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Prove that this equation is wrong.
Hint: You can't.
>>
>>9406727
Brainlet here what does this equation mean in stupid form?
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>>9406727
doesn't converge in the 2-adics, so it's technically wrong there.

[eqn]\lim_{n \to \infty} \left|\frac{1}{2^n} \right|_2 =\lim_{n \to \infty} 2^n[/eqn]
>>
>>9406729
it means 0.5 + 0.25 + 0.125, etc, all the way up to infinity.

A good 5 minute explanation of sigma notation for people who are clueless as to how it works is here: https://www.mathsisfun.com/algebra/sigma-notation.html
>>
>using the axiom of infinity

found the error
>>
>>9406727
Can I use induction on this shit?
1/2^n+1= 1/2^n + 1/2
1+1/2 = 3/2 =\= 1
QED xD
>>
>>9406727
Well the first term is a half and the rest of the terms don't feel like they get close to a half so it's probably wrong :^)
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>>9406734
so if a series doesn't converge for the n-adic numbers it's divergent? I think i understand how the adics work
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>>9406727
Fuck off, Zeno.
>>
>>9406727

Correct me if I'm wrong but with abs convergence test doesn't this come out to be 1/2?
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>>9406757
That's obviously wrong because it's 1/2 + some other positive stuff.
>>
>>9406751
if a series doesn't converge it's divergent full stop
>>
>>9406762
Wtf is the retardation in this thread.
The sum in ops picture is convergent (to 1)
>>
>>9406780
your the retard you need to work on you're reading comprehension
>>
>>9406727
Zenos paradox is itself a paradox. In his achilles and tortoise example he maps multiple factors that allow for a finite result, and in practice use finite inputs as well. The tortoise gets a head start but moves at half the speed of achilles. When achilles reaches where the tortoise started at 0.5 the race track, the tortoise has made it to 0.75. When achilles reaches 0.75, the tortoise has teached 0.875, so on until achilles reaches the tortoise.

The multiple factors are time, distance, and a surmised end or point at why achilles and the tortoise meet. In fact the graph for the ewuation shows divergent values above 1. Taking altogether, it's just retard bullshit to take it as proof of infinite work sums. Taken without the time or limit distance factor, there would always be a smaller slice of the line between achilles and the tortoise, therefore never reaching 1. Taken with the time and limit distance factors, they reach 1 at some moment and then achilles diverges by doubling the distance between himself and the tortoise behind him with every reference frame, 1 to 2 to 4 to 8 to 16 to 32 to 64 to 128 and so on until achilles as run off the face of the earth and travelled across the universe towards infinite distance.

Zeno was a fucking brainlet and everyone who treats him other than the fact will also be a brainlet. Infinite sum is stupid gay stuff.
>>
Daily reminder that a discrete spacetime solves Zeno's paradox.
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>>9406806
When achilles reaches where the tortoise started at 0.5 the race track, the tortoise has made it to 0.75. When achilles reaches 0.75, the tortoise has teached 0.875, so on until achilles reaches the tortoise.
> there would always be a smaller slice of the line between achilles and the tortoise
> never reaching 1
>>
>>9406727
[math]0.111..._2=1[/math]
>>
>>9406814
Daily reminder that Calc I solves Zeno's paradox.
>>
>>9406806
This just means that the application of infinite sums only works for finite things, but even then it would have to be extended to either defining or getting as a result the actual limit point at which a convergence occurs.
Applied arbatrarily to numbers and variables without finite limits is not the correct way to use it. For example it's easy to mechanically prove [math]\sum_{n=1}^{\infty} \frac{9}{10^{n}} = 1[/math] if and only if we define a finite decimal accuracy limit, which as a product would also define the n limit at which the decimal limit was reached, which for this problem would just be decimals+1. For 10 decimals of accuracy, on the 11th iteration it would round up to 1 from 0.999999999 at the 10th, so truthfully
[math]\sum_{n=1}^{X} \frac{9}{10^{n}} = 1 with 10 decimal accuracy[/math], but written more simply lacking necessary information would instead result
[math]\sum_{n=1}^{\infty} \frac{9}{10^{n}} = 0.\bar{9}[/math]
>>
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>>9406806
Yes, the sort of 'stupid gay stuff' that shares itself with one other infinity descriptor: "I'm not stupid or gay enough to think of a way to ultimately not be a super-immortal quasi-dimensionality."

