Presumably the table is going to have an infinitely sharp edge. The rounded-corner illustration of the table edge at 1:00, while pretty, may not be the best choice to illustrate this. You could say, of course, that the true edge of the table is just where the curve begins.
Tehom Yes, the image bothers me because it's wrong. I'm pretty sure that the mass with it's center of mass not touching a solid surface doesn't look down and say "I'm not supported but it's okay because there's this fancy little curve further down so I'll just pretend I'm supported by that instead."
TimGB Just place the center of mass even with the edge of the table. The infinitely long brick would stretch infinitely in both directions keeping the center... in the center. Trick 2: turn off gravity. I mean, while we're just doing _whatever_, might as well. Glue the bricks together so they don't float away.
Matthew Fennell an infinitly long brick wouldn't have a centre of mass or would at least never topple, because if the brick just slightly tilts, the points far away would have to travel faster than the speed of light, wich sadly isn't possible
There is a (very) related riddle that I like. There is a mathematical ant on a mathematical rubber band. The rubber band is 100m long and the ant starts on an end of the band and wants to reach the other end. Each day, the ant travel 1m. But each night, when the ant sleeps, an evil mathematician pull the band so that it is 100m longer. To see what happens, at the end of the first day, the ant traveled 1m and the band is 100m. But the next morning, the band is 200m and since both the part before and the part after the band are now longer, the ant has traveled 2m. At the end of the second day, the ant has traveled 3m but the next morning, the band is 300m long and the ant has traveled 4.5m. 295.5m to go… The question is will the ant ever reach is final destination ? And if so, in how many days ? The key to the answer is to see that on the n-th day, the band is (n+1)*100 m long, and thus on the n-th day, the ant travels 1/100*1/(n+1) of the band. So after n days, the ant will have traveled 1/100*(1/2+1/3+1/4+...+1/(n+1)) th of the band. But since the harmonic series diverges, this quantity will eventually be bigger that 1. And when this happens, this means that the ant will have crossed the whole band.
Well technically it's still possible to do it on a table with a curved edge except for one thing they were using the wrong balance point. The correct balance point would be where the base of support ends ie the moment that the surface shifts from being tangential to Earth's surface to curving towards the surface past that point it wasn't supporting the block anymore and the very first block would have rotated and flipped off the table as a result as they got the balance point wrong!
In regards to Shor's Algorithm causing world economic collapse: no worries! We already have elliptic curve cryptography. There is no currently known algorithm for breaking it on a quantum computer. By the way, I would love to see a future episode on elliptic curve cryptography!
In your first description, where you compare the proportion of the block to the right and left of the table edge, you actually determine the median position of the mass, not the mean position of the mass. This isn't necessarily the center of mass. Once a block is completely to the right of the table, it will be 100% to the right no matter how far away it is, but the CG will continue to change if it goes farther to the right. In the second approach, where you consider point masses, you're actually considering the distance and computing the first moment, so that is the center of mass.
Kind of off-topic/request: Could you make an episode on angles and rotation in general? covering degrees, radians and mainly: quaternions (which nobody seem to understand/explain). And also: why 360 degrees and not 100 ? ;) I love your channel, even if I struggle with it (me being an artist). Keep it up!
Our modern computers actually have the same issue though it's generally quite uncommon - cosmic rays will flip bits in memory and storage and cause strange effects, so modern cpus usually have error checking
A bridge is a structure that connects two or more separate edges. If the structure is infinitely long, by definition, there is no second edge for it to connect to. (How can it stop somewhere if it doesn't stop?) What you have isn't an infinite bridge but an infinite diving board.
And it's a good thing. If the maximum overhang went as the sum of natural numbers, and you had an infinite number of blocks, and you put the edge of the top block any further out than -1/12 - that means it's still entirely on the table by 1/12 of a block length - it would fall off.
-1/12 isn't the sum of all natural numbers. That's infinite. -1/12 is just the point on the Reimann Zeta function which would correspond to n= -1 if the function was actually defined there. Which it isn't.
About new forms of encryption, Physics Girl has a video on quantum encryption, which is truly unbreakable, just currently inconvenient, it needs to jump similar hurdles to the quantum computers themselves.
with a rounded corner like shown if the center of mass is at the edge of the table how long till the blocks are on the floor assume a standard table height
Thought: it seems that the limiting factor here is the number of dimensions we're building in: this is a one-dimensional bridge (we measure how far 'out' it goes) in a two-dimensional space (we 'stack' blocks) and we get a 1/3(n) growth rate. What happens if we build the largest two-dimensional bridge (i.e. we measure its area, not its distance) in a three-dimensional space? What's its optimal growth rate? At does the pattern extend to an x-dimensional bridge in an (x+1)-dimensional space.
The "x = 1/2n" at 5:10 looks disconcertingly similar to certain viral maths problems that occasionally suck the internet into arguments about rules of precedence.
I noticed that .5 * (HS1[I am denoting the harmonic series as (HSn) where n is equal to how many terms the partial harmonic series has]) =.5 while Ln(1) = 0, this is a pretty sizable difference. I went on to check for n= 2 and n= 10 they indicated two things, 1) that the Harmonic series is consistently an over approximation, and 2) that the difference decreases. Are my intuitions correct? That would mean a sequence described by A(n)= (Ln(n) - HSn) would converge to 0 as n approaches infinity. If that is true then would the series described by that sequence converge as well? if so what value would it converge to?
