From homework we got that Sum [0;inf] of 2^n = -1. Sum [0;inf] of y^k = 1/(1-y). It converges on (-1;1), but has unique analytic continuation given by formula: 1/(1-y). And plugging in y=2 we get -1. So everything fits.

@notfancy2000

Жыл бұрын

Of course ∑i≥0 2ⁱ = -1, that's its dyadic representation.

8:44 Remember when Mathologer got mad at Numberphile because of that expression? 😂 14:10 Homework 14:54 Good Place To Stop

@Noam_.Menashe

Жыл бұрын

He got more mad about 1+2+3+4... this he said was a cesaro sum.

@goodplacetostop2973

Жыл бұрын

@@Noam_.Menashe He got mad at both, yeah. 1-1+1-1… was part of the Numberphile proof for 1+2+3+4… = -1/12

@themathsgeek8528

Жыл бұрын

@@goodplacetostop2973 lol yes I remember

@MichaelPennMath

Жыл бұрын

I would be psyched if this video spurred some "response" content!!

Here's another way to look at the problem. Let D denote the derivative operator with respect to x. Then, y' + y = f(x) can be written as (D+1)y = f(x). "Inverting" this operator gives us y = 1/(D+1) f(x). We can formally expand the operator 1/(D+1) as a power series in D, giving us the particular solution y_p = D^0f(x) - D^1f(x) + D^2f(x) - ... = f(x) - f'(x) + f''(x) - ..., as shown in the video. There are several problems with this approach, however, but the most important is that the operator D+1 is NOT invertible. For example, the function exp(-x) is in the nullspace. If we quotient out the nullspace, then perhaps we can justify inverting the operator D+1 against its range (i.e., taking a psuedo-inverse). Indeed, if we choose a function space for which the operator D is a bounded linear operator, then the Neumann series D^0-D^1+D^2-... converges in that function space and represents the pseudo-inverse of D+1. Unfortunately, function spaces that make D a bounded operator are few and far between. (For most reasonable function spaces, D is an unbounded operator.) For your examples where the particular solution formula worked, you happened to pick a function in a nice enough function space for which D+1 could be inverted. In the cases where the particular solution formula did not work, you fell outside the set of "nice enough" function spaces.

@ryancreedon5100

Жыл бұрын

@@angelmendez-rivera351 Not sure I see how this fixes the situation, as (exp(-x)Dexp(x)) acting on exp(-x) still equals zero. The operator still has a nullspace. I'm not familiar with the concept of left/right-cancellability, but from what I can tell by a quick Google search, D is right-cancellable if and only if it is surjective. For D defined over reasonable function spaces, I'm willing to bet this is true. However, if I'm not mistaken, you also need D to be one-to-one in order for D to be right-invertible. Clearly this isn't the case, as the derivative of any constant function is zero.

Awesome video!! I wrote about example 2 (called Grandi's series) years ago in my blog MathematicalADD while writing about the "integration by parts table method". In essence, looking at the integral of e^(2x)dx, we do integration by parts but, against logic, choose D = e^(x) and I = e^x. Then the table method for IBP has all entries being e^x, so we get that the integral of e^(2x) is equal to an alternating sum of e^(2x). This means (1/2)e^(2x) = (1 - 1 + 1 - ...)e^(2x), so the alternating sum is 1/2. I pondered in that post (from 2017...) what other strange sums you can get from this method, and you've shown another here!! Very cool! EDIT: This is awesome! To be clearer, you can connect what you're doing with this "infinite integration by parts" by looking at the integral of f(x)e^x and doing IBP with D = f(x) and I = e^x. This gives Int(f(x)e^xdx) = e^x[Sum((-1)^nf^(n)(x))], the exact sum you were looking at! Then if Int(f(x)e^x) evaluates easily in the normal way, you get these kinds of sums! Ex) If f(x) = xe^x, you get xe^(2x)/2 - e^(2x)/4 = Sum ((-1)^n(xe^x + ne^x))e^x, so e^(2x)[x/2 - 1/4] = e^(2x)Sum((-1)^n(x+n)), x/2 - 1/4 = xSum((-1)^n) + Sum((-1)^n * n) which, matching up coefficients, gives the two results you found! EDIT 2: Thank you, Michael! This clears up the general method (multiply by e^x) that elluded me for years! Your videos are always absolutely wonderful

@sk8erJG95

Жыл бұрын

@@angelmendez-rivera351 The integrating factor is a great explanation of this, thank you! It makes me wonder - back then, I looked at the general integration of f(x)g(x) which is the sum of (-1)^nD^n(f)I^(n+1)(g) by IBP. This only gave sums that made sense sometimes, like f = g = e^x. And as we discovered, whenever g = e^x, this will work because of the integrating factor and Michael's diff eq. But for f = e^x and I = 1, it did not work and gave -1 = 0. So a general question: Will this infinite IBP work to give an interesting value to a divergent series only when g(x) is the integrating factor (or inverse?) of some linear diff eq with righthand side f(x)?

