The most memorable part was when you giggle, and my wife in the other room says "You're watching that math guy again?" As always, thank you for expanding my knowledge base.
@@teleny2 Gen. Turgidson: "'Strangelove'? That ain't no kraut name." Aide: "His original name was 'Merkwürdigliebe'. He changed when he became a citizen." Gen. Turgidson: "Huh. Strange."
Yes. Simple, indeed: In order for a fraction to be an integer, the prime factors of the denominator must form a subset of the prime factors of the numerator; but, in odd/even, the denominator always has the prime factor: 2, which the numerator never does, in odd/even; thus, P(denominator)* is never a subset of P(numerator); and thus, odd/even can never be an integer, in disguise 😌. *P = Prime factors.
Undergraduate mathematician here. The better I get at math, the more I appreciate your videos. These videos give a great visual experience which is generally not taught in proof courses. My favorite chapter was probably Chapter 5, reminded me of some of the concepts discussed in my analysis course.
Ramanujan summation is powerful but it is also very picky. For example if I add 1+2+3+4+... and 0+1+2+3+4+... under Ramanujan summation they would have completely different values! Even if I add the numbers in a different order they would have different values. A small price to pay given the ability to sum divergent series.
@@456MrPeople It's not so strange that order of summation changes the sum... It might happen even for convergent series. :-) Well... Actually it *does* happen for convergent series, except the absolutely convergent ones. :-)
My wife viewed this lecture,I made her,and just called you the biggest nerd on the planet.But that is good for she has been calling me the biggest one for 37 years I gladly pass the title over to you.I thoroughly enjoyed it and love your enthusiasm.I'm self-studying figurate numbers and would enjoy any lectures on this subject matter.Thank you
As an adult who barely survived "New Math" back in the 60s, I grew to *hate* math with a purple passion, though I loved it with an equal passion. I gave up, finally, in high school at algebra 1, with the only "C" I've ever received in all my school years. I guess they were trying to tell me that math is not my shtick. Today, that hatred has melted away and my love and curiosity shine again. I never miss any of your videos. I love your humor and your enthusiasm!! The most memorable part? The searching for and recognition of patterns. That is so delicious!
In a similar vein, a mathematician advertises a lottery in which the prize is an infinite amount of money. Lots of people pay for tickets, but when the winning ticket is announced, the mathematician explains the mode of payment: "$1 this week, $1/2 next week, $1/3 the week after, ..."
@@johnchessant3012 Given the common assumptions about compound interest and the time value of money, this prize can actually be funded with a finite amount of money. (The assumption is that $1 today is worth $e^(rt) at time t for some constant r.) A neat puzzle is to figure out how many "now dollars" that prize is worth. (Hint 1: If the prize was $1 every week, and assuming a realistically small interest rate like 1/52% per week, you would only need about $5200.50 to fund the prize.) (Hint 2: I don't know how to solve this by hand. I cheated and used WolframAlpha.) (Hint 3: It's surprisingly small! Less than $10.)
The most memorable part for me might be the idea of that gamma value: especially the super quick visual proof that it had to be less than one by sliding over all the blue regions to the left
Yeah that was really mind-blowing. Also, to answer Mathologer's question, Gamma is more than half because each time we slide over the blue part, there is a corresponding white part, but the blue part has a "belly", or it bulges into the white part, so they're not equally divided triangles. There are infinitely many blue-half/white-half pairs, and in each the blue part has a "belly" so adding the area of all the blue "halfs" should yield a sum slightly more than half. This is just a visual approximation though, I don't know how to prove how much more than half it is.
Your teaching style is just so good! I think it's a combination of the interesting topics, your smooth as heck animations, giggles, and the quick glances you give at the end of each chapter to summarize (it's especially nice for note-taking!). Not even to mention the fact that you don't give direct answers to questions you bring up, but instead direct the viewer to introductory terms and topics to look up and gain knowledge themselves. I wish I could attend one of your lectures, but until then this will have to do!
Most memorable part: derivation of γ. As in high school we learn about the approximation of the area under the 1/x curve but not many actually focus on the 'negligible part of the area' which in fact adds up to something trivial to the whole field of number series. 24:23 Sinple proof for γ>0.5: All the tiny little bits of that blue areas are a curved shape. By connecting the two ends of that curve line we can see each part is made up of a triangle and a curved shape. The total area of those infinitely many triangles equals to 0.5 so the total area of the blue sharpest be greater than 0.5.
@@youssefm1 it’s basically because the largest non-integer in the series is 1/2 and every subsequent one is half again, so the first few get you very close .5 and every one after that is less and less and therefor as the sum gets closer to infinity the area above the curve gets closer to .5 but never over. This is mainly because there are an even more infinite set of fractions between 1/2 and 1/∞ than integers between 1 and ∞
@@Ohhelmno , thanks. My son made me realise that the sum of the vertical lines (heights) of all the triangles = 1 so the areas of the triangles (being half the area of the rectangle of that height) = 0.5 and since the blue part was larger than the triangle, its area > 0.5.
The most memorable was the optimal towers, as I always thought that the leaning tower of lire was the best way to stack overhangs. It looked so perfect that I never questioned if there was a better way to do it!
Most memorable part: In university Mathologer apparently came up with an original finiteness proof for Kempner's series, and the grader failed the homework because they couldn't be bothered to check a solution that was different from the one on the answer sheet.
I had a similar experience in topology in undergrad. I did an unconventional proof and even my professor didn't understand it but he found another professor who said it was correct.
24:30 It's "obvious" because 1/x is concave, meaning between any two points the graph is below the secant line connecting those two points. Dividing the 1x1 square into rectangles in the obvious way, the blue areas include more than half of each rectangle and hence more than half of the 1x1 square.
The secant lines partition blue triangles as a lower bound for gamma, triangles add up as a telescoping sum 1/2*((1/1-1/2)+(1/2-1/3)+...-1/n) = 1/2*(1-1/n) = 1/2 in the limit
Most Memorable : Every seconds of this video. I couldn't choose a single thing. I am sure that this is the best video I have ever watched in my life related to anything. Thank you so much Mathologer.
Wonderful video as always. The more videos I watch the more I'm convinced that Euler must've been a time-travelling Mathologer viewer who really wanted to look smart by appearing in every video
24:35 You can construct a right triangles out of the corners of each blue region. The base of each is 1 unit while the height is 1/n - 1/(n+1). The sum of the areas of these triangles yields a lower bound for γ. We can see that this area is (1/2)*(1 - 1/2) + (1/2)*(1/2 - 1/3) + (1/2)*(1/3 - 1/4) + ... which is a telescoping series so we can cancel everything except 1/2*1, so 1/2 is a lower bound for γ
That's it. Of course, you can also just skip the algebra :) Having said that it's nice in itself that all this corresponds to a telescoping sum when you turn it into algebra.
