Not -1/12
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- Опубликовано: 9 сен 2024
- Yes -1/12, Ramanujan Summation: • the most famous Ramanu...
Sum of all n^2, • how Ramanujan did 1^2+...
How about let's get it to be -1/9: • I Got Put in the Math ...
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blackpenredpen | 曹老師
In a nutshell, the sum of all the positive integers is the friends we made along the way
Nice.
Conquering infinity with the power of friendship.
We lost friends
So we lost one that toxic friend out of the 8 we had made.
Perfect
Wow so that means you have -1/8 friends
Because infinity is involved, the number can represent anything. It’s like indeterminate form 0/0 and just dividing by 0 in general. Like how there’s one false proof that gets 1=2 by dividing by 0.
1(0)=2(0)
Divide zeros out of equation: 1(0)/0=2(0)/0
Bing bang boom you got a paradox: 1=2
@@axbs4863 8 year olds: Thank you for the forbidden spell, magic man
@@axbs4863 To further obfuscate (which I'd wager is the "proof" OP was thinking of):
Let x = y
xy = y^2
xy - x^2 = y^2 - x^2
x(y - x) = (y + x)(y - x)
x = y + x
Plug in x = y
x = x + x
x = 2x
1 = 2
@@capsey_ this math on meth
@@YOM2_UB Redditor located
The golden rules to be adhered to when dealing with divergent series are:
1) Do not use brackets
2) Do not remove any zero
3) Do not shuffle around more than a finite number of terms
why not
because of this shit
@@footlover9416 Because once you change the order of the series, you effectively change the series by messing with the terms like this.
@@glumbortango7182 Correct, but *blackpenredpen* did NOT change any order, and his brackets are also NOT causing the false proof.
The real issue with his proof is that
1 + 2 + 3 + ... = S
whatever is S, you need to prove that it is a number(Real number or Complex number or anything on which the rules of algebra, that he was using, actually works!)
And what if S is not a number, or maybe infinity, which means he was actually doing this:
1 + 9 * ∞ = ∞
adding - (9 * ∞) on both sides of equation
1 + 9 * ∞ - (9 * ∞) = ∞ - (9 * ∞)
since 9 * ∞ should still be ∞, therefore,
1 + ∞ - ∞ = ∞ - ∞
He created this indeterminate situation (∞ - ∞)
which he reduced to:
1 = - 8 * ∞
This is the only problem with his proof. Not brackets, or order of infinitely long sequence!
And to anyone thinking, it should be equal to zero; No it does not! since ∞ * ∞ = ∞, and n * ∞ = ∞, where n is a Real number. You may use these two statements to justify ∞ - ∞ = ANYTHING. (∞ - ∞) is a situation which tells *nothing* about the resultant value, which is why it is called indeterminate.
@@AceptronYou cant really say ♾ * ♾ = ♾, because infinity is not EQUAL to any value, it doesnt make sense to use ‘=‘ when talking about infinity
this man deserves an award for being the master of the pen switch
I agree :)
Yeah like how is he holding a pen and writing with the other one lol
He should make it his account name fr
I like this as a counterexample to the numberphile method of extracting -1/12
Numberphile did several videos explaining how it's divergent, how the summation isn't accurate, and how the Zeta function works to produce -1/12. I don't get why people like to make fun of Numberphile for this, since they're no worse than this channel in this regard.
Joshua Hillerup People do it because they like having reasons for criticizing math RUclipsrs, even if there legitimately are no valid criticisms with regards to a specific topic. That is why people do it.
Brian talks about He is wrong on purpose. He intentionally did something wrong just to prove the point that it is wrong because it produces contradictions. But he knows it’s wrong. It’s a teaching method.
Brian talks about Also, the sum being equal to -1/12 doesn’t have to do with it being divergent or not, since the two concepts refer to two different things altogether. There is no contradiction between saying the partial sums don’t have a limit and the sum has a value of -1/12
Brian talks about Well, I’m not denying that. I was just clarifying the distinction.
The vertex of the parabola y=x(x+1)/2 is at y=-1/8 LOL
And the area under the x axis (-1 to 0) is -1/12
👏
Fucc
Mind = Blown 😮
pi=3
e=2
0.01=0
sinx=x
*ok hand emoji*
How can the area be negative lol
Infinity is so interesting
It's called playing with non absolute convergent series, riemann has a theorem that any permutation of a series that is semi convergent(that is not absolutely convergent) can lead to any answer
But this isn’t even partially convergent, it’s fully divergent. Would this still apply
@@blackmamba1261 since the theorem is for semi convergent it also aplies to divergents,like black pen showed you,you can get any number by using tricks like adding series ,or replacing terms,numberphile does similar tricks and gets the -1/12,you can get a finite value for a divergent even though its wrong,this video was for comedic pourposes,when you are replacing terms you are basically changing the series(if its absolutely convergent the series will still be the same no matter what operations you do to it)
@@MA-bm9jz I am sorry but no matter the permutation you use, this series will always be infinity. The theorem you talk about is not the same thing, when you rearrange a semi convergent series, the new series is (semi) convergent and the value we give to it is a true value, it is the limit of the partial sum sequence.
@@MA-bm9jz More, your reasoning isn't correct, it is not because the theorem is for semi convergent series that it works also for divergent series. There is no including relationship between the sets of divergent and semi convergent series.