Funny, at 6:36 there is a graphic of ln(x) with negative terms still in the background. And why aren't the first row and column fully shown in the graph on top? It looks strange to me
Actually, to secure against quantum computers using shor's algorithm all you have to do is double the bit length of the encryption key. The methods we have are fine. They're still just as secure against quantum computers, quantum computers would just be faster at cracking them. I believe shor's algorithm takes sqrt time compared to a classic computer and doubling the number if bits just so happens to square the time required to crack it.
nice! Question though, at what rate would every new bottom block have to "grow in length" in order to make the growth off of the table consistent with each step? my intuition tells me it would be some "inverse" of the natural log function, but i'm not a math buff. P.S. I am very curious what your tattoo is of, if it is indeed a tattoo on your right shoulder. Extra Credit for anyone, because google proves to be useless no matter how I phrase this question, what is the Term for when you multiply 1*2*3*4*5*6*......? i believe it's written like "x!", ie- 5!=120 Thanks for any answers.
wouldn't this particular series converge when the distance x reached the planck length since you're talking about physical blocks made of atoms... at least I assumed you were
I've been told that the answer is no, but I keep on wondering if there's something in between "very slow growing" series that converge greater than any finite number, and the fastest growing series that do converge at a finite number. It feels like you should be able to create a series that at some part has a variable x, where if x is greater than some number the series does not converge at a finite number, and where x is less than some number it converges at a finite number, (or maybe there's actually a range of numbers), so that if you plug in the exact number (or range) it does something "in between", even though I don't know what that "in between" would mean.
for the first blocks there is an easier way to think about it... because both block are the same length and overlap half there length it means that the parts not overlapping have the same length, so they balance out, and in order to get the center in the middle you then just have to divide the overlapping length by 2...
Instead of setting up unknown x, it's easier just to use the centre of mass of the top block as the origin. Then you can easily right down the CM of the whole bridge as sum(1/k)
I have an issue with the solution to this problem presented here, as well as in other places on the net. I hope you can clarify it for me. Both my work and your solution agree on the first 3 terms, namely that the bridge extends by 1/2, 3/4, and 11/12 of a brick's length over the edge. But your solution for the 4th brick has the whole bridge extending by 25/24, while I'm pretty sure it's 19/18. Start by representing every brick with 36 squares on graph paper, or 36 symbols in a text editor. You can follow along here: pastebin.com/QTy0bsJB Hang the first brick by 18 spaces over the second (1/2 of 36), the first and second together by 27 spaces over the third (3/4 of 36) and the first three over the fourth by 33 (11/12 of 36). Now draw a line at position 38 of the four brick assembly and count the total number of symbols to the left and to the right of it. You will get exactly 72 symbols on each side. Since the four brick assembly is a rigid structure, per construction, and every symbol has the same mass (1/36 of the mass of a brick) it follows that position 38 is the center of mass of the entire structure, by definition: half the mass is to the left of it, half is to the right. But position 38 is 19/18 of 36, not 25/24! That's why I believe the common solution to this problem, which you presented here, is wrong. Either that, or I'm making a mistake that I fail to see.
I don't have a picture anymore, but around 15 years ago we were visiting my grandmother's house, my little sister and I found a box of old lego bricks and running out of ideas eventually built a suspension bridge out of plain old bricks and flat pieces. We used knitting yarn for the cable, with the vertical bits held in place between connected bricks of the deck,. It spanned almost a meter between the towers when unsupported the lego block surface couldn't hold together more than a few cm and would collapse under its own weight. It was really nice to see how the bridge was hanging itself up, using the towers and the cable to convert convert the downward force of weight into an inwards compression between the anchors at the ends. When completed, it held hold several kgs of additional weight. Legos are very strong in compression as studied extensively on youtube...
Why did you use log(n) to approximate H_n? The approximation H_n = log(n+½) is much more accurate. The error of log(n) is O(1/n), compared to the much better error for log(n+½), which is O(1/n²)
If 1/1+1/2+1/3+1/4...+1/n is infinite, but 1/1+1/2+1/4+1/8+1/16+...+1/n² is finite. Is there an infinite sum between those sums which is just at the edge of being infinite or finite?
Erdmännelchen The harmonic series is right on the border of being finite. For a series of elements 1/n^p, if p is greater then 1 then the series converges. If 0 < p
*Answer 1a*. 1/(n*log n) diverges and is between 1/n and 1/n^2. *Answer 1b*. 1/(n*log(n)*log(log(n))) diverges and is between 1/(n*log(n)) and 1/n^2. *Answer 1c*. ... continue in this fashion. *Answer 2*. Suppose you have a series of positive numbers x1+x2+x3+x4+... that diverges. Then you can find another series of positive numbers y1+y2+y3+... that diverges `infinitely slower'. By infinitely slower, I mean that the ratio (yn/xn) tends to zero as n tends to infinite. *One proof of Answer 2*: We know x1+x2+x3+x4+... diverges. This means that the sum eventually exceeds 1, and that it eventually exceeds 2, and eventually exceeds 3, and so on. For the sake of illustration and ease of notation, imagine it takes 100 terms to exceed 1, and the next 900 terms to exceed 2, and the next 9000 terms to exceed 3, and so on. Then we define the new series y like this: x_1,x_2,x_3,...x_100, (x_101)/2, (x_102)/2,...,(x_1000)/2, (x_1001)/3,(x_1002)/3,..., (x_10000)/3, (x_10001)/4,... Need to check two things: this new series diverges and that the ratio y_n/x_n tends to zero. *y_n/x_n tends to zero*: y_n/x_n starts off as 1, then after 100 terms the ratio becomes 1/2, then after 1000 terms the ratio becomes 1/3, and after 10000 terms becomes 1/4 and so on, getting closer and closer to zero. *y_n diverges*: By construction, y_1+...+y_100 >1. And y_1+...+y_1000 > 1 + 1/2. And y_1 + ... + y_10000 > 1 + 1/2 + 1/3. Continuing in this fashion, the right hand side is becoming the 1/n series, which we know diverges, and hence sum of y_n diverges. *Answer 3*. Naturally, your question actually still stands. Is there an edge? Sure, using Answer 1 or 2 repeatedly, we can create infinitely many (countably infinite) diverging series, each much slower than the previous one. But can you go beyond that? The answer is yes, using some kind of diagonal argument. You can find a series which diverges and which is slower than all the series constructed previously. And now we can keep on repeating. There is a technical sense in which this process has to end, which involves having to talk about ordinals! So it looks like you asked a deceptively simple yet highly interesting question - I am glad the next video in the series commented on your comment!