24:40 the sum of all the triangles that lower-approximate the blue areas is: 1/2*(1*(1 - 1/2) + 1*(1/2 - 1/3 )+ 1*(1/3 -... = = 1/2*(1 - 1/2 + 1/2 - 1/3 + 1/3 -... = 1/2*(1)= 1/2
@@vik24oct1991 yes, that's why he said lower approximate. The left over portions are convex. So thinking of them as triangles, there's some area left. That's why the total area is greater than 1/2.
@@sadkritx6200 That was my point , you don't need to calculate the sum of the area , if you prove that in each part the curves are convex then that implies that at the sum of the leftover is greater than half, no matter how the parts are divided.
The most memorable proof is the original proof of the harmonic series' divergence simply for the fact that this probably the only proof I could present to my year 10 math class and most of them would understand it.
The fact that the number of fractions summing to one in that doubled every time hinted at the logarithmic relationship, although I was thinking log base 2.
I liked A LOT that the sum of the “exactly 100 zeros series” is greater than the “no 9s series”! It is almost unbelievable. I need to check the paper. 🤣
First thought: Isn't the "exactly 100 zeros" series a subseries of the "no 9s" series? Second thought: No, what about the term 1/9e100? Third thought: So ... when you are dealing with *all* integers you thin out more by banning 9s than by *only* requiring 100 zeros. Mind blown. Fourth thought: Take a random billion-digit number. It will almost always have more than 100 zeros (you expect about 10 million of them, just 99 is very rare). So it's almost not a constraint at all. Although apparently enough to force convergence. On the other hand, almost no billion-digit numbers will have no nines (the probably is something like (9/10)^1e9 ~= 0 of randomly grabbing one). And almost all integers are bigger than just a billion digits. Mind now thoroughly blown.
In my opinion Kempner's proof at the end was the most fun since it's one of those proofs where one (or at least I) would have no idea where to even start, but once you see it, one suddenly realizes how obvious it is ;). Thanks for the very interesting lecture :)
@@MasterHigure I don't think so. For example, consider the sequence x_0 = 9/4 and for all n > 0: x_n = 1+(1/3)^n; form a series by summing these terms. The terms are ever decreasing, the series is divergent, and never hits any integers. Yet the partial sums never come closer than 1/4 to any integer, which it hits at the very first element a_0 alone.
24:30 (γ>1/2 :) It is equal to proving that the blue region is strictly greater than white region in that 1square unit box... since 1/x is concave up in (0,oo).. (Means a line formed by joining any two points on the curve (chord) will lie above the curve in that region) In those each small rectangles inside the 1unit box , the curve of each blue region (which is part of 1/x graph) will lie below the chord (here diagonol of that rectangle) As blue area crosses diagonals of each of these small rectangles (whose area is actually 1/(n) -1/(n+1) ) , it is greater than half the area of these rectangle... And adding up all thsoe rectangle gives area 1...and adding up all these small blue region is our "γ" So it is greater than half the area of 1. ie: γ>1/2. -----------------------------------------------
The variations on the Harmonic series were definitely my favourite - who even thought to ask such a strange question as "What's the Harmonic Series, but if you remove all the terms with a nine in them?" It would never have occurred to me to ask a question like that.
Thank you very much for another excellent demonstration of the amazing beauty of mathematics! I love the 700 years old divergence proof. Also, the unbelievably slow pace of divergence is absolutely amazing.
I have to say you are one of my favorite RUclipsrs! And that is saying something.... Most youtubers shy away from the math, but not you. Your visual proofs are brilliant and will span through the ages, I thank you because I have genuinely been looking for this content for years. Out of the bottom of my heart thank you, I needed this...
The most memorable for me has to be Kempner's proof, just due to how counterintuitive it is after seeing so many divergent series, but how intuitive the proof is.
Why is that optimal stack optimal? Those 3 bricks on the top right look like you could extend them more to the left and thereby push the whole center of gravity to the left and thereby the tower to the right
@@maze7474 moving a few bricks would necessarily shift the entire stack's center of gravity by a smaller distance. Since the blocks you're proposing to shift include the rightmost one, you'd lose more overhang than you'd gain.
@@maze7474 those 3 blocks have centre of mass at the middle of them which is at the edge of the block on which they are sitting, so if you try to shift left they will fall.
My favorite part was when you revealed that the sum of the 100 zeroes series is greater than the sum of the no 9 series. Absolutely mind-blowing. In truth, my favorite part was the entire video you just made me pick :) Thank you!!!
When you get to that age when you want your students to tell you a bedtime story of the old days via math proofs. It would be gracious of us to do so just like when we were little kids asking mommy for a bed time story. Hmm are tests care work? 👋🕊️
Peter Ustinov relates that his teacher gave him zero marks when he answered "Rimsky-Korsakov" to the question "Name one Russian composer." The correct answer was Tchaikovsky.
Most memorable: being invited to take a moment and post why it might be obvious that gamma is greater than 0.5 and then doing it. Hmm... why is it obvious that gamma is greater than 0.5? Well it didn't seem obvious... But imagine the blue bits were triangular; then there would be equal parts blue and white in the unit square on the left i.e. a gamma of 0.5. But the blue parts are convex, they each take up more than half of their rectangles and together take up more than half of the square.
Marble: Most memorable Idea: No integers among the partial sums. Your excitement is always a great feature not your presentations. And why you're my favorite Mathematics RUclipsr!
Most memorable: every positive number having its own infinite sum. It’s very obvious afterwards, but I would never believe it without your explanation. Thank you for all the interesting videos!
@@takyc7883 I remember that there is no such thing as a slowest diverging series. for example 1/n diverges, 1/(n*ln(n)) diverges, 1/(n*ln(n)*ln(ln(n))) diverges, and so on. As always, there's a math stack exchange thread talking about this topic: math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series
@@takyc7883 No. Using the ideas from chapter 6, you can actually show there is a subseries of the harmonic series that diverges as slowly as you would like. Simply take a function f(x) that diverges at a rate slower than the natural logarithm. At each integer, we will choose an entry from the harmonic series which is smaller than the one we had chosen previously. First, choose the largest entry of the harmonic series that is smaller than f(1). If this is not possible, choose the smallest entry of the harmonic series with is larger. Next, choose the largest entry possible so that the partial sum so far (just the first term and this one) is less than f(2). Again, if this is not possible, choose the smallest entry possible so that the partial sum is larger than f(2). Continue in this way and you will make a series whose rate of divergence is the same as f(x).
The most memorable is your voice... the giggle you make when telling us wonderful facts. Have a wonderful life. Stay safe... ✌️
4 года назад+3
The most impressive part in my opinion was the fact that the 100 zeros sequence converges to a bigger sum than the no 9s sequence. Greetings from Germany by the way and keep up that great work. It is always a pleasure diving into your mathematical discoveries!
If extended hugely, I guess that patterns of absent blocks will create pleasing curves.. I recall doing this with kids bored of 'Jehinga'. old uk duffer here :)
i thought it was beautiful also. Glad I'm not alone. Modern concepts of "beauty" are overfocused on symmetry. Observe more natural structures to appreciate the beauty in the "misshapen" and the perfection in the "imperfect"
Love your work, thank you for sharing so much knowledge & quality content. Generally RUclipsrs run for views, nothing more than that. You never click baits or anything that ever had made me regret clicking. You are just awesome.
