Cannons and Sparrows - Numberphile 

It's not that hard to understand you take the intervals of separation and now I'm lost

lol

It really wasn't too hard... when he said this is a topology proof, not a calculus one, because the perimeter must be continuous but not necessarily smooth, I immediately knew what he meant. Because calculus can't deal with those corners. They're not part of a function. Cause functions don't have corners. But shapes do. And the intermediate value theorem applies to shapes.

"Let's write a program to generate all possible real conceptually random homosedastic normally distributed data values. From there run the simulations with varying degrees of noise...."

Got a real laugh out of me. Had that exact feeling trying to help my young brother with his algebra. Didn't remember how to solve it the they wanted him to and I was like hmm well just find the area under the curve with this neat trick involving infinity. The look reminded me that whoops.. This is algebra.

Olaf Doschke "Mit Kanonen auf Spatzen feuern"...

That seems more akin to the English proverb "a bird in the hand is worth two in the bush".

This is completely unrelated to the video and possibly of interest to nobody but me, but the exact same proverb exists in Polish - as in the literal translation is identical. I found this interesting because it seems so rare to find a proverb that's the same in two or more languages; usually you'd find something with the same meaning, but phrased in a completely different way. Came here for the maths, learned a linguistic fun fact while the video was buffering. This is shaping up to be a fairly productive evening.

The second proverb isn't related by similar meaning, just also being about sparrows...:)

+Olaf Doschke Oh, I see.

Swallow, not Sparrow.

SirJMO, The Ministry of Truth has been notified of the historical error in the Monty Python document. Winston Smith has been assigned to review said document, and correct this and any other errors that can be discovered.

SirJMO whoosh

super funny man, you should do stand up comedy. Nobody ever has made this Monty Python reference about sparrows.

TrasherBiner Screw you. that was funny

That was the first thing I noticed in the thumbnail; this book made me the math lover I am now

B Spits that is why I started watching.

Isn't it crazier knowing that it doesn't work for prime products? It's almost like encryption in reverse. We know that two primes equal a product that can't be deconstructed. But we don't know what the product looks like. As opposed to seeing a product and brute force decomposing into prime factors. I've been drinking. Pardon my french

@@hunterbelch2524 This might still work for all numbers. We know for certain it can be done for prime powers, but we don't know for certain if it can be done for any other numbers. That is not the same as knowing for certain we _can't_ do it for other numbers.

Hold on a second - you _could_ do it, it's not like it's that easy

And if the edges are cutvy they are even circumferences !

mokopa not dyslexic but I can tell it could really annoy or confuse someone dyslexic

nex I was looking for this comment. Thank you for looking it up!

The question is not "Can we be sure they have the same perimeter?". The question is "Can we assert that it doesn't exist a certain way of cutting Q1 in R1 and R2, and Q2 in R3 and R4, so that R1 has a the same perimeter of R3?" That is the assumption we want to prove or disproof.

And in the case of R1,R2,R3 and R4, it actually always exists a way of cutting them.

The whole 22 minutes is math.... and besides, don’t I decide what my videos are about!?

It's different - wait until you see the result!

only because the proof for two pieces uses the same idea. The problems are not related in any way, this is much harder.

Of

Benjamin Wessel Dative "me"

+Czeckie They are related in some way, since they both use continuos changes between two numbers.

Will Turner which hannah fry?

The ham sandwitch one? I have a feeling that the ham sandwitch theorem video was a prep for this one.

This is really good point. Because it's easy to see that end points of line move continuously.

but, as soon as you modify the sub polygons to adjust the areas, the perimeters will change. with weighted points, you guarantee that the polygons will keep the same area no matter where you move them.

+Luis Pulido Jonathan was (probably) talking about the case where you want to split the polygon into 2 parts using a single line. Not about splitting any into more than 2 parts.

Ahsim Nreiziev indeed

it's the same case I guess. Imagine you start with 2 perimeters equal, how can you balance the areas without breaking the perimeters? this basically means that if you start with 2 equal perimeters then you have a line that cuts the polygon creating 2 points (A and B). If you then want to modify the areas, you have to modify A or B or both! breaking entirely both perimeters (to actually try to maintain the perimeters, you HAVE to move both, moving one only will break both perimeters). So we end up in the same conclusion as having already a line dividing the shape in 2 equal areas and moving the points to find the equal perimeters by moving both points, which basically means rotating the line.