You can try to change value of exponentiential: www.wolframalpha.com/input/?i=sum+of+1%2Fn%5E1.5 and try to approach 1: www.wolframalpha.com/input/?i=sum+of+1%2Fn%5E1.01 it seems to work until you're at 1 ...
You're actually asking about the value of the Riemann Zeta function : en.wikipedia.org/wiki/Riemann_zeta_function at the value s = 1 and at s = 2 This function is defined for all complex z such that Re(z) > 1 =, hence the series converges for all real numbers in ]1;+infinity[ . The function also has a pole at s= 1 since the series goes to infinity. So the value at the edge of being infinite would be something like 1+ds where ds is an infinetly small positive quantity.
I know this is an old video, but in case anyone sees this: what if you have one block of unit length, one of 1/2 length, one of 1/3 length... what is the furthest you can extend the bridge now? Is it still infinite?
I don't think it's enough to just compare the mass on the table to the mass off the table. The position of the mass matters as well. Imagine you have 2 bricks, one on each end of a long massless stick. Now you can place this contraption with one brick sticking just off the table and one well onto the table, or with one sticking far off the table and the other close to the edge. In both cases the mass off the table is equal to the mass on the table, but in one case it's very stable and in the other case it will fall right off the table.
If you had as many bricks as particles in the observable universe, your old bridge would be about 92 cm long (since 0.5 x ln(10^80) = 92,1...) and the new one would circle the earth about 6x10^16 times.
This is, of course, the Riemen Zeta function for s=1. For complex s, it has non-trivial zeroes only when the real part is 1/2. I have discovered a truly marvelous proof of this which this RUclips comment has too few characters to contain.
The fastest growing one is just an infinite mass at somewhere back of the table supported by one block at the edge and a horizontal stick with however length you want xD I'm just an engineer, trying to build simpler solutions hahaha.
That is so strange! I tried to build a bridge out of coins. Find out I need 86.05 grams to balance hanging 36.25 grams. That is only 42 percent efficiency, but due to the math and logic it HAS to be 100% - what you hang is what you balance it with, precisely. But I suspect it's not working. Coin on the edge lifts up the next one, nick is big enough to make it slide down. The 2nd coin cause a smaller gap, the 3-d - even smaller. And the fifth coin have no noticeable movements, but it does not hold all the previous anymore. So I had to add tower over the center to prevent this tiny motion/pitching. And it has to be quite heavy. =( No quick bridges so soon
I vote physicist. Physicists work with idealizations, and there's a lot of silly stuff you can tweak to idealize this, like... buoyant forces (put it in heavier air) static friction with the table (bricks are made of tape that sticks to the table) air flow (blow on the bottom of the bridge so it stays up) gravity (put another Earth-mass object on the other side of the bridge so it doesn't fall toward Earth) the weight of the duct tape (have it lighter than air to where it'll float upward with the same force that the tower would fall downward) Wingardium (Leviosa) thermal effects (heat the bridge so it expands until its density lowers enough to float) electromagnetism (think magnet trains) fluid effects (pressure within the bricks to make them extend like those long balloons do)
char whick But that's only if the brick is infinitely strong. Otherwise, it'll bend downward under its own weight as it stretches further outward. Physicists can make those idealizations, while engineers can explore them to find a system's physical limits, but an engineer would lose the job pretty quickly if that was the kind of answer the engineer gave to an employer.
If you want to build the longest possible bridge wouldn't it be better to use a cantilever where you fuse the bricks making up the bridge and balance the structure with a counter weight? The bridge could be expanded by fusing in new pieces and sliding them under said count weight via a mechanism designed to shift and angle the counter weight. Thus allowing the bridge to go on infinitely assuming we have a infinite supply of a infinitly strong material that can be welded and has a finite weight.
The situation assumes no welding- that is, the blocks have to be held together by gravity alone. If this were the case and it was infinitely strong, there's nothing stopping us from welding them end to end and placing it like one large first block extending infinitely far in both directions, which is kind of boring.
"The harmonic series has never been observed to diverge". It's like my own statement that an infinite iteration has not yet ever occurred (that is, if you believe the universe had some beginning, be it the big bang, the hand of God, or whatever).
I would like to point out that even "classical" computers are not 100% accurate. So there is in fact a small chance that 2+2 will equal 5. That's what programmers call "soft errors" and are caused by cosmic rays or thermal noise flipping random bits in your computer. There are even attacks utilizing such errors if they happen inside an URL.
They also ignored forces between blocks and the fact that the force of gravity on each block would decrease as the bridge stretched farther from Earth. Frankly I doubt they've carried out an environmental impact assessment or even considered how to fund this boondoggle. This is why I love pure math!😃
+Karma Arachnid Indeed there are of course a whole bunch of issues preventing it being an engineering solution. Other factors would be the fact that eventually the weight of the blocks will be so great the bottom blocks would be crushed into dust even if you use infinitely strong blocks you also can't have loose blocks stacked to infinite height. Above 36,000km the top blocks would be at a height where the centripetal force needed to keep it rotating at 0.26 rad/hr would exceed the local value of g causing the blocks to drift upwards from inertia. Fun to think about but physics has to go spoil it heh.
You can't make an infinite bridge even with blocks with perfect center of masses because the further you go you will eventually be far enough from earth that other gravitational forces will be higher than that of earths and some blocks 1000's of kilometers away will float away into space.
To make a bridge you first need to know how big infinity is to set the foundation blocks up.. so it fails. the blocks will need to be infinitely close to the edge to span an infinite gap.
