Ramanujan Summation
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- Опубліковано 28 вер 2024
- The third video in a series about Ramanujan.This one is about Ramanujan Summation.
Here's the wikipedia page for further reading: en.wikipedia.o...
Euler-Maclaurin Formula en.wikipedia.o...
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Here is an example of divergent summation, that hopefully shows its usefulness, and shows that it is rigorous and consistent with traditional summation.
Imagine a series c_n that can be split into two other series a_n and b_n as follows:
sum c_n = sum (a_n + b_n) = sum a_n + sum b_n
Using this we can work out sum c_n from the values of sum a_n and sum b_n. This is standard stuff when working with finite series, or convergent series.
It is also possible for a convergent series to be the sum of two divergent series. So in the above, c_n is convergent and a_n, b_n are divergent.
In that case, we can still work out sum c_n from the values of sum a_n and sum b_n, but now you have to use divergent summation.
This only works if the divergent summation method is regular (gives the same answer as convergent summation when applied to c_n) and linear (so sum (a_n + b_n) = sum a_n + sum b_n)
A specific example is sum(n) = 1 + 2 + 3 + ... and sum((1-n^3)/n^2) = 0 - 7/4 - 26/9 - ....
These are both divergent series but, by Dirichlet regularization, their divergent sums are sum(n) = -1/12 and sum((1-n^3)/n^2) = (1 + 2pi^2)/12.
Finally sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2) as expected.
Some series are are harder to sum than others. So there are levels of summation that you can use. The levels are something like this:
Finite series: Can be added together, multiplied, rearranged, as expected.
Convergent series: Has all the properties of finite series, except a sequence of partial sums does not end with the value of the series. Instead, the limit of the sequence is used as the sum of the series. Example: geometric series with decreasing terms 1 + 1/2 + 1/4 + ... = 2
Conditionally convergent series: Has all the properties of convergent series, but if you rearrange the terms you get different answers. Example: 1 - 1/2 + 1/3 - 1/4 + .... = ln(2)
Divergent series: Would go to infinity by definition of convergent series. Various other methods can be applied to give a value. Some divergent series are harder to give a value than others. See below. Any divergent summation methods needs to agree with the limit when applied to convergent series (i.e. regular).
Divergent series Cesaro summation: Can still be added and multiplied like convergent series (i.e. linear). Example: Grandi's series 1 - 1 + 1 - 1 + ... = 1/2
Divergent series Euler summation: A method of analytic continuation. Still can be added and multiplied as expected (linear). Example: geometric series with increasing terms 1 + 2 + 4 + 8 +... = -1
Divergent series Borel summation: Can give a value to harder series but still agrees with previous methods. Loses a property known as stability, where removing a term from the series does not simply subtract its value from the total sum. Adding and multiplying (linearity) still exists.
Divergent series Ramanujan summation: Still linear. Can be used on the most stubborn divergent series, but depends on your choice of a parameter. I called the parameter 'a' in the video. When a is taken as infinity this method agrees with convergent sums. Examples when a=0: 1+1+1+1+... = -1/2. 1+2+3+4+...=-1/12. Example when a=1: 1 + 1/2 + 1/3 + 1/4 + ... = 0.5772... the euler-mascheroni constant.
Zeta Function continuation: A method of analytic continuation. You will continuously approach the value, and agrees with Ramanujan summation. Example: 1^-s + 2^-s + 3^-s + ... = B_(s+1)/(s+1)
For s=0 we get 1+1+1+... = -1/2 and for s=-1 we get 1+2+3+4+... = -1/12.
Some may remember the numberphile video where Tony Padilla manipulates series to write 1+2+3+... in terms of Grandi's series, and got the -1/12 answer. Strictly speaking, Tony was manipulating the Riemann zeta function, then as a last step you take s=-1. Tony explains that method here: www.nottingham....
Dirichlet regularization: This takes zeta continuation one step further and can be used for series of the form sum f(n)n^-s. This is how I got an answer sum((1-n^3)/n^2) = (1 + 2pi^2)/12 in the example sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2).
As you can see, convergence isn't synonymous with sum. Convergence is just one method of summation out of many. But the idea is that these different methods fit together so it makes sense to call these methods sums.
I can't get over how much this guy looks like the guy from numberphile. Is he his brother?
+Marcel Robitaille not entirely sure if this was intentional, but you made me laugh. Thank you
+InShadowsLinger I know it's him
+Marcel Robitaille I suspected so, I just erred on the side of caution. It is funny comment either way.
+InShadowsLinger he mentions it in another video. It wasn't an original joke.
@@srednualotbus3090
This IS him! James Grime. Get over it. It’s not funny.
"Infinite weirdness" is going to be my go-to answer now whenever I come up against something tough in maths. Thanks for that!
+Peg Y EMBRACE THE INFINITE WEIRDNESS!!! Words to live by!
I hit the largest lottery of all time, but when I went to get paid they said "You're not going to like it.", then gave me a check for -$1/12.
Isn't that mean. You must pay them $1/12 ?
I've watched probably half a dozen videos on this topic over the past few days from the likes of Numberphile, Mathologer, and others and I must say having the formulas clearly laid out like that has made grasping this -1/12 business so much easier. Thanks for the video, it's definitely one of the best on the subject I've seen.
You are the mathy-est mathematician I've ever laid eyes on. The passion and unadulterated joy you have for math is so abundant that I believe it rubs off on the viewers. Everyone should have a math teacher like you. Thanks for making these videos.
correct..you said it
But he's a theoretical physicist
When it comes to infinite series, whether they converge or diverge, I think the language of saying what they are "equal" to is part of the problem for some people. If you look at the simple geometric series 1+1/2+1/4+1/8+... we say that is equal to 2. But, if you look at the list of partial sums of that series, at no point will you find the number 2. We consider it "equal" to 2 because we invented a rule that basically says (in this case, and paraphrasing for simplicity) the smallest number that is greater than all partial sums is the actual answer.
