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This video features Dr James Grime: www.singingbanana.com
His KZhead channel: / singingbanana
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Some papers about the Goat Problem...
Return of the Grazing Goat in n Dimensions: www.jstor.org/stable/2686558
A Closed-Form Solution to the Geometric Goat Problem: doi.org/10.1007/s00283-020-09...
The Grazing Goat in n Dimensions: www.tandfonline.com/doi/abs/1...
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I love how mathematicians casually talk about goats grazing in 5 dimensions whilst frowning upon tangible real-world numerical answers...
Engineer stops listening after hearing that it's about 1.15
A physicist would probably just approximate the problem by a harmonic oscillator. And outrageously, it would probably work, somehow.
@@zlosliwa_menda a harmonic oscillator moving in n dimensions and then taking the limit as n approaches 5.
Math is less about reality and more about the beauty of equations and patterns. I love that about it
@@happy_labs 1, take it or leave it.
I love Dr Grime . His smile is infectious and it just makes me excited to learn more .
There's a reason his channel is called "singing banana"!
@@Irondragon1945 It's because early on in his KZhead career his metamorphosis from sentient banana person to normal human person hadn't yet completed.
Yo his handwriting has punctuation :) such an expressive person
yeah, he doesn't deserve such a name.
Right?! Dude just absolutely lives and breathes mathematics :)
15:35 it actually is an important problem! I had to use it for my research in biology! Basically it was to calculate how the effusion of a substance in a circular arena affects animals and I stumbled across it online when I realized how difficult it was to calculate by hand, really great stuff!
lol that's kinda awesome! love it when stuff like that happens :)
The goat problem might be important, but finding a *closed form* solution wasn't. In any real-world application, a finite number of significant figures will do.
sea slugs? 🐌 :p
Goat droppings?
I was also thinking surely it has some use in physics, or computer game physics, where proximity radius is used a lot in collision and LOD etc
I think Grimes is one of the best people featured on this channel Every video is a joy to watch
He sure went a long way from his time on The Simpsons
Very glad he made it back safe from his Caldonian expedition.
New information era scientist??
Wow! Potato is here!
"Grimes"
James Grime is one of my most favorite personalities on Numberphile. You guys really feel like a friend :D and I would definitely recognize you guys in public!
hopefully you've checked out his personal channel, singingbanana! (if i recall correctly?) go and support him!
singingbanana is great indeed
Great educator. Genuinely excited by math.
For those interested in the trig: 1) Area of arc part (swept by tight tether) A1 = r^2 α/2 = 2α cos^2(α/2) = α (1+ cos(α)) 2) Area (swept by circle radius over the part of the circle that goat can visit) A2 = π-α 3) Overlap (four equal right-angled triangles) A3 = 2cos(α/2)sin(α/2) = sin(α) So we have to solve A1 + A2 - A3 = π/2, or α (1+ cos(α)) + π-α - sin(α) = π/2 which simplifies to α cos(α) - sin(α) + π/2 = 0 This gives α ≈ 1.90569572930988... (radians) r ≈ 1.15872847301812...
α actually have many solutions, but we are only looking for 0 ≤ α ≤ π/2.
@@earldominic3467 that would be 0
Duh!
Couple of friends of mine wrote a paper years ago on a generalisation of this problem and its connection to optimal siting of a radar jammer, or nodes in a mesh network to avoid mutual interference. It was called "On Goats and Jammers" and the technique used there was to split the problem into two integrals, one for the real part of the problem and one for the imaginary part (Shepherd and van Eetvelt, Bulletin of the IMA, May 95). The abstract says "The technique is a generalisation of the classical “goat eating a circular field” problem, which is resolved in passing".
Awesome!
Are you saying this was (potentially) solved decades prior? I can't find the paper online (the ResearchGate page has nothing on it). A link to the journal archive would be appreciated.
@@YjDe-qe8xt Researchgate: "On goats and jammers", S. j. Shepherd and Peter van Eetvelt, University of Bradford, January 1995.