It's like being Christian and seeing the sign for that whole 'shared belief' concept.
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>>9406827
oh god it's the shitlatex masturbator again
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>>9406793
>of course the mouth breather can't spell
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>>9406827
http://m.wolframalpha.com/input/?i=N%5Bsum%5B1%2F%282%5En%29%2C+%7Bn%2C+1%2C+118%7D%5D%2C+35%5D
This would suffice to solve achilles and the turtle if the smallest measurement rererence frame was a planck length with 35 decimals of accuracy. At n=118, 1/2^n converges to 1 with planck length accuracy. At n=117, its just 0.999...
>>
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>>9406852
>just going around proving zeno's paradoxes wrong on a chinese pottery bulletin board in the middle of the night on new years
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>>9406866
fuck you it's noon here
>>
>>9406852
>>9406866
>只是繞來繞去證明芝諾悖論錯在中國陶公告牌在新年內半夜

*SLAM DUNK*
>>
>>9406740
but
[math]\frac{1}{2^n}+1 \neq \frac{1}{2^n}+\frac{1}{2}[/math]
>>
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>>9406729
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>>9406873
Isn't that what I said? You can't use induction on this shit anyway, can you?
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>>9406729
It means you should go back to your board
>>
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>>9406852
>>9406827
>>9406866
holy
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>>9407363
Name one (1) intelligent Republican.
>>
>>9407525
I'm conservative republican and have proven one of zeno's paradox solutions/assumptions incorrect.
>>
now THIS is real deal.
[eqn]
\sum\limits_{n=1}^ \infty \frac{1}{\frac{1}{n}} = -\frac{1}{12}
[/eqn]
>>
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>>9407588
that equation tends towards infinity
>>
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>>9407602
>>
>>9407602
but the value is -1/4pi
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>>9407371
No you cant.
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>>9407683
No, the value is a positive number that incrementally adds up.
>>
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[math] \sum_{n=0}^m x^n = \dfrac{1-x^{n+1}}{1-x} [/math]

so for |x|<1

[math] \lim_{m\to \infty} \sum_{n=0}^m x^n = \dfrac{1}{1-x} [/math]

Now the special case x=1/2 gives

[math] 1 + \lim_{m\to \infty} \sum_{n=1}^m \dfrac{1}{2^n} = \dfrac{1}{1-1/2} [/math]
>>
>>9407723
>finite operations are the same are "infinite operations"
aww cute
>>
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>>9407733
http://m.wolframalpha.com/input/?i=sum%5B1%2F%281%2Fn%29%2C+%7Bn%2C1%2Cinfinity%7D%5D

n:1 = 1/(1/1) = 1 +
n:2 = 1/(1/2) = 2 + 1
n:3 = 1/(1/3) = 3.3 + 2 + 1
n:4 = 1/(1/4) = 4 + 3.3 + 2 + 1
...
[math]\rightarrow \infty[/math]
>>
>>9407745
limit =/= value
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>>9407825
take your shit in street pajeet ideologies elsewhere please. We're trying to do math here.
>>
>>9406727
[eqn]S = \sum_{n=1}^{\infty} \frac{1}{2^n}=\frac{1}{2}+\frac{1}{2}S \iff S = 1[/eqn]
QED
Also algebra is universal and Cauchy was a faggot and a /pol/tard.
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>>9407850
Are you retarded?
>>
File: image.jpg (1.22 MB, 3264x2448)
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I Know im late but to anyone still wondering how the original result comes to be here goes...
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>>9407873
are you?
>>
>>9407875
adding to that, i don't understand why its a paradox.... i didn't even knew zeno existed we just had that series in a calc homework
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>>9407825
t. I Don't Understand [math]\mathbb{R}[/math]
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>>9407369
something doesn't seem about this. you give the square an area of 1 before filling in the triangles, so it's like you're starting with the solution you're trying to prove
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>>9407905
the reasoning is that you can fit all the triangles in the unit area square I think.
>>
>>9407905
>>9407918
Its a good representation of the problem described by zeno as he compared multiple finitist factors like time and distance resulting in a finite answer of 1, but both his problem and that image arent actually [math]\sum_{n=1}^{\infty}\frac{1}{2^{n}}[/math]
Since neither require "infinite" work to reach the result >>9406852

Where >>9406827 also describes that the least limit required to solve the problem is the amount of decimal places + 1, which if n=infinity, the decimal accuracy would have to be infinity-1, which in any practical usage would simply define infinite work and summation is not actually being done.
>>
>>9407962
Required to solve a similiar problem*
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>>9407901
>infinity is in R
o boy
>>
>>9407962
If n=[math]\infty[/math] and it results 1 then how is it wrong
>>
>>9408001
It's wrong by misunderstanding the usage of infinite sum. It's only 1 at n=infinity IF a decimal limit of infinity-1 was provided. Decimal limits are important because its how subdivision into smaller and smaller units work. If the decimal limit for the sum 9/(10^n) were 35 places corresponding with the planck length being [math]1.616229 × 10^{-35}[/math], n needs only reach 118 to be considered converging. Just over a hundred iterations, not a thousand or a million or a billion nevermind infinity.