It's an arbitrary rule but people tend to be okay with accepting it (except sometimes when you use it to show that 0.999... = 1, then some people freak out). But, if you start to come up with specialized rules that let you get finite results for divergent series, people start to abandon ship. In the convergent example, at least you can see that the partial sums approach 2. In the sum of natural numbers, the partial sums just keep getting larger, and never get negative, so it makes sense to find the result puzzling. However, when various completely different methods, each developed by different people working independently and for different purposes, come up with the same result for this series, you have to start paying attention and realizing there must be something to it.
The sum of all natural numbers is not "equal" (in a traditional sense) to -1/12, but that value does somehow represent something meaningful about that series (Frenkel's "golden nugget"), and it shouldn't be ignored, despite how strange it seems.
+Arkalius80 This is a good comment.
You should be making videos about this guy when you're dead, James... no series about the man who knew infinity should ever be finite! ;)
Arkalius80 We could say it tends to, Limit of it tends to -1/12
No, it is not approaching -1/12.
+Math and UA-cam, for convergent series the limit is the thing being tended to. The limit doesn't tend to anything, it is a fixed value - it is a number - a constant.
Another reason to love Ramanujan: He finally got you to make new videos!
I understood exactly 0 of this video of the Ramanujan series.
+Ms. Chanandler Bong That's fine, I only understood -1/12 th of it.
+Shuaib Ahmed Syed Gilani witty lol
so all of it?
I expected from a guy who supposedly works in statistical analysis and data reconfiguration
You can say you understand more than 1+2+3+4+5+...
please continue this Ramanujan video series , It's very interesting (y)
Of course a Hardy would say that.
Oh my gosh. I watched this video 4 years ago and understood nothing, not even the integral. But now, I understand almost all of that! I love Euler Maclaurin summation and the Bernoulli numbers!
Fantastic. I remember the difference between starting university and ending university and suddenly realising I had learnt a lot.
(Unrelated to anything math(s) I think we should get James a better camera and microphone.. Just to show our appreciation. :)
Adding an infinite set of rational numbers and resulting in an irrational number isn't so surprising when you consider the contribution of each digit individually. For example:
3.0 + 0.1 + 0.04 + 0.001 + 0.0005 + 0.00009 + .... = 3.14159...
I've seen this sum discussed several times before, but your explanation is far more understandable, so thank you. Pointing out that generalization of formal derivations can be key to progress in mathematics is most helpful. Also, this time I was strongly reminded of renormalization in quantum mechanics. Is there something the two methods have in common?
I once added up all the natural numbers. I started with one, and spent an entire day finding the first partial sum (it was 1, fyi). I then spent the next two days adding two to my previous answer, getting the result 3. I then spent three days adding three, four adding four, etc... until I had added all the natural numbers. Sure enough, it really is -1/12. Interestingly, I finished my addition process 1/12 of a day prior to even starting it! Strange but true.
+Daniel Bliss For the record, this show is awesome, and Ramanujan is among the greatest geniuses who ever lived. That being said, the inescapable obviousness of the fact that the sum of the natural numbers is infinity makes me unable to fully accept these explanations. In some cases, the sum really must be considered infinite, and the -1/12 answer has no meaning. How can we tell when Ramanujan summation applies and when the classic infinite result applies?
+Daniel Bliss Great comment :P
+wowsa0 Thanks ;-)
That's some playful logic.
Daniel. In his full work, Ramanujan does state the conditions under which the result is meaningful.
For those who want to look deeper, +Mathologer made a really good (but really long) video explaining this result
and mathloger person arrogant and couldn't watch entier video...
It's not crazy, it's just not really summation in the normal sense of the word and if you present the topic as if it is, of course people will be confused.
Too many pop-maths videos doing the same thing. And then people get mad because it (correctly) doesn't agree with tuition. It's very frustrating - why not just carefully explain from the outset that assigning numbers to series in this way is useful, and extends the usual summation of series, but of course shouldn't be interpreted in the usual sense of summing a series? It gets more views this way I guess...
jamma246 Because the fact that it doesn’t agree with intuition doesn’t matter at all. Most of mathematics is counterintuitive. Saying that it isn’t really equal to summation is not really accurate either. We can prove these results. There even are theorems about this. The fact that Ramanujan summation simplified into the traditional value of a convergent sum whenever the series being summed is convergent should tell you something about this: this simply extends the domain that summation can cover, it doesn’t change summation.
jamma246 Also, these aren’t mere pop math videos. Again, these methods have been rigorously tested for centuries. It is an entire field of mathematics.
We have been indoctrinated from an early age to interpret the notion of summation to have a certain meaning. However, that particular interpretation of summation is only valid when a finite number of terms is involved. There are alternative ways of interpreting the notion of summation which allows values to be assigned to a summation represented by an infinite series. Depending on which interpretation we choose, we can define how we assign a value to a particular infinite series. In standard calculus, we define a summation represented by an infinite series as the limit of its partial sums. This definition only allows values to be assigned to a class of infinite series for which such a limit exists and they are known as "convergent" series.
A Ramanujan summation reinterprets/generalises the notion of summation and defines it to allow values to be assigned to a much wider class of infinite series, including convergent series, in a way that's consistent with the other definitions of summation. There is no fundamental difference between assigning values to convergent or divergent infinite series.
"Embrace the infinite weirdness."
The philosophy of the Singing Banana.
8:22 "you just have to embrace the infinite weirdness". Indeed! Summing rational numbers infinitely to produce an irrational number is an infinite weirdness. Thanks for the great videos!
I just imagine it as the numbers getting so freaking big that they just wrap around and go back through the negatives on the other side lol
+KessaWitdaFro that moment when the numbers just think "nope, screw this".
+KessaWitdaFro that does happen on many computers!
+Jack Mott if you're a shit programmer
+Marcel Robitaille Well that certainly could be the case f the problem domain is such that not allowing overflow is important. However there are problem domains where the performance of arithmetic operations might be more important than accuracy at the edge cases. In those cases a good programmers might well use unchecked arithmetic and allow occasional overflow to be possible. An example would be high scores in video games. Further, there are many problem domains where the wrap around behavior of an unchecked native type is actually desired to compute the desired result. Many ecryption algorithms make use of overflow on purpose, for instance.
