I Just Can't Believe what the Math is Telling Me

The Grandi's Series history is quite famous. The sum 1-1+1-1+1-1 to infinity should clearly diverge with no dispute. However, this 1-1+1-1+1-1 infinite series has a very controversial answer:
1-1+1-1 = 1/2 ?! How could this be?
Here's my option on the Grandi's Series answer and a Grandi's Series proof. Enjoy!
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

[@BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON KZbin) kzbin.info/www/bejne/aZTdmJl-irGNedU

[@marcosoekmono3009]

The engineers were right all along! Taking the average does give the true answer.

[@BriTheMathGuy]

*True* might be up for debate. I think the argument is very compelling through!

[@livedandletdie]

The average of 1 is 1 so 1 is 1, statement holds true.

[@uberless1]

There's a reason egineers have to take an ethics class and physicists and mathematicians do not...

[@mihailmilev9909]

@@uberless1 lmfao

[@manumano3887]

The engineers are always right .

[@BCQM_BCQM]

I like the fact that most different approaches lead us to the same ridiculous answer.

[@swapnil72]

Not always

[@vascomanteigas9433]

Analytic Continuation explains this.

[@shambhav9534]

The simple logic method doesn't.

[@BCQM_BCQM]

@@shambhav9534 so I said "most"

[@shambhav9534]

I think these ridiculous methods are assuming that the series is convergent. Which of course it isn't.

[@henrytang2203]

I think maths gets weird at infinity. As an engineer, I'm happy to accept the idea of limits, calculus and converging sums. But I give up on trying to justify values for divergent sums.

[@kazedcat]

Infinity > Logic. It is a very powerful and useful concept that is why we extract as much consistency out of it as we can but if you go deep enough into infinities then logic is washed away.

[@kasuha]

Mathologer did a good lesson on this one a few years ago. It's all fun and games but you can run into contradictions pretty easily so it's better to be very careful about it. Most important observation is that convergence of analytic continuation does not imply convergence of the original expression. That process may only work one way.

[@VikramSingh-qb4gw]

True

[@BriTheMathGuy]

I am a beginner at analytic continuation. To me, it doesn't really seem like the analytic continuation is even the same function we started with (that, or it's more a piecewise function on a different domain). The Taylor Series we used to analytically continue 1/(1+x) does NOT equal 1-1+1-... when we substitute x=1, yet it does seem like there's some sort of link between the two. I would really appreciate if someone could explain this a little bit better and give some insight into the validity of this method. Regardless, I hope you enjoyed the video! If you didn't enjoy it, make sure to give it a dislike! :)

[@a2izzard]

Hi

[@Rodhern]

To me it feels a little like this. We have some function, f, that we would like to examine at a given argument, say x. We then go and say "If f is really benign we *restrict* our search of good candidate approximations/substitutions for f to this *one* function g (with emphasis on argument x); all other candidates are bound to be useless. It, g, is the only one that is nice enough that we believe it might work". Then we remember that f is not really all that benign, and we change up our search strategy and say "There are no good candidates out there, but if we *must extend* our search to at least one candidate anyway, surely we should pick the only one that cannot be discarded from the search in the benign case".

[@matih1234]

i think that the series doesnt converge because lim an = lim (-1)^n =\= 0 what is necessary condition

[@shambhav9534]

I think you've stumbled upon a method to convert a divergent series into a convergent one by force which has created this weirdness.

[@aashsyed1277]

how is this comment two days ago?"????????????????

[@Liesmith424]

"Any math which uses letters is witchcraft." --George Washington

[@Bodyknock]

Mathologer and 3blue1brown have also both done videos on “sums” of divergent series that are pretty great. Basically you’re essentially extending the usual definitions of “sums” to assign a number that makes sense in a variety of contexts to these divergent sums. The famous 1+2+3+... = -1/12 for example makes no sense normally but, if you add a formal way to assign a number to that divergent sum, then that -1/12 makes sense and tells you something useful about the series itself. Likewise saying that 1-1+1-1+... = 1/2 doesn’t make sense in the usual settings, but when you extend the concept of summation so that you can assign a value to that series then 1/2 pops out as a sensible value. It’s a bit like how square roots of negative numbers make no sense in the Reals, but if you extend the number system to define and assign complex numbers to square roots of negative numbers then the resulting numbers make sense in that context. Or when you analytically extend the definition of factorials to be the gamma function, you have a new function that still gives you the normal values for factorials but also gives you useful values for other numbers. Mathologer - kzbin.info/www/bejne/oJSug3qmgs6Jitk 3blue1brown - kzbin.info/www/bejne/jnenfmSfpZp7mrc

[@ldx8492]

Exactly, and not only that, it has been repeatedly shown that imaginary numbers do play a huge role in physics and physical laws (e.g. Schrodinger Equation). Why shouldn't divergent series do the same? In fact they do, especially in particle physics

[@benjaminojeda8094]

@@ldx8492 what

[@tomkerruish2982]

Also, Flammable Maths did a video on infinity factorial equaling the square root of 2 pi. (Sorry, don't know how to imbed a link using my phone.)

[@daphenomenalz4100]

Ramanujan was just out of this world, how did he figure out that equals -1/12 0_0 as he was the first one

[@Bodyknock]

@@daphenomenalz4100 And he died at only 32, just imagine the things he could have done if he had lived longer!

[@ethanbartiromo2888]

Ahhh, Cesaro sums, the annoyance of my Reals class

[@MithicSpirit]

Very well done! Thank you for not oversimplifying like that one Numberphile video.

[@BriTheMathGuy]

Glad you enjoyed it!

[@SytheZN]

The answer 1/2 makes sense to me from a logical standpoint since for the finite series the sum is dependent on the mod 2 result of the number of terms. Given that infinity is neither (or both, depending on your frame of reference) odd nor even, and the convergence point is by definition a single value, the only way that it can represent both of them simultaneously is by their midpoint. By extension, for any series with a repeating pattern of sums the convergence point should be the sum of the vales in the repeating set over the number of values.

[@joetigani7269]

Think about it this way: the sum can't be 0, because there will always be an additional 1 to add. The sum cannot be 1 either, because there will always be an additional 1 to subtract. It bounces back and forth between 1 and 0 forever. Since the sum can't be 0 or 1, and because those are the only two values that the function will ever take, the only logical conclusion is that the infinite sum (or, in other words, our attempt to represent this infinite series with a single number) must be exactly halfway in between these two values, or 1/2.

[@sergeyromanov2116]

The only logical conclusion is that it diverges.

[@joetigani7269]

@@sergeyromanov2116 But the math tells you the answer is 1/2. That's what I am getting at--laying out a logical conclusion for what the math tells us

[@sergeyromanov2116]

@@joetigani7269 The math tells you that the series diverges and there is no sum. That some methods of assigning numbers to such series are misleadingly called "x sums" (Cesaro sum, etc. - you can never omit the naming element and just say "a sum") is neither here, nor there.

[@joetigani7269]

@@sergeyromanov2116 Apologies for my huge mistake. I guess the video is mislabeled then (I can't believe what the math is telling me) if the math doesn't tell us the answer is 1/2.

