For more information about openings at Mat-X go to www.matx.com/michaelpenn
@kubilayaytemiz72747 ай бұрын
At 14:52, multiplying with s+1/s+1 for s = -1 gives us 0/0 which is indeterminate.
@adb0126 ай бұрын
Yes, but that step would still be valid for s=-2. Does it mean that the sum of all squared integers is zero? Legit?
@kubilayaytemiz72746 ай бұрын
@@adb012 To obtain a similar equation for s = -2, we have to multiply with s+2/s+2 which is indeterminate also. To validate myself, you can see at 23:09, after doing I.B.P, we have Gamma(s+3). Only way to get Gamma(s+3) from Gamma(s+2) is multiplying the integral with s+2/s+2. And this will be the case for every s. Your conclusion is not valid I'm afraid.
My genuine reaction when the sum of all the integer squares is a "trivial" result and very obviously equal to 0.
@vascomanteigas94337 ай бұрын
zeta(-2)=0. The best formula is to use Bernoulli Numbers. For all integers n>0, then zeta(-n+1) = B_(n+1)/n
@spiderjerusalem40097 ай бұрын
@@vascomanteigas9433 Any book/articles to learn about bernoulli/euler numbers?
@MarcinSzyniszewski6 ай бұрын
Proof: it's trivial. 👍
@PowerUpStudio_9 күн бұрын
@@vascomanteigas9433 i think the actual formula is zeta(-n)=(-1)^n*B_(n+1)/(n+1) for natural numbers n>=0
@goodplacetostop29737 ай бұрын
25:58 Ah, when Mathologer got mad at Numberphile… Those were the days
@NotBroihon7 ай бұрын
And rightfully so.
@l.w.paradis21087 ай бұрын
@@NotBroihon 💯
@Alan-zf2tt7 ай бұрын
Em hmm I don't know and think context is important. In Math approach infinity goes on to infinitely many infinities In science approach quanta do things discretely yet there comes a place where quantum mechanics needs -1/12 to conform to observed behaviors. Each has its own context and I think those contexts need to be explored and explained in a way that satisfies both math and science parrticularly quantum observables
@NotBroihon7 ай бұрын
@@Alan-zf2tt that has absolutely nothing to do with what Mathologer rightfully criticized Numberphile for.
@Alan-zf2tt7 ай бұрын
@@NotBroihon I think what I am trying to say is that it is science and not a war. It deserves explanation and we can agree it seems controversial - for at least ten years
@giovanninuno79637 ай бұрын
Swapped summation and integral without proving convergence?
@jimallysonnevado39737 ай бұрын
i think this step is barely not allowed😂
@nHans7 ай бұрын
1:03 - Nice, I learnt a new mathematical term today: *_barely not allowed._* 🤣
@ericbischoff94447 ай бұрын
5:19 the odd terms sum starts at 3, but on left side of = it starts at 1. I think n should start at 0 for odd terms. The error is canceled at 6:20.
@williamperez-hernandez39687 ай бұрын
He used an incorrect expression for the odd numbers. Using 2n-1, allows to begin with n=1. Thus the sum over all naturals is recovered.
@aryakaranjkar9397 ай бұрын
Yeah, or instead it should be 2n-1 instead of 2n+1 .
@mikederleres43007 ай бұрын
2 sign mistakes, which nicely cancel out: At 16:20, on du At 18:00, on the integral
@f5673-t1h7 ай бұрын
It's been 10 years. I was in college taking calc 2 when that video dropped. I remember a classmate bringing up how it was "proved" that 1+2+3+... = -1/12, and our professor was just confused.
@kazedcat7 ай бұрын
Yeah Terence Tao has proven it. To be very precise he has proven that the sum of all natural numbers can be -1/12 if you extend the definition of partial sum into a weighted partial sum.
@l.w.paradis21087 ай бұрын
@kazedcat In other words, if you redefine the operation of summation itself. Well, then.
@kazedcat7 ай бұрын
@@l.w.paradis2108 Partial sum is axiomatic so there is no proof that piecewise summation is the only valid way to do infinite sum. This is similar to the fifth postulate and rejecting it leads to Non Euclidean geometry.