0:52 this is wrong, you can clearly see the edge is rounded. 2:21 the = should be a >. this is a shoddy remake of an old paradox popularized in the 70s by martin gardner. I was already thinking of 8:50 the way you described it but that was not the original question. A pity you didn't explain why it's n^(1/3) and I don't think it's true in general. The argument that a normal triangular pyramid would topple is not explained. The parabolic thing only works if you place groups of blocks at the same time, so a triangular version would work just as well: .xxx. xxxx .xxx. ..xx.. ...x... after which you place at the same time the top left and top right to complete the top layer along with a block on each to hold them in place. This is linear, so wins from your n^(1/3). Therefore I suspect there is some not explicitly mentioned constraint I either missed, or you failed to mention.
It can't be infinite, because if nothing else, eventually you will place a block and find out you can extend it only 1 more Planck length. And then you can no longer continue decreasing the length that you extend the next blocks.
Wouldn't a diamond shape bridge not work (you know: 1,2,3,4,5,4,3,2,1) It's balanced, and uses X² bricks (X being 5 in the example) to extend X/2 (each of the lower layers extend 1/2; all the layers higher then the middle are balast so it doesn't collapse That means a root(N)/2 progress, no?
The problem is that quantum computers will exist as standalone devices before they can be built into the infrastructure. The first quantum computers will be room sized or bigger. This will create a period of time where hackers will have potential access to quantum computers long before they are small enough to be built into the infrastructure of the internet. In this period there will be no communication that cannot be "easily" cracked. Quantum computers will likely be our downfall unless we can find some other means of protecting data that doesn't involved quantum mechanics.
Wouldn't just build a diamond shape give you a n^(1/2)? Like This － － － － The center of gravity will always stay center. And twice the bridge span leads to twice the width and twice the height so four times the coin number.
I always love these videos, but contents aside for one moment... If Kelsey (or someone else who knows) reads this; what's your shoulder tattoo? I always think of it as an octopus, but I'm sure it's probably not. Been bugging me for months.
You used a kind of 'greedy' way of stacking subsequent bricks, where each brick was placed as far as it could go under the previous conditions, i.e. the first one was placed 1/2 a distance away from the table, and then because of those conditions already, the next brick could only be placed 1/4 and so on. How do you know that this 'greedy' method of placing bricks is the optimum method for stacking one brick at a time? What if you placed bricks taking into account the moments of the future bricks? Could you use some sort of optimisation mathematics like dynamic programming to know how far to place each subsequent brick, taking into account future bricks?
it felt nice to know i was right about the power of the parabolic bridge, by gust guesstimating, good to see my master in physics at work more to the point, could you make a video on cnoidal (Jacobi elliptic functions) i had and still have a really hard time understating and explaining them, or about non harmonic waves in general, i keep trying to find an easy way to explain these to friends and family
About the comment on internet cryptography breaking down: As long as there are problems that are easy to state and have an easy way to check a solution, but which are very hard to solve, there will be potential for cryptographic algorithms, right?
Presumably the table is going to have an infinitely sharp edge. The rounded-corner illustration of the table edge at 1:00, while pretty, may not be the best choice to illustrate this.
You could say, of course, that the true edge of the table is just where the curve begins.
Tehom Yes, the image bothers me because it's wrong. I'm pretty sure that the mass with it's center of mass not touching a solid surface doesn't look down and say "I'm not supported but it's okay because there's this fancy little curve further down so I'll just pretend I'm supported by that instead."
one of the best channels on RUclips :)
I have watched 3 of her videos, this being the third, and have already subscribed and hit the notification bell.
Can beat all this.
Just use an infinitely long brick.
Matthew Fennell But would it be stable? Where would its centre of mass be; would it even have a centre of mass?
TimGB
Just place the center of mass even with the edge of the table. The infinitely long brick would stretch infinitely in both directions keeping the center... in the center.
Trick 2: turn off gravity. I mean, while we're just doing _whatever_, might as well. Glue the bricks together so they don't float away.
Matthew Fennell an infinitly long brick wouldn't have a centre of mass or would at least never topple, because if the brick just slightly tilts, the points far away would have to travel faster than the speed of light, wich sadly isn't possible
You'll need an infinitely long table. I win.
Then you don't even have a bridge. You're disqualified.
There is a (very) related riddle that I like.
There is a mathematical ant on a mathematical rubber band. The rubber band is 100m long and the ant starts on an end of the band and wants to reach the other end. Each day, the ant travel 1m. But each night, when the ant sleeps, an evil mathematician pull the band so that it is 100m longer.
To see what happens, at the end of the first day, the ant traveled 1m and the band is 100m. But the next morning, the band is 200m and since both the part before and the part after the band are now longer, the ant has traveled 2m.
At the end of the second day, the ant has traveled 3m but the next morning, the band is 300m long and the ant has traveled 4.5m. 295.5m to go…
The question is will the ant ever reach is final destination ? And if so, in how many days ?
The key to the answer is to see that on the n-th day, the band is (n+1)*100 m long, and thus on the n-th day, the ant travels 1/100*1/(n+1) of the band. So after n days, the ant will have traveled 1/100*(1/2+1/3+1/4+...+1/(n+1)) th of the band. But since the harmonic series diverges, this quantity will eventually be bigger that 1. And when this happens, this means that the ant will have crossed the whole band.
I like the way that this introduces subdivision, and limits by using physical representations
I watch the episodes twice. once to listen to her words explaining mathematics in that voice and again to see her hand expressions!
Love your videos because you always put references in the description
the Q&A at the end is a really neat concept!
a table with a curved top edge doesn't seem like a good choice for this discussion...
So I'm not the only one who was irritated by that?
KinoftheFlames agreed
Well technically it's still possible to do it on a table with a curved edge except for one thing they were using the wrong balance point. The correct balance point would be where the base of support ends ie the moment that the surface shifts from being tangential to Earth's surface to curving towards the surface past that point it wasn't supporting the block anymore and the very first block would have rotated and flipped off the table as a result as they got the balance point wrong!
Proof that the CG/Animators did not listen to her at all
implying that a non curved edge exists outside a mathematical model.