[@sounakbhattacharya9646]

Thank you. You explained in a very simple way which enabled me to understand it very quickly.

[@anonymous_4276]

We are just calling it summation but in fact doing something different. With odd number of terms we get 0, with even number of terms we get 1. Hence limit as n approaches infinity of this series doesn't exist. By changing and modifying definitions to suit our needs we can make it possible for the series summation to equal 1/2. But then we aren't even summing the damn sequence. We have changed the definition of summation.

[@angelmendez-rivera351]

The "traditional summation" is not summation either, though. So your point is moot. In fact, there is no such a thing as "summing a sequence". The arithmetic operation of "summing", by definition, can only be applicable to finite sets. Since sequences are functions from the set N to other sets, and N is an infinite set, it makes no sense to talk about "summing the set". This is also why it does not even make sense to talk about commutativity or associativity: those properties by definition assume finitely many elements. They are meaningless in the context of infinitary operations. Sure, people do call it "summation", but this is little more than just abuse of lanuage. A mathematician would not actually call it that, and would never describe it as such in their works, not a modern mathematician, anyway.

[@Rodhern]

@@angelmendez-rivera351 Do your inner voice tell you what "summation" intuitively might be or not be? Or are you able to silence that kind of distractions? Given a set of non-negative real numbers augmented with 'positive infinity', I get the feeling that a supremum (upper limit) to the sum of finite subsets is almost like doing (proper) "summation". A quick example of course tells me that I am doing something new, strange and/or wrong, because, why is it all of a sudden possible to get a result of 'positive infinity' even when none of the summands are(?).

[@angelmendez-rivera351]

@@Rodhern I think you are missing the point of my objection. It does not actually matter whether we are working on a topological extension of the real numbers, complex numbers... it does not even matter if you choose to talk about surreal numbers here. If you perform an arithmetic operation followed by a non-arithmetic operation, then the composite operation is not itself an arithmetic operation. Trying to pretend that it is, regardless, is what I am taking issue with here.

[@angelmendez-rivera351]

@@Rodhern As in, if I evaluate a function f at a point, then I evaluate another function g at the output of f, what I am evaluating is g°f, not some "extension" of f.

[@angelmendez-rivera351]

@UC9dnnW1McvAfT0MD2CsStnQ *ok, we're not summing but taking the limit of a sum as the number of terms approach infinity* Yes, but taking the limits of the sums requires ordering those sums in a very specific way. So it is not just an issue of it being about it technically being a limit of sums rather than itself a sum. It is the fact that the particular arrangement of the elements in the limit-sum is arbitrary, and changing the arrangement will change its value. *But at the very least we are taking the limit of the sum.* This is very unconvincing. What you are essentially doing is taking an operation f, then applying a completely distinct operation g, and then claiming that the composite operation g°f is actually f itself or merely a "generalization" of f, neither of which is true. And then when someone else does the exact same thing, but just using h instead of g, you complain that this is _not_ applying f at all, it is doing something different. Well, yes, it is something different, but so is the thing you are doing. Besides, I know what you are trying to get at. You want to argue in favor the idea that the only natural "generalization" of addition from finite sets to infinite sets is via limits of partial sums. But as I explained, using limits is completely arbitrary, as you also impose a specific, arbitrary order on the set you are adding. Consider this: rather than looking at the limit of initial partial sums, let us look instead at every sum f[i(1)] + f[i(2)] + ••• + f[i(n)], where i(1), i(2), ..., i(n) are natural numbers, not necessarily consecutive, not necessarily with i(1) starting at 0, and not in any particular order. Very importantly, each sum always has finitely many summands. Now, the set of every such sum can be called S. You can look at sup(S) and inf(S), the supremum and infimum respectively. If sup(S) = inf(S), then you could very well call this the "sum of f", if you want. Notice something interesting: if this gets applied to the set of powers of 1/2, you still get 2 as the result, and if you apply this to Grandi's series, you actually have that sup(S) = +♾ and inf(S) = -♾. However, if you have the sequence (1, -1/2, 1/3, -1/4,...), the alternating harmonic sequence, then while lim s[f] (n -> ♾) does exist, you still get sup(S) = +♾ and inf(S) = -♾ in this case. *gasp* So these two ideas differ, and my idea can easily be argued to be the more natural generalization, but it is not the idea that you are spousing as "the unique, obvious faithful generalization". This proves my point: there is no unique generalization that is the most natural and intuitive. Undermining the role that the limit plays does not make it more intuitive, and the fact that any confusion and debate exists at all proves that the relationship is superficial at best. It only merely looks like initial partial sums are the natural generalization if you choose to think about it for 4 seconds, rather than to think about it for 10 minutes. This is why no mathematician nowadays ever talks about "infinite sums" in their works as if the abuse of language was justified by intuition: because it is not. This is also why the "summation" notation for series that people like using on the Internet is not used by mathematicians anymore: because it is misleading, and they know it. *But then the answer 1/2 is hardly even related to summation.* I disagree. Cèsaro "summation" very obviously does use summation in its definition, even if the composite operation itself is not summation. When you evaluate the limit of partial sums, you evaluate lim s[f] (n -> ♾). With Cèsaro summation, you evaluate lim g·s[f] (n -> ♾), where g(n) = 1/n. It is the limit of weighted partial sums. It is only different from the limit of partial sums you want to work with in the weighting. You use the weight g(n) = 1, Cèsaro uses the weight g(n) = 1/n. The definitions are otherwise identical. And there are other alternative, equivalent definitions that make the relationship to summation more obvious, even if, at the end of the day, it still is not summation. But then again, evaluating lim s[f] (n -> ♾) is not summation either. *I just don't think the usage of the word "summation" here makes sense. I think it causes confusion.* I completely agree with you. Where we disagree is that I think using the word "summation" when evaluating lim s[f] (n -> ♾) is also nonsensical, and it obviously is also confusing, as proven by the very existence of this debate, whereas you do not think so.

[@nqa1893]

Pls you can make a video explaining the zeta function

[@rogerkearns8094]

Yes, I agree - in particular, how to evaluate the (seemingly unbreachable) functional equation for arbitrary values of z.

[@ramirezramirez475]

1+2+3+4+... flashbacks

[@captainyoyo99]

I have another example where this sum shows up, which I find pretty interesting. The integral of sin(x)*cos(x) is 1/2sin^2(x). But if you perform integration by parts you find that the integral of sin(x)*cos(x) is equal to sin^2(x) - that integral. So if you continue the integration by parts and factor out sin^2(x) you get the sum (-1)^n times sin^2(x) and therefore the relation that this sum is equal to 1/2. Edit: I ignored constants while integrating since this makes no difference in my argument.

[@PunmasterSTP]

Woe, this video has a grande amount of information! Again, I really appreciate your delving into the concepts and touching on so many of them in each video. I’m hooked!

[@sidharthshanu6554]

tis is what i will answer when they ask me to find Schrödinger's cat in maths

[@qbot8382]

I really like that he's probably the only KZbinr who asks his viewers to dislike the video if they don't like it.