@@davidgillies620 Doesn't work. When you try to download the paper all you get is a photo of the second author. The paper doesn't appear to be digitised anywhere else either unless it's in some obscure archive. It'd be neat to give those guys credit if they really got to the solution first. Maybe you could ask the authors to upload again to ResearchGate?
@@YjDe-qe8xt I'm afraid I've lost touch, this being thirty years ago now.
James Grime being friends with Graham Jameson is almost as impressive as the goat situation
Graham Jameson being friends with the charismatic and brilliant James Grime is even more impressive.
This will probably be said later in the video, but it just dawned on me that r tends to sqrt(2) in high dimensions because the volume of high-dimensional hyperballs is increasingly concentrated near the surface (a fact I probably learned from another Numberphile video), and r=sqrt(2) always halves the surface exactly.
Lovely intuition
Oh! That's a really clever observation
Ooooh, That's fair enough. I was worried about how sqrt(2) would always be halving the surface as there would always be some excess volume, but I suppose that would tend to 0 as more and more volume became concentrated far from the centre.
ima need to reread this later. i read that 4 times and didnt understand it.
So, the new challenge is to solve it in 1 dimension.
0.5
@@jk-kf7cvActually, 1. The _radius_ was 1, not the diameter.
@@LunizIsGlacey ohsh*it you’re right because the length would be 2 in this case😅
@@jk-kf7cv Ye lol haha
@@jk-kf7cv give this man a fields medal
That practice explanation at the end is so important, people always complain about money being spent on research that yields nothing or random seemingly useless knowledge but the researchers have to learn, improve process and tools somehow. Satisfying curiosity is important to help people focus, also tiny findings may help someone else with their process in the future.
a theme in mathematics and scientific research is figuring out something seemingly random and useless only to find it get used 10000 years later to solve even more advanced problems
Yes, and those researchers are also lecturers too. Even if their research is completely unimportant, you need the researcher to be invested in their field so they remain there and their skills are kept alive by new students.
Science has 2 directions with math Either the problem has been posed and solved before or found applications Or science came across an equation that math hasn’t solved or considered like Fresnel integral
There are an unbelievable amount of "pointless" problems that ended up having unexpectedly applicable solutions.
The most famous example was Maxwell equations of electromagnetism.
There are very few people I've been watching on youtube longer than Dr. Grime. It's always a treat to see him pop up here.
I appreciate the response at 15:22. Me trying to explain why I “waste time” programming things that are fun but don’t matter to anyone but me
"Because I enjoy it" is never a waste of time. We all need to remember that we're humans, not money printing robots. Fun is an important part of the human experience, even if it's not profitable.
I always love seeing Dr. Grime on this channel! ❤️
All we need to do is construct a collapsible Peaucellier-Lipkin linkage and tether the goat to that. Then the boundry of it's constraint will be a straight line instead of an arc, and figuring out the necessary length will be easy
Halfway thru this video I started working on the tether system (pickets, ropes, pulleys, cams, etc.) that would constrain the second goat to allow them the other half of the grass without infringing on the first goat's share. 🤔🤨🤯
@@JupiterThunder 🤣
Been following Dr. Grime on Numberphile for years and it’s always a delight to see his enthusiasm. I’ve been away from recreational maths because full time job gets in the way, but this video reminds me of those puzzle cracking days, which were awesome. And it’s also really really nice to see Dr Grime not changing a bit in his passion talking about maths in an accessible way to the general public.
I was taught this problem at school, and I think I recall that I was told that it was solvable exactly using only secondary school maths we had already learned. We spent the entire lesson trying to work it out, and it's not left my mind for half my life.
A great question to humble anyone. I thought this was easy until I actually tried it. Looks like there's still a lot to learn.
Yeah it is fascinating which joy and knowledge can be hiding behind such a simple looking problem. Oh an what I immediately noticed was that the goat didn't sound particularly healthy. Which, by the way, is completely normal for "experimental goats" and especially mathematical experimental goats:P
Imagine finding out you didn't actually know everything. Truly humbling.