"Some infinite series are finite." Hopefully not this one :D
I think she said, some infinte series converge to a finite value...
Maybe PBS Infinite series is actually finite and will not last forever.
This comment hasn't aged very well...
@@galladeguy123 :(
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In regards to Shor's Algorithm causing world economic collapse: no worries! We already have elliptic curve cryptography. There is no currently known algorithm for breaking it on a quantum computer. By the way, I would love to see a future episode on elliptic curve cryptography!
Very interesting, I prefer the classic bridge with the (1/2)ln n overhang.
It's so counterintuitive that it can overhang arbitrarily large distances.
Finally! A video I understand! I even know about the big O too!
In your first description, where you compare the proportion of the block to the right and left of the table edge, you actually determine the median position of the mass, not the mean position of the mass. This isn't necessarily the center of mass. Once a block is completely to the right of the table, it will be 100% to the right no matter how far away it is, but the CG will continue to change if it goes farther to the right.
In the second approach, where you consider point masses, you're actually considering the distance and computing the first moment, so that is the center of mass.
Kind of off-topic/request: Could you make an episode on angles and rotation in general? covering degrees, radians and mainly: quaternions (which nobody seem to understand/explain). And also: why 360 degrees and not 100 ? ;) I love your channel, even if I struggle with it (me being an artist). Keep it up!
I love the "buts" and "wells." And, great content.
Our modern computers actually have the same issue though it's generally quite uncommon - cosmic rays will flip bits in memory and storage and cause strange effects, so modern cpus usually have error checking
Watched a 2:30min ad because I love this channel and want to support it
A bridge is a structure that connects two or more separate edges. If the structure is infinitely long, by definition, there is no second edge for it to connect to. (How can it stop somewhere if it doesn't stop?) What you have isn't an infinite bridge but an infinite diving board.
but if it's infinite, how do you even find the edge to dive off of it?
CogniVision Good point. Maybe it's just an infinte balcony, then?
+Noel Goetowski Good point, but, uh, don't you need to get to the end of a diving board to make a dive?
You just need to go infinitely fast to get there.
They should just call it an infinite road and make it with yellow bricks.
So this is what happens when you have -1/12 blocks
Yeah I realize it doesn't make sense I just needed to make a -1/12 joke
And it's a good thing. If the maximum overhang went as the sum of natural numbers, and you had an infinite number of blocks, and you put the edge of the top block any further out than -1/12 - that means it's still entirely on the table by 1/12 of a block length - it would fall off.
-1/24, you need to divide it by two first.
You are right, I missed the factor of 1/2. So you can get the farthest edge of a block almost to the edge of the table before they all fall off.
-1/12 isn't the sum of all natural numbers. That's infinite. -1/12 is just the point on the Reimann Zeta function which would correspond to n= -1 if the function was actually defined there. Which it isn't.
You're the best! I love your videos.
About new forms of encryption, Physics Girl has a video on quantum encryption, which is truly unbreakable, just currently inconvenient, it needs to jump similar hurdles to the quantum computers themselves.
with a rounded corner like shown if the center of mass is at the edge of the table how long till the blocks are on the floor assume a standard table height
you teach very well!!!!! thanks!
"Btw it's an example of an Infinite Series" roll credits!
do the blocks have friction? might be able to do some interesting things with friction
Thought: it seems that the limiting factor here is the number of dimensions we're building in: this is a one-dimensional bridge (we measure how far 'out' it goes) in a two-dimensional space (we 'stack' blocks) and we get a 1/3(n) growth rate. What happens if we build the largest two-dimensional bridge (i.e. we measure its area, not its distance) in a three-dimensional space? What's its optimal growth rate? At does the pattern extend to an x-dimensional bridge in an (x+1)-dimensional space.
What happens if you stack infinitely many blocks? the table collapses under its weight.
The "x = 1/2n" at 5:10 looks disconcertingly similar to certain viral maths problems that occasionally suck the internet into arguments about rules of precedence.
i am with you, they should really write fraction as simple fractures and not in the format like here
it drives me nuts seeing that
I noticed that .5 * (HS1[I am denoting the harmonic series as (HSn) where n is equal to how many terms the partial harmonic series has]) =.5 while Ln(1) = 0, this is a pretty sizable difference. I went on to check for n= 2 and n= 10 they indicated two things, 1) that the Harmonic series is consistently an over approximation, and 2) that the difference decreases. Are my intuitions correct? That would mean a sequence described by A(n)= (Ln(n) - HSn) would converge to 0 as n approaches infinity. If that is true then would the series described by that sequence converge as well? if so what value would it converge to?
Funny, at 6:36 there is a graphic of ln(x) with negative terms still in the background.
And why aren't the first row and column fully shown in the graph on top? It looks strange to me
9:45 "The new bridge is growing exponentially faster"... is that true? do the rates of growth of n^1/3 and ln(n) have an exponential relationship?
Actually, to secure against quantum computers using shor's algorithm all you have to do is double the bit length of the encryption key. The methods we have are fine. They're still just as secure against quantum computers, quantum computers would just be faster at cracking them. I believe shor's algorithm takes sqrt time compared to a classic computer and doubling the number if bits just so happens to square the time required to crack it.
nice! Question though, at what rate would every new bottom block have to "grow in length" in order to make the growth off of the table consistent with each step? my intuition tells me it would be some "inverse" of the natural log function, but i'm not a math buff.
P.S. I am very curious what your tattoo is of, if it is indeed a tattoo on your right shoulder.
Extra Credit for anyone, because google proves to be useless no matter how I phrase this question, what is the Term for when you multiply 1*2*3*4*5*6*......? i believe it's written like "x!", ie- 5!=120
Thanks for any answers.
wouldn't this particular series converge when the distance x reached the planck length since you're talking about physical blocks made of atoms... at least I assumed you were
I've been told that the answer is no, but I keep on wondering if there's something in between "very slow growing" series that converge greater than any finite number, and the fastest growing series that do converge at a finite number. It feels like you should be able to create a series that at some part has a variable x, where if x is greater than some number the series does not converge at a finite number, and where x is less than some number it converges at a finite number, (or maybe there's actually a range of numbers), so that if you plug in the exact number (or range) it does something "in between", even though I don't know what that "in between" would mean.