This was probably one of the best video that I have seen on this topic. And the geometric limit of Gamma between 1/2 and 1 and the no-nine series proof was something I learnt for the first time.
Most memorable: the optimized leaning tower! Although it was very messy, I think there's a lot of beauty in the fact that the most optimal arrangement of bricks is such a mess. It reminds me of how an extremely simple physical system (like a double pendulum) can result in chaos!
Good point. The problem with I have that solution is perhaps that the original concept stipulated or assumed that the tower would lean only in one direction (which is what towers do). The recent innovative solution doesn't involve a "tower" at all. This is the problem with the "thinking outside the box" cliche. I tell you to connect the dots within a box using 2 lines and then you connect them by drawing lines outside the box. Wow, that's impressive. It's like telling your boss you can solve a problem by spending 10X the given budget.
My vote definetely goes to Kempner's proof, it is extremely elegant, since the concepts used are individualy simple, such as the calculation of numbers without nine or geometric series, but when cleverly combined they form this amazing result. Besides that, great video as always. Edit: typo
The most memorable part for me was seeing the 700 year old proof.I I didn't pay much attention to math in school, despite the fact that I learned it easily. Math class was always boring to me because once I learned and understood an how an equation worked, I had no interest in running the same equation again with different numbers dozens of times for homework. My teachers never made the subject as interesting and engaging as the many resources I have found since graduation. But now that I have finally found an interest in the topic, I realize that I am woefully undereducated on anything past basic high school algebra. This is the reason why I liked seeing the simple proof. It was so easy that I could have done it myself with my current skills! That gives me hope that, while I may have MUCH to learn, there are still intricate problems that even I can grasp. Thank you for another great video, and have a nice day!
That’s my favourite / most memorable part, too; because I managed to prove it, for myself, and find the pattern, for the nth partial sum: Σ(n) = (a(n-1)*n+(n-1)!)/n!, where a(n-1) = the numerator of the (n-1)th partial sum. 🙂
Definitely the most impressive part is the animated Kempner's proof, I've expected something extremely complicated and yet the whole thing was "nice and smooth".
Dear Mathologer, I am so pleased whenever I run across one of your videos. As for my vote for the portion that impressed me the most, it would have to be the leaning tower of Lire. There is something so lovely in its orderliness, that I sense my head bowing, much like the old Frenchman, Oresme. Thank you for another interesting and entertaining video on the beauties of math.
I wish I knew more. But the more I watch and try to learn, the more time I've used up getting nowhere. The dedication the geniuses have to mathematics and physics is astounding. And it is not done for reward other than the pursuit of knowledge. And that is as beautiful as the proofs the geniuses present. Thank you for the videos.
A bit late on the lower bound for gamma, but... You can take every blue piece, place it into a rectangle of dimensions 1 x 1/(2^n), and split that rectangle in half with a diagonal from the top left to the bottom right. If you were to take the upper triangle from every one of these divided rectangles, you would get an area of one half of the square. Because every piece has a convex curve, it will stick slightly outside of the upper half of its rectangle. This means that every piece has an area greater than half of the rectangle, and the sum of all the pieces is greater than one half of the square. Because the square is 1x1, the area of the blue pieces is greater than 1/2.
@@Mathologer - Is it true to say that in the "No 'n's" series where we intuit that the sum converges, the sum of all the removed terms containing 'n' must itself be infinite? You have a series summing to infinity minus another series. If the thing you subtract is itself finite then you would still have an infinite series left over, ergo the subtracted series must itself sum to infinity for the remaining series to converge. Not sure if that's simply obvious or if it's also an "AHA" moment.
11:50 Arrange the bricks in a two layer formation where, on the bottom layer, you place a single block with its center of mass on the cliff's edge (akin to a single-block maximum overhang position). Then, on the second layer, place both remaining blocks with their centers of mass aligned with each of the bottom block's edges. If done correctly, the block placed over the cliff should create precisely a 2 unit overhang (assuming all blocks are 2 units long), with the other brick on the same layer acting as a counterweight. We would need to assume all blocks weight exactly the same, have perfectly equal shapes and there are no external forces aside from gravity acting on the system. Below I'll try to make a small ASCII schematic to illustrate the formation ________ ________ ________ -------------| Cliff |
solution to there bricks with overhang of 2 units: place one brick with overhang of 1. then place another bring on top of this one all the way to the right with its own overhang of 1 unit. clearly this will fall. now place the third brick to the left of the second, making the top layer 4 units (2 bricks) long, and the bottom layer centered around it. Done.
Mathloger THE VERY BEST MATH AND LOGIC TEACHER ON THE PLANET. I missed him already when I was a school boy without knowing him. I enjoy every clip and its so well presented and easy to understand. He*'s a NATURAL AND A PRO... go on bro
Most Memorable: The fact that it is possible to *arrange* the bricks on the table such that the last brick can be as far as the size of *observable universe* from the table, and yet be perfectly balanced!!!🤯🤯🤯
I have to vote for the Kempner's proof animation, it was simply stunning to see such a seemingly complex problem; being broken down into techniques that a school student could understand 👏
11:55, simple: brick one on the edge, bricks two and three on either edge of brick one. 14:48 LOL, I love the editing style! This plus the proof is enough to make chapter 3 my favorite. 24:44 all the curves are "bulging", which is to say that they have a positive second derivative, so they take up more area than a series of triangles with the same start and end points
I was already... not comfortable, but let's say "resigned"... to the fact that there exist very slowly divergent series; but the fact that there are very slowly CONVERGENT series, whose sum is impossible to approximate computationally within a reasonable margin of error, like the no nines series... was a shock!
For the challenge at 20:24: I found a certain sequence of reciprocals (don't get too excited, I only have finitely many terms and don't even know how to extend it without using trial and error) that 1) has ever-decreasing terms (like the true harmonic series), and 2) when you calculate its partial sums it hits every integer from 1 to 3. It'll be under the "read more" button so I don't spoil it! 1 + 1/2 + 1/3 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + 1/16 + 1/18 + 1/20 + 1/24 + 1/25 + 1/30 + 1/36 + 1/48 + 1/60 + 1/100 + 1/140 + 1/180 is my sequence (the sequence adds to 3, first four terms add to 2, and first term is 1). I'm quite worried about how I can extend this to the integer 4, since my largest possible term available is 1/181 and my largest term with many divisors is probably 1/189 or 1/192 - which means it'll take a hard lower limit of 200ish additional terms to write it and almost certainly it will also take many more.