I actually had a similar kind of problem spring up with my job recently. My work was planning on retro-fitting one of its vessels with two new cranes, but they wanted to know the overlap of their work envelopes because both cranes sometimes need to work together. Each crane cost about £250K so they needed to know if it was worth it. I remember wondering why I didn't know exactly how to calculate the overlap of two circles & decided I had better things to be doing than doing geometry for an hour...
When I was 14, our teacher (best I've ever had) gave us a similar problem, only backwards: If the radius of the circle is 8m and the goat's rope is 6m, what percentage of the circle can the goat go graze? And yes, that's solvable.
With the goat being fastened to the fence again (and not the center)?
@@wullxz you can work out the area eaten just by putting the rope length into the trig you've calculated for the area, then you just divide area eaten by total area of the circle to get a fraction. This is exactly how you could go about approximating the rope needed for 50% - it's going to be more than 4 and less than 5. Just keep narrowing it down - more or less than 4.5? 4.25?
It is way easier to calculate. That is how approximation works: you are guessing what to use instead of the 6m, so the answer would be close to Pi/2
.... you were 14? What a teacher that must have been, to give you a shot.
That's a contour integral symbol not specifically a complex integral symbol. AFAIK there isn't a special symbol for a complex integral.
This is true, but at the same time complex integrals are almost always contour integrals, and using the shorthand "the circle means it's a complex integral" seems reasonable in the context of a divulgative video that isn't even about integration
It is a complex integral here though
The usual notation that suggests that one is dealing with a complex integral is the use of “z” as the variable of integration.
I think you're suffering from Mann-Gell amnesia.
I love how happy James is giving his friends a shout out
That's impressive. I don't understand the question "Why did he do that?" Why wouldn't he do it? It's cool.
That final formula was stunning. Been a while since I saw some math really outside my understanding - gonna have to investigate those complex integrals. Thanks Dr. Grime!
Always nice to see the Singing Banana back on the channel!
There is still a demonstrable lack of both singing and bananas. I want my money back.
I thought that the answer was something along the lines of "First you stay with the goat while the wolf brings the cabbage across the river..."
I love how James intuited the square root of two answer. Just shows that he thinks in higher dimensions. ❤️
His three years as a maths undergraduate were clearly well spent.
I love James, what an inspirational maths man. Been watching his videos since he first starting uploading
1:01 That is either a really sick goat or Chewbacca.
I worked this out in Geogebra several years ago. I could only approximate, just like the recent solution. It taught me a lot.
James is always so happy, it makes me very ready to learn
a little problem: the exact solution to those integrals would involve the residue theorem, which requires the poles (zeros of the denominator) of the function. setting sin z - z*cos z - pi/2 = 0 we get sin z - z*cos z = pi/2, which is the same equation we started with, slightly rearranged. maybe there's a better way to evaluate those integrals that i'm not seeing, but complex integrals are intrinsically connected to those poles in the integral domain, so i feel like whichever way we look at it, we have to solve this nasty equation.
That's interesting, as my main question after watching this video is "OK, but how are those integrals that I don't know how to do, a better answer to the problem than that equation I don't know how to solve?" And you seem to be saying that, actually, it isn't.
@@beeble2003 it is a closed form answer. It is better that just numerical approximation in that it contains a full solution. It's just not practical to compute. But i guess you could make a decent asymptotic analysis of it from it
@@your-mom-irl Expressions including integrals are usually not considered closed-form solutions.
Our maths teacher, when we asked him, did find a solution by resolving integrals (real numbers only) in French "maths sup" class. Quite computational, but he found a solution. That was about 30 years ago. I still remember the sketch of his computations: he divided horizontally the grazed area into two. Both left circle segment and right circle segment are curves of which we know the equation (but we don't know the intermediate bound of the integrals). The problem then is to compute the integral under these curves, and equate it to a fourth of the area of the field, so to find the absciss of the common bound. I do not remember if he found a closed form. Now I watch this video, I think not, but it was long ago and I could not sware.