As @ViperDaniel points out both -1/12 and -1/8 are related to the sequence 1+2+3+4+... If you graph in the equation (x+1)x/2 (which is the continuous version of the sequence above) you get a parabola. The signed area under the x axis is -1/12, and the minimum point has a height of -1/8. These two numbers are NOT arbitrary, they pop right out from plotting the equation. Saying the word "equals" in these situations is troublesome without giving a very specific context.
Never thought to investigate the link with the sum formula, thanks for this! Any thoughts on the significance of the value of the intercepts or the definite integral value?
integrals rediscovered)))
I did not get a parabola. I got a right triangle.
The function for counting is simply f(x)=x so that when x is at 1 y is at 1 or when x=3 y=3 and so on...
All you are doing is solving for the area of a triangle.
Furthermore, if you start counting from a point that isnt 0, say for instance, 31, and you count to 49 all you need to do now is find the area of a rectangle and then the triangle that sits at the top.
So from 31 to 49 the sum is...
31*31=area of the rectangle=961
49-31=19= the amount of times you had to +1(count) from 31 to 49.
((19+1)19)/2=area of the triangle=190
Addd both areas to get the sum of 31+32+33....+49=1152
This is also hooks law in physics.
*numberphile intensifies*
Is mathologer still claiming that 0.999... = 1? I felt so sorry for them.
@@Deibler666 yes, 0.(9) is APPROXIMATELY 1, and what's wrong with it? (1/3)X3=0.(3)X3=0.(9)=1
Approximately? It is or it is not? For me is both and none, ’cause they don't even agree if 0.999... is an actual number. I'm sorry for them who argue any side.
@@Deibler666 I wrote an equation in my previous comment which is the easiest proof that 0.(9) is 1. It is considered that 0.(9) is 1, because the endless 0.9999999... is just so close that in fact there's no difference between it and 1
@@Deibler666 Yeah, though it is a bit contradicting, but there isn't much point in wasting too much time for it 😂😂
Thanks blackpenredpen! You're the first person to get me legitimately interested in maths. All my math teachers in school were too slow and boring. But you actually make it interesting.
Thank you!!
There seems to be some jargon among a lot of math teachers that math is *_serious business,_* so it is very refreshing when someone talks about math in a more lighthearted way.
I am a big fan of the mechanics book "Introduction To Classical Mechanics" by David Morin specifically for that reason - that book feels sort of as if it were written for children, and yet it brings up college level topics like the Lagrangian Method and space mechanics, and that's definitely a very interesting contrast.
@@blackpenredpen "And they were stuck thanking each other forever, infinitely, or S times"
@@blackpenredpenstupid you substrate ∞ from ∞ and it's super sum not random sum
Make your school hire him immediately.
This is kind of interesting in the general case. You always get -1/8.
In general, 2n+1 consecutive integers will sum to a value equal to a multiple of (2n+1)^2, so long as the middle value is a multiple of 2n+1.
Ex.
n = 1: (3k-1)+3k+(3k+1) = 9k
n = 2: (5k-2)+(5k-1)+5k+(5k+1)+(5k+2) = 25k
or in general,
[(2n+1)k - (n)] + [(2n+1)k - (n-1)] + ... + [(2n+1)k - 0] + ... + [(2n+1)k - (n-1)] + [(2n+1)k - (n)] = (2n+1)(2n+1)k = (2n+1)^2 k
This allows for grouping of every (2n+1) terms after the first n terms have been summed, resulting in the nth triangular number.
Ex.
n = 1: (1) + (9)(1+2+3+...) = S
n = 2: (1+2) + (25)(1+2+3+...) = S
or in general, using the formula for the nth triangular number,
(n)(n+1)/2 + [(2n+1)^2](1+2+3+...) = S
Noting 1+2+3+... is just S, and some further simplification:
(n)(n+1)/2 + [(2n+1)^2](S) = S
[n^2 +n]/2 = S - [(2n+1)^2](S)
[n^2 +n]/2 = S(1 - [(2n+1)^2])
[n^2 +n]/[2*(1 - [(2n+1)^2])] = S
[n^2 +n]/[2*(1-[4n^2 +4n +1])] = S
[n^2 +n]/[2*(-4n^2 -4n)] = S
[n^2 +n]/[-8n^2 -8n)] = S
Finally:
-1/8 = S, no matter the choice of n. Very neat and strange.
How much time dyu have ?
oh i just asked that. very cool!
[n^2 + n]/[-8n^2 - 8n]=S
[n*(n+1)]/[-8n*(n+1)]=S *simplify the n*
[n+1]/[-8*(n+1)]=S *simplify the (n+1)*
-1/8=S
You're correct! But when n=0 the answer diverges because you can't simplified because you can't divided by 0
cause you are changing the series,check riemann's theorem for semi convergent series,ovoiusly this is wrong,when yur at an exam, you write +inifnity
dviscusi1 beautiful
You need to prove that S exists in the first place. The fact that it gives the result in these examples proves that it does not. Merely writing S on the board does not prove it exists.
history buff03
Hope you know that this is just for fun : )
blackpenredpen I know it, but do others? It is instructive for math students to know this (I too used to teach math, and I used to use examples like this to show students that a symbol has to stand for something "real" in the math world.)
@@blackpenredpen Thank you for this. I makes math fun to see these interesting examples. And if people don't get that it's for fun, they will eventually. Don't stop doing what you're doing.