Applied truly arbitrarily on numbers defined without a decimal or accuracy limit, the result should instead be nothing but [math]0.\overline{9}[/math] if we take the overline to define infinite arbitrary repitition as the result of an infinite arbitrary sum, since without a definition of accuracy limit with the only direction being [math]n\rightarrow\infty[/math] means no concatination occurs. If instead our smallest measure of time and distance was 1/10th, the sum 9/(10^n) = 1 at n=2 since 0.9 would be followed by 0.99 which would appear to be a huge chunk of a divisional increment we can't fathom, so that second 9 rounds up to 10, the first 9+1 = 0, resulting 1.0

The usage of [math]\sum_{n=1}[/math] is taught and used poorly for lack of understanding the limit is either something you should be supplying to find a decimal accuracy limit, or solving as a variable when supplied with a decimal accuracy limit.
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>>9408032
>the limit is either something you should be given to solve a decimal accuracy limit, or solving as a variable when supplied with a decimal accuracy limit.
fixed
>>
>>9408037
>>9408032
[math]\sum_{n=1}^{\stackrel{x}{d.10^{-35}}\frac{9}{10^{n}}[/math]
or
[math]\sum_{n=1}^{\stackrel{472}{d.x}}\frac{9}{10^{n}}[/math]
would both work as example ways to write the questions if the intent was to produce an answer indicative of convergence.
>>
>>9408044
>>9408052
>>9408054
[math]\sum_{n=1}^{\stackrel{472}{d.x}}\frac{9}{10^{n}}[/math] or
like
[math]\sum_{n=1}^{\stackrel{x}{d.35}}\frac{9}{10^{n}}[/math]
>>
>>9408032
>wall of text
why do all religious nuts have verbal diarrhea
>>
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>>9408069
>my cant unambled to read well so u r dum
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>>9408052
>>9408060
oh look you missed a bus again
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>>9408052
Might have handled the stackrel fraction but the problem is 4chan's spacing. One sum followed by another in seperate math/math blocks without text after the first block as well as a line break will mess it up.
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>>9406821
Spoken like a true engineer who doesn't understand mathematical at all.
>>
>>9406821

Daily reminder that Zeno's paradox isn't a paradox at all if the arrow lands at Zeno's foot.
>>
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>>9408142
>look up zenos arrow
>"motion is impossible because time is just countless instances of freeze frames"
goddamn what the hell was this guy's problem?
>>
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>>9408148
His brain was two sizes too small.
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>>9407973
I never said that, brainlet. Infinity is not where the other anon failed to grasp the set of reals.
You don't need infinity to deal with limits. That's what that whole epsilon-delta thing is for. (=
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>>9407905
It's a visual representiation of the fact that:

Given any ε>0, there exists a halving-step N such that after it (n>N) the difference between the filled area and the area of the square(which is 1) is less that ε.

That's what [math] \sum\limits_{i=1}^{\infty}\frac{1}{2^i} = 1 [/math] means:
[math] \forall \varepsilon >0 : \exists N : \forall n : n>N \implies |\sum\limits_{i=1}^{n}\frac{1}{2^i} - 1|<\varepsilon [/math]
>>
>>9408128
Spoken like a loser who failed Calc I.
>>
>>9408184
Not sure but it looks like the only practical usage of epsilon delta is for finding the zero of a wave function. Or working to redundantly prove a given non-infinite limit alongside a given result and finding the solution for the limit via epsilons and deltas which doesn't translate to hard numbers.

Its much easier to produce results through sigma sums if you just provide the sense and scale of required outputs via non-infinite limit or non-infinite decimal accuracy. In the case of 1/2^n with n to infinity, the answer would never even equal [math]0.\overline{9}[/math] because the summation always adds on an arbitrary halved amount of various whole and different integers, unlike for example 9/10^n which produces nothing but more 9's.

There is a very slipperly slope of brainlet logic derived from shit education surrounding these problems. First its [math]0.\overline{9} = 1[/math], then it's [math]0.999........99764243367532 = 0.\overline{9} = 1[/math], and soon enough it'll just be [math]3=\infty[/math]