[@jacobpotts7954]

SomeOrdinaryGamers does that too

[@audigamer8261]

Linus Tech Tips

[@samucabrabo]

I really dislike that he basically invented math drama. Every video he makes is some exaggeration or misinterpretation about mathematics.

[@dqrksun]

@@samucabrabo example?

[@abdulmasaiev9024]

I've seen this often. And KZbin sees this and goes "hey, it's engagement".

[@captainhd9741]

Out of everything in this video I was most fascinated by the idea of defining different ways for summation in an infinite sum. Why do we have to stick with the limit of partial sums? Like you mentioned in the video, the other method which (forgot his name) used was to take the average instead of a limit. It makes me wonder, “What other types of summations can we define other than the standard one?”

[@angelmendez-rivera351]

Firstly, we need to get rid of the idea that what we are doing in any of these cases is "summation". These are operations on sequences that mimic summation, but they are not summation. With that out of the way: secondly, the reason that the limit of the sequence of partial sums of another sequence is so ubiquitous, to the point that schools incorrectly present it as _the_ sum of a sequence, is because Taylor's theorem, which is arguably the most important theorem in all of analysis, uses this idea. Specifically, you have a sequence of Taylor polynomials, which by definition, are partial sums of the sequence f(n, x, α) = g(n, α)·(x - α)^n. The limit of the sequence of Taylor polynomials is equal to the function of interest in some region, a region centered at α. As such, these Taylor polynomials constitute the "best approximations of nth degree" of the function in a region centered at α. This is what Taylor's theorem claims, in a nutshell, and this theorem is incredibly overpowered, and is the cornerstone of mathematical analysis itself. Without Taylor's theorem, there is no analysis, no calculus, no higher mathematics. Why? Because polynomial functions are such simple functions, and the fact that there is a systematic, formal theory about when substituting complicated functions we want to work with, with polynomial functions, is allowable, makes for a very robust method that has uses in every application you can think of.

[@captainhd9741]

@@angelmendez-rivera351 Hello Angel! Nice to see you again. Thanks for this long and thorough explanation.

[@daphenomenalz4100]

In school, I once did a question in which there was limit to infinite and after few minutes my small brother saw my question Nd told that the answer was the average of the original range of the function. And we tried to apply it on few more questions and it worked. I don't remember the question, but after that we were kinda curious why that happened. This seems like an explanation to those problems 👀

[@nickharland9207]

Doesn't work. the sum of ((-1)^n)/n contains terms between -1 and 1/2 but the sum is -ln(2), which is not the average of -1 and 1/2 which is -1/4.

[@TheLordrain12]

General tests of divergance is always telling the truth. The one thing happens when you use these kind of formulas and get 1/2 means that "if" this series did converge, it wouldn't be anything other than 1/2.

[@davidjohns2582]

It is interesting that if instead you look at -1 + 1 - 1 + 1 -1 + ... then you end up at a sum of -1/2 (using the same reasoning). So although it is an infinite series, the sum result depends on which is your starting value (either +1 or -1)

[@Eric-wp9um]

Hey buddy. One thing I don't understand here is that when you plug in x=1 into the new series you came up with you don't get \sum (-1)^n, it's something like \sum (-1/7)^n (with some constant out in front). It seems like it's just two different series with two different intervals of convergence, so we can't just input whatever numbers we want from the new interval of convergence.

[@aahaanchawla5393]

Also, you forgot that if you input -1 as the power and 1 as x in the binomial theorem, you get 1/2 = 1 + 1 - 1 + 1...

[@angelmendez-rivera351]

Actually, not in the binomial theorem, but in Newton's binomial series, of which the binomial theorem is a special case.

[@aahaanchawla5393]

@@angelmendez-rivera351 Yeah That is what I meant to say

[@pedrosso0]

@@angelmendez-rivera351 hm?

[@oni8337]

|x|

[@EdKolis]

There's a computer game called Space Empires 5 that has a buggy scripting language where this sum is somehow any integer depending on how long you write it!

[@TM_Makeover]

If you have a basket and you put 1 apple and take out that apple, again put the apple and again take out it, and repeat this process forever where does the amount of apples in the basket approach or how many apples will be there in it at the end... Well this is the same question as that of yours... This is an infinite process which do not have a end. So there could be 1, 0 or the question is simply wrong. In other equations like this The sum approach some value Like Zeta function 1 + 1/2 + 1/4 + 1/8... Here this Zeta(2) converges to 2 Or so many other series or summations Either they converge or diverge you can find the answer by taking limits in case they converge This is also an infinite process It does not have a end but the values do converge If you put 1 apple then a 1/2 then a quarter you will see that the total number of apples are approaching 2 In your question the value neither converge not diverge it is... Kind of meaningless or idk Sorry Not that much time and iq

[@jadegrace1312]

The issue, is that you found two series that equal the same expression, but in different domains. If you evaluate the series at a value of x (in this case, x=1), but that value of x is in the set where the series equals that expression for one series, and but the other, then the two series aren't equal at that value of x. I hope that makes sense.

[@kjl3080]

That’s literally what analytic continuation means, extending the domain of a function beyond what is normally valid

[@angelmendez-rivera351]

Here is an idea to consider: the sequence f : N -> R with f(n) = (-1)^n has a sequence of partial sums equal to (s[f])(n) = [1 + (-1)^n]/2 = 1/2 + f(n)/2. So lim s[f](n) (n -> ♾) exists if and only if lim f (n -> ♾) exists. Of course, we know that lim f (n -> ♾) does not exist, so lim s[f] (n -> ♾) does not exist either. However, if lim f (n -> ♾) = 0 were true, then lim s[f] (n -> ♾) = 1/2 would be true as well. This motivates the concept of regularization. Rather than evaluating lim s[f] (n -> ♾), it may be interesting to look at some other linear operator of sequences instead. This linear operator, call it L, should coincide with lim f (n -> ♾) whenever it exists, but it should have the added property that L(f) = 0 if f(n) = A·sin(B·n + C). Ideally, it should also be a continuous operator. Then L(s[f]) would be the quantity of interest. Just a fun experiment. I believe at least one such operator L does exist.

[@p8a1]

quick solution - 1-1+1-1+1-1... = x 1-x = x you can see that subbing in x on lefthand side an infinite number of times will give us the original series = to x. solve for x 2x = 1 x = 1/2 tada, no complicated math

[@Termenz1]

This is exactly like dividing by zero. You don't get a defined number unless you decide to specify certain moments. Just as that sequence has 1 or 0 as the answer depending on the last number (which is infinity), dividing by zero is dividing by pure nothingness, which doesn't exist as well as make any sense. You could say that x/0 can be any number between negative and positive infinities, but that doesn't accomplish much.

[@JoeCMath]

I really like how you do your animations!!! Well done!

[@BriTheMathGuy]

Thank you so much 😀

[@elliottmanley5182]

For a long time I've been wanting to understand more about analytic continuation. This vid is the most accessible introduction to it yet. Thank you.

[@BriTheMathGuy]

Glad it was helpful!