Really interesting video, Michael. Thanks.

Wow, this is the kind of stuff that made me fall in love with math in the first place! I love the connection between topics. I would love to see a video on the nitty-gritty of convergence, distributions, analytic continuation, etc that goes into this sort of "sneaky" solution!

@leif1075

Жыл бұрын

Don't you find the notation needlessly confusing..what is the difference between y and f(×) in the initial equatiom..can anyone tell??

@Blackfir333

Жыл бұрын

@@leif1075 They're both (different) functions of x

@leif1075

Жыл бұрын

@@Blackfir333 don't you agree he should've used different notation then since I'd think most ppl.will thinkny equals f(×)?

@ayylmao2410

Жыл бұрын

@@leif1075 u will get along eventually

@RabbidSloth

Жыл бұрын

@@leif1075 There is ambiguity there, so it can be frustrating if this level of maths is relatively high for you, or you're not familiar with the common notation. The pedantic part of me would have liked him to define y

Thank you, Michael! It's an amazing connection between differential equations and divergent series. On another hand, as far as I know, first differential equations were solved with series by Newton. So, I think, these topics deep linked originally.

In physics, and as pointed out by others in the comments, the particular solution would almost always be obtained in terms of a convolution of a homogeneous solution (subject to some boundary conditions) with the "source" f(x). This is the method of Greens functions, where you first solve the equation for a Dirac delta source \delta(x), which is found by matching homogeneous solutions with the boundary condition obtained from integrating the equation for the delta source around the origin and then shrinking that region to zero together with some assumptions of continuity. In operator form here we have (D+1)G=1 such that the Greens operator G can be viewed as the formal inverse. The Greens function and its generalizations is one of the central objects in modern theoretical physics.

Hi, Home work : what would be the equation to get the well known : 1+2+3+4+... = -1 / 12

@riadsouissi

Жыл бұрын

Good question. Tried some functions that produce sum(n), such as e^-x or xe^-x but the resulting y solution from the differential equation does not match the particular solution (sum((-1)^n*f(x)(n)). Also tried different functional equations, such as y''+2y'+y=f(x) which has as a particular solution sum((-1)^n*(n+1)*f(x)(n)) but same issue, cannot get hold of an expression that matches sum(n) with a value from the solution of the differential equation. Same with sum(1), not obvious to get.

@FranzBiscuit

Жыл бұрын

Start with the generating function, f(x) = (x*(x+1))/2 = 1/2x^2 + 1/2x. Then f'(x) = x + 1/2, and f''(x) = 1. So Yp(x) = (1/2x^2 + 1/2x) - (x + 1/2) + (1) = 1/2x^2 - 1/2x + 1/2, and Yp'(x) = x - 1/2. And indeed, Yp(x) + Yp'(x) = 1/2x^2 + 1/2x = f(x).

Thank you, very intersting!

Great video! I'll try the homework problem and come back later with some more feedback. Thank you.

Interesting video. I really appreciate the continuity of the video.

Does someone know which font is being used by teacher Michael for making his videos thumbnail?

this can be used to get the infamous sum of natural numbers equals -1/12, by assuming the alternating sum does equal -1/4, with a bit of sketchy manipulation you can find the alternating sum equals 3s where s is the sum of naturals

Better than guessing a solution is to multiply both sides by exp(x). Then LHS is derivative of y.exp(x). Integrate both sides and divide by exp(x) to get y=exp(-x)*integtal of f(x).exp(x).dx. Sorted.

WOW. It blows my mind

14:15 we get 1+2+2²+2³+......=-1

Is it possible to use this method for f(x) = x^x?

Oha. Feeling strong Mathologer "Numberphile-debunking" vibes...

4:03 How did you come up with the original claim? What types of computation did you do to get it?

really fun!