@l.w.paradis21087 ай бұрын
@@kazedcat That's fine, but it's not the same thing. Why is it that in geometry, we never confuse Euclidean with Lobachevskyan, etc.?
@kazedcat7 ай бұрын
@@l.w.paradis2108 Because you don't deal with geometry in your everyday life. When was the time you had to compute the area of a parking lot to buy groceries. Only Google Maps developers have to deal with the difference between a flat map and a spherical map.
@__christopher__7 ай бұрын
In the partial integration step: x^s is0 for x=0 only if s>0. Which precludes not only inserting s=-1, but even taking the limit of s to -1.
@fable43157 ай бұрын
I think the sketchy step is 10:50. Swapping an infinite summation and integration is probably only under some assumptions possible, for example I think one condition should be that the sum under the integral converges?
@FranzBiscuit7 ай бұрын
Nice! Another way to look at it is as the indefinite integral of the generating function of the sequence which produces the sum-of-integers, ie. (n(n + 1))/2 = (n^2)/2 + n/2, which is of course -1/12.
@Noam_.Menashe7 ай бұрын
My favourite way to "prove" this is with derivatives of geometric sums. You start with 1/(1+x)=sum (-x)^n, after derivation it becomes 1/(1+x)^2=sum n*(-x)^n-1, putting in x=1 we get 1/4=sum n*(-1)^n-1 and finally by isolating even terms and dividing by two we get -1/12=sum n I liked the multiplying and dividing by s+1, then setting s=-1. If the integral converges it converges.
@galveston89297 ай бұрын
how do you justify exchanging the limit and the infinite sum?
@MichaelRothwell17 ай бұрын
@@galveston8929 by the limit I imagine you mean the limit as x→1, given that the sum is only true for |x|
@rrr00bb17 ай бұрын
i like doing it this way too. I use recursion so that you use nothing more than high-school math. it really only requires taking the definition of "a=b" more literally; like a computer does. With recursion, you can avoid all infinite sums. This is good, because you don't end up arguing about what you can and cannot do with infinite sums. kzbin.info/www/bejne/oHvJn55urcSLnaM
@coreyyanofsky7 ай бұрын
although other functions that "look like" 1 + 2 + 3 + ... when analytically continued can "assign" arbitrary values to the series, the value -1/12 is special because in quantum physics when the series 1 + 2 + 3 + ... occurs only the value -1/12 gives answers that agree with observation there has been recent work on how regularization of divergent sums and integrals in quantum theory works; i won't pretend to understand sentences such as "...regulators that allow the infinite series of natural numbers to converge towards -1/12 are intimately connected to the preservation of gauge invariance at one loop in a wide class of non-abelian gauge theories coupled to an arbitrary number of Dirac fermions" but if this topic is of interest to you, dear reader, you can find more information in the arXiv preprint "Smoothed asymptotics: from number theory to QFT" by Antonio Padilla and Robert G. C. Smith
@petterituovinem84127 ай бұрын
Is the sketchy step changing the order of summation and integration
@janeknowakowski57327 ай бұрын
Yes, i think so
@vincentbutton59267 ай бұрын
I missed that happening at about 10m20s in. Thanks for pointing it out.
@davidcroft957 ай бұрын
Not quite sure if it's a problem (or one of the problems, maybe), but surely the problem is in the next board, while solving the "zeta integral": he multiply (and divided) by s+1, which is zero in this setup
@jagatiello69007 ай бұрын
@@davidcroft95 At 14:47 he's exploiting the fact that gamma(s+1) has a simple pole at s=-1, which is why he then says the (s+1) in the denominator is absorbed by using the functional equation for gamma.
@davidcroft957 ай бұрын
@@jagatiello6900 he still divided by zero
@gerryiles39257 ай бұрын
When you wrote out the result of the first integration by parts step at 18:03 you missed the negative sign but then the error got undone in the second integration by parts...
@bloodyadaku7 ай бұрын
When you did integration by parts you skipped out on a negative, but you skipped out on the negative twice so it ended up cancelling out.
