sat Nov 25 21:46:30 2017 No.9321709 >>9321630Why?
 >> Anonymous Sat Nov 25 21:49:03 2017 No.9321720 File: 13 KB, 210x240, skeeter-valentine-doug-8.55.jpg [View same] [iqdb] [saucenao] [google] [report] >>9321545Lets take your example and apply ㅌ to it0.999...=x+ㅌ9.999...=10x +10ㅌ10x-x +(10ㅌ-ㅌ) =9.999...-0.999...9x+9ㅌ=9x+ㅌ=10.999...+ㅌ=1Where your original problem makes a fault is here at>10x-x=9.999...-0.999...>9x=9>x=1>0.999...=1x = 0.999[910x = 9.999[9But 10x-x, 9x, does not equal 9.09x = 8.999[1Your 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 0.999[9+0.999[9_____________1.999[8if 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
 >> Anonymous Sat Nov 25 21:58:00 2017 No.9321744 >>9321630>requires the intentional inclusion or addition of another abstract value to rival the concept of ∞>>9321672>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.
 >> Anonymous Sat Nov 25 22:17:03 2017 No.9321793 >>9321744Fuck 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 = 10.999...+ㅌ = 1excluding 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.
 >> Anonymous Sat Nov 25 23:08:44 2017 No.9321926 >>9319258>game theory harder than differential equations What a load of shit
 >> Anonymous Sat Nov 25 23:11:02 2017 No.9321932 >>9321793Infinite decimal expansions have rigorously defined meanings, they are equivalent to infinite summations.$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]  >> Anonymous Sat Nov 25 23:12:29 2017 No.9321936 >>9321926It 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.  >> Anonymous Sat Nov 25 23:12:29 2017 No.9321937 >>9321793>>9321932Fuck[math]0.\overline{9} \equiv \sum_{k=1}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k$$0.\overline{9} = \sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k - 9$$0.\overline{9} = 9 \left( \frac{1}{1-\frac{1}{10}} \right) - 9$$0.\overline{9} = 1$
 >> Anonymous Sat Nov 25 23:15:56 2017 No.9321948 >>9321630\displaystyle\begin{align*} \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}}} \\&\vdots\end{align*}\\ \displaystyle\Rightarrow 0.\overline{3} = \frac{1}{3}
 >> Anonymous Sat Nov 25 23:18:04 2017 No.9321954 >>9321936But 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
 >> Anonymous Sun Nov 26 14:34:39 2017 No.9323334 >>9321948The 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 = 1.0if 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 = 1Your 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 via0.333[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?cont.
 >> Anonymous Sun Nov 26 14:50:00 2017 No.9323361 >>9323334>>9321948that 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[2Performing 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.
 >> Anonymous Sun Nov 26 15:15:26 2017 No.9323411 >>9323334I thought you were trolling yesterday, but this is getting desperate.
 >> Anonymous Sun Nov 26 15:37:57 2017 No.9323463 >>9323361This 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[2But 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.
 >> Anonymous Sun Nov 26 15:41:15 2017 No.9323475 File: 66 KB, 600x471, 1511085854684.jpg [View same] [iqdb] [saucenao] [google] [report] >>9323411Nah, 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.
 >> Anonymous Sun Nov 26 15:57:17 2017 No.9323523 >>9323475You need to get your head around irrational numbers, buddy.
 >> Anonymous Sun Nov 26 16:32:13 2017 No.9323625 >>9319321How did you get from this:>s-4s=(1+2+3+4+...)-(4+8+12+16+...)to this:>-3s=1-2+3-4+...?Shouldn't it be -3s = 1-4+2-8+3-12+... ?
 >> Anonymous Sun Nov 26 16:40:01 2017 No.9323660 >>9323625See >>9321505I 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.
 >> Anonymous Sun Nov 26 16:47:48 2017 No.9323688 >>9323625In 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.
 >> Anonymous Sun Nov 26 18:22:10 2017 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.
 >> Anonymous Sun Nov 26 18:28:14 2017 No.9323921   >>9323361>uncountable>>9321948 isn't uncountableevery line = 1line #1 = 1line #10 = 1line #982376487236823498 = 1at infinity, it still is 1
 >> Anonymous Sun Nov 26 18:30:27 2017 No.9323929   >>9323361>uncountable>>9321948 (You) isn't uncountableevery line is exactly 1/3line #1 = 1/3line #10 = 1/3line #982376487236823498 = 1/3at infinity, it still is 1/3
 >> Anonymous Sun Nov 26 18:32:44 2017 No.9323938 >>9323361>uncountable>>9321948 isn't uncountableevery line is exactly 1/3line #1 = 1/3line #10 = 1/3line #982376487236823498 = 1/3at infinity, it still is 1/3
 >> Anonymous Sun Nov 26 18:34:43 2017 No.9323945
 >> Anonymous Sun Nov 26 18:35:40 2017 No.9323950 File: 7 KB, 645x773, Thatface20110725-22047-wlaopv.png [View same] [iqdb] [saucenao] [google] [report] >>93239000.999... 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.