Draw the line somewhere bitte.
>>
It's funny how people expect real numbers to behave like real stuff, ignoring the absurdity of their definitions. I blame primary teachers.
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>>9408248
I believe that not even you are understanding what you are saying desu.
>>
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>>9408435
(You)'s don't mean as much when you don't work for them. Put a little effort in next time.
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>>9407602
>>9407588
You are both correct. "The series diverges" and "the series equals -1/12" are not contradictory, just as "the alternating harmonic series does not converge absolutely" and "the series converges to ln(2)" are not contradictory. You're just extending the concept of a series in a unique way.
>>
>>9408435
[math]\sum_{n=1}^{n \rightarrow \infty} \frac{1}{2^{n}} < 0.\bar{9} < 1[/math].
.
.
[math]\sum_{n=1}^{n \rightarrow \infty} \frac{9}{10^{n}} = 0.\bar{9} \neq 1[/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow x}{d.35}} \frac{9}{10^{n}} = 1, x=118 [/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow 118}{d.x}} \frac{9}{10^{n}} = 1, x=35 [/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow x}{d.1}} \frac{9}{10^{n}} = 1, x=2[/math]
>>
>>9408498
where d is a placeholder for decimal point accuracy cause latex was having trouble displaying exponents in the small stackrel field but i'll try one more time:

[math]\sum_{n=1}^{n \rightarrow 118}{10^{-35}} \frac{9}{10^{n}} = 1[/math]
>>
>>9406727
Infinity doesn't exist. Done.
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>>9408515
[math]\sum_{n=1}^{\stackrel{n \rightarrow 118}{10^{-35}}} \frac{9}{10^{n}} = 1[/math]
>>
>>9408528
Ye that looks like ass. d dot works better.
>>
The sheer amount of retardation here is unbelievable. There are maybe 3 people here who understand math and the rest are brainlets.
>>
>>9408528
>>9408544
[math]\sum_{n=1}^{\stackrel{10^{-35}}{n \rightarrow 118}} \frac{9}{10^{n}} = 1[/math]
>>
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>>9408593
Looks better
>>
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>>9407373
This is my board
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>>9407373
>>9407525
>>9407588
>>9406727
>>9408593
The sum of e^n from 8 to 1 is -negrative e^(8+1)

=e^9
>>
>>9408743

It would be negtive e^7 because its acturally e^(8-1)

So it would look like

1 E n=8 of e^n = -e^7
>>
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>>9408593
>plank length is [math]10^{-35}[/math] metres as the smallest unit where things matter
>0.000000000000000000000000000000000001m
>planck time is [math]10^{-44}[/math] seconds as the the time it takes the speed of light to travel a plank length
>0.00000000000000000000000000000000000000000001s
>[math]\frac{0.00000000000000000000000000000000001}{0.00000000000000000000000000000000000000000001} = \frac{1pt}{1000000000pl} = \frac{0.0000000001pt}{1pl}[/math]
am i retarded or is this retarded
>>
>>9408777

dude nice trips
>>
>>9408777

No because the speed of light is a given whereas planks constant is imagninary
>>
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I don't accept that numbers exist so axiomatically your argument is invalid.
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>>9408498
Ok, but how is this relevant?
>>
>>9408973
it's not ok and isn't relevant
>>
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>>9407745
and you can add an infinite number of rational numbers together and get an irrational at the end. All of the partial sums are rational since the rationals are closed under addition but putting infinity in makes everything go to shit.

so what I'm saying is your post is wrong
>>
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>>9408526
>the axioms are hard, get rid of them
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>>9409116
Nah, keep your street shitting pajeet math to yourself, you brainlet subhuman.
>>
>>9408973
That yoy can find the exact finite result of a sigma sum via a sigma sum without unrelated delta epsilon fudged retard limit stuff by plugging in one more factor is what makes it relevant.
There is no such thing as infinite work leading to a rational number, and you only need do some finite work to reach any desired result.
>>
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>>9409774
[math]\frac{9}{10} + \sum_{n=2}^{\stackrel{d.10^{-n+1}}{n \rightarrow \infty}} \frac{9}{10^{n}} = 1 \neq 0.\bar{9}[/math]
>>
>>9409781
>>9409781
[math]\stackrel{d.10^{(-n+1)}}{\big[ \frac{9}{10} + \sum_{n=2}^{n \rightarrow \infty} \frac{9}{10^{n}}\big]} = 1 \neq 0.\bar{9}[/math]
>>
>>9409743
>you can find the exact result of a sigma sum