I'm not your teacher, but I found solution 40 years ago. I hate trygonometry so i use analytical geometry with integrals of y = sqrt(1-x^2) which is arctg(x) :D irc. I think it was always pretty solvalble problem.
@@jarosawmaruszewski1678 I think that was the approach of my former teacher too. BTW, arctg() is a kind of trigonometric function, isn't it? :-P
@@jarosawmaruszewski1678 I too discovered a truly marvelous proof of this, which the KZhead comments are too narrow to contain.
it's nice that you remember that
@@jamaloney1 Fermat
If this was in fact taught at naval academies I have a suggestion why. This is a wonderful illustration of 'picking the right tool for the job' or why you should always consider alternative solutions if the original plan becomes too complicated. If the goal is to have the goat graze half of the field, the easiest solution would be to ditch the rope and just build a fence :)
An alternate suggestion: the US Naval Academy's mascot is a goat, and a math prof thought the problem would be à propos
Better still buy a second goat and let them work it out between themselves. According to the British Goat Society, tethering "is the worst form of management". Another site (thefreerangelife) states: "Do not get just one goat. Ever. They will be sad, depressed, and unhealthy and probably quite loud as they call out for some company." "Each goat should be provided with at least a quarter of an acre of space." (Source unknown, but they mean a UK acre, a quarter of which would be just under 1,012m².) You can figure out for yourself how much a fence would cost - the nearest approximation I can find is exorbitant.
I'm dubious about the claim that it was taught in "US naval academies". Either Dr Grimes mangled it in the telling, or it's an urban legend. There's only one US Naval Academy.
Is seems like it's related to pursuit problem or a search of an area.
I was given the goat problem by a lecturer at college many years ago and never thought about using angles as the starting point as such. I got given the problem as I had solved the ladder & wall problem fairly quickly. Thanks for the answer.
The long running US radio program Car Talk posed this problem: semi-trucks, aka lorries, have cylindrical fuel tanks oriented horizontally. A caller wanted to know where to put the marks on a dipstick to be able to measure 1/4, 1/2, and 3/4 levels in the tank. Both of the hosts, being MIT graduates, say, "No problem!". And after a few minutes they start to realize this one may be a bit tricky...
I got the chills when Dr. Grime said "it tends tooo.... the square root of two *drops mic*"
9:15 "Polynomials will have an exact solution" - Galois is freaking out!
Something something x^5 + x - 1
Funny enough, the formula shown is degree 4, thus it does have a closed form solution. You can just put solve 3r^4-8r^3+8=0 into Wolfram Alpha, tap "exact forms" and you're done.
@Milan_Openfeint "If it is a polynomial then that will have an exact answer " This is the exact phrase; which is wrong
@@TheKnowledgeNook777exact meaning “an answer that can be expressed as a formula”
@@fulltimeslackerii8229 No; exact solution of a polynomial means the answer involves only +,-,*÷ and taking n-th roots operations performed on coefficients of the polynomial
"Here's the answer!", that was classic James Grime Gold 😂
HEY!! The 1-dimensional case (a line) is pretty easy to solve... r = 1/2 exactly. Now, where's my Fields Medal ? 🙂 Edit to add: Have to be careful how the line is defined in terms of "radius". r = 1 exactly if the line is 2 units long.
What about the 0-dimensional case?
@@Kumagoro42 0 dimensions lacks the meaning of length and thus the question can’t even be asked.
@@Kumagoro42 No grass so "no"
However, the very concept of a soft rope is impossible in 1 dimensions. Let the length of the line be equal to 1. Unless matter within the rope is destroyed, the linear goat would be forcibly fixed at whichever point on the line the rope terminates at, since no extra dimensions exist to accommodate any extra "slack" of rope. Therefore, no matter what r is equal to, the effective length that the linear goat can travel along the 1 dimension is 0. Then again, the rope would be infinitely thin. So thin as to not be able to exert any tension force on the linear goat. In this case, the goat can travel anywhere along the line, with the effective length being 1. Either the goat can travel all of the line, or none of it. There is no half. As the old saying goes "do, or do not. There is no try".