It's a running meme in the community.
@@historybuff0393 People know. For one thing, the whole "Infinity is -1/12" is a running gag in the math community, but also, the only people who would ever have to use this kind of math, or who would use this kind of math, are going to also be aware of how superfluous these "proofs" are.
Me: Everyday, I save my money with an increasing amount from $1, $2, $3, $4, and so on. I will be rich in the future!
My savings: *Let me show you how saving is equal to debt*
If you can show that debt is equal to saving, then you have either spoken with my wife or you work for the government.
Numberphile: 1+2+3+4... = -1/12 because Ramanujan.
BlackPenRedPen: HOLD MY BEER
: )))))
@@ViratKohli-jj3wj matlab kuch bhi
This is wrong
U don't know whether infinity is a multiple of 3
@@ViratKohli-jj3wj you are right bro. his result is very useful and has important applications. but he saw infinity in a different way.
@@ddm1912 nikal lavde
You can continue the pattern indefinitely and get -1/8 each time. 3 + 25s = s | 6 + 49s = s | 10 + 81s = s
This is fantastic. Most Ramanujan videos are confusing or shallow, whereas your Ramanujan content is challenging yet crystal clear. Thank you very much!
The sum is divergent, but if you consider it in the context of the Riemann Zeta function, the answer should be -1/12. But, of course, just as with 0/0, looking at it from different angles might give you different values, but if not equal, the sum is at least somehow related to -1/12, as seen with the analytic continuation of the zeta function.
Cats are related to dogs. That doesn't mean cats are dogs, does it? To say that the sum of positive integers converges to -1/12 is simply wrong. If you assume that it converges on anything, you can make that thing any number you wish. Ex falso sequitur quod libet.
@@MrCmon113 At the very least it converges to infinity right?
@taxtro please go study the Riemann zeta function. It will make sense to you then.
@@chrisbagshaw5676 please understand that zeta(-1) is not 1+2+3+4.... and its a completely different definition. The connection is right but strictly speaking he is right not you.
Hassan Akhtar he’s actually not right. We do not say things “converge” to infinity.
Srinivasa Ramanujan , Not every legend lives long. He lived BEYOND !!
Yeah Well put..
It's a divergent series, OK
Now it will focus in negative number line
This series is getting answer as
- 1/12. U are changing the concavity of series by adding them as pairs of 3 and 25 thus focusing the series on - 1/8, it changes the magnitude of infinity by pairing in 3 and 25
Neither of those, the series diverges
Marcovnikov Chung That has nothing to do with this.
this is humor, -1/8,-1/12 none of them are the answers, internet is really mad about this
Once you grasp the fact that infinity is infinitely larger than any number you can pin down... It might start to make a little more sense. Until then, enjoy the humor.
That is the point
@@bryandandoy885 you are right, but if we had to give an answer to this question, it would be -1/12 by application of the zeta function's extension
Can you try and find out what sqrt(2) factorial is!
JZ Animates
I like this question! However, I think we can only approximate the answer. Are you okay with that?
blackpenredpen Yes that is fine! I just wanted to see how the question would be approached.
@@jzanimates2352 Try it with Π function or Γ function.
thats the integral from 0 to infinity of x to the root 2 times e to the x with respect to x.
@@UjwalAroor e^-x
I am seriously concerned about all those -1/12-Videos regarding the infinite sum of all natural numbers.....I fear that some non-math-people start to think that this sum actually HAS a finite value ....
Nice video though, showing that assuming this divergent sum to be finite leads to multiple contradicting solutions...
I am a maths enthusiast, quite young. Can you tell me is anything wrong in the vudeo. Some say sum of all positibe integers is -1/12 and here it is -1/8. I m kinda confused
@@raghavsingh3381 infinitiy is so tricky) especially infinite sums, Ramanujan way to solve this was based on guesses of 1-1+1-1+1-1+... and 1-2+3-4+5-6+7-8+.. sums. This gave us -1/12, btw which can be proved by integral sum (another video) or even Rimman zeta function. Here is -1/8 and it's likely to be true as well)) The main point is that it's impossible to get the right answer, because people now cannot manage with infinite sums and these two results visualise the different approaches to simplify and just make an assumption how to counter infinite sums. just take it like a joke) Nobody pretends to find the only right solution, and... We don't know whether it is the only)
@@yuriiafonin1412 oh thanks man. For the help. I was confused for a long time. Since i could not believe that sum of infinite terms cam be a finite number. Thanks for the explanation.
I've seen supposedly Maths competent people (they claimed to be graduate level students) argue that it is a valid ststement.
@@raghavsingh3381 I view these as proofs by contradiction. The conclusions are clearly nonsensical, in addition to being contradictory, therefore the proposition that S converges to a definite number (as least using these arguments) is false.
Do you think you could find a way to make it equal to pi?
When he does S-9S=-8S, this isn't algebraically supported as strictly speaking true. S-9S would necessarily be conditionally convergent and subject to the Riemann rearrangement theorem, so he is saying that we should let S-9S be arranged in such a way to obtain -8S. This is really just a declaration, so you could just make it equal to S/pi instead, but you won't be able to trick people that don't know this into thinking that you did it legitimately. I don't think you'd be able to do that with this method that writes S as it's partial sum plus some odd square times itself, because it's constant at -1/8. You'd have to come up with some completely different method that tricks people into thinking it's legitimate, which I imagine would be pretty hard. This method worked so well because the partial sum is (x(x+1))/2 and the rest is (2x+1)^2, and 1-(2x+1)^2=-4(x(x+1)); the x's cancel and it becomes constant.