 >> Anonymous Sun Nov 26 18:40:36 2017 No.9323961 File: 34 KB, 601x508, e7db9e4e33592de521b23e955084488e.jpg [View same] [iqdb] [saucenao] [google] [report] >>93239500.999... approaching 1 is like living while thinking there is an after life.
 >> Anonymous Sun Nov 26 18:50:37 2017 No.9323982 >>9319219wtf the limit of that sum is infinite, how does this make any sense?
 >> Anonymous Mon Nov 27 01:38:03 2017 No.9324720 >>9321948This 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
 >> Anonymous Mon Nov 27 01:42:35 2017 No.9324725 >>9323982It makes sense in a different system than the one you’re used to.
 >> Anonymous Mon Nov 27 01:50:36 2017 No.9324730 >>9319338>1=.999... by dividing 1 by 3.1= 3/31/3=.333r1/3 + 1/3rd +1/3rd = 3/3rd.333r+.333r+.333r= .999rtherefor .999r=1does this upset you
 >> Anonymous Mon Nov 27 01:51:45 2017 No.9324734 >>9324720if it works, it worksno matter what bs you say
 >> Anonymous Mon Nov 27 01:58:22 2017 No.9324738 >>9321176the truth is 1 is rounding of .999r9
 >> Anonymous Mon Nov 27 02:00:42 2017 No.9324740   >>9324734 > 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
 >> Anonymous Mon Nov 27 02:03:37 2017 No.9324741 >>9324734Thats 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
 >> Anonymous Mon Nov 27 02:04:33 2017 No.9324743 >>9324720>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
 >> Anonymous Mon Nov 27 02:06:17 2017 No.9324744 >>9324741>it doesn't workwhat line is faulty
 >> Anonymous Mon Nov 27 02:06:45 2017 No.9324747 >>9324743You think a broken clock being right for two seconds a day suffices as an accurate clock. Plug 1/7 into that equation shitstick.
 >> Anonymous Mon Nov 27 02:12:45 2017 No.9324750 >>9324747nah I'm good you admitted it's right
 >> Anonymous Mon Nov 27 02:15:36 2017 No.9324752 >>9323660can someone take a look at >>9323660 and>>9321505 >>9319850maybe >>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.
 >> Anonymous Mon Nov 27 02:16:03 2017 No.9324754 File: 33 KB, 600x700, Pool.jpg [View same] [iqdb] [saucenao] [google] [report] >>9324750Jesus, how delusional are you. I've called it wrong multiple times now.How fucking sad is it to exist as you?
 >> Anonymous Mon Nov 27 02:16:43 2017 No.9324756
 >> Anonymous Mon Nov 27 02:24:47 2017 No.9324774 >>9324756The 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
 >> Anonymous Mon Nov 27 02:39:00 2017 No.9324788 >>9324774>The method is falsewell it's fucking easy to point out the faulty line then, isn't it
 >> Anonymous Mon Nov 27 02:46:20 2017 No.9324792 File: 142 KB, 617x347, 1509768568403.png [View same] [iqdb] [saucenao] [google] [report] >>9324788EVERY LINE IS FAULTY YOU STUPID PRICK, THE METHOD ENTIRELY IS WRONG.THE TOP LINE, FOR EXAMPLE, 0.3+ 1/30THAT IS {0.3} + 0.0333[3, AND EVERY LINE THERAFTER JUST ADDS ONTO THE PROBLEM FROM {0.33} TO {0.3333}O K ?you are injecting 3's into the answer then adding on an infinite repeating 3 decimal.1/30 = 0.0333[31/300 = 0.00333[31/3000 = 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
 >> Anonymous Mon Nov 27 02:51:05 2017 No.9324802 >>9324792omg we went over this already>>9324743
 >> Anonymous Mon Nov 27 02:59:09 2017 No.9324810 >>9319219does it hurt you that much that someone with only a fraction of resources compared to you accomplished something you couldn't in multiple lifetimes?
 >> Anonymous Mon Nov 27 03:01:37 2017 No.9324814 File: 73 KB, 334x319, 1508566033111-1.jpg [View same] [iqdb] [saucenao] [google] [report] >>93248021/n = ( ((floor(10/N))/10) + (1/(10×n)) )1/3 = (3/10 + 1/30) = 0.{3}333[31/7 = (1/10 + 1/70) = 0.{1}142857[142857When you plug in 1/7, you get 0.1142857...meanwhile, 1/7 = 0.142857...Real 1/7 = 0.142857your 1/7 = 0.114285do you comprende
 >> Anonymous Mon Nov 27 03:02:58 2017 No.9324817 >>9324814yeah your method is shit is seems
 >> Anonymous Mon Nov 27 03:04:35 2017 No.9324822 File: 7 KB, 420x420, b36.png [View same] [iqdb] [saucenao] [google] [report] >>9324817You should invest in bitcoin
 >> Anonymous Mon Nov 27 03:07:09 2017 No.9324823 >>9324792>you are injecting1 = 1/2 + 1/2not "injecting" anything, just cutting it up
 >> Anonymous Mon Nov 27 03:14:53 2017 No.9324829 >>93248230.5 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
 >> Anonymous Mon Nov 27 03:15:39 2017 No.9324832 File: 10 KB, 604x658, NegativeZeta.gif [View same] [iqdb] [saucenao] [google] [report] >>9319219It's black magic.
 >> Anonymous Mon Nov 27 03:21:16 2017 No.9324843 >>9324829you 10-based retard
 >> Anonymous Mon Nov 27 03:22:55 2017 No.9324844   Here's your fucking 7, snow-white\displaystyle\begin{align*} \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}}} \\&\vdots\end{align*}\\ \displaystyle\Rightarrow 0.\overline{142857} = \frac{1}{7}