I think the radius doesn't matter in 1 dimension because it has always an area of 0. Or it has infinite answers or the area should be compactified like the 10 dimensions in string theory.
We know James is the G.O.A.T in Numberphile.
I remember solving this for a programming contest. Of course, you only needed to compute an approximation (up to 5 decimals or so), and you didn't need much math since you could just do binary search.
Man I do love James And I was wondering why this video got so many more views than recent vids And going through the comments suprised to see how many James appreciators there are
If you squint your eyes a bit, the complex integral symbol looks like a Treble clef
It doesn't matter what length rope, the goat will eat it. The tether needs to be a chain length ;-)
Having worked out the length of the rope is only a part of the problem. The length of the goat's neck (distance from collar to front of teeth) needs to be added (or subtracted, if you prefer)
I seem to recall hearing about something similar, but it involved miniature golf. Or maybe baseball.
I was given this problem about 50 years ago, with a 50m radius field. After several days of complex trig equations I came up with an equality which I tried to solve iteratively by hand. I came up with (from memory) an answer of 57.18m. The person that gave me the problem hadn’t been able to solve it and didn’t know the answer so I never knew if it was even approximately correct.
Whenever you see a variable inside and outside a trig function together you know you're in trouble.
Although the 3-d answer was messy, it was so satisfying to know that it is infact an exact answer.
Came late to the video, but I just love any video with James. Thank you!
It's a insanity question.
FWIW, we had an exact answer from the start. It was the solution to π/2 = r² acos(r/2) + acos(1 - r²/2) - r/2 √(4 - r²). That solution can be found numerically, and it is "exact" in the sense that the solution to this equation also solves the original problem exactly; no approximations were made in the analysis. This is also exactly the kind of "easy answer" James is talking about with respect to polynomial equations. For instance, the equation x^5 - x - 1 = 0 has one real solution (and four distinct non-real solutions), but it cannot be represented by an elementary expression. The best way to write the answer is "the real solution to x^5 - x - 1 = 0." You can introduce new functions like the hypergeometric function to create an expression for this solution, but it's not a general method; if we move up to x^6 - x - 1 = 0, the solutions can no longer be represented with the hypergeometric function. There is no general way to solve these that is significantly better than just inventing a function that solves polynomial equations by definition. And this same problem arises in higher-dimensional versions of the goat problem. What we have now is a _closed-form_ solution to the goat problem. It's no more exact than the original, nor is it any easier to compute. It still can only be found numerically and only to finite precision. So it's no more or less "exact," just a different way of writing it, but it's nice in that r can be represented with a single mathematical expression. For what it's worth, whether this actually counts as a closed form is also debatable, since expressions involving integrals are usually by definition not considered closed. Traditionally, a closed-form solution used only a finite number of operations. In fact, this is the first example I have found that describes an integral expression as a closed form. In any case, this is the first time anyone has successfully written any mathematical expression _at all_ that exactly evaluates to the solution in question without making up new functions specifically for the problem at hand.
I know it's a bit of a debate as to what qualifies as 'closed form' but I'm kind of surprised that an unevaluated integral does.
@@TheEternalVortex42 It's particularly disappointing when you realize that the process of evaluating these contour integrals amounts to solving the original equation but with extra steps.
Well said.
I thought there was a way to calculate the areas separated by a chord, and this is two overlapping circles that share a chord, so my first thought was to calculate the grazing area as the sum of the appropriate portions of each circle
That quartic equation for the "bird in a cage" reminds me about one of the first topics that baffled me when I was younger: a general formula for the cubic equation. There's also one for the quartic equation, but not from fifth onwards. I think explaining about it would make for a couple of nice videos about Polynomials and Group Theory.