I notice that y = xe^x - e^x is another particular solution of of y' + y = xe^x then if i attempt to match with other particular solution then : infinite sum of (-1)^n = 1 and infinite sum of n(-1)^n = -1 .... crazy

Why would the infinite sum defining y_p converge? Don't we need more restrictions on f for that?

Hey do you have any advice for new math creators like me?

And that's a good place to start

1 - 1 + 1 - 1 + 1 - ... =x x = 1 - 1(1 - 1 + 1 - 1...) = 1 - x x = 1/2 Actually, that would work for any geometric series, except for 1 + 1 + 1 + 1...

@NotBroihon

Жыл бұрын

But it's wrong, you're inducing results by assuming convergence.

@oni8337

Жыл бұрын

@@NotBroihon yeah since |r|

Narrator: “As it turned out, it was *not* a good place to stop after all”

Working with divergent series like that feels so wrong.

I haven't done the homework, but it seems like it would give you the sum of 2^n, which I think will be 1/(1-2)=-1 even though it's clearly divergent. Edit: It's definitely a particular solution. y=-1*e^(-2x). Also using f(x)=1/x will result in the sum of factorials *x^-(n+1), since the nth derivative is (-1)^n*n!/x^(n+1). The particular solution is e^-x*Ei(x), so when "plugging" in x=1 you get that the sum of factorials is Ei(1)/e=0.6917....

As a Colorado long time viewer, thanks for representing!! Love your videos, especially since I’m a “frog” more than a “bird”

If there are "other summation methods" that give the desired result, wouldn't it make sense to use them for defining y_p? Well I guess the point is to show that the relationship carries over to this particular differential equation, so you can use it for studying alternative summations.

Edit: nvm I got it, Im thinking of a power series where the first term is a constant so differentiates to 0 always. Shouldn't Yp' summation index start at n=1 at 2:25 , making Yp' + Yp = f(x)-f'(x).

If the equation is y'+ry=f(x), then the particular solution is the same but with f(rx) instead (by the chain rule). Now that I think about it, I think r can be any function and it would be f(the integral of r wrt x). Correct me if I'm wrong, but it seems like it works

@pajrc1234

Жыл бұрын

Actually nevermind that only works when the forcing function is f(int of r wrt x) in either case

So what is the missing convergence concept for these series? How about the mean of a distribution?

@manun7105

Жыл бұрын

It's not convergence but the concept of divergent series' summation 😉

@ojasdeshpande7296

Жыл бұрын

Check out Mathologer's video on this

10:43 you do not have the right to distribute segma over + unless you prove that the 2 spareted series are convergent .

I didn't read all the comments, so this might have been pointed out by others. The series 1-1+1-1+1-... is Abel summable and the sum is exactly 1/2. If I remember correctly, none of the Cesaro methods can make this sum be become summable.

@wyboo2019

Жыл бұрын

this series (Grandi's Series) is actually Cesaro Summable. the sequence of the means of the partial sums is 1/1, 1/2, 2/3, 2/4, 3/5, 3/6, 4/7, 4/8, ... which definitely approaches 1/2 so it is Cesaro Summable

Your hairs now looks so good! 👍👌

Heresy! Ignite the stake!

the thing is all the manipulation done on the series are valid as long as the series converges, if you choose a f such that the series diverges, the manipulation are not valid anymore

@farfa2937

Жыл бұрын

Everything converges if you're brave enough

@disgorgeengorge

Жыл бұрын

Everything converges if you wait long enough

It diverges Differential equation is linear Integrating factor is e^{x} but variation of parameter also can be used

You remind me a US senator when you blank.

The answer to the homework is 1-2+4-8+16...=1/3.

Right off the bat confusion..doesn't y equal f(×)?? Then it'd just be y prime equals zero..why notnuse different notation?

preforming algebra on infinite sums can lead to these inconsistencies. factoring out e**x from that alternative sum of 1s is where the math goes awry.

I really got confused why a divergent series converge to sth...

@ConManAU

Жыл бұрын

The answer is that it doesn’t, which is why Michael was careful to put quotation marks around the equality. The solution method he derived at the start of the video only works if the infinite sum is convergent, but there’s a kind of continuity argument you can make that says that if the infinite sum *could* have a meaningful value, then it would probably be the one derived from this method. It’s a bit like how r+r^2+r^3+…=1/(1-r) only if |r|

@vascomanteigas9433

Жыл бұрын

Analytical continuation.