 >> Anonymous Mon Nov 27 03:24:28 2017 No.9324846 File: 60 KB, 1024x558, 1509533757591m.jpg [View same] [iqdb] [saucenao] [google] [report] >>9324843>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"
 >> Anonymous Mon Nov 27 03:31:22 2017 No.9324850 >>9324846you and your stupid floorsjust point out which line in >>9321948 fails
 >> Anonymous Mon Nov 27 03:32:42 2017 No.9324853   here is the 1/7 case\displaystyle\begin{align*} \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}} \\&\vdots\end{align*}\\ \displaystyle\Rightarrow 0.\overline{142857} = \frac{1}{7}
 >> Anonymous Mon Nov 27 03:38:37 2017 No.9324857 \displaystyle\begin{align*} \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}} \\&\vdots\end{align*}\\ \displaystyle\Rightarrow 0.\overline{142857} = \frac{1}{7}
 >> Anonymous Mon Nov 27 03:46:06 2017 No.9324867 >>9324829in base 3one-third is "0.1"Exactly.So now what, is 10 magic because that's what jesus gave us?
 >> Anonymous Mon Nov 27 04:24:12 2017 No.9324912 File: 20 KB, 306x306, 79e0f76f22f429dc6a79ebbe8f3b484d.jpg [View same] [iqdb] [saucenao] [google] [report] >>9324857desperate
 >> Anonymous Mon Nov 27 04:28:02 2017 No.9324917
 >> Anonymous Mon Nov 27 04:52:10 2017 No.9324951 File: 110 KB, 318x275, 1508479619038.png [View same] [iqdb] [saucenao] [google] [report] >>9319219>infinityNo such thing.
 >> Anonymous Mon Nov 27 04:53:53 2017 No.9324954 >>9320901>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?
 >> Anonymous Mon Nov 27 04:55:42 2017 No.9324958 >>9319219>never taken a real math course in my life the thread
 >> Anonymous Mon Nov 27 04:56:33 2017 No.9324960 >>9324958>real mathIn what way are "numbers" real math?
 >> Anonymous Mon Nov 27 04:57:59 2017 No.9324961 >>9324951hares overtake tortoises erry day
 >> Anonymous Mon Nov 27 05:00:28 2017 No.9324967 >>9324961Your IQ must be at least 30 to post in these threads.
 >> Anonymous Mon Nov 27 05:09:34 2017 No.9324977 File: 38 KB, 645x729, 1509035922690.png [View same] [iqdb] [saucenao] [google] [report] >>9324867And whats 1/2 as a decimal in base-3?
 >> Anonymous Mon Nov 27 05:10:52 2017 No.9324979 >>9324958I 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.
 >> Anonymous Mon Nov 27 05:11:43 2017 No.9324980 >>9324979Especially when they claim that "analysis" or anything related to it is "real math".
 >> Anonymous Mon Nov 27 05:14:32 2017 No.9324984 >>9321793>infinites>"nullify the infinite sequence"Engineer spotted.
 >> Anonymous Mon Nov 27 05:19:03 2017 No.9324988 >>93249770.111...
 >> Anonymous Mon Nov 27 05:19:53 2017 No.9324989 File: 82 KB, 645x729, 1506487478785[1].png [View same] [iqdb] [saucenao] [google] [report] >>9321793>epsilon is just like, a really really small number but not zero dudeplease, please, brainlet, give us your definition of "epsilon".
 >> Anonymous Mon Nov 27 05:24:18 2017 No.9324996 File: 485 KB, 1050x1080, 1495467753515.png [View same] [iqdb] [saucenao] [google] [report] >>9324988>...No such thing.
 >> Anonymous Mon Nov 27 05:27:10 2017 No.9324999 >>9324996I guess you've solved it.
 >> Anonymous Mon Nov 27 05:28:17 2017 No.9325001 >>9324999Solved what exactly?
 >> Anonymous Mon Nov 27 05:28:46 2017 No.9325002 >>9324977deci is 10, retard
 >> Anonymous Mon Nov 27 05:29:13 2017 No.9325003 >>9325001Math
 >> Anonymous Mon Nov 27 05:30:03 2017 No.9325006 >>9325003But what I replied to has nothing to do with math.
 >> Anonymous Mon Nov 27 05:32:04 2017 No.9325009 >>9325006Well you solved it, anon, you deserve it.
 >> Anonymous Mon Nov 27 05:33:26 2017 No.9325010 >>9324967ever heard of Zeno?
 >> Anonymous Mon Nov 27 05:36:03 2017 No.9325013 >>9325010Interesting. Could you point out the paper which proves that this so-called "paradox" is actually a paradox?
 >> Anonymous Mon Nov 27 05:37:37 2017 No.9325014 File: 973 KB, 312x213, x.gif [View same] [iqdb] [saucenao] [google] [report] >>93249961/2 doesn't exist!!!
 >> Anonymous Mon Nov 27 05:38:22 2017 No.9325016 >>9325014>1/2I don't see the "..." here, is this a typo?
 >> Anonymous Mon Nov 27 05:38:42 2017 No.9325018 >>9325013it isn'tbecause infinities are everywhere
 >> Anonymous Mon Nov 27 05:39:20 2017 No.9325020 >>9319258You got BTFO by reddit boihttps://www.reddit.com/r/math/comments/1rkc0r/found_this_on_4chan_what_do_you_think/
 >> Anonymous Mon Nov 27 05:40:40 2017 No.9325023 >>9325018>because "infinities" are everywhereI 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.
 >> Anonymous Mon Nov 27 05:41:55 2017 No.9325025 >>9325023>I thoughtkek
 >> Anonymous Mon Nov 27 05:44:38 2017 No.9325027 >>9325025>"kek"That's pretty low of you.