[@tomdekler9280]

Why do non-convergent sums even have an answer in the first place? If we're not approaching anything, then even if the answer is somewhat intuitive, it's still not doing anything, right?

[@RetjeeeTv]

This is crazy

[@BriTheMathGuy]

Right?!

[@tomkerruish2982]

That doesn't mean it's wrong (cf. quantum mechanics).

[@donovanb8555]

@@tomkerruish2982 or is it?

[@SavaConrad]

We want however the value of the sum itself not the analytic continuation of the function. We want a way for 1 - 1 + 1 … to have a value, not for a function to represented differently

[@Thepiecat]

Since (-1)^n is equivalent to cos(π*n), can we say that the sum of cos(nπ) also = 1/2? Can we make any conclusions about the integral of cos(x) from 0 to ∞?

[@angelmendez-rivera351]

Well, lim cos(π·n) (n -> ♾) does not exist in the standard topology. However, in a coarser topology, you may be able to have it so that lim cos(π·n) (n -> ♾) = 0, and this is really the only other possible result you can even have. Another idea is to regularize the function cos(π·n) systematically, such that the limit of the regularization is 0.

[@vascomanteigas9433]

Integral renormalization are a direct lemma from analytical continuation. Take from your example: Integral(cos(x),x,0,inf) Take the Laplace transform of cosine by definition: L(cos(x),x,t)=integral(exp(-x*t)*cos(x),x,0,inf) Then the original integral is the Laplace transform for t=0. The Laplace transform are a straightforward integral by parts two times: L(cos(x),x,t) = integral(t*exp(-x*t)*sin(x),x,0,inf) = t - t^2*integral(exp(-x*t)*cos(x),x,0,inf) = t -t^2*L(cos(x),x,t) Finally, L(cos(x),x,t)=t/(t^2+1) Which means integral(cos(x),x,0,inf)= L(cos(x),x,0) = 0

[@angelmendez-rivera351]

@@vascomanteigas9433 This uses the same concept as Cèsaro summation, but it applies to functions with a connected domain, rather than a discrete domain. It is fascinating. Underlying both Cèsaro summation and this Laplace transform regularization is the idea of a weighted limit, by which a weighted limit of a simple sinusoidal function is 0, and this is dilation-invariant and shift-invariant. So in both cases, you have weightedlim cos(x) (x -> ♾) = 0.

[@bluexer9198]

How do (-1)^n is equivalent to cos(π*n)????

[@p_square]

@@bluexer9198 (-1)^n = (e^{i*pi})^n = cos(n*pi) for n > 0

[@fox.6497]

Well in my opinion, 1-1+1-1… does equal to 1/2. Because if we think that 1-1+1-1… equals to zero because 1-1=0 that would mean that the last numbers of the série is 1-1. But, it’s an infinite série so there is not “last numbers of the série”. In my opinion, it depends of at which speed does the série converges. Because if we randomly look at a number in the série, it’s either +1 or -1. If we found +1 then the result is 1 but if we found -1 the result is -1. And since we have technically the same chance to found 1 than -1. For me, it makes sens that the result is 1/2. (It’s obviously only my opinion, sorry if I wasn’t very clear I’m not English and I’m not mathematician).

[@le1incognito]

Я не математик, но интересно что интуиция подсказала мне что ответом в подобном бесконечном ряду должна быть половина единицы, если мы начинаем ряд с единицы и, соответственно, -0,5, если начинаем ряд с -1. Бесконечность ряда порождает его неопределённость, которая, в свою очередь переводит итоговое значение в категорию вероятностных. А оно уже выдаёт компромиссный усреднённый результат двух значений, удовлетворяющий оба вероятных значения. Желая получить общий результат обсуждаемого бесконечного ряда мы как бы получаем эффект сглаживания меандра при взгляде на него с отдаления.

[@le1incognito]

Интересно что по факту число является и вероятностным значением, свойства которого проявляются при отсутствии чётких рамок (задействования бесконечности) в поставленном вопросе.

[@le1incognito]

Другая интересная мысль для меня в том, что по факту ряд, описывающий синус, должен начинаться с половинного значения крайнего расстояния между пиками по оси Y. Для меандра это будет 1-2+2+2-2+2-2.... . И тут интересным ответом будет то, что по идее мы опять не получим 0 :) Только если не сделаем ряд конечным и завершающимся на -2 :)

[@irokosalei5133]

It doesn't matter if it fits exactly the definition of convergence but that 1/2 is a meaningful value regarding this sommation.

[@afancybirb]

If you add up an infinite amount of zeros, that’s the same as 0 times infinity, which is indeterminate. So using (1-1) + (1-1) = 0 * infinity which is indeterminate. And using 1 + (-1 + 1)…. Is the same as 1 + (0 * infinity) and so that also uses an indeterminate. That’s why neither of those parenthesis groupings actually imply that it equals 0 or 1, you just rewrote it to make it LOOK like it equals 0 or 1.

[@joeblog2672]

It's like quantum probability from the infinite series perspective. Think of Schroedinger's cat. Dead or alive? You only find out when you open the box! In the infinite perspective, the box is never opened thus the cat is considered to be in a quantum flux of alive and dead simultaneously. Therefore to give some greater meaning to this flux state, we assign a 50% probability to either state being correct at any time. Thus the sum can be looked upon as 1/2 or -1/2 depending on how you start the series.

[@rubychakraborty2751]

If noticed, grandi series is the infinite-series expansion of the base-10 logarithm of 5..........

[@kofidwirahsclass4662]

Thank you so much.you have helped me and my brother

[@Fperm-t6g]

i was waiting eta function. -eta(0)

[@thecarman3693]

Try looking at this with a real world application. Let a light being turned on equal 1 and being turned off equal to -1. Flicking the light on and off with equal amounts of time for each condition over time will give an amount of total light equivalent to 1/2 the light of it being continuously on.

[@kazuhoshiinoue2695]

The Grandi Series is also approachable using Binomial Theorem (1+x)^n with n = -1 and x = 1. It's shown in Veritasium's video about how Newton broke the Binomial Theorem and computed pi using integrals.

[@angelmendez-rivera351]

This is not really the binomial theorem, but Newton's generalization of it.

[@kazuhoshiinoue2695]

I didn't say it IS the binomial theorem. But it is what you get when you blindly plug in n = -1 into the Binomial Theorem of (1+x)^n like what Newton did, which turns out to be equal to 1 - x + x^2 - x^3 + x^4 - x^5 + ... then just plug in x = 1.

[@Ytinasniiable]

Still dont really understand any of this, but i do find it rather interesting that seeing this after watching a bunch of stuff on infinity, my brain, looking at the problem (1-1+1-1...=?) Immediately went "one half, because infinity is weird"

[@erichlf]

You can't use partial sums if it doesn't converge, so it only is 1/2 if it converges, which it doesn't.

[@tommyproductions891]

I feel like the analytic continuation doesn’t make sense here, as it’s obviously not true for x=2 for example

[@KarmaPeny]

Shouldn't we be concerned that there are so many different tests for convergence, and that a series can be said to converge if we use one test and diverge if we use another? All this 'converges to' lark sounds like a right mess. It appears to confirm that 'real numbers' are no good; this is just another absurdity to add to the infinitely long list of problems with real numbers (as no doubt Prof. Wildberger will tell you).

