The best A – A ≠ 0 paradox

This video is about a new stunning visual resolution of a very pretty and important paradox that I stumbled across while I was preparing the last video on logarithms.

00:00 Intro
00:56 Paradox
03:52 Visual sum = ln(2)
07:58 Pi
11:00 Gelfond's number
14:22 Pi exactly
17:35 Riemann's rearrangement theorem
22:40 Thanks!

Riemann rearrangement theorem.

Gelfond's number
e^π being approximate equal to 20 + π may not be a complete coincidence after all:
@mathfromalphatoomega
There's actually a sort-of-explanation for why e^π is roughly π+20. If you take the sum of (8πk^2-2)e^(-πk^2), it ends up being exactly 1 (using some Jacobi theta function identities). The first term is by far the largest, so that gives (8π-2)e^(-π)≈1, or e^π≈8π-2. Then using the estimate π≈22/7, we get e^π≈π+(7π-2)≈π+20.  I wouldn't be surprised if it was already published somewhere, but I haven't been able to find it anywhere. I was working on some problems involving modular forms and I tried differentiating the theta function identity θ(-1/τ)=√(τ/i)*θ(τ). That gave a similar identity for the power series Σk^2 e^(πik^2τ). It turned out that setting τ=i allowed one to find the exact value of that sum.

Ratio of the number of positive and negative terms
It's interesting to look at the patterns of positive & negative terms when rearranging to Pi. We always only use one negative term before we switch. The first ten terms on the positive side are: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220,... If you look at the differences between terms, you get: 22, 23, 23, 23, 23, 24, 23, 23, 23, 23, 23, 23, 23, 24,...
The reason for this is that Gelfond's number is approximately equal to 23. It turns out that if an arrangement of our series has the sum pi, then the ratio of the numbers of positive to negative terms in the finite partial sums of the series converges to Gelfond's number. This is just one step up from what I said about us being able to get arbitrarily close to pi by turning truncations of the decimal expansion of Gelfond's number into fractions. Similarly for other target numbers. For example, to predict what the repeating pattern for e is, you just have to calculate e^e :)

@penguincute3564 thus ln(0) = negative infinity (referring to +0/1-)

Bug report: At the 1:18 mark, I say minus one sixth when I should have just said one sixth.

Music: Silhouettes---only-piano by Muted

Enjoy!

Burkard

[Mathologer]

Eddie suggested that I ask.the keen among you the following nice question (first watch the video): How many different ways are there to rearrange a conditionally convergent series to get the sum π? Yes, of course, infinitely many. The real question is whether there are countably infinitely many or uncountably infinitely many ways?

[alphafound]

This demonstrates something non-math people don't get: Infinity is full of trap doors, subtleties, and other frustrations. The early infinity theorists like Cantor nearly lost their mind over this kind thing.

[左括號]

I really love how there's subtitles for every video since I'm still learning English
Thanks for the great content

[markjames]

It's interesting to look at the patterns of positive & negative terms when rearranging to Pi. The first 10 terms on the positive side are: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220. If you look at the differences between terms, you get:
22,
23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 24,
23, 23, 23, 23, 23, 23, 23, 24 ...



You get similar almost repeating patterns when your target is e, or the golden ratio.

I think that what's going on here is that ln(23) is close to Pi, so we are very close to the fixed 23 positive 1 negative ratio.

[FFELICEI]

An Italian Math and CS Senior Lecturer here. Just want to share that, as it happened to the Mathologer, when my professor did the Riemann rearrangements theorem in Real Analysis I, as a freshman, was totally upset and amazed by this counterintuitive result. Congrats to The @Mathologer whose videos always make us see the things we know under new and interesting perspectives.

[MathFromAlphaToOmega]

There's actually a sort-of-explanation for why e^π is roughly π+20. If you take the sum of (8πk^2-2)e^(-πk^2), it ends up being exactly 1 (using some Jacobi theta function identities). The first term is by far the largest, so that gives (8π-2)e^(-π)≈1, or e^π≈8π-2. Then using the estimate π≈22/7, we get e^π≈π+(7π-2)≈π+20.

[julienarpin]

This channel has brought me intellectual ecstasy for years

[ABruckner]

As soon as you moved the negative fractions below the top line, my first instinct was "Wait...isn't the top part 'outpacing' the bottom part?" Then I lost confidence when you collapsed them, lining up all pos and neg, lol. I was like "but, but, but...." Anyway, I love that stuff!

[НикитаДёмочкин-йж]

It is always a pleasure to watch your vids. Not only because these are great educational videos, but also because your voice and wordings make them even better

[RebelKeithy]

I think it's easy to get distracted by the fact that there is a matching negative for every positive term in the sequence. A similar paradox makes it more intuitive what's wrong with rearranging terms.
∞ = 1 + 1 + 1 +...
∞ = (2 - 1) + (2 - 1) + (2 - 1) +...
∞ = (1 + 1 - 1) + (1 + 1 - 1) +...
∞ = 1 + 1 - 1 + 1 + 1 - 1 +...
Then we can pull out positive and negative terms.
1 + 1 + 1...
- 1 - 1 - 1...
So every +1 is canceled by a - 1.
You can even create a mapping from the nth positive 1 to the n*2 negative term, so every positive term has a negative to cancel it.
This, to me, intuitively shows why you can't add infinite sums by rearranging terms. You need to look at how it grows as you add terms.

[marom]

with the last two videos this channel has outdone itself. I have seen and re-watched them several times and as an amateur and enthusiast I believe that they are the two best calculus lessons I have attended. So illuminating and profound, they hold together all those details that leave one confused in a school course and which here instead receive the right attention and are explained with incredible ease. Bravo! ❤

[obscurity]

My daughter (ninth grade) just sent me this video with the message, “this is so interesting!”

Thank you for the proud dad moment, Mathologer!

[henridelagardere]

Squeezing and stretching the snake, that sounds like lots of fun, and the result is quite beautiful indeed.

[danieljudah]

In case you are wondering, the notification for this video works for me. With crazy YouTube algorithms many creators are talking about these days we need these notifications.

[yummyyayyay]

You've shown me the true beauty in math. Your videos are truly intellectually stimulating

[ProfAmeen]

Your videos do have a tremendous impact on me, making me wanna attend you at Monash and enjoy the rest being of my life in that kind of Maths you’re reciting to us in every single dope video of yours!

[agostinhooliveira]

Fantastic stuff!!!
Actually I found a pattern in your videos.
With each new video, the length of your channel's supporter list at the end grows enough to conclude that the length of subsequent videos approaches infinity!

[tomking]

This video was basically the final week of my very first analysis course at uni, and you explained it brilliantly. Maths is one of those things that never makes sense the first time, but then becomes crystal clear the second time.

One extra thing that could have been in this video was a bit more on why the positive and negative terms of a conditionally convergent series sum to infinity, because it’s not obvious in general unlike the other key fact about them tending to zero. *Edit* Thinking about it a second time, I’m not sure if you could do that without a full mathematical proof, and it’s at least well-known for the harmonic series, so maybe it was best left unexplained.

[ufogrindizer]

There is something weirdly relaxing and also beautiful watching the animations and the number somehow forming! ❤

[mathboy]

That ln(2) trick is really something special. Thanks for showing it.