@deadfish37896 күн бұрын
Really love the step-by-step approach which leaves the bad steps as late as possible. Where I would like more detail is, is this the only possible analytic continuation to a region containing -1, and if so, why? Could we use other tricks to get a different result?
@CoreyKatouli7 ай бұрын
Excellent video as always
@s46237 ай бұрын
4:57 wouldn't 2n+1 starting at n=1 gives you 3? Where did the leading 1 term go? (Should have been 2n-1)
@l.w.paradis21087 ай бұрын
NOW I know why ChatGPT is suddenly making mistakes in simple mathematics, and refusing to show its work! (Seriously, there's already a major paper on it.)
@Jack-e7i8s5 ай бұрын
When you used the formula for geom. series, the geom. series only works if the absolute value of the common ratio is strictly less than 1, though it will not be if x = 0.
@JackPullen-ParadoxАй бұрын
Did you separate the sun of the odd and even numbers into the sum of the odd numbers plus the sum of the even numbers? Do you have a similar solution for the odd sum and the even sum? Because neither of these sums converge ordinarily. If we assume that the sun of the natural numbers converges, does that imply that the sum of the odd numbers converge and that the sum of the even numbers converge? Would that imply that the sum of any collection of natural numbers converge? I wouldn't expect that 2+ 2 + 2 +... converges, but 1 + 3 + 5 + 7 + ... might. Or, any sum in which elements are selected only once from any set of natural numbers might converge. My question is can you justify the convergence of the two sums independently from the convergence of the total, or do you want to declare the total convergent?
@deadfish37896 күн бұрын
At that step, wasn't he still assuming that Re(s)>1? In which case, both sums do converge. He only extends to the rest of the complex plane after finding a function which agrees with the infinite sum on that region.
@JackPullen-Paradox5 күн бұрын
@@deadfish3789 Yes, in terms of the Zeta function, the odds converge and the evens converge, and things like 2+2+2+... do not converge. So, all is self-consistent. But the fact that the sum of the odd natural numbers and the even natural numbers via the Zeta function is not -1/12 may present an inconsistency of importance for certain uses.
@MasterHigure7 ай бұрын
I would be pedantic and call 1+2+...=-1/12 an *equality* not an equation.
@rdubeau7 ай бұрын
At 11:40 you substitute t=nx but accept infinity as your upper limit of integration, same as before the sub. I can see this maybe being “barely not allowed.”
@speedbird83267 ай бұрын
OMG i could have sworn @22:51 he was going to say "And that's a good place to stop" but he kept going !!!!!!
@vascomanteigas94337 ай бұрын
The reasoning was to use the Fermi-Dirac integral and apply integration by parts to find the result.
@Kratos_legends7 ай бұрын
Wooowww😮 you are the best teacher that I have seen in my whole life ❤❤ your skill of explanation is so great 👍
@gp-ht7ug7 ай бұрын
At 02:36 the formula is the sum of alternating signs?
@glenneric15 ай бұрын
To prove it is wrong I just go 'der'.
@KipIngram7 ай бұрын
6:27 - Whoa; hold it. Your sums from 1 to infinity were for (2*n+1) and (2*n). That starts at 2. But you replaced it with a sum that starts at 1.
@djfahed30027 ай бұрын
15:24 Once you multiplied and divided by s+1, you assumed that s!=-1 thus you lost the privelege of replacing s by -1
@dneary7 ай бұрын
@MichaelPennMath I saw you in Nina Williams's China Beach documentary! I know you from math videos, but apparently my son's know you more from climbing stuff!
@vascomanteigas94337 ай бұрын
Abel-Plana formula goes brrrr... 😂
@the_nuwarrior5 ай бұрын
it can be proved from hankel integral over C
@ruffifuffler87117 ай бұрын
So, you sail a boat past all the natural #s, using all the previous #s to pick up the next one, but, at some point the boat sinks, and reaches equilibrium at -12 feet? Just goes to show you that most numbers aren't worth a dam when their daid.!