Kelsey, have you heard of the Mathematics Genealogy project? I looked it up and you are adviser-descended from Isaac Newton!
That's pretty neat! Thanks!
What is that? It sounds cool.
wait... you have a Debian tatoo? you are getting even more awesome every time
The correct approach is, not have the blocks at 1/2 way point but offset by a delta and then take the limit as the delta approaches zero.
Cool scar on your shoulder. What's that from. Bet there is a cool story
for the first blocks there is an easier way to think about it...
because both block are the same length and overlap half there length it means that the parts not overlapping have the same length, so they balance out, and in order to get the center in the middle you then just have to divide the overlapping length by 2...
Instead of setting up unknown x, it's easier just to use the centre of mass of the top block as the origin. Then you can easily right down the CM of the whole bridge as sum(1/k)
I have an issue with the solution to this problem presented here, as well as in other places on the net. I hope you can clarify it for me.
Both my work and your solution agree on the first 3 terms, namely that the bridge extends by 1/2, 3/4, and 11/12 of a brick's length over the edge. But your solution for the 4th brick has the whole bridge extending by 25/24, while I'm pretty sure it's 19/18.
Start by representing every brick with 36 squares on graph paper, or 36 symbols in a text editor. You can follow along here: pastebin.com/QTy0bsJB
Hang the first brick by 18 spaces over the second (1/2 of 36), the first and second together by 27 spaces over the third (3/4 of 36) and the first three over the fourth by 33 (11/12 of 36). Now draw a line at position 38 of the four brick assembly and count the total number of symbols to the left and to the right of it. You will get exactly 72 symbols on each side.
Since the four brick assembly is a rigid structure, per construction, and every symbol has the same mass (1/36 of the mass of a brick) it follows that position 38 is the center of mass of the entire structure, by definition: half the mass is to the left of it, half is to the right. But position 38 is 19/18 of 36, not 25/24!
That's why I believe the common solution to this problem, which you presented here, is wrong. Either that, or I'm making a mistake that I fail to see.
I don't have a picture anymore, but around 15 years ago we were visiting my grandmother's house, my little sister and I found a box of old lego bricks and running out of ideas eventually built a suspension bridge out of plain old bricks and flat pieces. We used knitting yarn for the cable, with the vertical bits held in place between connected bricks of the deck,. It spanned almost a meter between the towers when unsupported the lego block surface couldn't hold together more than a few cm and would collapse under its own weight. It was really nice to see how the bridge was hanging itself up, using the towers and the cable to convert convert the downward force of weight into an inwards compression between the anchors at the ends. When completed, it held hold several kgs of additional weight. Legos are very strong in compression as studied extensively on youtube...
You need a better microphone at the place where you record answers to comments. Other than that, cool video as always :-)
Kelsey Houston-Edwards is an excellent example of an "Infinite Series."
8:50 Finally, I thought they'd never get to this.
Why did you use log(n) to approximate H_n? The approximation H_n = log(n+½) is much more accurate. The error of log(n) is O(1/n), compared to the much better error for log(n+½), which is O(1/n²)
If 1/1+1/2+1/3+1/4...+1/n is infinite, but 1/1+1/2+1/4+1/8+1/16+...+1/n² is finite. Is there an infinite sum between those sums which is just at the edge of being infinite or finite?
Erdmännelchen The harmonic series is right on the border of being finite. For a series of elements 1/n^p, if p is greater then 1 then the series converges. If 0 < p
*Answer 1a*. 1/(n*log n) diverges and is between 1/n and 1/n^2.
*Answer 1b*. 1/(n*log(n)*log(log(n))) diverges and is between 1/(n*log(n)) and 1/n^2.
*Answer 1c*. ... continue in this fashion.
*Answer 2*. Suppose you have a series of positive numbers x1+x2+x3+x4+... that diverges. Then you can find another series of positive numbers y1+y2+y3+... that diverges `infinitely slower'. By infinitely slower, I mean that the ratio (yn/xn) tends to zero as n tends to infinite.
*One proof of Answer 2*: We know x1+x2+x3+x4+... diverges. This means that the sum eventually exceeds 1, and that it eventually exceeds 2, and eventually exceeds 3, and so on. For the sake of illustration and ease of notation, imagine it takes 100 terms to exceed 1, and the next 900 terms to exceed 2, and the next 9000 terms to exceed 3, and so on. Then we define the new series y like this:
x_1,x_2,x_3,...x_100, (x_101)/2, (x_102)/2,...,(x_1000)/2, (x_1001)/3,(x_1002)/3,..., (x_10000)/3, (x_10001)/4,...
Need to check two things: this new series diverges and that the ratio y_n/x_n tends to zero.
*y_n/x_n tends to zero*: y_n/x_n starts off as 1, then after 100 terms the ratio becomes 1/2, then after 1000 terms the ratio becomes 1/3, and after 10000 terms becomes 1/4 and so on, getting closer and closer to zero.
*y_n diverges*: By construction, y_1+...+y_100 >1. And y_1+...+y_1000 > 1 + 1/2. And y_1 + ... + y_10000 > 1 + 1/2 + 1/3. Continuing in this fashion, the right hand side is becoming the 1/n series, which we know diverges, and hence sum of y_n diverges.
*Answer 3*. Naturally, your question actually still stands. Is there an edge? Sure, using Answer 1 or 2 repeatedly, we can create infinitely many (countably infinite) diverging series, each much slower than the previous one. But can you go beyond that? The answer is yes, using some kind of diagonal argument. You can find a series which diverges and which is slower than all the series constructed previously. And now we can keep on repeating. There is a technical sense in which this process has to end, which involves having to talk about ordinals!