Differential equation has infinite particular solutions, it is not correct to equate particular solution to another particular solution without claim that both of them are the same solution

I think we need to talk about that infinite distributive law that is applied on the divergent series. Seems to me that it is the root of all of this.

im confused as always when i see them

I have big issues with many of the various summations that come under the name of analytic continuation. We know that integers are closed under addition. However many integers we add we won't get a fraction. So I am sceptical of purported sums that come to -1/12 and so on. Likewise the sun of naturals is closed under addition, yet by analytic continuation it is suggested that the sum of 2^n comes to -1, which is not a natural number. I would _really_ appreciate a video on exactly why do many otherwise competent mathematicians think these results have any validity at all. In terms accessible to a physicist (a prolific user of maths but no way a pure mathematician) why is it reasonable to accept that a sum of an infinite series can defy the normal rule of closure? I do have one semi-plausible point of entry when thinking about this. We know that the sum of a series of rationals can converge to an irrational (think of the Taylor series for sine or exp). However while such series do escape the rational range of the input they end up with an irrational that is defined to lie between two arbitrarily close rationals. What blows my mind about -1/12 or -1 is that both these results fall onto the number line in a section outside the envelope of the normal range. My mind blows a fuse at that point ... Explaining that would make a great video, even though you might feel it's not at quite the same level as your other number theory ones.

@Simarillio

Жыл бұрын

This might be an issue of philosophy/interpretation rather than mathematics. After analytic continuation, the Riemann zeta function is no longer "the sum of 1/n^s", but a function which agrees with that on the restriction Re(s) > 1. However, we are naturally attracted to assigning values to things, and since we don't have a better value to assign to the sum of the integers, why not assign zeta(-1), which "looks like" the sum of the integers *if you pretend you are using the definition for Re(s)>1*? This viewpoint is maybe attractive if you are mainly working with the zeta function - then the sum 1/n^s is a "special case" of zeta, the thing you are really studying. Of course, it is also reasonable to not assign that value if you care more about the standard definition of sums. The reason that your mind blows a fuse is because it has some good intuition about how "standard" sums behave, and rebels when a definition that disagrees with that intuition appears. This is usually a good thing - we typically try to define mathematical objects so that they behave in the ways that we have come to expect. Context matters, and often disappears when people make provocative statements!

@user-jc2lz6jb2e

Жыл бұрын

I'll be a bit meta here before I get to the actual point: If I write something like "1+1", these symbols have no inherent meaning. I have to GIVE them meaning by defining them. So we would have to start by defining "+" and give it meaning. If you look up the definition of addition on real numbers, you will see it is only defined on TWO real numbers at a time. So "1+1" would be defined, but "1+2+3" wouldn't be (however, "(1+2)+3" WOULD be defined, because it's clear in what order we'd like to resolve the additions). We'd then go on and prove that addition has associativity, so it doesn't matter in what order we resolve the additions. Then we DEFINE what something like "1+2+3" means, and in general, finite sums. With me so far? I hope you can see where I'm going My actual point is this: infinite sums have NO inherent meaning. We have to GIVE meaning to them. One way to do that is with convergence of partial sums in the reals, but that's not the only way. Moreover, there are many sums that would be left undefined by this. Sure, it's "intuitive", but who says this is the ONLY valid way? There are other ways to define infinite sums. One similar way is convergence of partial sums in the 2-adics (though you'd be dealing with 2-adic numbers rather than reals). So while a sum like 1+2+4+8+16+... of powers of 2 would be undefined bg convergence of partial sums in the reals, it would be defined in the 2-adics. And the result they would converge to in the 2-adics would be -1. Similarly, the sum 1+1/2+1/4+1/8+1/16+... would be 2 in the reals, but would be undefined in the 2-adics. The point is there are many ways to define infinite sums, and none of them is more "valid" than the other; they're all valid. Analytic continuation is valid. Convergence is valid. All of them. You may not think that the convergence in the 2-adics is valid and it's all made up math mumbo jumbo, but we actually use it in computers. Just as we use approximations to represent real numbers on computers, we use approximations of 2-adics. If you look at any integer data type, it only has a finite number of bits, so you're actually working modulo a power of 2. If you look at the representation of -1 in any integer data type, you will see it's just a string of ones in binary, meaning it's a sum of powers of 2. And if you have more bits, you'll just have more powers of 2 in -1's representation. So really it's just like the 2-adics, where -1 is a sum of powers of 2. So it's all "valid". Just define what you mean by an infinite sum, and it's all good. Different contexts will have different meanings, so just be clear about which you're using.