 >> Anonymous Mon Nov 27 05:45:43 2017 No.9325029 File: 53 KB, 403x448, 1509935607777.png [View same] [iqdb] [saucenao] [google] [report] >>93250020.1 is 0.333[30.2 is 0.666[60.1222[2 is 0.407[407but 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 0.5this tard math seems mighty familiar
 >> Anonymous Mon Nov 27 05:46:54 2017 No.9325031 >>9325029>...>higher thanEngineer spotted.
 >> Anonymous Mon Nov 27 05:47:25 2017 No.9325032 >>9325029There is no rounding going on, stop this and go read a book.
 >> Anonymous Mon Nov 27 05:49:44 2017 No.9325033 File: 33 KB, 500x500, 1506140326977.jpg [View same] [iqdb] [saucenao] [google] [report] What is 1/3 in base 12?
 >> Anonymous Mon Nov 27 05:51:18 2017 No.9325035 >>93250330.4
 >> Anonymous Mon Nov 27 05:56:00 2017 No.9325038 File: 89 KB, 363x475, 1504229489395.png [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Mon Nov 27 05:58:04 2017 No.9325040 >>9325020>>9325038>r*dditYour kind isn't welcome here.
 >> Anonymous Mon Nov 27 06:20:58 2017 No.9325054 >>93249770.120 1 2 3 4 50 1 2 10 11 120.5 ~> 0.12
 >> Anonymous Mon Nov 27 06:33:14 2017 No.9325063 >>9325054$0.12_3 \times 2_3 > 1_3$
 >> Anonymous Mon Nov 27 06:53:51 2017 No.9325075 >>9325027top kek
 >> Anonymous Mon Nov 27 07:06:22 2017 No.9325082 >>9319408your math is retard , u are 100% american
 >> Anonymous Mon Nov 27 08:33:35 2017 No.9325180 File: 13 KB, 389x500, x.png [View same] [iqdb] [saucenao] [google] [report] >>9325029>0.1222[2 is 0.407[407christ, this idiot can't even count
 >> Anonymous Mon Nov 27 14:03:16 2017 No.9325796 File: 55 KB, 600x719, db782446bcb491d356c77e13bd4b98a0.jpg [View same] [iqdb] [saucenao] [google] [report] >>9319321$1=\sum_{k=1}^{\infty} k-\sum_{k=2}^{\infty} k=\infty-\infty=0$
 >> Anonymous Mon Nov 27 14:34:26 2017 No.9325837 0.1000003= 0.333(...) = 1/30.111(...)3 = 0.500000= 1/20.2000003= 0.666(...) = 2/30.222(...)3 = 1.000000= 1/1
 >> Anonymous Mon Nov 27 14:39:08 2017 No.9325844 File: 7 KB, 211x239, stump wojak.png [View same] [iqdb] [saucenao] [google] [report] >>9319258>strokes theorem
 >> Anonymous Mon Nov 27 15:15:27 2017 No.9325921 File: 71 KB, 731x720, 1508571001225.jpg [View same] [iqdb] [saucenao] [google] [report] >>9325796>$\infty$No such thing.
 >> Anonymous Mon Nov 27 19:57:34 2017 No.9326537 >>9324984Well 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 80,001there 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.
 >> Anonymous Mon Nov 27 21:12:56 2017 No.9326673 >>9326537>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.
 >> Anonymous Mon Nov 27 21:20:48 2017 No.9326694 >>9326673MUH ABSTRACTMUH MENTAL MATHimagine 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.
 >> Anonymous Mon Nov 27 21:22:19 2017 No.9326698 >>9324979Well, I'm in grad school right now at arguably the best institution in my country for my field (operator algebras).
 >> Anonymous Mon Nov 27 21:22:38 2017 No.9326699 >>9326694abstraction doesn't mean>hurr I did these mindless calculations in my head
 >> Anonymous Mon Nov 27 21:26:58 2017 No.9326710 >>9326694>MUHStopped reading right there. Take your subhuman engineer self somewhere else if you can't handle the most basic abstractions.
 >> Anonymous Mon Nov 27 21:40:00 2017 No.9326741 x1 = 0.99...9x2 = 0.99...9 - 0.00...9x2 = 0.99...90 \\ truncate the zerox2 = 0.99...9 \\ same number of 9's as beforetherefor 0.00...9 = 0x1 = x2
 >> Anonymous Mon Nov 27 21:41:47 2017 No.9326746 >>9326698That 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.
 >> Anonymous Mon Nov 27 22:12:05 2017 No.9326826 none of you fucking brainlets know anything about advanced topics in divergent series. Neck yourselves.
 >> Anonymous Tue Nov 28 00:40:23 2017 No.9327169 >>9326710Yet 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.
 >> Anonymous Tue Nov 28 01:08:24 2017 No.9327226 >>93267410.1 = 1/10^10.01 = 1/10^20.001 = 1/10^30.000...1 = 1/10^inf = 0
 >> Anonymous Tue Nov 28 01:13:20 2017 No.9327232 >>9327169>Any attempt to write1 = 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 + 10/10000and so on, ad infinitum.No slow creeping up towards one, each line already =1
 >> Anonymous Tue Nov 28 01:32:50 2017 No.9327255 File: 245 KB, 1063x1063, 1508010693769.png [View same] [iqdb] [saucenao] [google] [report] >>9327232>ad infinitum
 >> Anonymous Tue Nov 28 01:50:02 2017 No.9327274 >>9319261oosh, that's a niche meme
 >> Anonymous Tue Nov 28 02:58:16 2017 No.9327361 File: 90 KB, 785x629, x.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Tue Nov 28 04:54:31 2017 No.9327511 File: 963 KB, 990x1146, 85a.png [View same] [iqdb] [saucenao] [google] [report] >>9327361You cannot do anything infinitely, and if you could it would be probably be a waste of your time.
 >> Anonymous Tue Nov 28 05:04:52 2017 No.9327518 >>9327169>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.