So it looks like you asked a deceptively simple yet highly interesting question - I am glad the next video in the series commented on your comment!
You can try to change value of exponentiential:
www.wolframalpha.com/input/?i=sum+of+1%2Fn%5E1.5
and try to approach 1:
www.wolframalpha.com/input/?i=sum+of+1%2Fn%5E1.01
it seems to work until you're at 1 ...
Christophe Andre. There are more things you can do than change the exponential. E.g. the sum of 1/n*log(n) diverges and is in between 1/n and 1/n^2.
You're actually asking about the value of the Riemann Zeta function : en.wikipedia.org/wiki/Riemann_zeta_function
at the value s = 1 and at s = 2
This function is defined for all complex z such that Re(z) > 1 =, hence the series converges for all real numbers in ]1;+infinity[ . The function also has a pole at s= 1 since the series goes to infinity.
So the value at the edge of being infinite would be something like 1+ds where ds is an infinetly small positive quantity.
Welp, time to go out and by a shit ton of Jenga bricks
I know this is an old video, but in case anyone sees this: what if you have one block of unit length, one of 1/2 length, one of 1/3 length... what is the furthest you can extend the bridge now? Is it still infinite?
That is a cool question. I will have to think about it.
I don't think it's enough to just compare the mass on the table to the mass off the table. The position of the mass matters as well. Imagine you have 2 bricks, one on each end of a long massless stick. Now you can place this contraption with one brick sticking just off the table and one well onto the table, or with one sticking far off the table and the other close to the edge. In both cases the mass off the table is equal to the mass on the table, but in one case it's very stable and in the other case it will fall right off the table.
Shriek is like a parabolic stack. They have layers.
Alright, wait here. I need to get some infinitely thin building blocks and an infinitely strong table.
If you had as many bricks as particles in the observable universe, your old bridge would be about 92 cm long (since 0.5 x ln(10^80) = 92,1...) and the new one would circle the earth about 6x10^16 times.
The rabbit run halfway then relax, halfway and relax. The turtle run continously so she wins
So how long can we build that one-gram-block bridge with the mass of the Observable universe? with and without dark matter.
Sometimes when she frowns in a particular way, π appears on her forehead. Ever noticed?
I'm in love with you! lol
This is, of course, the Riemen Zeta function for s=1. For complex s, it has non-trivial zeroes only when the real part is 1/2. I have discovered a truly marvelous proof of this which this RUclips comment has too few characters to contain.
share it on google disk or something
loljk
and then he died. lol
Would you not be able to reach out further if you turned the bricks at an angle?
11:11 Make that music during question time available to the masses, for lord's sake PLEASE
The fastest growing one is just an infinite mass at somewhere back of the table supported by one block at the edge and a horizontal stick with however length you want xD I'm just an engineer, trying to build simpler solutions hahaha.
That is so strange! I tried to build a bridge out of coins. Find out I need 86.05 grams to balance hanging 36.25 grams. That is only 42 percent efficiency, but due to the math and logic it HAS to be 100% - what you hang is what you balance it with, precisely. But I suspect it's not working.
Coin on the edge lifts up the next one, nick is big enough to make it slide down. The 2nd coin cause a smaller gap, the 3-d - even smaller. And the fifth coin have no noticeable movements, but it does not hold all the previous anymore. So I had to add tower over the center to prevent this tiny motion/pitching. And it has to be quite heavy. =( No quick bridges so soon
This problem was extra credit for my final in physics.
Nice 100k subs.
where can I find the proof for ln(n) =~ 1 + 1/2 + 1/3 ... + 1/n ??????
what if the blocks we used were of increasing or decreasing weights (or masses), that would have a different effect on the center of mass.
There are already encryption algorithms in existence that are hardened against quantum computers.
Make the bricks have the same density as air. voila. now you just need duct tape
Found the Physicist.
actually we found the engineer
I vote physicist. Physicists work with idealizations, and there's a lot of silly stuff you can tweak to idealize this, like...
buoyant forces (put it in heavier air)
static friction with the table (bricks are made of tape that sticks to the table)
air flow (blow on the bottom of the bridge so it stays up)
gravity (put another Earth-mass object on the other side of the bridge so it doesn't fall toward Earth)
the weight of the duct tape (have it lighter than air to where it'll float upward with the same force that the tower would fall downward)
Wingardium (Leviosa)
thermal effects (heat the bridge so it expands until its density lowers enough to float)
electromagnetism (think magnet trains)
fluid effects (pressure within the bricks to make them extend like those long balloons do)
Free Diugh The engineer answer is "an infinitely long bridge is easy, just use an infinitely heavy counterweight"
char whick But that's only if the brick is infinitely strong. Otherwise, it'll bend downward under its own weight as it stretches further outward. Physicists can make those idealizations, while engineers can explore them to find a system's physical limits, but an engineer would lose the job pretty quickly if that was the kind of answer the engineer gave to an employer.
Can we all just stop and appreciate how her dress looks like a harmonically & parabolically stacked bridge. Wow.
If you want to build the longest possible bridge wouldn't it be better to use a cantilever where you fuse the bricks making up the bridge and balance the structure with a counter weight? The bridge could be expanded by fusing in new pieces and sliding them under said count weight via a mechanism designed to shift and angle the counter weight. Thus allowing the bridge to go on infinitely assuming we have a infinite supply of a infinitly strong material that can be welded and has a finite weight.
The situation assumes no welding- that is, the blocks have to be held together by gravity alone. If this were the case and it was infinitely strong, there's nothing stopping us from welding them end to end and placing it like one large first block extending infinitely far in both directions, which is kind of boring.
Could you do a video on POINCARE CONJECTURE & its solution? Topology is one of my interest in Mathematics, also #ilovepbs
Ignore gravity and all of your problems will be solved.