@KipIngram7 ай бұрын
I understand why this upsets people. Most people realize that they don't know "high math," but they do at least think they understand adding integers. Then this comes along and more or less says to them "you know NOTHING." The thing to understand here is that the -1/12 does NOT come from actually adding up these numbers. You can write an expression that is equivalent to this series when its parameters are set a certain way, but that expression itself isn't defined everywhere. You can then use analytic continuation to carry it from the region where it IS defined to "everywhere else," and it is THAT function that has the value -1/12 at a key point. It's not the same function - you've replaced it with a different one at that point. In some ways it's a "cheat," though in some ways it's perfectly valid and actually has physical applications.
@gp-ht7ug7 ай бұрын
If I’m not wrong there is a new Numberphile video on this matter
@thes72744737 ай бұрын
Two videos, actually. While watching the first one I was like "Yes, a redemption arc!" as they had a guest explaining analytic continuation. Then they posted another video where they doubled down and "proved" that the sum is actually equal to -1/12 by totally changing the problem.
@l.w.paradis21087 ай бұрын
I hope not.
@kazedcat7 ай бұрын
@@thes7274473It's not changing the problem but just changing the axiom of a partial sum. There is no solid reason why a partial sum needs to be the way it is. This is similar to changing the fifth postulate to derive Non Euclidean geometry there is no valid reason why the fifth postulate must only be true.
@NotBroihon7 ай бұрын
Probably Numberphile's worst video. Completely unmathematical and basically just clickbait.
@l.w.paradis21087 ай бұрын
@@NotBroihon It was despicable.
@Mohanchous7 ай бұрын
1. Why the Riemann zeta function? Couldn't other functions be used to deruve a value for 1+2+3+...? If so, would other functions imply the same -1/12 evaluation? 2. Is there a number system, a variant or extension of the reals in which 1+2+3+... is convergent?
@debtanaysarkar97447 ай бұрын
Answer to ur first question, there's another function which was devised by Ramanujan himself to prove this result, and guess what, that uses complex analysis 💀
@MuffinsAPlenty7 ай бұрын
Great questions! 1. Good question! Because your concerns are valid. We can develop other series-defined functions which "evaluate" to 1+2+3+4+... outside of their region of convergence, and which analytically continue to assign values _other than_ -1/12 to input which "evaluates" to the series 1+2+3+4+... [See appendix B of this comment.] The only examples I've seen, however, are pretty contrived and aren't useful. So the question then becomes how can we justify that the Riemann-Zeta function is the best choice to use. 2. No. And this is because convergence of the sequence of partial sums leads to a _stable, linear_ summation method. And it is impossible for 1+2+3+4+... to be assigned a value under any _stable, linear_ summation method. [See appendix A of this comment.] *APPENDIX A* A "summation method" is a function S with inputs being (some) infinite series and outputs being numbers. A summation method S assigns a value to a series ∑aᵢ if S(∑aᵢ) is defined. A summation method is considered to be _linear_ if S is a linear function: 1. If S(∑aᵢ) and S(∑bᵢ) are defined, then S(∑(aᵢ+bᵢ)) is defined and has the value S(∑aᵢ)+S(∑bᵢ) (essentially, term-wise addition of adding series is consistent with adding the sums of the series). 