Very nice idea to package it all into one series. When people play with this they usually use completely different sums to make the integers. When you do that it's fairly easy to show that you can get all positive integers, in fact you can make all positive rational numbers. Have to ponder what's possible with your approach :)
I really enjoyed it. I guessed no 9s has finite sum by thinking about binary numbers with no 0s. that's 1/1 + 1/11 + 1/111 + 1/1111 + .... < 1 + 1/10 + 1/100 + 1/1000 + ... = 10 (=2). I most enjoyed the Leaning Tower of Liire, because they didn't teach it in my calculus course and it's simple and beautiful. Thank you Bar
Proof that γ>1/2 (24:30): Replace all the blue parts with right triangles inside of them and the resulting sum is 1/2 (visually this is obvious, writing out the sum isn’t too bad either).
It is really interesting how eulars comes into all these different formulas. Personal favourite was the very neat proof at the end that the sum was less than 80
Me too. The untidy structure called to mind Sabine Hossenfelder’s “Lost in Math: How Beauty Leads Physics Astray”. Physicists are addicted to symmetries. Dirac thought beauty was the most important (and convincing) feature of any formula in mathematical physics.
yeah, i just went for the odds and said: well it is finite, because maybe something will happen, that i cannot see now, so in contradiction i am still alive :) But... just to ask for clarity: if that grid exists with 9s, wouldt it exist with all other numbers >0 ? what about the 0 ?
@@MrTiti The grid is the "same" with every digit, including 0. In the first 10 numbers, 1/10 contain a (insert digit) or 10% In the first 100 numbers, 1/10 for every "ten" + 9/100 (the ones starting with your digit, like 31 32 33...), or 19% containing your digit. In the first 1000 numbers we have 19 every 100, + the ones with the first digit (300, 301, 302...), or 271/1000 or 27.1% total. And so on... Every time the % of numbers containing the chosen digit keeps increasing, reaching almost 100%. Works for all 10 digits (0, 1, 2... 9).
Euler mascheroni constant being greater than 0.5 can be seen in the figure at 25:13 by noticing that the blue convex triangles contain straight triangles whose area sum to 1/2. Great video Mathologer.😉
It's odd that while the terms get smaller and smaller, the sum doesn't converge. You look at every book in the tower being less and less over the edge and think "there must be a point where the top book moves so little that it's impossible to notice no matter how many books you pile, and the new books barely go over the edge for even microns" and at the same time think that it will always go further as far as you want it to go.
Thanks for another amazing, informative, extremely clear and well made presentation. The fact that the series explodes to infinity is one thing, how slow it happens and how many terms are needed for just a tiny increase makes my head spin.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
The most memorable part was when you giggle, and my wife in the other room says "You're watching that math guy again?" As always, thank you for expanding my knowledge base.
😂😂😂
When she hears all of the profanity, she knows you're watching Flammable Maths!
Same over here :)
His giggling always sounds like Dr. Strangelove to me. Man, Peter Sellers was a great actor.
@@teleny2 Gen. Turgidson: "'Strangelove'? That ain't no kraut name."
Aide: "His original name was 'Merkwürdigliebe'. He changed when he became a citizen."
Gen. Turgidson: "Huh. Strange."
The most suprising part for me was the "terrible aim" the fact that odd/even is never an integer is so simple yet i would have never thought about it
Yes. Simple, indeed: In order for a fraction to be an integer, the prime factors of the denominator must form a subset of the prime factors of the numerator; but, in odd/even, the denominator always has the prime factor: 2, which the numerator never does, in odd/even; thus, P(denominator)* is never a subset of P(numerator); and thus, odd/even can never be an integer, in disguise 😌.
*P = Prime factors.
Undergraduate mathematician here. The better I get at math, the more I appreciate your videos. These videos give a great visual experience which is generally not taught in proof courses.
My favorite chapter was probably Chapter 5, reminded me of some of the concepts discussed in my analysis course.
Any divergent series: *exists*
Ramanujan: Allow me to make it convergent.
Ramanujan summation is powerful but it is also very picky. For example if I add 1+2+3+4+... and 0+1+2+3+4+... under Ramanujan summation they would have completely different values! Even if I add the numbers in a different order they would have different values. A small price to pay given the ability to sum divergent series.
@@456MrPeople It's not so strange that order of summation changes the sum... It might happen even for convergent series. :-)
Well... Actually it *does* happen for convergent series, except the absolutely convergent ones. :-)
@@456MrPeople that's the normal problem of infinity
...to something that is not even close to where the series goes.
It would like to see how TREE(1)+TREE(2)+TREE(3)+... can be made convergent. 😉
My wife viewed this lecture,I made her,and just called you the biggest nerd on the planet.But that is good for she has been calling me the biggest one for 37 years I gladly pass the title over to you.I thoroughly enjoyed it and love your enthusiasm.I'm self-studying figurate numbers and would enjoy any lectures on this subject matter.Thank you
Most memorable part: me losing my life after failing the “no nines sum converges”
sure.. x2
I didn't lose my life at that part! I gamed the system, by already losing it way earlier on in the video! lmao
@@Torthrodhel t
I lost my life too!
If you had collected 1000 Coins (or more), in Japan; then, it’s no problem. 🙂
As an adult who barely survived "New Math" back in the 60s, I grew to *hate* math with a purple passion, though I loved it with an equal passion. I gave up, finally, in high school at algebra 1, with the only "C" I've ever received in all my school years. I guess they were trying to tell me that math is not my shtick. Today, that hatred has melted away and my love and curiosity shine again. I never miss any of your videos. I love your humor and your enthusiasm!! The most memorable part? The searching for and recognition of patterns. That is so delicious!
That's great, your comment made my day :)
"Are we there yet?"
"No just 1+1/2+1/3+1/4+... more minutes."
:)
In a similar vein, a mathematician advertises a lottery in which the prize is an infinite amount of money. Lots of people pay for tickets, but when the winning ticket is announced, the mathematician explains the mode of payment: "$1 this week, $1/2 next week, $1/3 the week after, ..."
@@johnchessant3012 Given the common assumptions about compound interest and the time value of money, this prize can actually be funded with a finite amount of money. (The assumption is that $1 today is worth $e^(rt) at time t for some constant r.)
A neat puzzle is to figure out how many "now dollars" that prize is worth.
(Hint 1: If the prize was $1 every week, and assuming a realistically small interest rate like 1/52% per week, you would only need about $5200.50 to fund the prize.)
(Hint 2: I don't know how to solve this by hand. I cheated and used WolframAlpha.)
(Hint 3: It's surprisingly small! Less than $10.)
i just realized that that means that they are pretty close to their destenation, about half a minute away
@@MGSchmahl In today's world, 1% is actually a high interest in USD: the Fed Funds effective rate is only 0.08% (annually).
Mathologer video series are definitely better than any Netflix series. They surprise me anytime.
With a small amount of effort one could probably get Mathologer onto Netflix. It's just filling in forms and checking video quality and whatnot.
Netflix? No comparison. Mathologer wins every time, and it's free.
Mathflix. The best series (Taylor, MacLaurin, armonic, ...)