"The harmonic series has never been observed to diverge".
It's like my own statement that an infinite iteration has not yet ever occurred
(that is, if you believe the universe had some beginning, be it the big bang, the hand of God, or whatever).
I would like to point out that even "classical" computers are not 100% accurate. So there is in fact a small chance that 2+2 will equal 5. That's what programmers call "soft errors" and are caused by cosmic rays or thermal noise flipping random bits in your computer. There are even attacks utilizing such errors if they happen inside an URL.
2:00 something tells me you're not going to take into account rotational inertia
They also ignored forces between blocks and the fact that the force of gravity on each block would decrease as the bridge stretched farther from Earth. Frankly I doubt they've carried out an environmental impact assessment or even considered how to fund this boondoggle. This is why I love pure math!😃
+Karma Arachnid Indeed there are of course a whole bunch of issues preventing it being an engineering solution. Other factors would be the fact that eventually the weight of the blocks will be so great the bottom blocks would be crushed into dust even if you use infinitely strong blocks you also can't have loose blocks stacked to infinite height. Above 36,000km the top blocks would be at a height where the centripetal force needed to keep it rotating at 0.26 rad/hr would exceed the local value of g causing the blocks to drift upwards from inertia.
Fun to think about but physics has to go spoil it heh.
What do you think this is???
A Physics channel??? ;)
Why so hostile?
Not hostile, just poking fun
Edit: that's why there is a ";)" at the end but text isn't alway good at showing tone
Quantum computer proof (or resistant) algorithms for cryptography have already been identified/created.
I love this channel, why hasn't she been invited to numberphile yet? Sad
Cool guy 49 I suspect that is a problem of geography.
You can't make an infinite bridge even with blocks with perfect center of masses because the further you go you will eventually be far enough from earth that other gravitational forces will be higher than that of earths and some blocks 1000's of kilometers away will float away into space.
To make a bridge you first need to know how big infinity is to set the foundation blocks up.. so it fails. the blocks will need to be infinitely close to the edge to span an infinite gap.
0:52 this is wrong, you can clearly see the edge is rounded.
2:21 the = should be a >.
this is a shoddy remake of an old paradox popularized in the 70s by martin gardner.
I was already thinking of 8:50 the way you described it but that was not the original question.
A pity you didn't explain why it's n^(1/3) and I don't think it's true in general. The argument that a normal triangular pyramid would topple is not explained.
The parabolic thing only works if you place groups of blocks at the same time, so a triangular version would work just as well:
.xxx.
xxxx
.xxx.
..xx..
...x...
after which you place at the same time the top left and top right to complete the top layer along with a block on each to hold them in place. This is linear, so wins from your n^(1/3).
Therefore I suspect there is some not explicitly mentioned constraint I either missed, or you failed to mention.
It can't be infinite, because if nothing else, eventually you will place a block and find out you can extend it only 1 more Planck length. And then you can no longer continue decreasing the length that you extend the next blocks.
John Plissken There is also the small problem of finding sufficiently many bricks.
Wouldn't a diamond shape bridge not work (you know: 1,2,3,4,5,4,3,2,1)
It's balanced, and uses X² bricks (X being 5 in the example) to extend X/2 (each of the lower layers extend 1/2; all the layers higher then the middle are balast so it doesn't collapse
That means a root(N)/2 progress, no?
heck, if we presume we can balance a brick on it's middle point, why wouldn't (1,2,3,4,5) work?
The infinite bridge will also be infinitely high so you'd better wear a space suit when you really want to cross it...
TIL "asymptotically" has no "p" and an extra syllable.
One limitation is that all the bricks would get crushed.
The problem is that quantum computers will exist as standalone devices before they can be built into the infrastructure. The first quantum computers will be room sized or bigger. This will create a period of time where hackers will have potential access to quantum computers long before they are small enough to be built into the infrastructure of the internet. In this period there will be no communication that cannot be "easily" cracked. Quantum computers will likely be our downfall unless we can find some other means of protecting data that doesn't involved quantum mechanics.
Correction: "Uri Zwick" doesn't have a y-glide at the beginning. "Yuri" is a totally separate name.
Or, you can build a circular bridge and start walking along the circle, BOOM YOU NEVER END #INFINITEPERPETUALMOTIONHAHAEINSTIENBEATYOLOLE=MCMYASS
'The normal method is very slow'
The speak more quickly!
If we can put 2 bricks on each step on the second bridge... How are the odd number of brick layers created in one step then ?
will try in Polybridge
So, once we've built our infinite bridge, can we cross it?
Wouldn't just build a diamond shape give you a n^(1/2)?
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The center of gravity will always stay center. And twice the bridge span leads to twice the width and twice the height so four times the coin number.
I always love these videos, but contents aside for one moment... If Kelsey (or someone else who knows) reads this; what's your shoulder tattoo? I always think of it as an octopus, but I'm sure it's probably not. Been bugging me for months.
You used a kind of 'greedy' way of stacking subsequent bricks, where each brick was placed as far as it could go under the previous conditions, i.e. the first one was placed 1/2 a distance away from the table, and then because of those conditions already, the next brick could only be placed 1/4 and so on. How do you know that this 'greedy' method of placing bricks is the optimum method for stacking one brick at a time? What if you placed bricks taking into account the moments of the future bricks? Could you use some sort of optimisation mathematics like dynamic programming to know how far to place each subsequent brick, taking into account future bricks?
it felt nice to know i was right about the power of the parabolic bridge, by gust guesstimating, good to see my master in physics at work
more to the point, could you make a video on cnoidal (Jacobi elliptic functions) i had and still have a really hard time understating and explaining them, or about non harmonic waves in general, i keep trying to find an easy way to explain these to friends and family
About the comment on internet cryptography breaking down: As long as there are problems that are easy to state and have an easy way to check a solution, but which are very hard to solve, there will be potential for cryptographic algorithms, right?
The table's edge is round lol