What is a sigma sum
>>
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>>9406727
[eqn]\sum_{n=1}^\infty 2^{-n} = \int_0^\infty 2^{-t}dt = 2^0-2^\infty = 1[/eqn]
>>
>>9409781
>>9409850
[math]\sum_{n=1}^{\stackrel{d.(n-1)}{n \rightarrow \infty}} \frac{9}{10^{n}}} = 1 \neq 0.\bar{9}[/math]

is there any place to preview latex rendering before posting?
>>
>>9409866
[math]\sum[/math]
this thing. Its the letter sigma.
>>
>>9409872
[math]\sum_{n=1}^{\stackrel{d.(n-1)}{n \rightarrow \infty}} \frac{9}{10^{n}} = 1 \neq 0.\bar{9}[/math]
>>
>>9409850
>>9409878
Yeah d dot ends up working better than an exponent formula, since [math]10^{1-1}[/math] isn't clear towards defining how many decimals of accuracy when the intended input is [math]d.(1-1)[/math]. The exponent implies 1 decimal accuracy but the d dot clearly says 0, to save against writing it even more complex. Needs the aesthetics and simplicity if its going to stick.
>>
>>9406727
>>9408248
[math]\sum_{n=1}^\infty \frac{1}{2^n}[/math] never even makes it to [math]0.\bar{9}[/math] let alone 1.
>>
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>>9409929
>doesn't understand convergence
>>
>>9409945
Convergence can only occur with finite parameters. n to infinity is an infinite parameter.
>>
>>
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>>9409975
>Convergence can only occur with finite parameters
https://en.wikipedia.org/wiki/Limit_of_a_sequence
https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit
>>
>>9409945
>>9409975
Zeno's achilles and the tortoise problem includes finite paramaters. Ita no big deal the tortoise moves at half the speed of achilles. Whether they're racing to a given goal line and the tortoise gets to start in the middle of the length or whether they're just running to see if zeno can catch the tortoise over an arbitrary distance but the tortoise gets a head start while achilles waits, both situations present finitist factors that defy the infinite equations derived from them.
in the first case, a finite track length must be determined so that is knowable where the middle of it is so the tortoise may start there. In the second case, a finite amount of time must pass as the alotted time given to the tortoise for the head start before achilles can start moving.
In either case, if we know half the distance of the track to know the full distance with achilles and the tortoise racing to a goal then they will meet and tie and converge at the goal to equal the finite 1, or if we know how much time has passed to suffice as a head start of achilles merely chasing to reach the tortoise then they will also meet at a finite convergence of 1 as if a 10 second head start would become 5 seconds, 2.5, 1.125, and even into milliseconds or microseconds or nanoseconds of accuracy up to even imaginary planck time reference frames, but it's safe to say that with a 10 second head start, achilles will have met the tortoise by the time 20 seconds have passed since the tortoise began.
There are definite finite factors here and the only way to truly dissociate them would be achilles giving a tortoise a head start and then never even attempting to chase or follow it as a mean way of telling the tortoise to fuck off somewhere else.

In reality there is no such thing as an infinite sum beyond looking for repeating decimals.
>>
>>9409992
I dont care about asinine justifications for why retards believe infinity is a number or how liberally they can stretch and smear the definition of convergence. I especially dont care about wikipedia cause its monitored and adminstrated by shitlords.
>>
>>9410034
so much hand waving it's a helicopter
>>
>>9410046
>i speng half my salary on tin foil hats
>>
>>9410051
I bet you vote liberal democrat given you're a projecting cuck too dumb to realize you're projecting.
>>
>>9409988
>waahhh, cutting off at some finite amount doesn't give me all nines.

but given any number of nines there exists an n such that the sum gets up to that amount of 9s. and then if you keep adding more terms you get even more 9s.
There isnt an upper bound on how many nines you _eventually_ get to.
>>
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>>9410056
>[math]0.992 = 0.\bar{9}[/math] cus um there no limit to um thing of 9's and stuff
overline repeating decimals are a concept of truncation so you dont blow your asshole out trying to do an infinite, unending amount of work to solve a single equation. We see the simple pattern of how the numbers come out repeating and we truncate to [math]0.\bar{9}[/math]. Even if a pattern exists to the numerals coming out of 1/2^n, you still end up with a number smaller than 0.9repeating and would probably be required to be written in a novel way like [math]0.\underline{9}_{\overline{1234567890}}[/math]
>>
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>>9410055
>>
S = 1/2 + 1/4 + 1/8 + ... = 1
(1/2 - S/2) + (1/4 - S/4) + ... = 0
(1/2 - 1/4 - 1/8 - 1/16 - ...) + (1/4 - 1/8 - 1/16 - ...) + ... = 0