I don't think nine would be happy to be called eleven. Also confusing.

Does it have to be 911 or will any other Sophie Germain prime do as well?

Nice one, Kirk!

Brain cells committing suicide

Numberphile is flexing a bit.

That should work also for all numbers n. For n=6, first divide the polygon into 2 parts and then divide each part into 3 smaller parts

It might not work so easily. Start with a polygon P, cut it in two pieces Q1 and Q2 with same area and same perimeter. If you cut Q1 in R1 and R2, and Q2 in R3 and R4 there is no reason we have the same perimeter for R1 and R3.

Yes, if the case n=2 works for any polygon you could just prove it recursively. Edit: Ops, Pif de Mestre is right, when dividing two different polygons the perimeter won't always be the same

I don't think it's so easy because you have to ensure that each sub-polygon has equal perimeter to each other. Imagine cutting a very wide rectangle, say with x=8 and y=2 in half lengthwise so you have two rectangles with x=4 and y=2. You take the left one and cut it in half like that again so it's x=2, y=2, p=8, a=4 each. The one on the right you cut the otherwise, so they're x= 4, y=1, p=10, a=4 each.

For this to work, we would need to prove that the perimeter of sub polygon in a solution depends only of the perimeter of the starting polygon

It hasn’t even been 22 minutes since upload

I technically wrote that comment just after midnight, so technically this one is going up "tomorrow." Now, a little bookkeeping. Imbed the original shape in the Cartesian plane and parametrize the dividing line L as Xcos(theta) + Ysin(theta) = K, with K varying for different theta. This allows a couple of things. Less urgently, it shows that the appropriate dividing line varies smoothly with direction; moment to moment, it rotates around the center of the shared boundary to maintain the same area, and this midpoint too must vary continuously because the shape was assumed to be convex. More importantly, this allows those of us living in three dimensions to visualize the solution space. Admittedly, UVZ-space is not anyone's first choice, but it's better than reusing X and Y. In this graph, U=cos(theta) and V=sin(theta). (I'm using U and V instead of theta partly because this means graphing points on a cyclic function only once, on the wall of a bounded cylinder.) So L is the line UX+VY=K, and the UVZ graph will relate to the portion of the shape where UX+VY >= K. Specifically, for every UV pair, consider all possible dissections of the half-shape into P convex shapes of equal area and perimeter. The area is fixed; take Z to be every possible value of the equal perimeters on that side of the line. The problem reduces to finding, or proving there exists, some value of theta such that Z(theta) overlaps with Z(theta + pi). This... *ought* to work, since we already chose P to be solved, but it's not certain. The set of perimeters Z is nonempty for each theta, but is it continuous for each theta? One might imagine Z satisfies only a narrow band around Z=1+theta/3 where theta ranges from 0 to 3pi. The solutions would have to have some other quality, like continuity of the upper bound of Z, or no tapering, or somesuch. And that's still assuming we're looking for a solution for 2P regions. Even if it works for all P, that only proves some even numbers. To truly construct all solutions, you need to be able to multiply by *odd* prime powers too, and even 3P carries no guarantee. Two thirds switching, even with their perimeter values varying smoothly enough for the case of 2P, might not match the remaining third when they match each other.

anon8109 you rotate it such that the areas remain the same. Not all rotations will do this but it is still possible. I don't think you would need to limit yourself to rotating it around a point.

Exactly, it's just a ratio so you would need to put it say, in the middle of a regular polygon or solving for that point on other polygon. Of course it only works for one line in most cases

You wouldn't pick a singular point around which to rotate the line. In a sense the line would "wander". But it can be said that for a polygon there is a like at angle x that bisects it into equal areas. This solution would require to prove that more rigorously and then rotate that line through multiple angles and comparing the resulting perimeters.

Right, for any given angle there must exist a line at that angle that divides the area evenly in two. The set of all such lines (for all angles) does not _necessarily_ contain a common point.

Or another way to look at it: Take any point on a shape you're trying to divide, you can draw a line from that point to another point on the shape to divide it in half. Therefore, you can move the line or rather the two points in a continuous fashion around the shape you're dividing.

Technically you understood the problem but didn't understand the solution.

I agree... but I suspect the definition and ITS explanation would take well over an hour.