(Seen in his t-shirt)
Exactly true 👍
The most memorable part for me might be the idea of that gamma value: especially the super quick visual proof that it had to be less than one by sliding over all the blue regions to the left
Yeah that was really mind-blowing. Also, to answer Mathologer's question, Gamma is more than half because each time we slide over the blue part, there is a corresponding white part, but the blue part has a "belly", or it bulges into the white part, so they're not equally divided triangles. There are infinitely many blue-half/white-half pairs, and in each the blue part has a "belly" so adding the area of all the blue "halfs" should yield a sum slightly more than half. This is just a visual approximation though, I don't know how to prove how much more than half it is.
@@chessnotchekrs Yeah, that was a fun one to just suddenly get (though, like he said, it was “obvious”).
Clearly, the highlight of the Euler-Mascheroni constant is a splendid part of the video...the sum of no 9's animation is very impressive.
Your teaching style is just so good! I think it's a combination of the interesting topics, your smooth as heck animations, giggles, and the quick glances you give at the end of each chapter to summarize (it's especially nice for note-taking!). Not even to mention the fact that you don't give direct answers to questions you bring up, but instead direct the viewer to introductory terms and topics to look up and gain knowledge themselves. I wish I could attend one of your lectures, but until then this will have to do!
Most memorable part: derivation of γ. As in high school we learn about the approximation of the area under the 1/x curve but not many actually focus on the 'negligible part of the area' which in fact adds up to something trivial to the whole field of number series.
24:23 Sinple proof for γ>0.5:
All the tiny little bits of that blue areas are a curved shape. By connecting the two ends of that curve line we can see each part is made up of a triangle and a curved shape. The total area of those infinitely many triangles equals to 0.5 so the total area of the blue sharpest be greater than 0.5.
That's it:)
Why is the total area of those triangles 0.5?
@@youssefm1 it’s basically because the largest non-integer in the series is 1/2 and every subsequent one is half again, so the first few get you very close .5 and every one after that is less and less and therefor as the sum gets closer to infinity the area above the curve gets closer to .5 but never over. This is mainly because there are an even more infinite set of fractions between 1/2 and 1/∞ than integers between 1 and ∞
@@Ohhelmno , thanks. My son made me realise that the sum of the vertical lines (heights) of all the triangles = 1 so the areas of the triangles (being half the area of the rectangle of that height) = 0.5 and since the blue part was larger than the triangle, its area > 0.5.
Don
The most memorable was the optimal towers, as I always thought that the leaning tower of lire was the best way to stack overhangs. It looked so perfect that I never questioned if there was a better way to do it!
Most memorable part: In university Mathologer apparently came up with an original finiteness proof for Kempner's series, and the grader failed the homework because they couldn't be bothered to check a solution that was different from the one on the answer sheet.
We've all had those graders.
[i
I had a similar experience in topology in undergrad. I did an unconventional proof and even my professor didn't understand it but he found another professor who said it was correct.
How often do you imagine an answer different from the answer sheet is actually correct?
@@jetzeschaafsma1211 In math, more often than you think.
24:30 It's "obvious" because 1/x is concave, meaning between any two points the graph is below the secant line connecting those two points. Dividing the 1x1 square into rectangles in the obvious way, the blue areas include more than half of each rectangle and hence more than half of the 1x1 square.
Exactly :)
Finally something I had seen myself with my very low level of maths
That makes sense! Good explanation, I got it without any visuals! Haha.
You mean convex. You triggered one of my pet peeves.
The secant lines partition blue triangles as a lower bound for gamma, triangles add up as a telescoping sum 1/2*((1/1-1/2)+(1/2-1/3)+...-1/n) = 1/2*(1-1/n) = 1/2 in the limit
For me, the "no integers" part was the most memorable, but honestly the whole video was of great quality (as expected).
Most Memorable : Every seconds of this video. I couldn't choose a single thing. I am sure that this is the best video I have ever watched in my life related to anything. Thank you so much Mathologer.
Wonderful video as always. The more videos I watch the more I'm convinced that Euler must've been a time-travelling Mathologer viewer who really wanted to look smart by appearing in every video
24:35 You can construct a right triangles out of the corners of each blue region. The base of each is 1 unit while the height is 1/n - 1/(n+1). The sum of the areas of these triangles yields a lower bound for γ. We can see that this area is (1/2)*(1 - 1/2) + (1/2)*(1/2 - 1/3) + (1/2)*(1/3 - 1/4) + ... which is a telescoping series so we can cancel everything except 1/2*1, so 1/2 is a lower bound for γ
That's it. Of course, you can also just skip the algebra :) Having said that it's nice in itself that all this corresponds to a telescoping sum when you turn it into algebra.
24:40 the sum of all the triangles that lower-approximate the blue areas is:
1/2*(1*(1 - 1/2) + 1*(1/2 - 1/3 )+ 1*(1/3 -... =
= 1/2*(1 - 1/2 + 1/2 - 1/3 + 1/3 -... = 1/2*(1)= 1/2
you have to also prove that the left over portion is greater than area of triangle.
@@vik24oct1991 yes, that's why he said lower approximate. The left over portions are convex. So thinking of them as triangles, there's some area left. That's why the total area is greater than 1/2.
@@sadkritx6200 That was my point , you don't need to calculate the sum of the area , if you prove that in each part the curves are convex then that implies that at the sum of the leftover is greater than half, no matter how the parts are divided.
The most memorable proof is the original proof of the harmonic series' divergence simply for the fact that this probably the only proof I could present to my year 10 math class and most of them would understand it.
Would be interesting what your kids would make of the animation of this proof :)
@@Mathologer Maybe I'll use it in my "Mathe AG". :)
The fact that the number of fractions summing to one in that doubled every time hinted at the logarithmic relationship, although I was thinking log base 2.
I liked A LOT that the sum of the “exactly 100 zeros series” is greater than the “no 9s series”! It is almost unbelievable. I need to check the paper. 🤣
It is intuitively plausible as the 100 zeros series includes "a lot" of terms with 9s that the no 9s series leaves out.
@@imacds I can't say that's intuitive to me 😂
First thought: Isn't the "exactly 100 zeros" series a subseries of the "no 9s" series?
Second thought: No, what about the term 1/9e100?
Third thought: So ... when you are dealing with *all* integers you thin out more by banning 9s than by *only* requiring 100 zeros. Mind blown.
Fourth thought: Take a random billion-digit number. It will almost always have more than 100 zeros (you expect about 10 million of them, just 99 is very rare). So it's almost not a constraint at all. Although apparently enough to force convergence. On the other hand, almost no billion-digit numbers will have no nines (the probably is something like (9/10)^1e9 ~= 0 of randomly grabbing one). And almost all integers are bigger than just a billion digits. Mind now thoroughly blown.
how was the paper?
This blew my mind!
In my opinion Kempner's proof at the end was the most fun since it's one of those proofs where one (or at least I) would have no idea where to even start, but once you see it, one suddenly realizes how obvious it is ;). Thanks for the very interesting lecture :)
The most memorable part was Tristan's fractal. Fractals are beautiful and they always show up when you expect them the least.