1/2 + 1/4 - 1/4 + 1/8 - 1/8 + 1/16 - 1/16 + ... = 0
1/2 = 0

the equation is wrong lol
>>
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>>9410160
I suppose you don't think >>9410051 is ad hominem huh.
>>
>>9410170
No, it criticizes the explanation, not the writer.
Not being AH doesn't mean it has to be polite.
>>
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>>9410179
ahuh...
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>>9410034
To elaborate so brainlets understand, zeno's achilles and tortoise representation of [math]\sum_{n=1}^{\infty}\frac{1}{2^n}[/math] presents finitely divisible factors of accuracy because the way the problem is presented. Yes, 1/2 is half of 1, but these are just numbers, not units. Planck time might have 44 decimals of accuracy, but that doesnt mean arbitrary numbers with more decimals cant be crafted, so the actual application of [math]\sum_{n=1}^{\infty}\frac{1}{2^n}[/math] merely on arbitrary numbers also implies arbitrary decimal accuracy. Achilles may reach the tortoise in finite time over a finite distance with a finite amount of n iterations, but the arbitrary infinite sum never reaches 1 because there are an uncountably infinite amount of decimal numbers between 0 and 1 that if plotted and graphed similarly to the achilles and tortoise problem would never present convergence and instead imply the acceleration path of achilles is actually parallel to the acceleration path of the tortoise if they were given infinite time to cover an infinite distance, which infinite time and infinite distance is what is directly prescribed by the offhand unrelated representation of [math]\sum_{n=1}^{\infty}\frac{1}{2^n}[/math] which can't converge and remains parallel to some degree of undulating accuracy to the effect of being written as [math]0.\underline{9}_{\frac{1}{2}}[/math] or [math]0.\underline{9}_{0.5}[/math] which gives the necessary info to interpret the number by a quick glance to know that a similar problem like
[math]\sum_{n=1}^{\infty}\frac{2}{3^n}[/math] would result in a larger number [math]0.\underline{9}_{\frac{2}{3}}[/math] or
[math]0.\underline{9}_{0.\bar{6}}[/math]

Higher math is absolutely out of control with the bullshit it tries to pull over people's eyes. Either define decimal accuracy required, define a non-infinite limit, or fuck off.
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>>9406757
Absolute convergence test is a test whether a series converges or not. It doesn't tell you anything about what it converges to, if it does converge
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>>9407525
Arnie
>>
>>9407726
this
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>>9407905
You don't start with a square that you partition. You start with a triangle, call its area "1/2", then attach a triangle to its side that is half as large, and attach another that is half again as large and so on, and only then do you end up with a square that visually represents the answer that the sum of all these numbers must be 1.
>>
>>9410321
Again, decent depiciton of achilles and the tortoise, but a bad depiction of [math]\sum_{n=1}^{\infty}\frac{1}{2^n}[/math]. Achilles and the tortoise race over a finite distance in finite time equivalent to a boxed area that can be finitely divided into halved triangles. The number sum without minimal time or minimal distance accuracy can instead be infinitely divided, where [math]\frac{1}{2^1}[/math] so no actual relation to the sum unlike the bounded square area filled with triangles. That the bounded example actually shows however is a finite amount of triangles to consume the last available pixel, which defies a limit set to infinity. The limit is instead the resolution boundary of the boxed triangle image.
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>>9410290
>Either define decimal accuracy required, define a non-infinite limit, or fuck off.
Every mathematician: "No. You fuck off."
>>
>>9410410
Post your university.
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>>9410034
>In reality there is no such thing as an infinite sum beyond looking for repeating decimals.
We're talking about math here bud, not "reality".
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>>9410290
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>>9410416
Yes, math, the thing you have to do. As in, it takes time to do it. As in its bounded by reality.
setting a limit to infinity then dancing around like a fruitcake trying to fuzzy what the fuck you actually mean by infinity when you're told an infinite amount of a thing cant be solved is not intelligent. Yea, yea, infinity like "not really infinity but if you do this and that we can prove that probably at some arbitrary point we could figure something out", whatever - just stop writing infinity. Accept your loss and move on.
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>you: infinity is not a number