I've missed you on this channel, James! Glad to see you back 😁
This was such a nice problem! I also had written down a small geometry problem, totally of no use but to pass time during a bus trip; but I couldn't solve the problem myself. It is on arXiv with id: 1903.09001 The problem is about n "lighthouses" which are circles with radius 1, placed around a common center, equidistant at n units away from the placement center. Consecutive lighthouses are separated by the same angle: 360/n which we denote as α. Each lighthouse "illuminates" facing towards the placement center with the same angle α, and the question asks the total amount of dark area behind the lighthouses. There were two variations, I solved one but got stuck on the other one. Take a crack at it if you would like!
I heard of this problem years ago when I was in school (probably 50 years ago) and I could never work it out. The internet finally gave me access to the brain power needed to solve it. Such a simple looking puzzle with a nasty twist. Thanks for this explanation of the solutions - very entertaining.
It's called insanity
At first I was really confused at why this is a hard problem and a Numberphile video because Im pretty sure we had this in a school exam. But then I realised that it was only approximated answer using trigonometry and the actual solved answer gets pretty damn hardcore. That exam was my first and only math exam which I got full points 36/36 :)
When i read the problem i paused the video and spent an hour solving to get 1.1587. I came back to the video all proud of myself and found out that wasn’t what you were looking for.
This is interesting! But actually, the integral formula for alpha isn't really all that mysterious if you've taken complex analysis. Basically, complex integrals, defined as summing along the contour analogous to real integrals, can also be evaluated by finding a specific number (called the "residue") associated to each of the function's poles enclosed by the contour. Notice that in the formula, the pole of both integrands is the solution to the equation sin z - z cos z = pi/2. The integrals are arranged so that when you do the residue calculations, you get the value of z at the pole, which is your solution. So, the bulk of the solution is to play with these integrals a different way to try to get a closed form.
I was very underwhelmed by the final "answer". Finding explicit zeroes to analytic functions is easy if you accept residues. I was expecting (well, hoping) for an elementary transcendental expression for the angle. Sure, this isn't a grade-school level answer, but it's certainly an undergrad-level one. I may actually assign this as a problem next time I teach complex analysis...
@@EAdano77 I was extremely underwhelmed too. The integral is itself a limit of a sum, so I don't see it as any better as presenting alpha in terms of some other limiting process (e.g. iterates generated by Newton' Raphson).
Square root of 2 felt very wrong for me, at least for a circle, because the picture that flashed in my mind was very simple: the rope end cuts the field border in two points that form a diameter, so there already is half of the field on one side of this diameter, so everything between the diameter and the arc formed by the end of the rope is too much. It feels like it is the same thing in higher dimensions, but I guess it means the contribution of this extra volume becomes smaller as the number of dimensions increases.
Yes, this was exactly my reaction too... square root of 2 feels very wrong because it guarantees the goat will be able reach the half-way point *at the boundary* of the field (regardless of the number of dimensions), and therefore necessarily be able to eat more than half.
This guy is the reason why I love this channel
I always enjoy it when Dr. Grime hosts
Complex analysis was one of my favorite topics in uni, sad to have forgotten it all now haha
Greatest of all time is in my view the members of numberphile team who always nail great problems
Never has an expression in a numberphile video caused such a physical aversion in me, and this one has _multiple_ of those!
I first came across this puzzle at an Open University Summer School in 1982. Of course I tried to solve it but after ending up with many, many terms such as sin(sin(θ)), I thought I'd just gone wrong. I was convinced there must be a straightforward, simple solution, even if it eluded me at the time. Now , 40 years on, I know. Thanks!
I like seeing new developments for old problems. Another one was the recent closed-form solution (well, AGM anyway) for the exact period of a pendulum. Did you cover that already?
“you think Alpha is grassy?” is now my favorite Dr. Grime quote
Aw man, math is so fascinating. I don't understand most of what was just happened, but it's fantastic that people can solve problems like these without having to procure a goat, a circular plot of grassy land, and a piece of rope.