 >> Anonymous Tue Nov 28 06:07:43 2017 No.9327583 >>9327511no need tothe process shows that increasing the decimals changes nothing
 >> Anonymous Tue Nov 28 06:56:27 2017 No.9327650 File: 24 KB, 800x450, IKIFEEL.jpg [View same] [iqdb] [saucenao] [google] [report] >>9327583> 0.99 = 0.9
 >> Anonymous Tue Nov 28 07:40:32 2017 No.9327698 >>9327650how about you actually read >>9327232protip: use your calculator
 >> Anonymous Tue Nov 28 10:37:30 2017 No.9327922 >>9319247>use bad notation>blame the playerthe 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.
 >> Anonymous Tue Nov 28 14:38:00 2017 No.9328452 >>9327169>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.
 >> Anonymous Tue Nov 28 14:51:33 2017 No.9328495 >>9321793>nigger>All i'm saying is>self-taught math>just okiedokiefound the urban redneck retard
 >> Anonymous Tue Nov 28 15:04:22 2017 No.9328533 File: 9 KB, 211x239, wojak headlet.png [View same] [iqdb] [saucenao] [google] [report] >>9320930>you cannot perform arithmetic on infinite sequences
 >> Anonymous Tue Nov 28 15:18:28 2017 No.9328555   File: 42 KB, 479x720, UvGfe7Y.jpg [View same] [iqdb] [saucenao] [google] [report] >>9328533You can't.Here's your fedora
 >> Anonymous Tue Nov 28 15:29:52 2017 No.9328581 9328555>reddit filenameSubhuman trash is not welcome here.
 >> Anonymous Tue Nov 28 16:04:29 2017 No.9328655 File: 15 KB, 221x250, 1510711638411.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Tue Nov 28 16:06:51 2017 No.9328661   File: 42 KB, 645x729, tfw no brain.png [View same] [iqdb] [saucenao] [google] [report] >>9328555(a_n)_n + (b_n)_n \equiv (a_n + b_n)_nhttps://en.wikipedia.org/wiki/Direct_product_of_groups#Infinite_direct_productshere you go brainlet
 >> Anonymous Tue Nov 28 16:08:26 2017 No.9328664 File: 42 KB, 645x729, tfw no brain.png [View same] [iqdb] [saucenao] [google] [report] >>9328555$(a_n)_n + (b_n)_n \equiv (a_n + b_n)_n$https://en.wikipedia.org/wiki/Direct_product_of_groups#Infinite_direct_productshere you go brainlet
 >> Anonymous Tue Nov 28 16:24:15 2017 No.9328692 File: 82 KB, 680x680, f5753870a40ccef114a6cb88e7f48531.jpg [View same] [iqdb] [saucenao] [google] [report] >>9328664It'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.
 >> Anonymous Tue Nov 28 16:57:33 2017 No.9328792 File: 18 KB, 211x239, wojak spastic.gif [View same] [iqdb] [saucenao] [google] [report] >>9319321>WTF is limit
 >> Anonymous Tue Nov 28 17:00:10 2017 No.9328807 >>9328692>truncating infinityThe 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?
 >> Anonymous Tue Nov 28 17:01:46 2017 No.9328811 >>9319289>doing subtraction via two's complementYou mean 2-adic numbers.
 >> Anonymous Tue Nov 28 17:27:36 2017 No.9328870 File: 90 KB, 848x480, 1494816445491.png [View same] [iqdb] [saucenao] [google] [report] >>9328692>"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.>infinityNo such thing.
 >> Anonymous Tue Nov 28 18:41:48 2017 No.9328991 >>9319219I 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
 >> Anonymous Tue Nov 28 18:46:20 2017 No.9329000 >>9328991it'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
 >> Anonymous Tue Nov 28 18:48:40 2017 No.9329004 File: 157 KB, 1024x576, 1470910645307.jpg [View same] [iqdb] [saucenao] [google] [report] >>9328991>...>I know what sums are
 >> Anonymous Tue Nov 28 23:17:57 2017 No.9329540 >>9328581fuck off back to >>>/pol/
 >> Anonymous Tue Nov 28 23:22:40 2017 No.9329549 >>9329540I don't like left-wing boards, sorry.
 >> Anonymous Wed Nov 29 00:12:39 2017 No.9329615 >>9326746If you could point to when I called anyone an idiot, that'd be enlightening.
 >> Anonymous Wed Nov 29 00:18:15 2017 No.9329620 File: 125 KB, 500x358, 1510270439396.png [View same] [iqdb] [saucenao] [google] [report] >>9328664Why was the post deleted
 >> Anonymous Wed Nov 29 01:27:03 2017 No.9329706 >>9328870>>infinity>No such thing.how does the hare catch the tortoise then?
 >> Anonymous Wed Nov 29 01:36:04 2017 No.9329721 >>9329706That's more of an engineering question, ask the subhumans.
 >> Anonymous Wed Nov 29 02:06:00 2017 No.9329755 >>9329721>t. I don't know shit
 >> Anonymous Wed Nov 29 02:32:23 2017 No.9329796 >>9329755How can anybody be this delusional?
 >> Anonymous Wed Nov 29 02:34:55 2017 No.9329798 >>9328692>This is like the gender studies of maths.Now that's an expression I don't hear very often.
 >> Anonymous Wed Nov 29 03:29:00 2017 No.9329861 >>9329796just answer the fucking question
 >> Anonymous Wed Nov 29 03:44:02 2017 No.9329879 >>9326537>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
 >> Anonymous Wed Nov 29 14:32:10 2017 No.9330732 >>9320665learn to fucking read, faggot