+Jack Mott touché
You look great in orange!
As always, this video is well appreciated ^^
When I was at school, I was good at every subject except Mathematics. This video confirmed I haven't improved since.
It only makes me upset because everyone says that the infinite sum "equals" -1/12. That is plainly and self-evidently false, unworthy of refutation. But what is being done here is finding a value which corresponds to a divergent sum. The value obviously cannot be the actual sum because it is divergent, and therefore cannot be summed; but a value can still be found which consistently corresponds to it in some way.
I am not happy with - 1 / 12. But thing is. Reality doesn't care if we are happy.
Well, the weird thing about mathematics is exactly that it's difficult to call it "reality", it's mostly all in our mind but it seems that our mind itself can blow our mind.
palmomki be careful how far down that rabbit hole you go as you'll end up proving that 1+1=2 is in fact invalid in reality
+Lucid Moses
Well, it isn't, necessarily. Add one ball of blu-tack to another ball of blu-tack and you end up with one ball of blu-tack. Mathematics only describes reality if you define which aspects of reality you are applying which bits of mathematics to.
Lucid Moses lol
@Phi6er bro its used to prove physics which is for reality ok so it is beyond us
When it comes to infinity, the method you try to calculate things shapes the resultif we made 1+2+3+4 ... to this : 1 + (1+1) + (1+1+1) ... we would get n*1, if we extend it to infinity, we get the answer infinity*1, which is easily, infinity.
Such a witty chap. Impresive
love this guy explanations!!! i watch him at first time in number phile than it was crazy too and really good way to bring people like maths..... I have question no related with ramanuja formula... one student ask me - teacher if every polinomium can be write like an x^n + a(n-1) x^(n-1)... bla bla... + ao x^0 ( this is used for example in partial fractions) what happens when x=0 the last term will be ao 0^0 and that is a indetermination so i cant answer this question.... can u explain this????
NOW i understand, when numberphile did the gold nugget video they didnt go into details into the constant so it wasnt clear, even mathlogger's video tried to simplify it too much, but your video went into enough details to make it clear
I really like the homemade feel of these videos, I fell like you're my teacher or something
Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. ...
I think what people don't understand is that -1/12 is not the limit of the series.
In fact, a divergent series... diverge.
We just have methods of assigning finite values to divergent series, which are not their limits by definition.
This article does in more depth in infinite series:
www.science4all.org/article/infinite-series/
Sure -1/12 is weird, counter intuitive and does not reflect anything that exists in real life. But that is also true for the imaginary number i. You can't have i apples in real life. And you can't have a voltage in a circuit that has an imaginary component but still you can use that to calculate alternating currents in circuits and get the correct results. The results will never have an imaginary component though. The same goes with the sum of all integers being equal to -1/12. It doesn't apply to anything that exists in reality but it helps simplify calculating wave functions that are the sum of every wavelength or some such weird string theory stuff.
If you think about it you can't have -4 apples either but the concept of negative apples is still immensely useful. We just need to figure out what a negative apple means before we can apply it.
It's always a matter of knowing the limitations of when something is applicable and on how you can apply it.
+Anders Öhlund Exactly. As I called it, solving problems through abstraction. The better you understand these ideas the less weird they seem.
+singingbanana As von Neumann put it, "In mathematics you don't understand things. You just get used to them."
That's a very beautiful rendition sir. I wish more people understand this instead of just rubbishing it as an 'exaggerated truth' or whatever appellation they use.
I love RamanujanWell explained!!
I absolutely loathe the fact that -1/12 is actually correct here. I study mathematics and my ex-roommate studied physics, so we had huge arguments over this answer. Alcohol had a part in it of course :P
Thanks for the entertaining video!
It's "correct" based on a redefinition of mathematic symbols that says it's correct -- it's nearly tautological.
if sum (1....infinity) = -1 / 12 then
infinity = - 0.5 ( 1 +- (1/3)^0.5)
given that sum ( 1 .... n) = n(n+1)/2
I don't have a problem with the Ramanujan sum. However I DO have a problem with the notation I see everywhere, where an divergent sum EQUALS a finite result.
I would be comfortable with a more rigorous notation such as the word "limit" (or lim), or an arrow, given the proper context, and perhaps an additional notation for Ramanujan or Borel or Cesarò summation or the zeta regularisation, to make things clear. As for the use of such magical results in Quantum Physics or String Theory, so-called normalisation magically turns infinite results into finite ones, but the justification of it isn't rigorous, even if it works.
I really like the video! I had no idea that Ramanujan made an entirely new method of summing divergent series to justify his answer. Kinda makes me wish I'd majored in something where I could make an excuse to take more abstract math courses so I could figure out the reasoning :P
he did it without formal education... and don't u know about sqrt of -1 (i) ? would u call this invented to prove something? Yes its imaginary but proves and used in lot of real things... Why would a guy like him try to justify a thing rather than finding true solution for it? and how about Euler and Rieman wo proved same?
Why I can't be like Ramanujan?
Nobody stops you.
+Haggard380
You are. You just need to BELIEVE!
Don't stop. Believing!
culwin whoa
+culwin this took a turn
+Haggard380 because ramanujan was possibly more skilled mathematically than the greatest mathematician currently alive (though he probably knew less about maths because we have learned things since when he was alive). you are stopped by the sheer improbability of you (or any one person individually) happening to be that naturally gifted.
i wouldnt dwell on it too much, nor attempt to achieve that. instead i would recommend using someone more relatable as a role model.
The good old 1+1=2 type summation is not true for 1+2+3+.... Notice the change that happens between a=0 and a=infinity. The integral gets removed because it diverges to infinity.
But imagine you left that divergent integral in. Then you would have, 1+2+3+...=infinity + -1/12. Yes, infinity + -1/12 = infinity, but what Ramanujan "summation" is saying is that if we have to assign a value to the sum 1+2+..., it should be the little name tag it comes with, in this case -1/12. It is not normal summation. You remove infinity from the answer to find it. And it turns out to be very useful in math and physics. Hope this helped even a little.