[@kazedcat]

The problem is the concept of testing for convergence. How do you know the outcome of an algorithm that does not stop. There will be algorithms that are predictable but there are some that is unpredictable. Then there are outliers that would seem unpredictable but they are actually predictable and there those that seem predictable but are actually unpredictable.

[@pedrosso0]

Then something has to be wrong with analytic continuations, because the sum clearly diverges... but what's wrong with it?

[@Lucaazade]

The word “continuation” or “extension” refers to a new thing, which agrees with the original thing where the original thing makes sense, and makes sense in some new places where the original doesn’t.

[@joeo3377]

Here's another way to think about it: Grandi's Series is a divergent sum (clearly, this is true), but it is divergent in a particular way which allows us to find many different values for it. The question is then, of the many values that we can assign to the sum, which one is the "best" to characterize it? In this case, 1/2 is the best way to characterize the sum, but we shouldn't say that 1/2 is the actual result of the sum.

[@chris-hu7tm]

Its equal to 1/2, your intuition doesnt help when youre dealing with infinity

[@OptimusPhillip]

I never understood infinite sums. Like, at first it's exactly what it sounds like, adding up an infinite number of addends. But then when you run into "sums" that shouldn't exist like this, it feels like mathematicians just say "the sum can be whatever I want it to be" and just pull something random out of their asses.

[@kjl3080]

There is only one possible value for this, as holomorphic functions only have one valid continuation based of off specific properties. It’s like saying (Yes, it diverges, but if it wasn’t, it can’t be any value other for 1/2 no matter how you compute it such that it is differentiable everywhere)

[@marcusscience23]

It's definitely 1/2. Don't see how people are confused. S = 1-1+1-1+1-1+... S+S= (1-1+1-1+1-1+...) +( 1-1+1-1+1-...) 2S = 1 S=1/2 S = (1-1+1-1+1-1+...)=1/2 QED.

[@nikolatopalov6887]

I still refuse to believe this

[@anonymous_4276]

You are not wrong. We are just calling it summation but in fact doing something different. With odd number of terms we get 0, with even number of terms we get 1. Hence limit as n approaches infinity of this series doesn't exist. By changing and modifying definitions to suit our needs we can make it possible for the series summation to equal 1/2. But then we aren't even summing the damn sequence. We have changed the definition of summation.

[@angelmendez-rivera351]

@@anonymous_4276 Taking a limit is not the same as taking a sum either, so as I pointed out in your own comment thread, your point is moot.

[@BriTheMathGuy]

You're not alone!

[@b1akk-b7p]

The step on 2:33 is wrong. Geometric series doesn’t not converge uniformly on the whole interval (-1,1). Thus, we cannot conclude that the limit as x approaches -1 is equal to Grandi’s series

[@angelmendez-rivera351]

You have completely missed the point of that argument.

[@moneygrowslikegrass]

Mathematicians getting lost in their mad rambling as their common sense silently weeps in the back of their mind

[@markjohnson9227]

If you take A=1-A then with that logic this expand should look like this A=1-1-1-1-1-1-1...............-1-A = (-infinity-A) therefore, 2A=-infinity or A=- Infinity, so the actual result will be negetive infinity, because its a infinity series.

[@Jotakumon]

At the end of the day, it's more a semantical argument than a mathematical one. What do we consider a sum? It's not an objective fact, it's a convention that we make. Much like 0.999....=1. Some might argue that this is true because we define 0.999... as the limit of a series, but that *is* the convention behind that notation, as much as 9 is a conventional notation for the natural number succeeding 8, but theoretically could mean anything, such as the equivalence class {3,9,15,...} in Z mod 6, or something silly like the number of particles that make up Mars. But we would argue that if I say 8+7=9, that this is wrong, because 8+7 and 9 have conventional meanings. I would have to specify that I'm working in Z mod 6. And I would say that the convention behind 1-1+1-1 -... or 1+2+3+.... is the limit of the series as seen in basic analysis. Which can either converge or diverge in the way we commonly talk about it. And thus conventionally, 1-1+1-1+... wouldn't be 1/2. Now we may change that in the future, but given what we intuitively perceive as a sum and what utility the conventional definition serves, and given that both utility and intuition tend to inform our conventions, our notations, and our language, I think we're still far away from needing Cesaro or Ramanujan summations as the conventional idea behind a "sum". Not that there is anything wrong with it.

[@Anikin3-]

uh huh

[@lightbearer313]

It can be shown that 0.999... = 1 without invoking limits: Let x =0.999..., then 10x = 9.999... 10x - x = 9x and 9.999... - 0.999.. = 9 9x = 9, therefore x = 1, but x = 0.999..., therefore 0.999... = 1.

[@sergiolucas38]

great video my dude :)

[@cmilkau]

"when x approaches 1, this sum approaches the alternating series". Not quite. It's a POINT-WISE convergence, that, is, each ELEMENT of the sequence approaches the corresponding element of the alternating sequence. However, this is actually less than "normal" convergence.* In particular, you cannot make any conclusions about lim f(x(n)) from f(lim x(n)) if x(x) converges only point-wise, even when f is continuous.** Consequently, the analytical continuation of 1/1-x doesnt say anything about the alternating series. It doesnt say more than taking parentheses (which is identical to choosing a suberies and gives you a cluster point but not necessarily a limit) or doing arithmetics (which makes no sense at all, when the limit is no number). Taking the average of the partial sums is the same as taking a weighted average of the cluster points, in this case 50%*0 + 50%*1. Doing arithmetics preserves this value just like it would an actual limit, so it's no surprise you get the same answer. There is a lot more to say but this comment is already too long. TL;DR mathematics is not magic, it is only surprising when you haven't understood the details yet. *) That would require all elements to converge at a common minimum speed, if you want, but in this case convergence is slower and slower for later elements. **) It may work if f has as stronger form of continuity, but this is also not the case here.

[@aaronl19]

1-1+1-1+1…. = 1/2 both seems like a ridiculous answer but also the only answer it could be

[@yodo9000]

What I wonder, is if these relaxed definitions of sums give invariant values under reordering. That seems pretty necessary to me.

[@BriTheMathGuy]

Good question!

[@nickharland9207]

They don't. However, technically the normal definition of infinite summation is not always invariant under reordering unless it's absolutely convergent. For example the sum of (-1)^n/n is -ln2, but can be re-ordered to get any number you want.

[@killmeister2271]

I love watching people try to comprehend infinity

[@vivekc8563]

Guido Grandi is like ariana grandi of maths

[@carlolrac7889]

Well, the summation can be calculated as 1/2 + 1/2*(-1)^n, meaning for infinity we don't know the answer, since it could be multiple answers, also, it would be very weird using Taylor series to input a limit that isn't one, the summation isn't even continuous. We also know that the formula of calculating a limit has to have an x bigger than zero and smaller than one, otherwise the summation of all 2^n would result in a negative answer, which doesn't fit the mathematical properties we know. So far it gives nonsensical answers to limits, because not every formula is made for every number. However, I think it's very interesting to try and expand these to every number, maybe these work in complex environments, but only with interest well get there!