 >> Anonymous Wed Nov 29 14:42:44 2017 No.9330755 File: 93 KB, 1194x884, ping pong anime screenshot.png [View same] [iqdb] [saucenao] [google] [report] >>9319330>arithmatic
 >> Anonymous Wed Nov 29 14:46:36 2017 No.9330762 >>9319514you're a retard
 >> Anonymous Wed Nov 29 14:55:12 2017 No.9330772 >>9320891>even if you had four orders of eternity to spend writingjust one eternity would suffice, faggot
 >> Anonymous Wed Nov 29 14:56:54 2017 No.9330775 >>9320891>literally trying to perform arithmetic on infinityhave you literally never heard of ordinal arithmetic you colossal waste of matter?https://en.wikipedia.org/wiki/Ordinal_arithmetic
 >> Anonymous Wed Nov 29 14:59:07 2017 No.9330779 >>9320901AC is independent of ZF, jackassi can reject it all day long and have valid results
 >> Anonymous Wed Nov 29 15:29:44 2017 No.9330827 >>9330779You can reject ZF all day long and have valid results.
 >> Anonymous Wed Nov 29 15:46:36 2017 No.9330859 >>9330827not within ZF
 >> Anonymous Wed Nov 29 15:51:51 2017 No.9330868 >>9330859>implying anyone mentioned doing something internally to ZFEngineer spotted.
 >> Anonymous Wed Nov 29 18:04:47 2017 No.9331079 >>9330762save us anon with the beacon of your cranium
 >> Anonymous Wed Nov 29 18:38:44 2017 No.9331149 File: 67 KB, 600x738, 600px-Omega-exp-omega-labeled.svg.png [View same] [iqdb] [saucenao] [google] [report] >>9330775This 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.Bravo.Math needs government regulation so retards can't get away with making up garbage like this.
 >> Anonymous Wed Nov 29 18:40:37 2017 No.9331154 >>9327232And yet you'll never be rid of that fraction on the end no matter how far you go.
 >> Anonymous Wed Nov 29 18:43:17 2017 No.9331162 >>9331154don't need tothe sum is already at 1your job is to prove that it ever deviates from 1
 >> Anonymous Wed Nov 29 18:45:31 2017 No.9331165 File: 81 KB, 315x360, 1511761761718.jpg [View same] [iqdb] [saucenao] [google] [report] >>9331162.999...[80,000]9~>shitmath1.000...[80,000]9.999... > 1.0
 >> Anonymous Wed Nov 29 18:47:10 2017 No.9331169 >>9331165try again after you've taken your medication
 >> Anonymous Wed Nov 29 18:47:20 2017 No.9331170 What is that in the image?
 >> Anonymous Wed Nov 29 18:51:41 2017 No.9331177 >>9330775>colossal waste of matterunderage edgelords like you have polluted this board
 >> Anonymous Wed Nov 29 18:52:30 2017 No.9331178 File: 18 KB, 384x384, 1487360360836.jpg [View same] [iqdb] [saucenao] [google] [report] >>9331169Its 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[...]9And 1.000[...]9 > 1.0I'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.
 >> Anonymous Wed Nov 29 18:55:38 2017 No.9331183 >>9331178>has to assume an endnope, at infinity there are infinite 9sand the last part is 1/inf = 0and all the time, the sum = 1
 >> Anonymous Wed Nov 29 19:03:09 2017 No.9331191 File: 175 KB, 600x600, 58b.png [View same] [iqdb] [saucenao] [google] [report] >>9331183> 1 / ∞ = 0Oh 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
 >> Anonymous Wed Nov 29 19:26:15 2017 No.9331226 >>9331191The 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.
 >> Anonymous Wed Nov 29 19:28:20 2017 No.9331228 >>9331149>brainlets can't into the best field of mathematics>set theory is appropriately reserved for those with the highest taste and intelligenceall is well in the world
 >> Anonymous Wed Nov 29 19:35:47 2017 No.9331238 >>93219481/3 = (3/10 + 1/30)... = 0.333r3/3 = (9/10 + 1/10) = 1.0 + (9/100 + 1/100) = 1.1 ... = 1.111r????
 >> Anonymous Wed Nov 29 19:37:42 2017 No.9331241 >>93312383/10 + 1/30= 9/30 + 1/30= 10/30 = 1/3congratz, you just went full retard
 >> Anonymous Wed Nov 29 19:39:31 2017 No.9331250 File: 102 KB, 300x256, 1486363850107.png [View same] [iqdb] [saucenao] [google] [report] >>9331241You wanna take another look at that post?
 >> Anonymous Wed Nov 29 19:43:53 2017 No.9331258
 >> Anonymous Wed Nov 29 19:50:44 2017 No.9331279 >>9319284Serious 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?
 >> Anonymous Thu Nov 30 00:45:59 2017 No.9331763 >>9331279real analysis and a GOOD grasp of topology (point-set)
 >> Anonymous Thu Nov 30 02:12:50 2017 No.9331889 >>9319346e is an infinite seriesso are the trig functionsso when you do sin(2) you're doing arithmetic with infinity that is generally considered beneficial
 >> Anonymous Thu Nov 30 02:18:46 2017 No.9331895 >>9331889but that's a sinjesus told me so
 >> Anonymous Thu Nov 30 07:58:23 2017 No.9332239 >>9331191>∞No such thing.>r*ddit imageFuck off.
 >> Anonymous Thu Nov 30 08:34:42 2017 No.9332272 >>9329861The question isn't math related so I don't give a fuck.
 >> Anonymous Thu Nov 30 09:24:00 2017 No.9332340 >>9332272>t. don't know a fuck
 >> Anonymous Thu Nov 30 17:20:17 2017 No.9333412 He's awesome
>>