_"Then you would have, 1+2+3+...=infinity + -1/12. Yes, infinity + -1/12 = infinity, "_
That's complete nonsense.
@@miloszforman6270 You deny infinity + -1/12 = infinity ?
@@jhonnyrock
Problem is, you are fiddling with undefined terms. I can't see any logical sense in what you were saying.
I love how you have your prime counter in the corner there
First, I stumble across Stand-up Maths, and now this channel. Do any of the rest of you Numberphile guys have other channels, or maybe your regular guests like Integer Sequence Guy and Klein Bottle Guy?
The initial premise that 1-1+1-1+1-1 is 1/2 is flawed. Averaging it out due to infinite terms is the mistake as the terms increment infinity IN PAIRS. With this obvious flaw in place you set yourself up for wackiness such as -1/12. You can NOT average it.
I see you have watched the first numberphile video. There are many methods to get these answers, you don't have to use 1-1+1-1+ at all. The method in this video is completely different. Also, I do not understand why you think the terms increase in pairs only.
The summation can only be 1 or 0 can't be a half, averaging it is the mistake that leads to the erroneous answer that all integers added up = -1/12. Just because Ramanujan made the error doesn't mean it is right... the best mathematicians often make errors.
It doesn't matter, you don't have to use 1-1+1-1.. at all to get to this result, they are many other methods to get it, which furthermore confirms that 1-1+1-1.. = 1/2
You can get the result 1/2 very easily from that sum, without having to average it.
Yeah the summation is 1 or 0 for finite series. One easy way wich is also explained in numberphile video is that you consider the infinte sum as S=1+1-1+1-1+-...
Then you add it to it's self and arrange the integers like below
S=1-1+1-1+1-1+...
+S= 1-1+1-1+1-...
2S=1
S=1/2
Our numbering system is not without flaws. There are work-arounds such as do not divide by zero and believing in i, the square root of -1. Summing all the integers to a negative number is a similar flaw. I don't think it ever manifests itself in real life so we should treat it as an artifact.
Matter is a derivative of virtual worlds. 12 resonant forms.
Next time please explain the Cauchy's general principle of convergence, I couldn't understand it🙏🙏🙏
I still confused for a simple reason. We have a serie, the addition between number should be exacte. We don't have a function with errors estimations between rectingals of integrals. The evaluation is -1/12 of R result, like in expirences but mathematecly not explaned.
First lesson in string theory. Literally first chapter in all the string theory maths books.
Is there anything interesting about the following?: The finite sum of natural numbers is (n/2)(n+1). The integral from -1 to 0 is -1/12 (just like the Ramanujan Sum for the relevant infinite sum). Also, if you find the finite sum of the first n square, cubes, etc of natural numbers then integrate them from -1 to 0, you also get the same as the Ramanujan Sum for the corresponding infinite sum . I assume this is a known or obvious result, but I haven't seen it mentioned anywhere.
Good question! It took me a while, but I figured it out. You can write the sum of the kth power of naturals as a sum of Bernoulli numbers (Wikipedia has all the identities). Once you integrate n, you may apply an identity that turns the sum of Bernoulli numbers into a single Bernoulli number.
On the right hand side, where you apply Ramanujan summation, you can equivalently calculate the Riemann-Zeta function at -k. This function looks complicated, but has simple outputs for integer k. In fact, it simply produces a Bernoulli number with a coefficient. The same Bernoulli number and coefficient obtained above. Thus proving your implied conjecture.
I can't fully follow the reasoning for this sum of integers, however I have a question about it. Since any integer can be split into a bunch of 1s, like 3 can be split into 1+1+1, is the sum of 1+1 repeating to infinity also equal to -1/12?
It is a good thought but the ruleset for working with infinit sums differs from much of the more "intuitive" way often learned in school. So the answer is no.
Even just changing the order of two numbers in the series 1 + 2 + 3 + ... will alter the result (1 + 2 + 4 + 3 + 5 + ... ≠ - 1 / 12).
The sum 1+1+1+... actually has a Ramanujan sum of -1/2. Loosely speaking, we have to stretch the concept of a sum to such a degree that we lose some of its properties. Even putting a 0 in front of 1+2+3+... to make 0+1+2+3+... yields a different Ramanujan sum (+5/12, if I'm not mistaken).
@@tomkerruish2982 thanks for the reply. It is funny seeing my original comment, as I was in high school at the time. Now I’ve graduated undergrad in math and starting my math PhD in 6 weeks. And yes I believe you are correct.
This is what that Numberphile video SHOULD have been.
Here's my interpretation.By most constructions of the real numbers, addition is a binary operation. To add more than two terms together, we must invent special definitions. a + b + c is defined as (a + b) + c, and thanks to the associative and commutative properties, we can add any order we like. a + b + c + d = (a + b + c) + d, as we just defined sums of three numbers, and so on. This method doesn't work with infinite sums, so we must find another way to define those. If the sequence (a, a + b, a + b + c, ...) converges, the traditional sum is defined as that. If it goes to infinity or minus infinity, the traditional sum defines it as whichever of those. And if it completely diverges, there is no traditional sum. This is a different definition of how to do infinite sums, where the sequence (a, a + b, ...) may not "tend to" the defined sum.
Can anyone explain why it's completely fine to plug in a=0 instead of a=infinity and still say that it represents the same thing? I get that plugging in infinity doesn't work in the first place because it'll diverge, but that just kind of confirms the suspicion that the sum doesn't exist in the first place.
The good old 1+1=2 type summation is not true for 1+2+3+.... Notice the change that happens between a=0 and a=infinity. The integral gets removed because it diverges to infinity as you mentioned.
But imagine you left that divergent integral in. Then you would have, 1+2+3+...=infinity + -1/12. Yes, infinity + -1/12 = infinity, but what Ramanujan "summation" is saying is that if we have to assign a value to the sum 1+2+..., it should be the little name tag it comes with, in this case -1/12. It is not normal summation. You remove infinity from the answer to find it. And it turns out to be very useful in math and physics. Hope this helped even a little.