[@jhonnyrock]

The true limit doesn't converge and neither does its partial sums so I'm not convinced. However, there definitely seems to be a link because the series is taken to equal 1/2 as part of the Riemann Zeta Function if I remember correctly.

[@joeo3377]

The sum is certainly divergent, but that doesn't mean we can't characterize it by some value. It turns out that the best value to characterize Grandi's series by is 1/2, but this is not the actual sum of the series because the series diverges. What we're seeing is that Grandi's series behaves like 1/2, in some sense.

[@ZiseGzu]

Riemann's Zeta function isn't defined at -1, where it would be equal to the series. The link most people see here is that the sequence of partial sums 1,0,1,0,1,0,... is what's known as almost convergent sequence, and, as a consequence of that property, most of the "reasonable" ways to generalize summation would bring up 1/2.

[@jhonnyrock]

@@ZiseGzu Mathematicians have extended the Riemann Zeta function to include values -1 and less than and have defined the series 1+1-1... to be 1/2 for the function at that value

[@ZiseGzu]

@@jhonnyrock Yes, there exists an extension. Analytic extension of a function is not the same function anymore. So the rule that sends x to sum of x^(-s) doesn't necessarily hold.

[@Default-s1l]

You did play around with alternating series

[@Furkan-yv5ew]

It still doesn't make sense to me. How do you explain 1+2+3+...=-1/12? Even the avarage of it absolutely doesn't converge to -1/12 since its always positive. Can you explain me why is it so amazing because it makes no sense to me? Where is it used? How to use that formula? Why should i know this? I can't make any proper conclusions using this equation.

[@TheOneMaddin]

Analytic continuation just gives you 1/(1-x) of course. So its not better than you just plugging in x=-1.

[@moldysandwich9947]

Could you please look at the domain of the function y=ln(sin(x))

[@stann8032]

Brooo. This was super cool

[@adamoksiuta4715]

This sum NEVER be equal 1/2! Why? Because (-1) to even exponent is equal 1 and (-1) to odd exponent is equal -1. So this sum, when n is even is equal 0 (you have many pairs like -1 +1; -1 +1=0 and 0 times anything is 0). When n is odd the sum is equal -1because (-1) to odd exponent is equal -1.

[@gustavowadaslopes2479]

Whenever you try to achieve continuous results out of a digital formula, you risk getting results that only woek if it were a continuous fomula.

[@manla8397]

Search for "The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation" Terry Tao. This should give you more insight into this type of series. (YT just keep deleting my comments. )

[@DokterrDanger]

You should also see the Ramanujan Summation which says that 1+2+3+4+5+6+7+8+... infinity it's gonna give you -1/12, using the same Cesàro Summation.

[@sergeyromanov2116]

You didn't bother watching the video.

[@geosalatast5715]

A way i can understand this is this; S=1-1+1-1... (1). So S=1-(1-1+1...) => S=1-S => S=1/2. Another way for me to look at this when i was younger was this; S=(1-1)+(1-1)+...=0 or S=1+(-1+1)+(-1+1)+...=1, so the average of 0 and 1 is 1/2.

[@clementboutaric3952]

I love how you are trying your hardest to make something converge when it doesn't converge by the hecking definition of convergence. Of course if you use different object you can get convergence. But the ceasaro sommation isn't the sum of (-1)^n. There nothing to be convinced of. Just math that are true wether you like it or not.

[@neilgerace355]

I'm not a mathematician, but I think analytical continuation is just another way of cheating.

[@BriTheMathGuy]

Maybe! 😂

[@angelmendez-rivera351]

Analytical continuation is how we do complex analysis, though. There is no complex analysis without analytic continuation.

[@ITHrealXD]

Funny thing: the title says 1-1+1-… which means youre saying it repeats after the “-“ sign, so youre saying 1-1+1-1-1+1-… which=-∞

[@debangshuchakraborty7077]

It is also used in the Ramanujan summation where the sum is taken as half which leads to the summation that sum of all natural number are -1/12 Make a video on that too

[@angelmendez-rivera351]

The Ramanujan "summation" of the sequence f(n) = n + 1 is indeed -1/12, but a rigorous demonstration of this, unlike the one used by Numberphile, does not actually use the Grandi series at all. This Ramanujan "summation" is also not a "summation" in any strict sense of the word. The "traditional" sense of the word "summation" in the context of series is also not actually a summation either. You cannot add infinitely many elements together. You can, however, transform a sequence into another sequence using partial sums, and then evaluate the limit of the latter sequence. For example, you cannot actually add all the natural powers of the reciprocal of 2 and conclude that the result is 2. Again, summing infinitely many things is nonsensical. You can, however, look at the sequence f(n) = (1/2)^n, evaluate its sequence of partial sums to (s[f])(n) = [1 - (1/2)^(n + 1)]/(1 - 1/2) = 2 - (1/2)^n, and then evaluate lim (s[f])(n) (n -> ♾). This can look like you are indeed evaluating a sum of infinitely many elements if you do not think about it hard enough, which is why people like to abuse language, and call this "the sum of the powers of 1/2", but it is not actually a sum: you transformed a sequence into another sequence, then you found the limit of the latter sequence. Similarly, Ramanujan "summation" involves applying some linear transformations to a sequence, different than the ones you use to get to the sequence of partial sums, and significantly more complicated, and you evaluate the limit of the output of those transformations. Again, it can look like you are indeed adding infinitely many elements from a "generalization" stand point if you are not thinking about it carefully, but ultimately, it is not.

[@debangshuchakraborty7077]

@@angelmendez-rivera351 you are 100% correct but you are looking at it way everyone does think a bit differently, however all the math is correct in the Ramanujan summation so can't this be an indication to future discoveries that might settle that infinity is not what we generally think it might be a point on number line to which every number oscillates or the numbers line might be a number curve or number wave so this is the indication to many things so we can't just deny the theory right now we just gotta dig deeper and find out

[@angelmendez-rivera351]