text:
Checksums (part I)

I once worked in a company specializing in embedded electronics for industrial applications. In one particular project, the device communicated through a RS-422 cable to the computer and seemed to return weird data once in a while, causing unwanted behavior in the control computer whose programming did not provide for this unexpected data. So I took upon myself to test the communication channel as it seemed that the on-board software was operating properly and did not contain serious bugs. I added a check-sum to the data packet and it turned out that some packets came in indeed corrupted despite the supposedly superior electrical characteristics of the RS-422 link.

After a few days’ work, I implemented the communication protocol that could detect and repair certain errors while reverting to a request to retransmit if the data was too damaged. I then started gathering statistics on error rate, number of retransmit, etc, and the spurious behavior on the controller’s side went away. My (metaphorically) pointy-haired boss opposed the modification because “we didn’t have any of these damn transmission errors until you put your fancy code in there“. Of course, this was an epic facepalm moment. I tried to explain that the errors have always been there, except that now they’re caught and repaired. Needless to say, it ended badly.

Notwithstanding this absurd episode, I kept using check-sum to validate data whenever no other layer of the protocol took care of transmission errors. So, this week, let us discuss check-sums and other error detection algorithms.

The simplest type of check-sum, the check sum simply adds all numbers (or bytes, or bits) and reports that sum as the check-sum. This, unfortunately, doesn’t give you much power for error detection. Indeed, it’s very easy to understand that simply computing the sum of a series of number doesn’t tell you much. There are many ways of getting a given sum in the presence of multiple errors. Even a simple inversion goes undetected.