Thank you for great presentation. I took a piece of paper and guesed that rope should be somewhere 1 + 1/(2pi). Did not expect so complicated solution.
I like misleading problems like these Always a good riddle for friends
At first glance, i don't understand the culprit. My solution would be to write double integrals for this area. The change in integrals is where the Goat Circle intersects the field circle, so this requires calculating this point X. I have two circles, one x^2+y^2 = 1 and second (x-1)^2+(y-1)^2 = r^2 . Let's consider only top half as we have symmetry along x axis. We get the intersect at x=1-0.5r^2. Now we write two double integrals : 1)From (1-r) to (1-0.5r^2) dx and from 0 to (sqrt(r^2-(x-1)^2) dy 2)From 1-0.5r^2 to 1 dx and from 0 to sqrt(1-x^2) dy The sum of those should equal to 0.25pi (half of semi circle (quater of the field). ... OH. Ok Integrating this is fine, but what comes after is a monster. We have a polynomial equation with r at degree of 3 and r inside inverse trig functions. Hah.
I watched the rest of the video, i might've switcher to polar coordinates :D. Don't know if i lost anything in my thinking
😳😶🌫️
Integration is likely how they approximate the solution
@@Macialao I've reached a mixed polynomial/trig equation every time I've tried to solve this. For me, switching to polar coordinates and doing the integral over the upper half using symmetry seems to give the simplest route. I always had to solve the equation by numerical methods giving an answer around 1.16, so I'm impressed that someone has found a closed form for the solution.
@@RexxSchneider I wonder what do they mean by going to complex numbers. Maybe they switched to Euler form, found out imaginary solution which might've been simpler and they figured out the real solution by working out the symmetries in complex plane.
I can't believe a seemingly easy problem can have such a complex, humongous and ultimately ridiculous exact answer. I guess after all the hard work, this problem really is the GOAT of all (easy) problems.
As I was watching, with the grazing goat problem. First thought I had, assume the rope is connected at 0 radians (directly to the right) of the unit circle. Draw a horizontal line down the center of the circle (y axis). Find a rope length where the area of the goat’s circle on the left side of the circle centerline matches the unshaded area on the right.
I feel the video would've been much more interesting if James went a little deeper in to how people got to the old approximation of a rather than just giving us the number.
What the heck does "Nick is my friend for other reasons" at 9:58 mean?? 😂😂😂
They probably play D&D together
Means he wasn't his lecturer lol.
We were both members of the Society of Lancaster University Jugglers.
@@nojameson Hi Nick! I didn't expect to get an answer, but that is awesome!
James Grime’s enthusiasm is wonderful. In life nothing is cooler than enthusiasm.
Tending to square root of 2 is super cool. From start to end of all dimensions from the two square two dimensions to the infinite dimensional circle.
I found it odd how much James danced around saying the words "closed-form solution" during the entirety of the video...opting instead for multiple rephrasings of the much more vague "exact answer"
Not to mention how James stressed that polynomials always have an "exacr answer"... Galois turning in his grave!
@@energyboat4682 I was looking for this comment
Props to the intro animator with the dots and then just few poops to change their being, changing the acronym to an actual word, goat. And just in a few passing seconds. I love it.
I love that answer about why the mathematician spent time working on the problem. Bravo.
the incredulity of "you think alpha is grassy?", I love James
So wait. The exact formula we got for the 2D case is a fraction of 2 integrals. But are those integrals "computable" to an exact formula? I know there are some integrals that are impossible to write down to a closed form. Do we know that this is not the case here?
Exactly what I wondered. It doesn't look easily computable which is what I would expect a closed form would be
A closed-form solution usually doesn't allow for integrals to be part of the solution. So I don't think is a closed-form solution, but it is an explicit (alpha = some expression without alpha) rather than implicit function which defines the angle, which is still an improvement.