@@jhonnyrock
It's still the same nonsense. It's nonsense to argue that you can get a number "infinity -1/12", and you simply subtract that infinity to get -1/12. That's outrageously stupid. Or you have to define stringently what you are doing, especially how you define addition and subtraction of infinite numbers. You did not do that, though. You're merely _postulating_ that this will work, without any evidence and proof, just coming from outer space.
@@miloszforman6270 I was not giving a proof. I was trying to make the very complex topic of Ramanujan Summation a little easier to grasp. If my explanation didn't work for YOU, that's fine. I don't think it's fair to call it "outrageously stupid" just because you didn't find it helpful or understood the point I was making
@@jhonnyrock
_"I was trying to make the very complex topic of Ramanujan Summation a little easier to grasp. "_
I can't understand how any such theory should be "easier to grasp" if you are wrapping it into esoteric bullshit which nobody can really understand. I know that Mr. James Grime of this video channel does that, as well as Prof. Tony Padilla in the "Numberphile" channel. "Mathologer" has made some very clear statements about this bs, and I'm convinced that he is right.
Padilla came out recently with another bs video of such a kind, this time presenting Terence Tao's "weighting function summation". I read some of Padilla's paper, which is aimed at his mathematical and physicist colleagues, so I can clearly see that he indeed knows his math. But why on Earth is he talking esoteric weirdness on his "Numberphile" channel? We have lots and lots of such bs going on in this world, with governments lying all day, and scientists obsequiously narrating things they do not believe in but are told to tell. Why should even mathematicians contribute to all this pernicious confusion?
In the audio book of "The Man Who Knew Infinity," they specifically mentioned that English people at Cambridge tended to mispronounce Ramanujan as Rama-NOO-jin. :)
What is the correct way to pronounce it?
Ciara Sookarry
Ruh-MON-a-jahn.
+EGarrett01 Yup. I did it on purpose. That was the first way I heard it so when I speak fast that's what I say. It would take longer to film if I keep correcting myself, and they use both interchangeably in the film.
+EGarrett01 more like rah - mah - nuu - juhn
Jishnu Viswanath
Emphasis needed.
i've seen tons of your videos and only now realised your screen name is singingbanana. kudos on that :P
Its easy to enough to understand convergent sums. That's standard calculus. This goes beyond convergent sums. to me, all we have to do is accept essentially that 1 - 1 + 1 - 1...= 1/2 in order to accept that 1+2+3+4...=-1/12. As long as you accept that basic premise, all of Ramanujan's logic flows nicely. I can't replicate it, but it does flow from it!! So, even though that first premise isn't accepted under standard calculus, the fact that he achieved a meaningful result in that the analytical continuation of the Riemann function at -1 equals -1/12, that means something to me. Who cares if it "really" equals -1/12. Isn't infinity supposed to be mysterious anyways?
Wait, why would you evaluate it at a=0 if the series diverges?
If anything, that would mean that the Ramanujan sum of a divergent series would be a number closely associated with that series, but it wouldn't be the actual answer to the problem... right?
So it would be a very different kind of answer compared to, say, 1+2+4+8+... = -1, which is what I always think about when it comes to negative sums of positive series.
+Eric Vilas Yes, Ramanujan summation is different to other divergent summation methods, and involves a choice of a. Some say a better choice for a is 1 rather than 0. But a=0 agrees with the Riemann Zeta Function method. You would have to look at the topic deeper than I have to learn all the pros and cons. There are degrees of difficulty with infinite sums, 1+2+3+4+... is one of the hardest.
Why does every professional mathematician and physicist who is smarter than me come to this same conclusion about the sum of natural numbers?
They must all be wrong.
Shawn Pitman there is a video about mathologer on that topic where he disagrees
i loled... people get jealous over dead people's work too...
Josh, and there is another video of numberphile where they explain it better.
except in that video they apply a value of 0.5 to the series 1-1+1-1+.... even though it is clearly not converging to anything.
Why do people who don't understand Ramanujan summation always state the sum of natural numbers being equal to -1/12 like it's a fact without qualification beforehand?
in the Ramanujan prove he used the fact that 1+2+3+... = c, where c is a constsnt, then he multiplied the equation by 4, so he got 4 c = 4 x (1+2+3+...), but this can be done only when c is a number, but c is infinity, then 4 c is also infinity and not 4 times infinity
I wish I had a math teacher like this kid...
Loving these videos on Ramanujan, what are your thoughts on mathematicians apology? Cheers
Hello singingbanana,
There is something that I have been wondering and although not directly related to Ramanujan, I have to ask since this video reminded me of it. I have been following Numberphile and your channel for quite some time and there was at one point this video on different types of infinity, or well larger and smaller infinities, but does that actually matter when devicing these formulas? Does it matter what "size of infinity" it is at all? Because it seemed to be a pretty big deal yet I fail to understand why it is a big deal. Surely infinity is infinity despite perhaps being larger or smaller than another infinite. I mean, it's infinite. I have tried to make sense of it thinking of it as an object and the items are denser but is that relevant somehow at all? I just don't get it.
This is the video: ua-cam.com/video/elvOZm0d4H0/v-deo.html
Hoping to not have made a fool of myself and kind regards,
War (sorry, I don't like using my real name online)
ok lets assume that 1+2+3+4+5+...=-1/12 .i d like to know the sum of 1+1/2+1/3+1/4+1/5+.... how much is it.the question is inevidable.if the second sum diverges and goes to infinity how is it possible the first sum not to lead there since ALL exept the first term are MUCH BIGGER than the second series?
And from substituting the result back into the formula s=n(n+1)/2 we get the result, that infinity is either equal to (1/sqrt (12))-1/2 or -(1/sqrt (12))-1/2. Seems legit :D
+Denascite Yup, but I don't think that you should be expecting to get infinity from solving a quadratic equation in this case. It's way more abstract than that.