A bit more protection is given by Luhn‘s algorithm. Luhn’s algorithm on a series of digits (let us consider only digits as it generalises to any radix anyway) starts with an initial sum of $s_0$, which may be 0 or salted. Every odd-position digit is added once to $s_0$, each even-position digit is added twice to $s_0$. For example, the digit series 6,4,2,5,1 would be summed up as $s=s_0+6+(2\times{}4)+2+(2\times{}5)+1$. Then, $s \mod{}\:10$ is returned as the check-sum. We could reformulate the operation as:

$\displaystyle s=s_0+\sum_{i=1}^n 2^{((i+1)\mod{}\:2)} x_i$

An example of this simple type of check-sum is the Canadian Social Insurance Number, whose numbers are chosen so that the check-sum is $\equiv 0 (\!\!\mod 10)$. Needless to say, the validity of the number as a social security number cannot be limited to the verification of the check-sum alone; it has to be checked against the national registry—it’s ridiculously easy to generate a SIN that passes the check-sum test.

The idea can be extended to using other multipliers than 1 and 2 in the sum. The ISBN, or International Standard Book Number, uses either 10 or 13 digits codes to uniquely identify a book. With the correct ISBN alone, you can order a book from your favorite bookshop without ambiguity. The ISBN-10 algorithm computes the sum:

$\displaystyle x_{10}=\left(\sum_{i=1}^9 i\:x_i\right)\mod\:11$

Which yield the last digit of the code (with X encoding 10). The ISBN-13, however, is a Luhn check-sum using 1 and 3 rather than 1 and 2.

Luhn’s method generalizes to any number of digits (and to other bases as well), but is not very strong. Much stronger than Luhn’s method, the CRC, or cyclical redundancy check, computes the “check-sum” as the remainder of the division of the message, considered as a big integer, by a divisor polynomial. There are number-theoretic reasons to call the divisor a “divisor polynomial” but let us not go there right now (if you’re really interested in the mathematics, see here). For all practical purpose, the divisor is a small integer. The divisor is chosen in order to give a remainder on 16, 32 or 64 bits. While the long division of a very big number (the message) by a small integer (the divisor) seems cumbersome, some efficient algorithms exist that will compute the CRC very efficiently, and without doing any arbitrary precision arithmetic.

However, if the CRC is much better than the simple check-sum and Luhn’s algorithm, it is not infallible. As it encodes the message’s structure on a very small number of bits compared to the message itself, it cannot possibly trap every errors. But it turns out that for a good divisor polynomial (there are a few standard ones), it will detect quite many errors. It detects all errors of exactly one bit. It will detect all errors of two bits if the errors are separated by more bits than the order of the polynomial; for a polynomial of order $n$, they must be $n$ bits apart. It will also detect all error bursts of $n$ bits or less. However, it doesn’t do well with long series of leading zero bits, as deletion and insertion of leading zeroes will go undetected.

The CRC is the preferred method of computing check-sums in a number of applications (from Zip files to Ethernet packets) but there’s a number of ad hoc check-sum algorithms out there, not all equally fantastic. For example, the Adler32 function is meant to be efficiently computed; but yields incomplete coverage for messages that are too short. I also presented an ad hoc hash function in a previous post that is reasonably good while amenable to efficient implementations1.

*
* *

Which brings me to the topic of check-sums and hashes. In UEID, I presented a few hash function families. I discussed secure hash functions like MD5 and SHA in the context of providing a unique descriptor to a piece of data. That’s also the goal of a check-sum. A hash function wants to yield a different value whenever the message changes, ever so lightly, which is also a the goal of a check-sum function.

It should be clear by now that a (good) check-sum and a (good) hash function are the same both the exact same thing: they map a series of bits (the message) onto a smaller series of bits (the check-sum/hash) in a way that captures as many changes as possible in the messages.

*
* *

It turns out that inventing a good check-sum/hash algorithm is extremely difficult—even Don Knuth got his ass bitten on this one2. Unless you want to invest a considerable amount of time developing and thoroughly testing your own check-sum/hash function, I would advice to use one that already exist, that have been examined by experts, and that has Open Source implementations available.

1 I tested its behavior in the context of hash table look-ups, and it exhibited a most cromulent behavior. See here (in French).

2 Donald Knuth &mdash The Art of Computer Programming, Volume 3: Sorting and Searching — Addison-Wesley, 1973.

One Response to Checksums (part I)

1. […] a number of different occasions, I briefly discussed Hash Functions, saying that if a hash function needn’t be very strong if […]

format: plain