@@debangshuchakraborty7077 This not a matter of what "infinity" is. This a matter of semantics. The phrase "sum of all natural numbers" is an incoherence, because "sum", by definition, refers to a finitary operation, and the set of natural numbers is not a finite set. You can argue that we should redefine "sum" so that having a "sum" of infinitely many elements is a coherent concept. Sure, but this is akin to redefining the English word "house" to mean something entirely different than what it is agreed-upon to mean, arguing that "my intuition tells me that this redefinition is a natural generalization", and then pretending that this is the actual definition, rather than specifying in-context that this is a redefinition. It is not an argument, just fascetious word games. That is not mathematics, that is semantics. Since we should agree that the definition of the mathematical consensus of "sum" necessarily only deals with finite sets, and since we should agree that there is no intuitive or uniquely useful redefinition with sufficient warrant, at least according to pedagogical research, in contrast to mere personal anecdote, it follows that we need to accept that there is no such a thing as the "sum" of all natural numbers. This is nonsense. So the fact to the matter is, Ramanujan "summation" does not prove that the "sum of all natural numbers" is -1/12, because the object that is being proven to have this value is not the sum of the elements of an infinite set. Everyone else who objects to this always responds with "the sum of all natural numbers diverges" or "the sum of all natural numbers is ♾", but they are wrong too, because firstly, there is no accepted definition of the symbol ♾, and it is not established that it is a number, let alone that it can be the result of a "sum", much less if the sum in question is an incoherence in the first place; secondly, for the reason I already explained, that the object in question is an incoherence. If you cannot sum infinitely many elements coherently, then the sum is not anything: not finite, but not infinite either. It is just undefined. It is a meaningless construct, again, because it is not a coherent object. For this reason, Ramanujan summation also does not prove -1/12 = ♾, or anything of the like. So it does not actually indicate anything regarding the infinite, or the symbol ♾, for that matter. This does not mean that Ramanujan summation is wrong. I am not dismissing the theory. I am simply telling you that Ramanujan summation is not what you think it is.

[@sergeyromanov2116]

@@debangshuchakraborty7077 what you wrote makes no sense.

[@brunesi]

That's interesting, the reasoning looks sound, and I liked it, but I believed the correct answer would be zero. To me it's very clear this series notation represents two parallel lines, whose sum, or composition, would be the center line, zero. It could also be seen as a discretized, infinite length, sin function, which in turn would again sum up to zero. It looks very intuitive to me. And it could not be rearranged in order to lead to 1/2. But well.

[@angelmendez-rivera351]

I think part of the issue with this topic, and part of what makes it to be confusing and controversial, is the myth that there is supposedly a "traditional summation method". I blame the school system for this. The fact to the matter is that none of these things are summation any recognizable sense of te word. Firstly, it is important to realize that the exploration of these divergent series are older than even the definition of limit itself. Grandi's series was already thought of as being 1/2 before calculus existed, so the idea that there is a "traditional" interpretation of summation that results in Grandi's series not having a value is an invention of modern schooling, not a hystorical fact. Secondly, and strictly speaking, summation is just a generalization of addition to arbitrary n-tuples of quantities, rather than just being restricted to addition of 2-tuples. This is because addition is associative. If I sum the 2-tuple ((x, y), z), I get the same total as when I sum (x, (y, z)). As such, I can use recursion to define a generalization of this binary operation and turn it into an n-ary operation, such that when the operation acts on the 3-tuple (x, y, z), it gives the same result. By the way, this is where the notion of "empty sum" and "empty product" come from: when this generalization is applied to the unique 0-tuple (), the result is 0, so 0 is said to be the empty sum. This is an important point, because since this generalization is recursive, it necessarily only works for n-tuples. It does not give you anything to sum sequences or summing over other infinite sets. In fact, you cannot generalize this to infinite sets faithfully, and if you were to try to generalize, then this generalization would most definitely not be unique. There are some linear operations you can perform on the vector space of sequences, and these operations sort of mimic summation when you look at them sideways, so mathematicians sometimes abused notation and language, and called them "sums", even though there is never a point where you are actually evaluating the addition of the infinitely many elements of the set, because again, you cannot. The outdated notation being used, "f(0) + f(1) + •••", does not help. Thankfully, no mathematician today ever actually uses this notation in their works, as the notation is inherently ambiguous and nonsensical, contrary to what some people here in the comments would like to say. So among modern mathematical works, there is no controversy to begin with, and no confusion as to what kind of operations are being applied to a sequence to get a result. This is why I find the idea of this topic being controversial, it particularly being a controversy only on the confines of the Internet and nowhere else, to be rather dumb. Anyway, as I was explaining, you can sort of mimick the idea of addition when working with infinite sets, but ultimately, it never is addition. It always boils down to taking a weighted sum of finitely many elements, and then taking the supremum and infimum of the set of all such weighted sums. This is also one way you define integrals: rather than staying limited to countable sets, you can "sum" over the real numbers, for example, as long as as "sum" does not actually refer to a sum, but instead refers to a supremum of sets of sums. With a sequence f, you can define the sequence s[f] by (s[f])(0) = f(0) and (s[f])(n + 1) = (s[f])(n) + f(n + 1) for every natural n. This is a linear transformation that has f |-> s[f]. Now, you can evaluate lim s[f] (n -> ♾), and get a result, and this process is quite useful, primarily because of Taylor's theorem: a sequence of Taylor polynomials, when having its limit taken, results in a function that is meant to be the "best approximation" (in a rigorous sense) of the original function with which the Taylor polynomials were produced. This also allows for a nice generalization of the ring of polynomials into the ring of formal power series. This makes this concept very useful, and for this reason, it gets introduced early on in calculus courses. However, because the education system is incapable of taking a step forward without also taking a step backward, calculus teachers always introduce this concept as being "the sum of f", with the added bonus that they always denote the concept with Sigma notation, rather than using the correct notation for it, which is misleading as hell. I am aware that this abuse of notation and language is meant to be an oversimplification to make the topic supposedly easier to teach to youngins, but the fact that this Internet "controversy" exists at all is definitive evidence that such a pedagogy does not work. It creates this false narrative that this is the "traditional summation method", all while ignoring that (1) there is no true summation happening here, just operations of sequences, of which only one single step includes summations of finitely many elements, (2) all the other "methods" are hystorically contemporaneous with this one. If you ask two expert mathematicians if "1 - 1 + 1 - 1 + •••" = 1/2 or it diverges, many of you would expect them to unanimously reply one way or the other, or maybe you believe there is a slim chance they will start to have a debate about this if they have differing opinions. But this is most definitely not what will happen. If given time to think carefully through the question, both mathematicians will agree that the question is nonsensical, because the symbol "1 - 1 + 1 - 1 + •••" has no meaning. If you try to fix it up by using some misused Sigma notation like modern schooling does, there may be somewhat more inclined to reply that it is divergent, but generally, they are still more likely to say that this is still not right, since Sigma notation is not even meant to be used with infinite sets, as already explained. English wordings would not help either. That is the real issue with how this topic gets discussed, really. This idea that you are "summing" infinitely many elements is not accurate, so when you try to present it on the Internet as if it were accurate anyway, controversy is bound to happen, especially when the notation accompanying this is abysmal. So overall, I do not think the question "what is 1 - 1 + 1 - 1 + ••• ?" makes sense to start with. If you ask me specifically what the limit of s[f] is, with f(n) = (-1)^n, then I agree that said limit does not exist, though there is something funny that happens if you were to regularize the sequence (-1)^n to have limit 0. If you ask what the limit of s[g] is, where g(n, x) = f(n)·x^n, and then evaluate this as x -> 1, then I agree that this value is 1/2.