2. If S(∑aᵢ) is defined an c is a constant, then S(∑caᵢ) is defined and has the value cS(∑aᵢ) (essentially, distribution/factoring of constants over an infinite series is consistent with multiplying the sum of the series by that constant). A summation method is considered to be _stable_ if the following holds: If S(a₁+a₂+a₃+...) is defined and a₀ is any number, then S(a₀+a₁+a₂+a₃+...) is defined and has a value of a₀+S(a₁+a₂+a₃+...). (essentially, you can add a term to the beginning of the series, and the sum changes by adding the same amount). Again, one can prove that convergence is both stable and linear. Let's now show that 1+2+3+4+... cannot be assigned a value under any stable, linear summation method. We will do this by contradiction. We will assume 1+2+3+4+... can be assigned a value under a stable, linear summation method, and we will arrive at the conclusion that 0 = 1. Assume S = 1 + 2 + 3 + 4 + ... Then using stability, we can add a 0 term at the beginning, which will make the new series have the value 0+S = S. S = 0 + 1 + 2 + 3 + ... Using linearity, we can term-wise subtract the bottom series from the top series, and it will result in a series having the value S-S = 0. 0 = 1 + 1 + 1 + 1 + ... Using stability, we can add a 1 term at the beginning, which will make the new series have a value of 1+0 = 1. 1 = 1 + 1 + 1 + 1 + ... However, this new series is identical to the previous series. So we have 0 = 1 + 1 + 1 + 1 + ... = 1 or 0 = 1. So we have arrived at a contradiction. This shows that 1 + 2 + 3 + 4 + ... cannot be assigned a value under any stable, linear summation method. And since convergence of the sequence of partial sums is always a stable linear summation method, 1 + 2 + 3 + 4 + ... cannot have a sum using convergence, no matter what "number system" one uses. *APPENDIX B* I can't take credit for this example. I obtained it from the Math Stack Exchange post entitled "Uniqueness of analytic continuation to assign sum values" asked on January 14, 2018. Let C be any complex number, let h(x) = (1−x) − C(1−x)^3, and consider the function: g(x) = 1/(1−x)^2 − h(x)/(1−x)^3. Note that 1/(1−x)^2 = d/dx[1/(1−x)] = d/dx[1+x+x^2+x^3+...] = 1+2x+3x^2+4x^3+..., and this converges absolutely on (−1,1) Additionally, −h(x)/(1−x)^3 = −h(x)/2 * d/dx[1/(1−x)^2] = −h(x)/2 * [2+6x+12x^2+20x^3+...] = −h(x) * [1+3x+6x^2+10x^3+...], and this converges absolutely on (−1,1). Adding together, we get (1−h(x)) + (2−3h(x))x + (3−6h(x))x^2 + (4−10h(x))x^3 + ... = 1(1−2h(x)/2) + 2(1−3h(x)/2)x + 3(1−4h(x)/2)x^2 + 4(1−5h(x)/2)x^3 + ... = the sum of n (1−(n+1)h(x)/2) x^(n−1) from n = 1 to infinity. Let's take f(x) = ∑n∙(1−(n+1)h(x)/2)∙x^(n−1), which is defined on (−1,1). Notice that h(1) = 0, so f(1) = ∑n∙(1)∙1^(n−1) = ∑n = 1 + 2 + 3 + 4 + ... So we have found a function f(x) so that f(1) is 1 + 2 + 3 + 4 + ... Now, we have already shown that on (−1,1), we have f(x) = g(x). On this same interval, g(x) = 1/(1−x)^2 − h(x)/(1−x)^3 = 1/(1−x)^2 − [(1−x) − C(1−x)^3]/(1−x)^3 = 1/(1−x)^2 − (1−x)/(1−x)^3 + C(1−x)^3/(1−x)^3 = 1/(1−x)^2 − 1/(1−x)^2 + C = C The constant function C is an analytic function; hence, j(x) = C is the analytic continuation of f(x). Therefore, we associate 1+2+3+4+... = f(1) with j(1) = C through this particular analytic continuation. But recall that C was _any_ complex number. Hence, we can use analytic continuation to associate the series 1+2+3+4+... with _any complex number we want._ The function f(x) here isn't particularly important, and there's no reason we should expect to use f(x) to assign a value to the series 1+2+3+4+... in any given context, but it shows that there should be more than merely saying "analytic continuation". One should justify why the Riemann zeta function is the particular function we want to use. And there are good reasons to believe that the Riemann zeta function is a good choice, but I believe people should do a better job of explaining that.
@MathFromAlphaToOmega7 ай бұрын
One kind of strange thing about the sum is that if you use the function f(z)=0/1^z+1/2^z+2/3^z+3/4^z+.... and continue it to z=0, you get f(z)=zeta(z-1)-zeta(z), so f(0)=5/12. That suggests that 0+1+2+3+4+5+...=5/12, so these kinds of arguments are very sensitive to the exact function you use.