Thought I had recently: If you make the (false) assumption from the start that the sum 1+2+3+... converges, then you are allowed to manipulate the series the way you did in the Numberphile video. If you do that, it all boils down to evaluating 1-1+1-1+... which can be done using Abel's theorem and the geometric sum 1/(1+x^2) (again, you assume the series converges).
That made it seem a little less mystical for me, because there aren't too many false assumptions you have to make. Just say, "suppose this DOES have a sum," and then you arrive at the value it must be.
This is probably the same result you get using analytic continuation of the zeta function, but thats a whole other can of worms for someone like myself who hasn't studied complex analysis.
I take it you are also against irrational and imaginary numbers?
It very much is the result obtained by analytic continuation of the zeta function. Good intuition!
I saw a video with proof/visualisation of how the number line could represented into these nested squares. The numbers would not be in order, but some algebra would still be conserved. This way making the 'convergence' towards -1/12 seem more natural. Does anyone know this video? i can't find it again
Are you talking about p-adic numbers?
There is something quite wrong with this paradox. If Ramanujan is working with numbers he is not working with the Infinite, he can't be working with the infinite because the Infinite has no numbers.
He can only work with numbers after finding the frame of the numbers.
You can't just write numbers if yo don't know where are you placing them.
So the sequence does not sum to -1/12, but that number is a related constant.
+Rupt
Yes. The thing is, infinite sums don't really work at all. All you can say is that by adding up more and more numbers, they approach a certain limit. You have to be precise about what approaching means. By doing it in a non-traditional fashion you might get the Ramanujan limit.
+Zardo Dhieldor right. We have to define what we actually mean by an infinite sum. The axioms we choose for finite summation are chosen based on what we find to be the most appropriate properties for addition - associativity, commutativity, that sort of thing. These axioms don't automatically extend to infinite sums so we have to define what we actually mean by an infinite sum - indeed, simply rearranging the terms can often change the value of an infinite sum, yet these series are still contained under the traditional definition of an infinite series. When we work with generalisations of the notion of series, what we're really doing is finding a consistent answer to more types of series so as to extend the definition.
+Zardo Dhieldor maybe we'll use a different metric than the absolute value metric and we'll get that 1+2+4+8+...=-1. Maybe we'll use cesaro summation to average the partial sums and we'll get that 1-1+1-1+...=1/2. Maybe we'll take the averages of the averages and get 1-2+3-4+=1/4. Under it all is the concept of analytic continuation - that is, we take a holomorphic function (that is, differentiable everywhere in the complex plane) which is only defined for a certain range of values in the plane (satisfying a few conditions - see the identity theorem) and we can extend it to a unique holomorphic function spanning the entire complex plane (except for maybe a few singularities).
Point is, when we have a power series defined on some finite non-zero radius we can find a unique holomorphic function which is equal to it everywhere on that radius _and_ is defined everywhere else on the plane. Then in some senses the holomorphic extension is still equal to the sum - the only thing that causes the series to diverge under normal conditions is simply because of the fact that we're representing it as a sum, when in reality it's a function which still has a defined value outside its usual radius of convergence. Hope this makes things more clear.
But the complex continuation thing doesn't work as easy as one might imagine. First the analytic continuation while (under reasonable circumstances) unique is not at all always existent. Secondly there exist power series converging on one disc with different partial sums converging to different functions on another disc. :( I haven't really found any consistent theory bringing these pseudo-conervence methods together.
I just wonder if this kind of abstraction goes any further. If we can find the Ramanujan summations of different series, which I think is found using summations, subtractions, and multiplications of summations, can we do even more complex functions with summations? What would that mean? Its almost like the Riemann sphere to me- first you have a complex plane, then you extend it even further into 3 dimensions...
Looking at -1/12, it does make one think that duodecimal would be the best "natural" base. The answer "minus point one" would be even more baffling.
What happened with numberphile?
I have a few questions: First, why do you just choose the constant _a_ depending on your function? You basically use two different summation methods for divergent and convergent functions!
Also, what happens if you take a different function that coincides with the first one on all the integers but not everywhere else? Wouldn't you get a different limit for the same sum?
+Zardo Dhieldor Yes. It would be different summation values depending on your choice of a. Other summation methods don't depend on a choice of a. If you get deeper into the topic you can find the nice properties of Ramanujan summation and how it relates to other summation methods. Unfortunately I only know the most basic definitions.
Thanks for the brain buster, sir. You mentioned the name of the person who did the numberphile video on the same subject, but, as I'm American, I can only listen so fast. Could you please repeat his name (preferably typing slowly so I can understand it.)
:D
Thanks again.
+rogerdotlee Edward Frenkel on numberphile
Thanks. I'll check it out.
it's lovely! -(1/12) is gonna be my favourite number now!
I will embrace the weirdness of the infinity! Thank you!
Ramanujan, Cantor and Riemann my infinite heroes.
"Ramanujan's answer for the sum of all integers is -1/12". Shouldn't that be natural numbers, not integers? And would the sum of all integers just be 0?
Is there a symbol or way of showing triangle numbers, similar to n! ?
Scott Hill I've seen it as T_n
triangle number = (n + 1 )C2
The function for 1+2+3+4+5+6+7+... is not x. Simply graph x and the answers to the equation at x=1 y=1 at x=2 y=3 at x=3 y=6 at x=4 y=10. With the function x then y=x and you have the 45 degree line. The sum of all integers is not a line.
If sum(n) to infinity = -1/12 then what about the necessary condition of convergence???
The condition of convergence is a part of calculus. Different tasks have different rules.
Does this sum really "add to" or "equal" -1/12, or is that simply a related constant?