[@angelmendez-rivera351]

As an unrelated side note: if I have some f for which (s[f])(n) converges to a real number S[f], then [s^2][f](n)/n also converges to S[f]. This is Cèsaro's theorem, and this is the reason that Cèsaro "summation" is useful. lim ([s^2][f])(n)/n (n -> ♾) may exist even when lim (s[f])(n) (n -> ♾) does not, but if the latter does exist, then the two are always equal. Meanwhile, Abel "summation" amounts to an exchange of limits, or also to interval closure. When you evaluate the limit of partials sums, you evaluate lim [lim s[f(n)·x^n] (x < 1, x -> 1)] (n -> ♾), because polynomial functions are continuous functions, so lim s[f(n)·x^n] (x < 1, x -> 1) = s[f(n)]. Abel's evaluation evaluates instead lim [lim s[f(n)·x^n] (n -> ♾)] (x < 1, x -> 1). This also amounts to interval closure, because if g(n, x) = s[f(n)·x^n] has radius of convergence (-1, 1), the sequence of partials sums is g(n, 1), while Abel's evaluation instead evaluates lim g(n, x) (x < 1, x -> 1). So if h(-1) = lim g(n, x) (x > -1, x -> -1), h(1) = lim g(n, x) (x < 1, x -> 1), and h(y) = lim g(n, x) (x -> y) otherwise, then h has domain [-1, 1] rather than (-1, 1), and coincides with g everywhere on (-1, 1). Of course, if g has an analytic continuation to almost everywhere in the complex plane, then this result is trivial.

[@gabemerritt3139]

I mean the answer alternates between 1 and 0 depending on whether infinity is even or odd. Such a concept is nonsensical for infinity. The average value is 1/2 though...

[@gbyt034]

What if the answer is X is greater than or equal to 0 and less than or equal to 1

[@redg349]

well technically, if the natural results of the sum adding the parenthesis could be 1 or 0, then 1/2, that is the middle point between the two, is a good result for me

[@SupremeSkeptic]

If the result doesn't make sense then it probably won't be useful in real life. The fact that a nonsensical value is outputted, means that there is something wrong with the method of outputting it.

[@MadKingIII]

I can't speak to the math portion, but 1/2 makes sense as it being either 1 or 0 would mean it would definitely end at some point, where there's either an odd number of terms or an even number of terms, which can't be true if its infinite. So the only answer its somehow 1/2.

[@moosewhisker8072]

I’m happy to accept that 1-1+1-1+…=0.5 in some other number systems, but to say that it’s also the case for the real numbers just isn’t true.

[@noAbbreviation]

Analytical functions: Meaningful continuation to weirdness

[@davidculp6266]

I'm going to call this Schroedinger's Series. The sum is both zero AND one until you peek at the "end", at which time it resolves to either zero OR one. That's my story and I'm sticking to it.

[@riddhibratabhaumik9796]

all you proved by doing analytical continuation was that the taylor series of 1/(1+x) around x=3/4 converges to 1/(1+x) when x € (-1,5/2). Now you equated the taylor series of 1/(1+x) around x=3/4 to summation (-x)^n, which is wrong to do as summation (-x)^n coverges to 1/(1+x) only when -1

[@GopaiCheems]

Out of context, the fact that integers sum to a fraction shouldn't be surprising XD, given that fractions sum to irrationals (and even transcendental) Great video! Once again I learnt something new about a topic about which I've read so much about. In the division by 0 video, it was the stereographic projection, and here it is the Taylor series centred elsewhere.

[@BriTheMathGuy]

Very glad you got something from it! Thanks for watching and have a great day!

[@angelmendez-rivera351]

Eh. I know what you are getting at, but I would rather be accurate and clarify that rational numbers do not "sum" to irrational numbers. These series are not "sums", not even in the "traditional method". A sum of rational numbers is rational, but here, we are not looking at sums of rational numbers. We are looking at the limit of a sequence, and this sequence is a linear transformation of another sequence that involves sums, but that does not make the limit of the former itself a sum. The rational numbers are not closed under limits of sequences. This is to say, the limit of a sequence of rational can be not a rational number. What this means is that the rational numbers are not complete. In a similar fashion, the integers are not complete either. Since every single one of these "methods" consists of taking the limit of some transformation of our sequence of interest, whatever this transformation is, it is no surprise that the result may be not an element of the set the terms of the sequence belong to. This is another reason why it is inappropriate to think of series as "sums" in any capacity, regardless of whether you are using the so-called "traditional" definition, or any of the alternatives.

[@GopaiCheems]

@@angelmendez-rivera351 yes, thanks for correcting the technical errors. What you said about rationals and integers not being complete and closed under limit of sequence, was exactly what I wanted to say. Thanks for clearing it up!

[@angelmendez-rivera351]

@Javier Concepción *When you sum fractions you get fractions with more decimals in these cases* Sometimes, yes, but this is an artifact of the notation that we choose, and of the fact that we work in base 9 + 1, instead of base 2 or base 9 or some other base. Also, this not make those fractions any less "fraction-y", for that matter: a sum of rational numbers is still a rational number. The rational numbers are closed under addition (by definition) in the exact same way that the whole numbers are closed under addition (also by definition). The property is not meaningfully different for both sets. *so if you sum infinite you logically get a number with infinite decimals* No, not quite. What you are failing to think about here is the fact that "infinite decimals" are specifically defined as limits of sequences of "finite decimals". This is key in understanding how they work. This is why 0.999... = 1. Also, as I have pointed out, at no moment of this calculation are you ever evaluating a sum of infinitely many rational numbers. All you are doing is taking a Cauchy sequence of rational numbers, finding its sequence of partial sums, and finding the limit of that sequence. *However, when you sum integers you are not actually getting decimals in any steps.* Again, this is an artifact of imperfect notation, and it has nothing to do with the arithmetic properties of these sets. One could just as easily do away with the idea of infinite decimals and declare them to be nonsense, and mathematics would still work just fine. You can axiomatically demand that the only positional digital-string representations of rational numbers that are allowed are those representations that terminate. Does that mean 1/3 is not a real number? It does not, but it does mean that 0.333... is not a valid symbol to use, because it stands for a string of digits that never terminates. With this new notation, you suddenly never have the problem you describe.

[@angelmendez-rivera351]

@Javier Concepción It seems I did misunderstand your argument, though the language and descriptions you used, in my opinion, were unhelpful, so I do not think I could have understood your argument the first time around regardless of how I tried. So thank you for the clarification. You are right. It is true that it is possible for a sequence of rational numbers to not converge to a rational number despite being a Cauchy sequence. This is not possible for the integers. This is because the set of rational numbers is dense, while the set of whole numbers is not, and every point of the set of whole numbers is an isolated point.

[@mathsman5219]

Time to extend the definitions.

[@tamazimuqeria6496]

Can you do calculus 3 things tips and trics i have exam in 5 days 😫

[@mohammedal-haddad2652]

That was like magic 🎩

[@BriTheMathGuy]

🪄

[@Pedritox0953]

Nice problem! Would be easier if we define the infinity odd or even... But this is always true??

[@ashes2ashes3333]

I think your comment is essentially correct. The math is telling you "the analytic continuation of the function that matches a geometric series on -1 to 1 is 1/1-x over all x except 1." It's not telling you that the value of 1/1-x now has anything to do with other series with x outside -1 to 1.