Most memorable: that the harmonic series narrowly misses all integers by ever shrinking margins
I agree that an infinite number of non intergers is quite amazing.
I mean, any diverging series with ever smaller terms will have ever shrinking margins (as long as it doesn't actually hit any integers).
@@MasterHigure I don't think so. For example, consider the sequence x_0 = 9/4 and for all n > 0: x_n = 1+(1/3)^n; form a series by summing these terms. The terms are ever decreasing, the series is divergent, and never hits any integers. Yet the partial sums never come closer than 1/4 to any integer, which it hits at the very first element a_0 alone.
@@landsgevaer You're right. The terms need to converge to 0. I done goofed.
It also managed to miss infinitely more and infinitely denser all irrational numbers as well. THAT seems even more impressive!
24:30 (γ>1/2 :)
It is equal to proving that the blue region is strictly greater than white region in that 1square unit box...
since 1/x is concave up in (0,oo)..
(Means a line formed by joining any two points on the curve (chord) will lie above the curve in that region)
In those each small rectangles inside the 1unit box , the curve of each blue region (which is part of 1/x graph) will lie below the chord (here diagonol of that rectangle)
As blue area crosses diagonals of each of these small rectangles (whose area is actually 1/(n) -1/(n+1) ) , it is greater than half the area of these rectangle...
And adding up all thsoe rectangle gives area 1...and adding up all these small blue region is our "γ"
So it is greater than half the area of 1.
ie: γ>1/2.
-----------------------------------------------
Thanks
basically their area is greater than their triangle counterparts and the triangle area is ½
The variations on the Harmonic series were definitely my favourite - who even thought to ask such a strange question as "What's the Harmonic Series, but if you remove all the terms with a nine in them?" It would never have occurred to me to ask a question like that.
It's like the "Bee movie, but without bees" type of memes, I guess it's just the human nature
Thank you very much for another excellent demonstration of the amazing beauty of mathematics! I love the 700 years old divergence proof. Also, the unbelievably slow pace of divergence is absolutely amazing.
Most memorable part: all
I’m just constantly being mind blown throughout the whole video
Same bro. The whole video was magnificent 👌
Most memorable: If my life depended on knowing if the sum of no nines series is finite I would not be alive
I have to say you are one of my favorite RUclipsrs! And that is saying something.... Most youtubers shy away from the math, but not you. Your visual proofs are brilliant and will span through the ages, I thank you because I have genuinely been looking for this content for years. Out of the bottom of my heart thank you, I needed this...
The most memorable for me has to be Kempner's proof, just due to how counterintuitive it is after seeing so many divergent series, but how intuitive the proof is.
The weirdest thing you showed is definitely the unusual optimal brick stacking pattern.
Why is that optimal stack optimal? Those 3 bricks on the top right look like you could extend them more to the left and thereby push the whole center of gravity to the left and thereby the tower to the right
@@maze7474 moving a few bricks would necessarily shift the entire stack's center of gravity by a smaller distance. Since the blocks you're proposing to shift include the rightmost one, you'd lose more overhang than you'd gain.
@@ramenandvitamins sorry, typo from my side.I meant top left, those 3 that are stacked exactly over each other
@@maze7474 I suspect they'd no longer suffice to hold down the second-rightmost block if they were moved any further left.
@@maze7474 those 3 blocks have centre of mass at the middle of them which is at the edge of the block on which they are sitting, so if you try to shift left they will fall.
My favorite part was when you revealed that the sum of the 100 zeroes series is greater than the sum of the no 9 series. Absolutely mind-blowing. In truth, my favorite part was the entire video you just made me pick :) Thank you!!!
I liked your evil mathematician back story, with the teacher refusing to grade the "wrong" proof.
I didn't like it. Second hand annoyance. grrr
When you get to that age when you want your students to tell you a bedtime story of the old days via math proofs. It would be gracious of us to do so just like when we were little kids asking mommy for a bed time story.
Hmm are tests care work?
👋🕊️
I relate to that experience.
Peter Ustinov relates that his teacher gave him zero marks when he answered "Rimsky-Korsakov" to the question "Name one Russian composer." The correct answer was Tchaikovsky.
Most memorable moment was the cat going "μ".
Has a cat the hacker-nature? "Mew...."
As a cat-purrson, I approve 😻😌👍🏻.
So much great content packed into 45 minutes! Something I’ll always remember will be that the no nines series converges, and how simple the proof was!
Wow, ich hätte nie geglaubt, dass etwas das mit Analysis zu tun hat auch Spaß machen kann... :) Sehr cool!
Most Memorable: getting the Mathologer seal of approval
Most memorable: being invited to take a moment and post why it might be obvious that gamma is greater than 0.5 and then doing it.
Hmm... why is it obvious that gamma is greater than 0.5? Well it didn't seem obvious...
But imagine the blue bits were triangular; then there would be equal parts blue and white in the unit square on the left i.e. a gamma of 0.5.
But the blue parts are convex, they each take up more than half of their rectangles and together take up more than half of the square.
Exactly how I pictured it!
It also makes it obvious that γ is much closer to ½ than it is to 1.
Fred
My thoughts, exactly 🎯! Articulated better, than I could have put it 😌👍🏻.
Marble: Most memorable Idea: No integers among the partial sums. Your excitement is always a great feature not your presentations. And why you're my favorite Mathematics RUclipsr!
You’re so good at starting simple and yet including stuff that’s interesting for the fully initiated! Great work!
Most memorable: every positive number having its own infinite sum. It’s very obvious afterwards, but I would never believe it without your explanation. Thank you for all the interesting videos!
The most memorable part is the connection of the 'γ' and the log() function to the harmonic series! Really amazing!!
They should have called it the 'Barely Divergent Series'
is t the slowest diverging series?
@@takyc7883 I remember that there is no such thing as a slowest diverging series. for example 1/n diverges, 1/(n*ln(n)) diverges, 1/(n*ln(n)*ln(ln(n))) diverges, and so on.
As always, there's a math stack exchange thread talking about this topic: math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series
@@takyc7883 No. Using the ideas from chapter 6, you can actually show there is a subseries of the harmonic series that diverges as slowly as you would like. Simply take a function f(x) that diverges at a rate slower than the natural logarithm. At each integer, we will choose an entry from the harmonic series which is smaller than the one we had chosen previously. First, choose the largest entry of the harmonic series that is smaller than f(1). If this is not possible, choose the smallest entry of the harmonic series with is larger. Next, choose the largest entry possible so that the partial sum so far (just the first term and this one) is less than f(2). Again, if this is not possible, choose the smallest entry possible so that the partial sum is larger than f(2). Continue in this way and you will make a series whose rate of divergence is the same as f(x).
The most memorable is your voice... the giggle you make when telling us wonderful facts. Have a wonderful life. Stay safe... ✌️
The most impressive part in my opinion was the fact that the 100 zeros sequence converges to a bigger sum than the no 9s sequence. Greetings from Germany by the way and keep up that great work. It is always a pleasure diving into your mathematical discoveries!