@vascomanteigas94337 ай бұрын
Because this renormalization only are valid using the Riemann Zeta Function.
@kazedcat7 ай бұрын
There is another way to get -1/12 using weighted partial sum.
@theartisticactuary7 ай бұрын
Must be swapping the order of summation and integration that's the problem. Everyone knows that you're only allowed to do stuff like that in physics. Do it in maths and you're always going to run into problems. This is why we have rules.
@Anonymous-zp4hb2 ай бұрын
barely not allowed "equals" allowed
@donach97 ай бұрын
Γ(s + 1) = Γ(0), is undefined. Isn't that the dodgy step?
@djttv7 ай бұрын
But, the sum of all positive integers does not equal -1/12
@MichaelRothwell17 ай бұрын
That's why he always wrote "=" and not =. Only some physicists and intellectually dishonest mathematicians actually believe the sum in a literal fashion.
@joseoncrack7 ай бұрын
@@MichaelRothwell1 Yes, judging from the current state of theoretical physics (/astrophysics), I would even dare say a bit more than "some physicists". 🙃
@JackPullen-ParadoxАй бұрын
It may be a question of what infinity means in the context. One would think that is would at least mean that the sum is an integer. and perhaps positive. But 1/0 might be any positive number. A different infinity. But here they are using a technique that is usually reliable and just assume that the sum of the natural numbers converges. Then the resulting sum should have be consistent within the mathematics they are using. Exactly what I mean by "consistent within the mathematics they are using" is not so easy to define.
@xaxuser50337 ай бұрын
i don't know why people are talking about multiplying by s+1/s+1 , or swapping integral and summation. The proof is perfectly fine and the value of zeta at -1 is indeed -1/12. The only problem is that the two expressions of zeta coincide at {s / s>1} and not in the real line. The first expression of zeta holds for s>1 , the other one holds for a bigger set, the two expressions coincide at a set which is not all the real line. And example of what i say : let f(x) = (x^2-1)/x-1 and g(x) x+1 f and g coincide in the real line except for x=1 where f is not even defined. We can't say then that f(1)=1 cz it doesn't even have sense, but we have still an equality nearly everywhere.
@debguha87277 ай бұрын
At 17.19 x^s is 0 only if s>0. So we cannot put s=-1 in the last integral.
@wolliwolfsen2917 ай бұрын
3:38 The Zeta function converges for s>1, but you evaluate it later for s=-1. So 1+2+3+4+5+6+7+… can not be equal to -1/12, that‘s where the „proof“ fails
@vladimirlinhart44867 ай бұрын
It is not equal in a sense of equivalence. It is an analytical continuation of dzéta function. Do not mix the result with the sum of the natural numbers.
@wolliwolfsen2917 ай бұрын
@@vladimirlinhart4486 of course zeta(-1)=-1/12 but the zeta(s) is defined by the dirichlet series (sum of 1/(n^s)) for Re(s)>1
@coreyyanofsky7 ай бұрын
of course this is true -- the series is divergent; however there is a certain well-defined sense in which this series is "equal to" -1/12 + ∞ you get there by (instead of looking at the limit of partial sums) smoothing the sum with a parameterized regulator function such that the terms of the (convergent) sum becomes closer and closer to those of the series 1 + 2 + 3 + ... as the parameter grows without bound and then doing asymptotic analysis on the parameter; the zeroth order term is -1/12 and the first order term diverges along with the parameter Terrence Tao wrote a blog post about this approach entitled "The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation"
@vascomanteigas94337 ай бұрын
You need to derive the Analytical Continuation of Riemann Zeta Function. My favorite Proof is to apply the keyhole contour integral of Bose-Einstein integral and it os derived. Also it is possible to deduce a Laurent Series for Zeta valid for all complex Numbers except the pole itself at x=1. Zeta(x) = 1/(x-1) + sum from n=0 to Infinity of Stietjes(n) * (-1)^n * (x-1)^n / n!. The Stietjes constants are a generalizarion of Euler-Mascheroni constant.