+MegaMrFroggy If you pull at that thread then convergent sums wouldn't be called sums because the definition of finite sums doesn't apply. Maybe they shouldn't be called sums, but you can see why they are. Fortunately the meaning of convergent sum has an exact definition, and the same is true for divergent sums.
it really sums to -1/12. think of it as: after the usual part of the sum goes to hell, all that it is left is that "gold nugget"
+singingbanana I am a VERY big fan of you and I love mathematics. But could you please explain it in simpler words because I'm just in 9th grade. I would really appreciate it if you replied. Thanks
I dislike the example that the sum of rationals can give an irrational if the sum is infinite. It's not magic it's that the rationals are not a complete set and that they're dense in the reals.
The set of divergent functions are not dense. You cannot define uniformly C0 functions on such sets that would result in unique results. One could define a different sommation that for convergent sequences gives that result and for divergent ones gives any number you desire. And it would be just as valid.
Hardy's book on divergent series explains well why -1/12 shows up often. If your sommation rule follow certain properties you get the result.
+sunetulajunge I think we're saying the same thing in different words. I have that book as well, it's here in fact. I don't know where -1/12 first shows up, are you thinking of linearity, stability etc in the introduction? You can do it that way. This video is about Ramanujan summation.
+sunetulajunge i think you mean that irrational numbers are by "definition" a limits of Cauchy sequence of rational ones, right? or what you think irrational numbers are ?
I am extremely confused by this, i have very limited knowledge of mathematics but what i understood form this video is that essentially this method is being 'misapplied' as because it gives a correct answer when the series is finite or convergent infinite series we are just willing to assume that it works for nonconvergent infinite series as well, but wouldnt that be like saying that since the Pythagoras theorem works on right-angle triangles it has to work on on all triangles or even on rectangles, no? I have looked into Cesàro summation, as well, to try and understand whether there is a different meaning to word sum and the best i got is that its the limit of a series as n approaches infinity which doesn't really help either. Even as a limit it doesnt work i mean literally the first value in the series is bigger than the supposed answer and subsequent values are added not taken away. the only way i can see this making any sense is that this value has some sort of relationship to the inner working of an infinite series but isnt the actual total value of the series which i can accept and would like to know if they have figured out what that relationship is or how they are determining that relationship, but to suggests its the total value of the series or the limit of the series I cant seem to comprehend it. I have been trying to understand this for quite some time, its unfortunate because i cant seem to find any resource that can explain this without all the confusing math terms and weird calculations/formulas or any that do more than just claim "the formula proves it".
The good old 1+1=2 type summation is not true for 1+2+3+.... Notice the change that happens between a=0 and a=infinity. The integral gets removed because it diverges to infinity as you mentioned.
But imagine you left that divergent integral in. Then you would have, 1+2+3+...=infinity + -1/12. Yes, infinity + -1/12 = infinity, but what Ramanujan "summation" is saying is that if we have to assign a value to the sum 1+2+..., it should be the little name tag it comes with, in this case -1/12. It is not normal summation. You remove infinity from the answer to find it. And it turns out to be very useful in math and physics. Hope this helped even a little.
Hey john yes it did help thank you very much for taking the time.
Though i am going to be honest i am a still a bit lost in how this could be used/applied, i think its just my lack of experience in mathematics that is causing my inability to fully appreciate the intricacies of this "summation" @@jhonnyrock
@@RD-gr1ru
Don't let youself be confused and misleaded by this @jhonnyrock bs.
is that series is convergent or divergent?
If a series tends toward a certain value (called the limit) it is convergent. If it tends toward infinity it is divergent, since it does not tend toward or reach a limit.
but the series tends to infinity so it has to be divergent so why the series has finite sum.you r just giving me the definition of cgt and dgt
There has to be a use for that number. I don't know what it could be, but there has to be one. Does a finite "nugget" exist for every infinite sum? If so, it could be useful.
Some infinite series are harder than others (there are definitely some with no value to them). The divergent sum definitely has meaning. In my description for the video I talk about how you can add two divergent series together to make a convergent series - and the value of the convergent series will be the values of the two divergent series added together.
Talk about interesting. Thank you. I hope someone's thinking about it.
Isn't Alef-Null the set of all positive integers? Does that mean that Alef-Null, in a weird sense equals = -1/12? Or something like it?
Hey, I said it was weird. I came up with it three minutes ago. I know it's crazy.
Anyway, I'm sure that's nonsense.
It appears in bosonic string theory, where it implies you need at least 26 dimensions for it to be consistent.
Good. I'm glad it's useful for someone.
I'm more of a LQG guy, but more than that, I'm a German major, so although I'm glad it's useful, I'm out of my depth.
I am a former number theorist doing my PhD in astrophysics. I just love your way of discussing mathematics and the passion that reflects. Although I know these, I love to come back to your videos again and again. The only bad thing about your videos is, these make me regret for leaving Mathematics. Thanks and Regards !!
I'm glad you liked it!
Thank you! I will embrace my inner, infinite weirdness.
I wonder if you can define a new metric on the real numbers such that n(n+1)/2 actually does approach -1/12 as n goes to infinity. I guess that still wouldn't explain Grandi's series.
A comment about the use of -1/12 in string theory: An exhaustive search of all the string theory textbooks at hand showed that there is just one main use of the -1/12 result, namely in determining the dimensionality D of space-time required for the bosonic string theory to be self-consistent. The divergence taken seriously produces D=2; using the -1/12 substitute (it's not an equality) produces D=26. This wrecked the formerly widely advertised expectation that the values of such universal physical parameters as 4-dimensionality should arise uniquely in string theory. It didn't take string theorists long to devise a kludge, namely to roll up the extra dimensions into an unobservable scale. Eventually they were forced to hypothesize a "multiverse" where most of physics' theoretical parameters are *not* uniquely determined by the theory. For criticism of this pursuit and other aspects of string theory as it is practiced, I refer you to Lee Smolin's book "The Trouble with Physics".
I am wondering, are there still formula's from Ramanujan not being understood today but shown to be correct?
I don't think that getting a negative number out of a infinite summation of positive numbers is remotely in the same realm of weird as getting an irrational number out of an infinite summation of rational numbers.
After all, aren't irrational numbers irrational on account of an infinite summation? ie. there's always another non-terminating number added on after all the numbers previous to it. pi = 3+1/10+4/100 etc. It's only irrational because this summation is infinite. Not 'infinite weirdness' at all IMO... it's actually intuitive.