The most memorable thing is how ugly the optimal leaning tower is
you mean beautiful
Some people say warthogs are ugly.
To me they are stunningly beautiful, with their faces resembling the Mandelbrot fractal.
If extended hugely, I guess that patterns of absent blocks will create pleasing curves.. I recall doing this with kids bored of 'Jehinga'. old uk duffer here :)
It was the most beautiful leaning tower.
i thought it was beautiful also. Glad I'm not alone. Modern concepts of "beauty" are overfocused on symmetry. Observe more natural structures to appreciate the beauty in the "misshapen" and the perfection in the "imperfect"
Love your work, thank you for sharing so much knowledge & quality content. Generally RUclipsrs run for views, nothing more than that. You never click baits or anything that ever had made me regret clicking. You are just awesome.
Most memorable: The most efficient overhanging structure being the weird configuration instead of an apparently more ordered one.
Most memorable: An overhanging structure with n=google bricks.
Exactly what I needed today
EDIT: My favourite part was the no 9-s proof. It is just simply elegant.
This was probably one of the best video that I have seen on this topic. And the geometric limit of Gamma between 1/2 and 1 and the no-nine series proof was something I learnt for the first time.
Most memorable: the optimized leaning tower! Although it was very messy, I think there's a lot of beauty in the fact that the most optimal arrangement of bricks is such a mess. It reminds me of how an extremely simple physical system (like a double pendulum) can result in chaos!
Good point. The problem with I have that solution is perhaps that the original concept stipulated or assumed that the tower would lean only in one direction (which is what towers do). The recent innovative solution doesn't involve a "tower" at all. This is the problem with the "thinking outside the box" cliche. I tell you to connect the dots within a box using 2 lines and then you connect them by drawing lines outside the box. Wow, that's impressive. It's like telling your boss you can solve a problem by spending 10X the given budget.
My vote definetely goes to Kempner's proof, it is extremely elegant, since the concepts used are individualy simple, such as the calculation of numbers without nine or geometric series, but when cleverly combined they form this amazing result.
Besides that, great video as always.
Edit: typo
I agree with you!
The most memorable part for me was seeing the 700 year old proof.I
I didn't pay much attention to math in school, despite the fact that I learned it easily. Math class was always boring to me because once I learned and understood an how an equation worked, I had no interest in running the same equation again with different numbers dozens of times for homework. My teachers never made the subject as interesting and engaging as the many resources I have found since graduation. But now that I have finally found an interest in the topic, I realize that I am woefully undereducated on anything past basic high school algebra.
This is the reason why I liked seeing the simple proof. It was so easy that I could have done it myself with my current skills! That gives me hope that, while I may have MUCH to learn, there are still intricate problems that even I can grasp. Thank you for another great video, and have a nice day!
My new favorite maths channel
Most memorable: The harmonic series misses all integers up to infinity
That got me too.
Yeah it must be ∞.438882647883976917983791870000364553678223... or something.
That’s my favourite / most memorable part, too; because I managed to prove it, for myself, and find the pattern, for the nth partial sum: Σ(n) = (a(n-1)*n+(n-1)!)/n!, where a(n-1) = the numerator of the (n-1)th partial sum. 🙂
Great stuff. It's always amazing how you manage to find such intuitive explanations. Most memorable is probably the "no 9s" visualization.
Most memorable part: the 100 zeros sum being larger than the no nines sum.
Definitely the most impressive part is the animated Kempner's proof, I've expected something extremely complicated and yet the whole thing was "nice and smooth".
Dear Mathologer,
I am so pleased whenever I run across one of your videos.
As for my vote for the portion that impressed me the most, it would have to be the leaning tower of Lire. There is something so lovely in its orderliness, that I sense my head bowing, much like the old Frenchman, Oresme. Thank you for another interesting and entertaining video on the beauties of math.
Math: '*exists*
Euler: "First!"
so how did math work before euler?
@@ashtonsmith1730 The same way it always did - just in a dark room. Euler just turned on the lights to a lot of rooms.
When you're investigating maths, if you inspect close, there are really small notes all over the place. "Euler has been here"!
@@ashtonsmith1730 Lots of wordy descriptions and cobbled-together notations from dozens of different people.
@@neomew you'll never kill the real Dumbledore hehe
"Is the no 9 series finite? You life depends on this!"
Me: suspicious, has to be finite!
"Believe it or not, it is finite!"
Me: YAY
PS this comment is a joke, I have heard about the no 9s series a long time ago
Survival squad for the win!
@@JM-us3fr i would reply to you but ive been executed since i got it wrong
The e^gamma equation really blew my mind. Good stuff!
Good Stuff Burkard! :)
POLSTER! :)
QWERTZ
Yay Oily Macaroni!
Omg the pufferfish meme hits differently now god bless, Papa
@@Daniel-ws9qu Qwert Zuiopü, der Gallertprinz aus der 2364. Dimension
Most memorable part: the visualization of the "no nines sum convergence"
What an awesome way to look at it.
I wish I knew more. But the more I watch and try to learn, the more time I've used up getting nowhere. The dedication the geniuses have to mathematics and physics is astounding. And it is not done for reward other than the pursuit of knowledge. And that is as beautiful as the proofs the geniuses present.
Thank you for the videos.
Most memorable: the fact that gamma is the Ramanujan summation of the harmonic series.
No way
You beat me to it.
A bit late on the lower bound for gamma, but...
You can take every blue piece, place it into a rectangle of dimensions 1 x 1/(2^n), and split that rectangle in half with a diagonal from the top left to the bottom right. If you were to take the upper triangle from every one of these divided rectangles, you would get an area of one half of the square.
Because every piece has a convex curve, it will stick slightly outside of the upper half of its rectangle. This means that every piece has an area greater than half of the rectangle, and the sum of all the pieces is greater than one half of the square. Because the square is 1x1, the area of the blue pieces is greater than 1/2.
That's it. Never too late to have a great AHA moment :)
@@Mathologer - Is it true to say that in the "No 'n's" series where we intuit that the sum converges, the sum of all the removed terms containing 'n' must itself be infinite? You have a series summing to infinity minus another series. If the thing you subtract is itself finite then you would still have an infinite series left over, ergo the subtracted series must itself sum to infinity for the remaining series to converge. Not sure if that's simply obvious or if it's also an "AHA" moment.
That is very much true 🎯👍🏻.
as for me its such a charm to see the Mathologer seal of approval, I really dont know, but it really made my day!! thanks for the great video!