You wrote it all or voice typing?
Yaa i can !!..
@@SlipperyTeeth idk it's hard for me to understand. Am an 11th grader.
@@SlipperyTeeth are you studying at varsity?
The Riemann/Ramanujan assignment of values to an infinite series works because it establishes a particular pattern of summation. When you group the elements of the sum differently, it's a different pattern and it is easy to get different values (not just -1/8) by grouping things differently. In particular, if you make manipulations that rely on the fact that there are an infinite number of terms and shift them arbitrarily, you'll get all sorts of answers other than infinity or -1/12, mostly because you're essentially adding infinity to one side or the other of the equation.
The "real life" proof of the Riemann/Ramanujan approach is as it applies to the sum of all cubes, which arises in calculating the Casimir effect, the method of derivation yielding a sum of all cubes which is divergent. Using the analytic continuation of the zeta function yields 1/120, which turns out to agree with the experimental answer. Therefore this method of assigning a value to an infinite sum has a real meaning. Remember that mathematicians used to think that sqrt(-1) was mathematically meaningless, too, until it was found to be extremely useful in physics and could be regarded as a rotation of 90 degrees.
The sum of all integers plays a role in String Theory, and that theory uses -1/12 as a valid result, but I don't regard it as the same kind of real life proof, because there are no experiments that can be done to verify it. For me, it is much more interesting when nature itself proves that the math is right.
perfect answer.. the same with sqrt(-1), this magical number i (commonly used in electromagnetics, radio communications , AC etc) and -1/12 are supposed to be tools for finding solutions and not to be proven true..
@@ardinhajihil5011 in simple sum is not defined it is only defined by analytic continuations of zeta function.
when you subtract inf with inf, you will have some interesting result. The same as dividing by zero, powering 1 by inf, powering inf with 0 and so on... Here S is supposedly inf (divergent series)
Okay... But are we substracting a more massive infinity with a less massive infinity, or...? :^)
I have always thought it was weird that it was -1/12 since it’s negative and a fraction so thanks for clearing it up. This makes so much more sense
Not really no 🙃
@@coursmaths138 this one is wrong
2 reasons:
1.he has chosen only one infinite series in your method and you did subtraction operation on the same infinite series that means he is subtracting infinity from infinity virtually which will give indeterminate value
Where as Ramanujan had chosen three different infinite series such that when he did subtraction operation on two infinite series he got a different infinite series so in his case there is no indeterminate kind of thing.
2. This does not satisfy the Euler Reimann function
@@kanishkverma706 I'm not saying that what did Blackpenredpen is correct. I'm saying that the 1st comment is wrong by saying " _This makes so much more sense_ ", because what did B doesn't prove the usual computation to be wrong and doesn't "explain" why -1/12 would be false.
How can you say he substracts "same" infinite (what does this even mean?) ?😳 How do you see in that computation something related to limit (i guess it's what you mean...) ?
What do you call "different infinities"?
2 doesn't mean anything. What does mean "satisfying a function"? 😮 Though you probably mean the Riemann Zeta function...
@@coursmaths138 yeah I meant that
S is infinite, so no matter what you do to it, multiple it or sum it up with another number.... You will always get S again.
Exactly. A divergent series such as this one has no finite limit (i.e., a real number) and therefore no finite sum. The -1/12 "sum" should be called, more accurately, the Ramanujan Manipulation or the Ramanujan Conjecture.
It depends on the definition of an infinite sum. The partial sum is only one way to define it.
This man's marker switching is ON POINT
Not -1/12, Chen Lu is always the answer
Coincidentally or mathematically I don't know, but the sum of x natural nos. is equal to [x(x+1)]/2 , and the minimum value of this quadratic equation is -1/8 or -0.125. Please explain why...
And the area under the x axis (-1 to 0) is -1/12
@@siddharth.4671 Lol same
OMG! WTF!
there is only one explanation to this: 👽
Wowza
I always like the "not (yet) listed" videos.
and as always: never mess with infinity ;)
Riemann's theorem says any non- absolute convergent series can be arranged in a manner in order get a desired no number. This series resembles that, though it's not convergent but has a special character of suppliing all natural numbers.
Therefore, - 1/12=-1/8. Q. E. D. :)
Brilliant
This the consequence of a more general property of divergent non-oscillating sums. Take, for instance, the sum from n = 1 onwards of n + m so that one has (m + 1) + (m + 2) + (m + 3) + •••. Using Ramanujan summation, or any other method which you find appropriate, you can evaluate this sum by using the linearity of summation, getting Sum(n + m) = Sum(n) + Sum(m) = -1/12 + -m/2. Now add every n = 1 to n = m. The result will be m^2/2 -1/12. Now we see that the summation (1 + 2 + ••• + m) + [(m + 1) + (m + 2) + ••• ] is not the same as the summation 1 + 2 + 3 + •••. This proves a statement about how association works with this class of divergent series. There is a more general statement that does the same, but with (n + m)^p for all natural p, such that one gets 1^p + ••• + m^p + Sum[(n + m)^p] = m^(p + 1)/(p + 1) + Sum(n^p). This can be proven fairly easily: one simply needs to use the binomial theorem to expand (n + m)^p into a polynomial of n, then use linearity on the polynomial to evaluate the sum of each monomial individually, using the identity Sum(n^s) = -B(s + 1)/(s + 1), where B(n) is nth Bernoulli number. This will give a finite sum. Then, to add 1^p + ••• + m^p, use Faulhaber’s formula. Combining both terms, one obtains a telescoping finite sum which will simplify to the right side of the equation.