>mathematicians: [math]\infty \geq 1×10^{6}[/math]
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>>9410449
lol
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>>9410449
Correct
https://en.wikipedia.org/wiki/Extended_real_number_line
>>
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>>9410429
this happens when lower monkeys learn language but can't understand logic and abstractions and trivial deductions from axioms
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>>9410860
Math tried true abstraction once and it was called [math]i[/math], as in [math]5i+4i = 9i[/math]. The difference there is that the result carries the same abstract compound of the input, meanwhile succaticians prescribe infinite abstract work to equate a finite "real" result. Dumb, very dumb stuff.
Every single practically engineered usage of higher math based on axioms of infinity end up obeying and prescribing finite values in their equations to derive finite results. It's only a flaw of math-centric maths, wacky abstract arithmetic for the sake of giving yourself something to do to tease your brain, that these concepts of infinity are maintained. Fourier transforms are done on finite values, electrical equations are done on finite values, everything that actually works derived from axioms of infinity simply works because they drop the whole infinity part.
>>
>>9410526
Yea
0.999999 exists but
>0.9999999
>oh fuck [math]10^{-7}[/math]?¿?¿?!?
>FUCK THAT SON TOO MANY NUMBERS, THAT SHIT IS 1.0
>>
>>9410880
Like when you evaluate limits right? But the thing doesn't actually make any sense without the infinite aspect first.
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>>9410886
Nah. Decimal accuracy limit is what is implied. Easy to produce a number that way. 15 decimal places is accurate for most anything.
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>>9410894
>Math
>Decimal accuracy limit is what is implied
Retard engineer detected
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>>9410894
>15 decimal places is accurate for most anything.
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>>9410963
>>9410964
>>
>>9410880
Not everything fourier is the fucking FFT. Analysis of, say, LRC circuits uses integration from -inf to inf.
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>>9410973
>bounded by sin and cos
SO CLOSE
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>>9411021
It's an integration across an infinite span of time so I don't see your point. Also the transform of the unit step, which is used all the time, go to +inf as frequency goes to 0 from the right.
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>>9407588
V := "for all"
E := "there exists"
/E := "does not exist"
Convergence of a sequence (an) to a value c is equivalent to:
V e > 0 E N: V n >= N, |an - c| < e.
A sequence (an) is divergent if /E c: an converges to c.
which is equivalent to:
V c E e > 0: V N E n >= N: |an - c| >= e.
The sum of all natural numbers is the limit of the sequence of its partial sums, if it exists.
so an = n*(n + 1)/2
let c be arbitrary
choose e = c
let N be arbitrary
choose n = max{N + 1, ceil(c)}
case c > 0:
we see that |n(n+1)/2 - c| >= |(c(c+1) + 2c)/2| = |(c^2 + 3c)/2| = |c^2/2 + 1,5c| >= c = e.
therefore we have shown that the sequence of partial sums of all natural numbers up to some n does not converge to any value, which means by definition that it diverges.
case c < 0 is trivial
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>>9406727
How can you prove something wrong when it is true?
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>>9406727
guys i'm so close just a little longer and ill have it...
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>>9406736
>never studied analysis
>doesn’t realize the definition of convergent sequences doesn’t require infinity
The only error was your father’s missing prophylactic
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>>9406729
Go back to your board.
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>>9410963
>>9410964

an interesting problem would be knowing how fast the 9s in 0.999... grow
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>>9409737
Im not wrong tho you brainlet nigger
>>9409872
in the reply box in the lop left corner
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>>9410034
So with 20 seconds being the entire duration of the chase to convergence where the tortoise was given 10 second head start, with an infinite amount of time [math]n\rightarrow\infty[/math], achilles would start giving chase at [math]n=\frac}\infty}{2}[/math] and they would converge at [math]n=\infty[/math]
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>>9412459
>[math]\frac{\infty}{2}[/math]
Infinity is not a motherfucking number
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>>9412464
[math]\frac{\infty}{2} \geq 500,000[/math]
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>>9412469
[math]\frac{\infty}{1000000} \geq 1[/math]
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>>9412482
>[math]\frac{\infty}{\infty} = 1[/math]
makes sense to me boyo. Read a book.
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>>9410290
so whats [math]0.\underline{9}_{0.\bar{9}}[/math] equal
>>
>>9412565
[math]0.\underline{9}_{0.\underline{9}_{0.\underline{9}_{0.\underline{9}}}}[/math]
>>
what the absolute fuck is happening in this thread
>>
[math]\sum_{n=1}^{\infty}\frac{2}{3^n} = 0.\underline{9}_{\frac{2}{3}}[/math]
.
.
[math]\sum_{n=1}^{\infty}\frac{9}{10^n} = 0.\overline{9}[/math]
.
.
[math]\sum_{n=1}^{\infty}\frac{15}{16^n} = 0.\overline{9}_{\frac{15}{16}}[/math]
.
.
i dont know what the fuck even but its self evident that our predecessors had a very shit understanding of numbers. Repeating overline is really insufficient by itself. There are entire layers of shit here that'd otherwise been covered by loose approximations to 1.
>>
>>9412743
Boy it'd be nice if 4chans latex didnt suck cock. Literally trips on on stuff outside of larex blocks.
Fuck. Retard board.
>>
>>9407905
>you give the square an area of 1 before filling in the triangles, so it's like you're starting with the solution you're trying to prove
The solution is not that "The square has area one"

The point of the pic is that that area can be perfectly covered by the triangles. No one is trying to prove the area of the square is one, that's a given.