11:50
Arrange the bricks in a two layer formation where, on the bottom layer, you place a single block with its center of mass on the cliff's edge (akin to a single-block maximum overhang position). Then, on the second layer, place both remaining blocks with their centers of mass aligned with each of the bottom block's edges. If done correctly, the block placed over the cliff should create precisely a 2 unit overhang (assuming all blocks are 2 units long), with the other brick on the same layer acting as a counterweight. We would need to assume all blocks weight exactly the same, have perfectly equal shapes and there are no external forces aside from gravity acting on the system. Below I'll try to make a small ASCII schematic to illustrate the formation
________ ________
________
-------------|
Cliff |
solution to there bricks with overhang of 2 units: place one brick with overhang of 1. then place another bring on top of this one all the way to the right with its own overhang of 1 unit. clearly this will fall. now place the third brick to the left of the second, making the top layer 4 units (2 bricks) long, and the bottom layer centered around it. Done.
Summary:
Cliff edge is x=0. Bricks are measured at the middle.
Layer 0: a brick at x=0
Layer 1: two bricks, at -1 and 1
Idea: Put the left upper coin before the overhanging one. It will not fall.
Mathloger THE VERY BEST MATH AND LOGIC TEACHER ON THE PLANET. I missed him already when I was a school boy without knowing him. I enjoy every clip and its so well presented and easy to understand. He*'s a NATURAL AND A PRO... go on bro
Most Memorable: The fact that it is possible to *arrange* the bricks on the table such that the last brick can be as far as the size of *observable universe* from the table, and yet be perfectly balanced!!!🤯🤯🤯
Pretty amazing, but don’t get the table from IKEA, they are bad quality.
Lol
@@channalbert they mess with the physics of balance
Imagine making a mistake and it all fall down
Wouldn't that mean that you would need a stack higher than the universe? Even if each layer was just one atom thick.
I have to vote for the Kempner's proof animation, it was simply stunning to see such a seemingly complex problem; being broken down into techniques that a school student could understand 👏
11:55, simple: brick one on the edge, bricks two and three on either edge of brick one.
14:48 LOL, I love the editing style! This plus the proof is enough to make chapter 3 my favorite.
24:44 all the curves are "bulging", which is to say that they have a positive second derivative, so they take up more area than a series of triangles with the same start and end points
I was already... not comfortable, but let's say "resigned"... to the fact that there exist very slowly divergent series; but the fact that there are very slowly CONVERGENT series, whose sum is impossible to approximate computationally within a reasonable margin of error, like the no nines series... was a shock!
Most memorable: The proof that the bishop came up with, beautiful simplicity
For the challenge at 20:24: I found a certain sequence of reciprocals (don't get too excited, I only have finitely many terms and don't even know how to extend it without using trial and error) that 1) has ever-decreasing terms (like the true harmonic series), and 2) when you calculate its partial sums it hits every integer from 1 to 3. It'll be under the "read more" button so I don't spoil it!
1 + 1/2 + 1/3 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + 1/16 + 1/18 + 1/20 + 1/24 + 1/25 + 1/30 + 1/36 + 1/48 + 1/60 + 1/100 + 1/140 + 1/180 is my sequence (the sequence adds to 3, first four terms add to 2, and first term is 1).
I'm quite worried about how I can extend this to the integer 4, since my largest possible term available is 1/181 and my largest term with many divisors is probably 1/189 or 1/192 - which means it'll take a hard lower limit of 200ish additional terms to write it and almost certainly it will also take many more.
Very nice idea to package it all into one series. When people play with this they usually use completely different sums to make the integers. When you do that it's fairly easy to show that you can get all positive integers, in fact you can make all positive rational numbers. Have to ponder what's possible with your approach :)
Most memorable: Chapter 1: "Let's assume that the grey bar does not weigh anything - thought experiment - we can do this - hehe" Top notch video!
It (the likes) was prime, I clicked and it remained prime ;p
I really enjoyed it. I guessed no 9s has finite sum by thinking about binary numbers with no 0s. that's 1/1 + 1/11 + 1/111 + 1/1111 + .... < 1 + 1/10 + 1/100 + 1/1000 + ... = 10 (=2).
I most enjoyed the Leaning Tower of Liire, because they didn't teach it in my calculus course and it's simple and beautiful.
Thank you
Bar
I thought the same kind of thing. What happens with the values if we change the basis. Is it possible to find an expression to link the values?
Great video, watched with my husband! My favourite part was the visualization of the no 9s series, thank you Tristan for the idea!
Proof that γ>1/2 (24:30): Replace all the blue parts with right triangles inside of them and the resulting sum is 1/2 (visually this is obvious, writing out the sum isn’t too bad either).
I found it too, but I watched the video 2 days later than you, ahh sigh
@@davidrheault7896 Good job!
I would have never related the armonic series with a leaning tower, amazing!
Danke für diese wundervollen Videos!
The fact that the harmonic series misses all the integers is beautiful to me!
Most memorable: fractal visualization of no-9's series being finite
Agreed! Simple and thus memorable
I really enjoy your work and every time you laugh, I laugh. Thank you.
It is really interesting how eulars comes into all these different formulas. Personal favourite was the very neat proof at the end that the sum was less than 80
I liked the crazy optimal overhang tower the most. Didn’t expect that at all.
Me too. The untidy structure called to mind Sabine Hossenfelder’s “Lost in Math: How Beauty Leads Physics Astray”. Physicists are addicted to symmetries. Dirac thought beauty was the most important (and convincing) feature of any formula in mathematical physics.
You have great math, but often you get beyond me. Love your stuff!
The most memorable part was me dying because I didn't know the no 9's series was convergent
yeah, i just went for the odds and said: well it is finite, because maybe something will happen, that i cannot see now, so in contradiction i am still alive :)
But... just to ask for clarity: if that grid exists with 9s, wouldt it exist with all other numbers >0 ?
what about the 0 ?
@@MrTiti The grid is the "same" with every digit, including 0.
In the first 10 numbers, 1/10 contain a (insert digit) or 10%
In the first 100 numbers, 1/10 for every "ten" + 9/100 (the ones starting with your digit, like 31 32 33...), or 19% containing your digit.
In the first 1000 numbers we have 19 every 100, + the ones with the first digit (300, 301, 302...), or 271/1000 or 27.1% total.
And so on...
Every time the % of numbers containing the chosen digit keeps increasing, reaching almost 100%. Works for all 10 digits (0, 1, 2... 9).
Love it! - I wish I had another lifetime for this stuff . .
Euler mascheroni constant being greater than 0.5 can be seen in the figure at 25:13 by noticing that the blue convex triangles contain straight triangles whose area sum to 1/2.
Great video Mathologer.😉
Most memorable: Sad looking portrait of Nicole Oresme along with the Leaning Tower of Lire
It's odd that while the terms get smaller and smaller, the sum doesn't converge. You look at every book in the tower being less and less over the edge and think "there must be a point where the top book moves so little that it's impossible to notice no matter how many books you pile, and the new books barely go over the edge for even microns" and at the same time think that it will always go further as far as you want it to go.
Thanks for another amazing, informative, extremely clear and well made presentation. The fact that the series explodes to infinity is one thing, how slow it happens and how many terms are needed for just a tiny increase makes my head spin.
28:50 Gotta love the Numberphile-burn 😅.
This is the reason why I'm getting a PHD in mathematics: the infinite beauty of the numbers.