Also, I assume this itself is a special case of an even more general theorem about grouping in general. For instance, what happens if one tries all different groupings. What if I group 1 & 2, 3 & 4, 5 & 6, and so on? In general, 2n - 1 + 2n = 4n - 1. Hence the sum 3 + 7 + 11 + ••• = Sum(4n - 1) = 4Sum(n) - Sum(1) = 4(-1/12) + 1/2 = 1/2 - 1/3 = 1/6, which is equivalent to -2(-1/12) = -2 - 4 - 6 - 8 - •••.
Ultimately, the theorem which describes what happens when a given grouping of addends is performed on the series will be equivalent to another theorem about the set partitions of the set of natural numbers. This is very interesting, since the value of the associator in this case is m^(p + 1)/(p + 1), which is the antiderivative of m^p with respect to m.
Another example could be 1 + (2 + 3) + (4 + 5 + 6) + (7 + 8 + 9 + 10) + ••• = 1 + 5 + 15 + 34 + ••• = Sum[(n^3 + n)/2] = (1/120)/2 + (-1/12)/2 = -9/240 = -3/80. Of course, it easy to understand that once you perform a grouping on a series, it no longer is the same series anymore, and obviously the value changes. This is, a series is defined by the set of elements over which it performs the summation. We can consider the grouping a type of operator on sets. Naturally, there exists some grouping G such that G{1, 2, 3, •••} = {1, 4, 9, 16, •••}, and we know the sums of the two to be different because the sets are different. Each set has a unique summation, though, so the theory is naturally consistent. The issue is that when we perform a grouping a finite set, the resulting sets is decreased in cardinality. In other words, the sum of two finite sets is invariant under grouping. For absolutely convergent series, the sets are now countably infinite, but the sums of the sets are invariant under grouping. In other words, we could declare that, if S is a natural number, then G(S) = S, and if S is a set, then G(S) = T, where T is another set. Then we say that ΣG(S) = GΣ(S). A demonstration would be for example the sum 1/2 + 1/4 + •••. The simplest non-trivial grouping is 1/2 + 1/4, 1/8 + 1/16, and so on, which gives the elements 3/4, 3/16, 3/64, etc. This sum simply (3/4)/(1 - 1/4) = 1. This sum was preserved under this grouping because the new sequence of partial sums simply has its argument scaled, and as such has equal limit. For conditionally convergent series, you no longer have sum preservation under a grouping transformation. Naturally, you would not have it for divergent sums either.
What's cool to me is that -1/8 works using not just 3 and 5 but every similarly fashioned odd number.
7? Yep. 11? Yep. 101? Yep.
I can put a link to my scribbled-out proof if anyone's interested, but it can be shown by leveraging Gauss' formula for summing consecutive numbers.
: ))))))))
using series is easy too, lim an = lim n that is infinity (this by the theorem, if the series converges, then lim an is equal to zero, if is not equal to zero, the series diverges)
An insane trolling level. :)
All infinities are equal" might sound like a classic goof in math circles, but don't let that fool you! This brain teaser is anything but ordinary. It's a real gem of a video, packed with fun and excitement. Watching it and working out its puzzles is a blast from start to finish.
I think its niether. Both proofs (the -1÷8 and -1÷12) make the assumption the series 1+2+3+4··· has a sum, but they don't prove that the series converges. If anything, I think you can prove by contradiction that since a convergent series cant have multiply sums, it is not converging.
trash can The fact that the sequence of partial sums does not converge does not prove that the sum has no value. The problem with the algebra of this video is that it ignores the important Riemann theorem of rearrangement, which is important even for convergent series.
You sum all number accordingly not in water process . Sum of all integer number is always -1/12 not -1/8 . -1/12 infinity gives by a great mathematician .
I like that jacket
Thanks.
If we will add 3 numbers and then take 9 common then the bracket will be s÷3 as there are n number of natural number
→1+3s=s
S=-1/2
Looking at the comments, I've realised that not many understand the brilliant satire here.
It diverges. Let’s just leave it at that so my brain doesn’t explode. Thanks.
It can't be -1/8 because you are indirectly saying that the last 2nd positive number of this infinite series will be a multiple of 3 (3k), just like in the infinite series of 1 -1 +1 -1 +1 -1 +..... You can't say it answer is 0 as you can't say that the last digit is -1 and everything cancels out so answer is 0, last digit can be +1 so answer will be 1. Similarly the infinite series in video can have it's last 2nd digit as (3k-1) or (3k+1).
Well said! This method makes an assumption about the last term of the series.
that is also the wrong way to think about it. there is no last digit or last term in an infinite series.
ThoughtGaze , i think you & he are pointing out the same problem.
Het Shah , Brilliant! I hope Prof.Red Pen sees your comment!
Appilesh , you say it well too!