On what planet however would multiple positive numbers be summed to a negative? What's the mechanism for that?
+Kevin Theriault The sum of the series is infinite. The value of -1/12 is the analytic continuation: basically, it is inserting an invalid number in an equation. The value of -1/12 does have some meaning, but is is not the "answer" in the normal sense.
"On what planet however would multiple positive numbers be summed to a negative?" say im at the equator and i travel east along the equator for 24,900 miles. in that situation the sum of the distance i traveled is 24,900 or -2 miles. not saying that is analogous, just answering the question.
Your summation is relative to what? You'll quickly notice relative to your origin not all of the vectors for this summation are positive....
blkgardner yeah I understand what the answer means that's not what I was saying. I'm just nitpicking the point that this answer is similar to getting an irrational number from an infinite summation. It really isn't in my opinion.
The irrational numbers are indeed another dense subset of the real numbers
What about the infinite sum -1 - 2 -3 -4 -5 -6 to oo ? does that equal +1/12 ? And then how do we explain it......as we travel through infinity and flip to negative infinity could we have shifted by +/-1/12? Am I a foot taller or shorter for example......?
+Martin Carpenter I'm not sure but I think it's not +1/12. That would be true for finite series and convergent series. It's also true for some divergent series. But some divergent series are stubborn and need more general summation methods, in that case you have to lose the intuitive things you expect from finite sums. 1+2+3+4+... is one of those stubborn series.
+mehfoos In fact, I've changed by mind. By Ramanujan summation -1-2-3-4-... = 1/12. Using f(x)=-x. I was thinking of Riemann Zeta Function continuation, which is not linear (but doesn't apply to that series).
However, there are levels of difficulty with infinite series. Convergent series can be added and multiplied as you expect. Some convergent series can be rearranged, some cannot. Some divergent series can be added and multiplied, some cannot.
The levels are something like this:
Finite series: Can be added together, multiplied, rearranged, as expected.
Convergent series: Has all the properties of finite series, except a sequence of partial sums does not end with the value of the series. Instead, the limit of the sequence is used as the sum of the series. Example: geometric series with decreasing terms 2 = 1 + 1/2 + 1/4 + ...
Conditionally convergent series: Has all the properties of convergent series, but if you rearrange the terms you get different answers. Example: ln(2) = 1 - 1/2 + 1/3 - 1/4 + ....
Divergent series: Would go to infinity by definition of convergent series. Various methods can be applied to give a value. Some divergent series are harder to give a value than others. See below. Any divergent summation methods needs to agree with the limit when applied to convergent series.
Divergent series Cesaro summation: Can still be added and multiplied like convergent series. Example: 1-1+1-1+...
Divergent series Euler summation: A method of analytic continuation. Still can be added an multiplied as expected. Example: geometric series with increasing terms -1 = 1 + 2 + 4 + 8 +...
Divergent series Borel summation: Can give a value to harder series but still agrees with previous methods. Loses the following property: remove a term from the series does not simply subtract its value from the total sum (stability). Adding and multiplying (linearity) still exists.
Divergent series Ramanujan summation: Can be used on the most stubborn divergent series. Example 1+1+1+1+... = -1/2. 1+2+3+4+...=-1/12. 1 + 1/2 + 1/3 + 1/4 + ... = 0.5772... the euler-mascheroni constant.
Zeta Function continuation: A method of analytic continuation. But nonlinear. Agrees with Ramanujan summation. Example: 1^s + 2^s + 3^s + ... = B_(s+1)/(s+1)
+mehfoos Yes they can conflict. It's better if they are compatible though - obviously.
+mehfoos I'm thinking on a rearrangement attempt that shows the 'wackyness' of infinite sums: sum all integers and then substract it from itself. You should get zero, right?:
S = (1+2+3+4+...) - (1+2+3+4+...) = -1/12 + 1/12 = 0
But you could try to rewrite all the terms differently:
S = 1+ (2-1) + (3-2) + (4-3) + ... + (n - (n-1)) +... = 1 + 1 + 1 + ... = -1/2
The 'conclusion' would be that 0 = -1/2 !!
Can't wait to go to college and be able to understand every formula shown in this vid.
06:16 Awh, I'm sure you could get the hang of it James! :)
i have a question. how can u imagine the highest number?? because u cant imagine it. every time u imagine a highest numbet there is always a possibility to get another greater number than that!! i would not believe the sumation of it.... xxxxxxx
-1/12 may be my favorite number.
yup, didn't comprehend a sentence of this. Of course, this isn't james' fault
Good job. Thanks for making the effort.
Madhava, buddhayan were the two ancient hindu mathematicians who even calculated summation of such series way before the western mathematicians discovered such summation series
So, you're saying it's not nonsense, and that there are several ways to get this result, and even that it's useful.
Yet. In common arithmetic it can't be right, as any 8-year old will tell you; the sum has to be larger than 1, and the sum of just whole numbers is never a fraction. So, what gives? What are the conditions for this to work?
We have been indoctrinated from an early age to interpret the notion of summation to have a certain meaning. However, that particular interpretation of summation is only valid when a finite number of terms is involved. There are alternative ways of interpreting the notion of summation which allows values to be assigned to a summation represented by an infinite series. Depending on which interpretation we choose, we can define how we assign a value to a particular infinite series. In standard calculus, we define a summation represented by an infinite series as the limit of its partial sums. This definition only allows values to be assigned to a class of infinite series for which such a limit exists and they are known as "convergent" series. In standard calculus, summation of "divergent" infinite series is undefined; just as the square root of -1 was once undefined. A Ramanujan summation reinterprets/generalises the notion of summation and defines it to allow values to be assigned to a much wider class of infinite series, including convergent series, in a way that's consistent with the other definitions of summation. There is no fundamental difference between assigning values to convergent or divergent infinite series.
Oh God. Great is thy creation. Finite is our understanding
best explanation of sum(1:∞) = -1/12 to date