When he says ‘space of configurations of 3 points’ he is just referring to all possible ways of choosing three distinct points on the figure. In mathematics ‘space’ can mean a lot of things. It’s true that in this case it’s not absolutely clear what kind of space he is talking about, but in general, space means a set of points with some additional structure (metric spaces, topological spaces, vector spaces...). In this case, the set of points would be the set of possible configurations. To each possible way of choosing three different points (a configuration) corresponds a point of the space. In this case it isn’t clear what kind of additional structure he has given to this set of points but is probably at least a topological space that is a way of giving the space a way of telling which functions are continuous by selecting some special subsets called open subsets.

A year after the fact, but this video used many cuts to almost completely skip the argument. Some of these cuts are obvious, but some might be lost in the way the video is put together. In any case, you are more than right that the video does not prove the theorem. Rather, the point seems to be to sort of demonstrate how complicated the mathematics behind the partial proof is. The idea is that a deep theorem is being used to prove results for something easily understood on the surface. Deep theorems practically by definition cannot be proved on this channel. But the idea of using a powerful result to prove a weak one is still sort of interesting, and I think that's where this video originally came from. As a small joke, I reposted a valid but silly proof from _American Mathematical Monthly_ in the comments to try to get the idea across. Maybe more to the point, imagine using the fundamental theorem of calculus to prove the area of a triangle. This is definitely not a great video on the channel. I feel like this one needs some help to be of any value at all.

It's not quite that they forget they're speaking to non-experts, it's more that they aren't sure how much non-experts know, so they sometimes assume incorrectly, either assuming too much or too little knowledge. That's something teachers in every field struggle with.

I like your close-to-perfect sentence!

Just like poetry, it's really just a cryptic way to say something.

There was a story about a guy who wanted to become a mathematician, but he wasn’t creative enough so he became a poet. Anyway this is one of the most complicated numberphile videos I’ve seen and I don’t understand it.

@@misterhat5823 true. And it shows that speaking poetically is easier.

@@connorcriss the story has to do with the legendary mathematician Hilbert. One of Hilberts students dropped out of math to pursue poetry. In response Hilbert said "He didn't have the imagination for maths".

Daniel Macculloch fr fr though

There is a zero percent chance of explaining cohomology off the cuff to a general audience. Even math undergraduates would find it very hard to understand.

And thus, because I say so these two sets are not equal and we have success! If you can't explain it then pick a different topic for a video. This taught me nothing.

If he explains high level Algebraic topology to you, you'll have a stroke! But you're right, he just asked us to believe a bunch of facts without giving a convincing argument.

It's a very dull point.

@retepaskab Suppose you want to plot a semi circle around the origin in an (x,y) graph. The way to do this is by using the function y = √(1-x²) Go ahead and plot it in Wolfram alpha to see what it looks like. Note that the plot only has positive y-values in its reach, due to the square root in the function. Now square both sides of this equation to get y² = (1-x²) This equation has the first equation as a solution, as well as the negative of that first equation. The latter plots the negative y-values, thus making another semi-circle. Rearranging the second equation yields x²+y²=1 Note that by adding variables we go into higher dimensions, so with 3 variables we're plotting on a sphere. Or a 3-dimensional circle, but the basic premise that sums of squares add up to a "circle" holds. Hope this helps!

@Moustaffa Nasaj ... That's half arsed backwards. y = ±√(1-x²) is the equation you get from rearranging x²+y²=1 in the first place. If you wanted to explain why x²+y²=1 is an equation of a circle just explain that - After applying the Pythagorean theorem to the coordinates (x,y) you find x²+y² = r² Where r is the distance to the origin (0,0) So, the equation x²+y²=1 is just the previous with r = 1 So the solutions to x²+y²=1 are all the points with a distance of 1 from the origin (0,0) This is, by definition, a circle with radius 1 centered at (0,0).

It must be so because the first value is equal to the row number. So it can only be divisible by p.

Fermat's last theorem itself is a classic example of killing flies with nukes. And the nukes are invented in the proof.

Interesting, then we have a classic teacher (boring as F) and a guy who makes a living making maths fun and lighthearted. Thanks for suggesting the better one of these two.

James Tisajokt it must be a convex polygon, so I would say no.

+LostName Ah! I forgot that bit.