You can’t do standard algebraic operations on S if S is infinite or, more generally, if it belongs to a set where those operations are not defined for S.
We know that sum 1+2+3+4+5+6+... is divergent and -1/12 have sense only then we using imaginary number i=sqrt(-1) Ramanujan summation, this video uses only real numbers. Still logic behind summation is amazing.
How u prove with complex number?
You can generalize what you did here. You will get:
k? = [1 - (2k+1)²]S
k(k+1)/2 = - 2k(2k+2)S
S= -1/8, for all integers k
Both are wrong, infinity is undefined.
Ordinary addition is a binary process . To add any quantity of numbers you have to add them in pairs first. But an infinite series is NOT a binary process , for that reason you cant group the numbers of an infinite series in order to add them , because you cannot include the last number in any group (because there is no last number!). This is why this paradox is happening.
2:45
IDK I can be wrong but S tends to infinity and S - 9S is infinity minus infinity which is indeterminate.
Good Point
I am just worried what to do when you can't deduce instantly that series is divergent, for example when it's geometric, then we shouldn't do any operations until we are sure it is convergent?
Why don't we do the general case instead. Add the first 2n numbers up, and then goup by (2n+1) numbers, so n(n+1)/2 + (2n+1)^2 S = S, and so S=-1/8.
If Ramanujan was alive today he might have another ans to it. RIP "GREATEST MATHEMATICIAN"
Why not both? -1/12 = -1/8 solves the dilemma.
It is none. Since they're divergent series, the Ramanujan Sum assigns them a value, but normal algebra cannot be used to redistribute the terms of a divergent series without changing the result. (If you take the series as it is, 1+2+3+4+5... without rearranging the terms, then it is - 1/12, if you come up with these new series then you get other results).
Well, its neither -1/8 nor - 1/12. You cannot group and factor infinite series like that, the rules that apply to finite series do not work for infinite ones. And also -1/12 is the result of plugging -1 into the analytic continuation of the riemann zetta function. While Riemann zeta function of - 1 would represent the sum of all whole numbers, it is not defined for negative numbers, we can only find the value of its analitic continuation which does not represent the sum of whole numbers anymore.
The interesting thing is, this method of calculation will always result in S=-1/8,
no matter how many consecutive numbers we group up together.
For each natural number n, we can pretend:
S=1+2+3+...=(1+2+...+n)+[(n+1)+(n+2)+...+(3n+1)]+[(3n+2)+(3n+3)+...+(5n+2)]+...
S=(n+1)n/2+(2n+1)^2+2*(2n+1)^2+3*(2n+1)^2+...
S=(n+1)n/2+(2n+1)^2S
[(2n+1)^2 -1]S=-(n+1)n/2
(2n+1+1)(2n+1-1)S=-(n+1)n/2
S=-(n+1)n/(2*(2n+2)*2n)
S=-1/8
This the consequence of a more general property of divergent non-oscillating sums. Take, for instance, the sum from n = 1 onwards of n + m so that one has (m + 1) + (m + 2) + (m + 3) + •••. Using Ramanujan summation, or any other method which you find appropriate - I merely chose Ramanujan summation because it is the method BPRP used in his previous video, and also because it is most familiar to people - you can evaluate this sum by using the linearity of summation, getting Sum(n + m) = Sum(n) + Sum(m) = -1/12 + -m/2. Now add every n = 1 to n = m. The result will be m^2/2 -1/12. Now we see that the summation (1 + 2 + ••• + m) + [(m + 1) + (m + 2) + ••• ] is not the same as the summation 1 + 2 + 3 + •••. This proves a statement about how association works with this class of divergent series. There is a more general statement that does the same, but with (n + m)^p for all natural p, such that one gets 1^p + ••• + m^p + Sum[(n + m)^p] = m^(p + 1)/(p + 1) + Sum(n^p). This is very interesting in itself. Also, I assume this itself is a special case of an even more general theorem about grouping in general.
can you come up with a method to end up at an arbitrary number?
I think you can't close your eyes or hide infinite with the symbol which has behavior like finite number. Because after all you are subtracting infinite from infinite which is a typical example of indeterminate form.
But i admit i do't know about math in deed.
Please don't take it seriously.
That's actually one of the better arguments against this I've seen.
Actually, the idea that 1+2+3+4+...=-1/12 is actually talking about the zeta function and extending the domain of the function. Specifically, we're trying to find out what is the zeta function equal to at -1/12.
It's like if we said, "If we have a right triangle with an angle of i, the ratio of the lengths adjacent/hypotenuse is about 1.54." The cosine function, when analyzed at i, does mean the ratio of the lengths at that angle. Similarly, the zeta function when analyzed at -1, doesn't mean 1+2+3+4+....
The Rational Thinker
You are right buddy.
But some people are taking analytical continuation in the primary school sense !
The thing about that though is that the method used to identify the analytical continuation of the Zeta Function is almost identical to the Numberphile method for obtaining 1+2+3+4+... = -1/12.
The primary difference between the two methods is that the analytical continuation maintains z as a variable while manipulating the series and then puts the value of z = -1 in after the manipulation, while Numberphile puts the value of z = -1 in first and then manipulates the series.