@AlcyonEldara

Жыл бұрын

A series IS NOT an addition. That's the problem, you are extrapolating a "finite" process into something else. Addition is commutative. Series aren't. They aren't even associative.

@jeroenvandorp

Жыл бұрын

You can’t naively sum a divergent series because it hasn’t got a sum. Not to say that the regularized “answers” are thus nonsense, because they’re not and will give you actual correct real life results if you bump into such a series like in quantum mechanics. Regularized answers give you information about something else, like the development of a series. Not to claim we know everything about it. The Numberphile video with Edward Frenkel at kzread.info/dash/bejne/YoOV3MRwebrgkqQ.html (only watch the first minute or so) gives you a small insight in what professional mathematicians think about this.

@trueriver1950

Жыл бұрын

@@user-jc2lz6jb2e looking just at the point you make about 2-adics. As you say modern computers use modulo 2^n for integer arithmetic. (I've been in the trade long enough to have worked with one's complement as well. I certainly accept that the sum of N powers of two (from n=0 to n= N-1) modulo 2^N is -1. But in letting N approach infinity analytic continuation drops the modulo from the expression. That, like some of the other intuitive issues I have, is a step too far for me to accept your claim that they are "just like" each other. But thanks for your thoughts... I enjoyed thinking about them.

This may have also philosophical and ethical implications if you ask me. I know you are speaking of math here, but... I try to see further. Why? Because in our daily life we deal with infinity notion (in disguise, so most of the time it isn't obvious), and we are force to take actions, on these kind of situations, and we end up with different results, exactly because we weren't aware of the infinity notion, and we obeyed the classical logic, and classical logic can't deal "well" with infinity.

@radupopescu9977

Жыл бұрын

@@angelmendez-rivera351 strictly-speaking - No. Lato-sensu - Yes. Try (if you want!) to make correlations. Math, and infinity notions is "very" present in everyday life, but in disguise, not in obvious way. Think only to Zeno-paradoxes. Or case by case situations, All happen because infinity notion in disguise. Case by case results are found in indeterminate case at limits, when 1^infinity for e.g. can have any solution, depending on the case you study. There are limits with indeterminate forms that have no solutions at all, e.g. lim of cos (z) (again reals numbers are so restrictive!!!). Most of the paradoxical situations in life (and I don't mean falsidical paradoxes, or semantic ones!!! - e.g. liar paradox!!!) appear because of infinity notions in disguise.

@radupopescu9977

Жыл бұрын

@@angelmendez-rivera351 Yes but continuity has in common infinity, because you can divide it ad infinitum.

@radupopescu9977

Жыл бұрын

Would you be so kind an take piece by piece and say where am I wrong? Does any indeterminate form has case by case resolutions? Is involved infinity notion in indeterminate form? Please point out where do you think am wrong.

Why would you do that... 1+2+3... +(-1-2-3...) = Inf - inf = undefined

Nope. No traction at all. Tires slipping from the first words. 😞

Sometimes it's good to joke. But I think you should announce this.

@hybmnzz2658

Жыл бұрын

He says in the video he's doing sketchy things so your wish is already granted. And who cares if it leads you to discovering cesaro sums.

@mathcanbeeasy

Жыл бұрын

@@angelmendez-rivera351 why you 2 take it so seriously? Even my comment was a joke. 😂

@hybmnzz2658

Жыл бұрын

@@mathcanbeeasy sorry for my hostility

Interestingly, if we apply this method to f(x)=P(x)e^(rx), where P(x) is some finite polynomial and take a particular solution via undetermined coefficients of Q(x)e^(rx), where Q(x) is some polynomial of the same degree, after solving for the coefficients of Q in terms of the coefficients of P we can equate the two expressions for y_p. After a whole lot of manipulation and some sketchy order switching of sums, we get sum_n=0^inf nCr(n,k)r^(n-k)=1/(1-r)^(k+1), where r!=1. This looks a little like the last formula in the section here (en.wikipedia.org/wiki/Ces%C3%A0ro_summation#(C,_%CE%B1)_summation) on Cesaro summation. I don't really know anything about this topic, but a follow up video explaining the link between whats going on here and these divergent summation methods would be greatly welcomed.