It's the equality which is illustrated
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>>9406727
>Prove that this equation is wrong.
The burden of proof is on you

see, I won like we do on /pol/
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>>9406735
>a real answer
What the fuck?
>>
$$ \forall n \in \mathbb{N}, n > -1 $$
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>>9412743
>>9412746
>i am math god and know everything better than all dem mathematicians
>can't figure out latex
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>>9413001
>imfuckingplying this was my fault and not 4chan's
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>>9413024
>I'm stupid so the machine is "guilty"

It's simply an iq test you didn't pass
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>>9407363
>side with
get back to your own tribe, /pol/itard
>>
"Sum from 1 to infinity" is ill defined as infinity is not a number. It should be more like

[eqn]\lim_{N\to\infty}\sum_{n=1}^{N} \frac{1}{2^n} = 1[/eqn]
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>>9413247
>>
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>>9413247
kek fuck.
>"Consider the following sequence: 1.79, 1.799, 1.7999,... It can be observed that the numbers are "approaching" 1.8, the limit of the sequence."
Math is broken down a beaten path of retardation and calculus was a mistake.
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>>9411288
use latex you fucking sperglord
>>
>>9413330
but I cant :(
>>
Holy shit, how uneducated do you have to be to not understand how infinity works in mathematics? Why do you even pretend you reason like a scientist when you have clearly never taken the time to open a book about analysis?
I hope you are all either underage fags or larping as brainlets, because 18-year-olds in my country can solve this without spewing bullshit like you do.
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>>9415574
Every thread like this is made by trolls just pretending to be retarded to bait people like you. I mean
> because 18-year-olds in my country can solve
Good fucking lord, how pretentious.

But still, I pity you so please let me give you the quick guide to retarded posts in /sci/:

In general terms, there are only three kinds of retarded posts:

Case 1: Fake retardation posts
This is when someone who is knowledgeable of math and science simply wants to troll. A distinguishing feature of these threads is that they will include use of mathematical notation and will be short in order to enhance comedy.

Case 2: /pol/ retardation
Anything that anyone from /pol/ does is usually retarded, you know that. So I will focus on you can distinguish Case 2 posts from Case 1 posts. And the rule is easy: These threads will not include mathematical notation, because /pol/tards typically dropped out of high school or failed to get into college (which they blame diversity for). This means the entire thing, despite being supposedly mathematical or scientific, will be written in common English. This is, of course, a red flag. Because this is the only board with Latex which means you have to be beyond infinite retardation in order to cripple your own message by not using appropriate notation. And who do we know is beyond infinite retardation? That's right, /pol/

Exercise: Is this thread case 1 or case 2?
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>>9415586
you have no latex so
fuck off back to >>>/pol/
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>>9415595
No, poor /pol/tard. Those rules are only valid for retarded posts. Not every post is retarded. I know this is a hard notion to comprehend for someone for /pol/ who must believe that retarded is the only kind of post that exists. But if you look outside of /pol/ there are, usually, non-retarded posts. You won't find them where you come from, but I still think you have to go back.
>>
>>9413247
It's almost as if "Sum from 1 to infinity" was a shorthand notation for something else. I wonder what that "something else" is...
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>>9415586
If you actually, unironically believe calculus, limits, and the use of infinity in math "make sense", you are a god forsaken brainlet who has only tricked himself into believing self-intelligence based on how much you didn't critcally think and just copied insteuctions verbatim to get good grades. Really dont need autism analysis posts like yours.
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>>9416036

Your troll is weak.
>>
>>9416200

>>9413001
>>
>>9416094

[math]x= \big( \sum_{n=1}^{\infty} \frac{9}{10^{n}} \big) = 0.\overline{9}_{\frac{9}{10}} \neq 1[/math] .
.
[math]y= \big( \sum_{n=1}^{\infty} \frac{1}{2^{n}} \big) = 0.\underline{9}_{\frac{1}{2}} \neq 1 [/math] .
.
[math]z= \big( \sum_{n=1}^{\infty} \frac{10}{11^{n}} \big) = 0.\overline{9}_{\frac{10}{11}} \neq 1 [/math] .
.
[math]w= N \big[ \big( \sum_{n=1, k=9}^{\infty} \frac{k}{k+1} \big), (n-1) \big] = 1 \neq 0.\overline{9}[/math] .
.
y < x < z < w=1
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>>9410449
>mathematicians can't handle anything past mega
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>>9416208
N's accuracy is outside the scope of the sum.
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>>9416315
[math]w= N \big[ \big( \sum_{n=1}^{g} \frac{k}{(k+1)^{n}} \big), (g-1)\big] = 1 > 0.\overline{9}; k\geq 9
[/math]

probably the least effort required to write a general infinite sum that can actually converge to 1 without
>m-muh approaching approximation
>>
This whole discussion is bullshit. 1 means its 1.
0,99999999999999999999999 etc is still not 1 even tho it's so close that in any practical sense it is the same number.

OPs formula shouldn't use the = sign imo

>>9416450
Now multiply that infinitely you say its the same as if you'd multiply 1 by infinity?
>>
>>9406756
Underrated post




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