Other than that the processes are pretty much identical.
chandan kar Not at all. What you people fail to understand is that the analytic continuation is not the only method for obtaining these results. There are multiple well-defined methods founded on independently rigorous theories which gives the result that 1 + 2 + 3 + ••• = -1/12. Y’all should stop trying to make this discussion about analytic continuation and instead genuinely try to understand the motivation and the rigorous foundation behind the theory of divergent summation, which draws from several other theories, as I stated.
The problem is that f(z)=∑1/n^z is only define when z≥1 so the way of find 1+2+3+....=-1/12 through zeta function is also wrong!
Gui Silva the whole analytic continuation thing is just a "what if"... that's why mathematicians are trying to free the domain of the Riemann zeta... it's like, what if we can plug -1 into it? That doesn't mean it's wrong
Everyone repeat with me:
1. A subsequence limit is only a cluster point of the full sequence.
2. The set derivative (cluster-point set) is not a number.
3. Infinity minus infinity is not zero.
How about a video explaining these and showing what's actually going on?
S = 1 + 9 S: we used grouping so actually the RHS is only a subsequence of S. Even if it were the full sequence, S "converges" (in compactification metric) to +infinity and subtracting infinity from that equation gives undefined = undefined, not 0 = 1 + 8 S.
reason for why the first one was -1/8 is because you had skipped an indeterminate form of infinity - infinity, notice that 1 + 9 * infinity = infinity and therefor satisfies the equation.
Imad Hamaidi No, because S is not infinity. You cannot have them be infinity because you cannot perform arithmetic with infinity.
@@angelmendez-rivera351 when I say infinity I mean the limit sense of infinity, and limits do have arithmetic properties!
Imad Hamaidi Yes, limits have arithmetic properties, but S is not a limit. The limit of the sequence of partial sums of the sequence of natural numbers listed in order, with the indicator function as the cut-off of summation, does indeed diverge, but it was never said S must be equal to this limit. A sum only equals this limit when the limit exists. If the sum must be defined as an arithmetic operation, then of course it cannot be the limit every time.
@@angelmendez-rivera351 S is the limit of the sequence S(n) = n*(n+1)/2 as n approaches infinity.
Imad Hamaidi No, it is not. That is simply not its definition. An arithmetic operation on real numbers cannot be equal to a limit. That is literally impossible if you understand what an arithmetic operation is. S is only the limit of S(n) when the limit exists, but it is not an identity for every single S that S = limit S(n). It simply is not. And if it, then go ahead that 2 + 2 = 4 using this identity. I’ll be waiting an eternity here -.-
Numberphile vs blackpenredpen.... where’s my popcorn
3:45 "I will first call this to BS"
Ladies and gentlemen, I think my brain just commited suicide...
None of the two answers is correct. the fault in your solutions arises when you do the subtractions (s -9s or s-25s) ; you subtracted infinity from infinity which is indeterminate.
Then s(infinity) is also indeterminate then how do you find ans
@@saicharanmadem267 You can't.
2:49 subtracting infinity from infinity is not allowed, since it is undefined.
I think this step is invaid
You are right, both of S is infinite and should not be subtracted together
This entertaining "solution" generalises for any natural number k.
(k/2)(k+1) + (2k+1)^2S = S [i.e. kth triangular number + (2k+1)^2 ]
=> S(4k^2 + 4k) = -((k^2+k)/2)
=> S = -(k^2+k) / 8(k^2+k) = -1/8
: ))))))
You can not group of numbers because we don't know whether the pair is ending with same no. of group or not .
Let's do it legitimately.
in your zero double factorial video, you have presented 2 answers for zero !! which were
sqrt(2/pi) and 1.
let's assume them then that sqrt(2/pi) = 1.
from where we get, pi=2. but we know previously we know pi=3,
hence 2=3.
from this video, also, we get
we get 1+2+3+......=-1/8 but previously it has been defined =-1/12
hence,
-1/8=-1/12
hence, 2=3
so, these two instances actually prove that 2=3.
Pi is not 3...
@@cosimobaldi03 mostly people round it off to 3. and i just made it up jokingly. dont take it seriously
Do not forget that π = e.
And e = 2
Fundamental theorem of engeneering
Mathematics is to generalise as much as possible and leave as little ambiguity as possible. In which of different domains/settings/contexts the question is posed is reflected in the answer. The largest domain rules.
S is undefined ( the sum diverges). Therefore all calculations after, using S, are not correct. And your examples are as good as -1/12... Well done!
Sir I wanted to tell you that sum of all the natural number is also equals = -1/18 . I can show the proof.
Bro u did a great mistake cuz if u r not getting ans as -1/12 then u can't solve questions of 26 dimension. Ya u heard right 26 dimension. U even can't solve problems of quantum mechanics and string theory. If u r correct then 4=5 as per many RUclipsrs
Or 0/0=2 or 0/0=1.
Interesting attempt on disproving it. The base of your disprove is the assumption of this is that the the "prove" for -1/12 is done, is based on arbitrary tricks with infinite series, so you just do a different arbitrary trick to show that you can get any number.
As it turns out, infinite sums aren't well understood yet, and there seem so be some operations that are legal (as in "don't change the result" or "lead to the right result") and others that aren't. We just haven't figured out the ruleset yet.
The recent discovery of smooth weighting functions for infinite sums stronly points towards -1/12 being the right answer if you read it as the equivalency class "-1/12 mod infinity".