@vikramanbaburaj5257 ай бұрын
While I appreciate your videos, I also find them to have little to no use in practical life.😅
@tenjan757 ай бұрын
Your April fools prank is 5 days late….
@Alan-zf2tt7 ай бұрын
I cannot help but wonder if Euler had troubles with his new math. Did notion of imaginary numbers unsettle people? It is rumored that Cantor had problems introducing notions of different infinities s0 maybe there is a thing happening there than needs solving to mathematical satisfaction? My own take is that there has to be a point, surface, event space where discrete math becomes continuous math, integer sums become integer integrals. And if it is a boundary is it a fuzzy boundary with epsilon like error bounds or a pointwise boundary like the circumscribing of a circle? In essence the three assumed bits/parts/partitions are: discrete, boundary, continuous seeming to correspond to quanta, boundary, continuity -shrug-
@mircoceccarelli66897 ай бұрын
1 + 2 + ... + n = n ( n + 1 ) ÷ 2 n € IN => n >/= 0 , n + 1 > 0 ( + )( + ) = ( + ) =/= ( - ) Ramanujan N.O.K. !!! 😊🤪👍👋
@JackPullen-ParadoxАй бұрын
This result looks similar to the results for some p-adic numbers. But Penn indicates that it has to do with complex variables. Two different approaches that lead in a similar direction are an encouragement.
@NarutoSSj67 ай бұрын
The annoying thing is that this divergent sum equals - 1/12 using all those sketchy methods but never anything else, like - 10 or whatever. If only one could find a way for it to equal somethings else than - 1/12 then i am happy to disregard it.
@StephTBM47 ай бұрын
There is a way to (falsely) prove that the sum of all integers (let's call it S) equal -1/8 ... Regroup 2+3+4 =9 5+6+7 =18 8+9+10=27 ... So that S=1+9(1+2+3+...) Then S=1+9S so 8S=-1 and S=-1/8 !
@StephTBM47 ай бұрын
So, continuing on false assumptions, it appears that -1/8=-1/12, thus one can "prove" that 0=1 😂😂 Never make computations on diverging series ...
Glad you're getting some sponsorship dollars out of the "AI" machine learning bubble anyway but it's on par with advertising for a crypto company
@galveston89297 ай бұрын
Somehow sloppy math. You cannot change the order of a infinite sum an integral without proper justification. Let's call this "a lucky accident" unless there's a better exposition using Ramanujan sum and analytic continuity.
@NoahPrentice7 ай бұрын
I think that was the "trick that's not allowed in one of the steps" he mentioned around 0:55.
@galveston89297 ай бұрын
@@NoahPrentice it's more of "carelessness" than a "trick".
@NoahPrentice7 ай бұрын
@@galveston8929 given the video was already 25 minutes long and not about moving limits across integrals, I think it's more likely that they intentionally decided not to delve into that detail, rather than mere carelessness. If you want a proof that goes into every detail, you can probably find a paper on the topic (or work through it yourself).
@davidcroft957 ай бұрын
Yeah, that's the point of the video. Congrats 🎉👏🏻
@galveston89297 ай бұрын
@@NoahPrentice There are plenty of examples where you may use a "trick" like this on a diverging sum and come up with an arbitrary finite value. Sufficient rigor must be carefully put into assigning a finite value to a diverging sum. KZbin is rife with clickbait videos about sum of all integers being -1/12, Mathologer's is the only one I've seen so far that treats the subject properly.
@nightmareintegral55937 ай бұрын
copied from flammable math
@randomguyontheinternet_697 ай бұрын
what are you talking about ? does he own this infinite series?
@jellyfrancis7 ай бұрын
@@randomguyontheinternet_69yo 😂💥💀
@nightmareintegral55937 ай бұрын
@@randomguyontheinternet_69 not but method is same XD
@nightmareintegral55937 ай бұрын
@@randomguyontheinternet_69 that s why this is copy past
@MuffinsAPlenty7 ай бұрын
@@nightmareintegral5593 Have you accused flammablemaths of copying Riemann?