I guess there must be a mistake. Bcz analetic continuation of a function is unique. And so ans should be -1/12. This is what I know from formal mathematicians here. But thy might also have done a mistake which even they don't know :p
Subramanya M There is a mistake. The mistake is intentional. It was done to show that certain algebraic manipulations are not valid.
The error is at 1-9s=s; s is infinite, then infinite-infinite is undetermined.
Both the solutions shows that Infinity is infinitely complicicated as well as Interesting.
Since the problem is well understood, that would make it less than infinitely complicated, right ;) Also, this was a joke for Numberphile-viewers ;)
You proved Ramanujan wrong??????????????
Because is wrong
The simple fact of the matter is that we accept infinities can have different sizes when we consider the Naturals and the Reals, but we miss this we setting an infinite series to a variable, and using substitution. So, the series that is being substituted in these equations is actually not S. Let’s call it T. Where T is definitely a smaller infinity than S and can be shown to be smaller than S by S=3T+1… or for the second equation, 5T+1, as the 3 or 5 is number of numbers we are reducing S by.
Can you make video on reimann hypothesis please
I have another way to put this. Since we took S to be sum of all natural numbers till infinity and attempt to calculate it, its obvious that it wont converge. So we have a diverging sum that tends to infinity.
Now in the method shown, we group them into 3 (or 5) numbers at a time and find a S in that which is also fine. The problem comes where we write 9S-S = -1 and then 8S = -1. Here for S which actually is (tends to) infinity, doing mathematical subtraction isn't allowed, its more like infinity minus infinity form which is not defined unless given constraints. This is the reason why we got such different answer.
Open to your views on this 👍
Top 10 Pranks That Went Too Far
The correct sums for the following divergent series mentioned in this video are:
1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12
1 + 9 + 18 + 27 + 36 + 45 + ... = 19/4
3 + 25 + 50 + 75 + 100 + 125 + ... = 161/12
-1/12 = -1/8
-12 = -8
4 = 3
By induction, all integer numbers (and rational too) are equal
*THANKS FOR BREAKING MATHS, DUDE*
Just like women claiming gender equality, it doesn't make sense *dab*
Absolutely mind blowing approach 😊
When the school system sees how many hours of sleep we need.
The basic error is to combine two series which are not at all the same.
given S = 1, 2, 3, 4, 5, 6, .... then
1, 9, 18, 27, ... is not S, and 3, 25, 50, 75 is not S
*M* *A* *D* *N* *E* *S* *S*
This is an infinite series that means you can't work out only a few numbers and can't make any decision based on those few numbers. Any infinite series means undefined. This is the secret of these strange results.
This is absolutely brilliant! Incidentally you also get -1/8 if you consider sets of 7 in integers e.g. 4 to 10 and 11 to 17 etc. I would love to see numberphile respond to this.
brilliant, yeah! well perhaps until the time when you see what you did agree on,
1 + 9×∞ = ∞ , "therefore" ∞ = -1/8 ; and
3 + 25×∞ = ∞ , therefore ∞ = -1/8 , &c,
all this in spite of our very sure knowledge that these 'infinities' certainly originated from sums of positive integers only .
is this sincere, integer, brilliant algebra?
is it reasonable, rational mathematics?
do you call this realistic calculating?
huh . shiverr .
S=1+2+3+4+5+6+...+n =(n^2+n)/2
When n tend to infinity S also tend to infinity.
That is the fact based on logic.
Me: infinity is interresting.
Him: No
It's the second to last step in both cases where you basically have
*∞ - ∞ = ∞ - ∞*
on both sides, which can be anything.
Again and again how sum of positive numbers is negative ???
İt cant be negative. That is so absurd result.
Fantastic brain teaser! Great job on always making fascinating videos!
To get S = 1 + 9S, you are using assotiativity wich doesn't hold for infinite sums and anyway, this equation has not 1 but 3 solutions because -Infinity and +infinity definitely are solutions. And indewd you could say that S diverges to +infinity ;)
At least that's what we would say at my level ahah
Othun s Gungnir The sum diverging to infinity does not make infinity a solution. Infinity cannot be a solution because it is not a number. Stating that sum diverges is a statement about its asymptotic behavior, not about its value.
@@angelmendez-rivera351 yeah but if you want to solve the equation in R U {+infinity, -infinity} (R with its boundaries included), +inf and -inf are solutions, even if it might not make a lot of sense to consider the solutions this set.
Sometimes it's ok not to mess with infinity because we are living in a finite space
The golden rules to be adhered to when dealing with divergent series are:
1) Do not use brackets
2) Do not remove any zero
(unless the divergent series is stable)
3) Do not shuffle around more than a finite number of terms
Not adhering to these rules yields incorrect results.
No disrespect but I genuinely hate it when people assume that infinite series are the same as if you cut out part of it, like for example in finding the derivative of x^x^x^x^x^x... This is as stupid as dividing by 0 and proving 1=2.
Dawn Ripper
Hope you know this was just for fun. : ) happy new year!
On ne peux pas mettre de parenthèse pour une somme infinie...
Tiens! Un francophone ! qui est le seul a avoir un commentaire véridique en plus !
Tu as raison, ça impliquerait que les derniers trois termes entourent un multiple de 3
C’est vrai, parce que le théorème de reconfiguration de sommation de Riemann dit que c’est impossible
When you know there is a mistake but everything makes sense