These are integrals of a complex function over a closed curve in the complex plane, which are usually extremely easy to compute: they always equal zero. The exception is if the function you are integrating has a spot inside the circle where it looks like 1/z at 0, a spot at which the function is discontinuous and goes to infinity, like 1/x does at zero. Those two integrals each have a spot like that, actually at the point alpha, where alpha is the angle from the original equation. So really, all they have done is taken the original equation for alpha and disguised it as two integrals, but this is a purely mechanical transformation that can write the root of any equation as two complex integrals like this.
Just tie a goat in the field and gradually make the rope longer. When the field is half eaten measure the rope. Quick Maff
I love the fact that you say the root 2 answer feels right - the human brain is quite amazing when it comes to things like that.
The fact that it tends to sqrt(2) when dimension gets big is quite funny because there's this thing I don't exactly recall perfectly called infinite norm which is the max of all the coordinates of a vector. If you define an unit circle with the infinite norm, you get a square of side 2 the diagonal of which is 2×sqrt(2) which almost links to the result !
The difference between an exact and an approximated solution is a little bit semantic here. Sometimes you can't write a solution in terms of sums and products and powers of rational numbers, but so what? You can't do that with pi either, we just happen to have a symbol for it. Otherwise you'd have to say cos(x) = -1 had no "exact" solution. But I can give a symbol for the solution to this problem and then claim I can solve it exactly by producing that symbol as the answer. The point is, if you can describe an algorithm that gives you a solution to arbitrary precision, that is the same thing as an exact solution.
Could anyone explain how he derived the equation at 4:01 ? it is killing me! lol
You can draw a line through the goat's tether point and the center of the circle and make a triangle out of that (length 2), one of the two arms (length r) of angle alpha, and a third line closing the shape. By Thales's theorem, the angle between the length-r line and the third line is 90°, so now you know two angles (90°, alpha/2) and one side length (2), which is enough to use the law of sines to find r.
First draw line connecting the vertex of alpha to the center of the circle. Then draw another line connecting the center of the circle to one of the other points on the circle. The first line bisects alpha, and the two lines both have length one, so the new triangle formed is isoceles. Draw a line that bisects the third triangle leg, r, and you've created two right triangles where cos alpha/2 = r/2. Let me know if you have any questions!
@@fauxpas4589 Thank you! makes sense!
My first immediate thought was, "Can't we just integrate this??"...but I had NO idea that complex integrals were a thing too!
it's cool how you can see the architecture of the quartic formula in that exact answer
Interesting variation: The goat is restricted to the _outside_ of the circle and can wrap its rope around to exactly reach the point opposite its anchor. What's the area of the region it can graze?
That's hilarious.
My question is whether you can use this same method of integrals for the higher dimensional problems too - is it a more general solution?
Yes. It's a basic application of the residue theorem.
I tried solving this myself while watching the video, using my school level math. Took me a couple hours I did come up with a different though very less elegant solution for the 2d case. Assuming the radius of the field is 1, and goat should only have access to half the field you simply have to solve: pi/2 = a - b + c - d where, a = (x^2)*arcsin( -x/2 ) + ( -x^3 / 2)*sqrt(1 - x^2/4) b = (x^2)*arcsin( -1 ) c = arcsin(1) d = arcsin( 1 - (x^2)/2 ) + (1 - (x^2)/2)*sqrt( 1 - (1 - (x^2)/2)^2 ) Which, when I plotted it into desmos turned out to be around x = 1.1587284.. so it was surprisingly accurate I thought. The way I produced this monstrous equation was by cutting the field in half to get two semi circles that can be written as functions. And realizing that by symmetry, if the semi circle overlaps with half the area of the other semi circle, then that's the same solution as the full circle. And then because these are now functions, I took an integral equation to get the overlap. Finding an equation for the intersection wasn't too bad, and the bounds were a little funky. The integral of square roots get the arcsins, and the funky bounds made the stuff inside the arcsins and square roots a little funky. Hence the funky, a,b,c,d above. P.s. I don't know if this is solvable exactly, it's probably also transcendental, this was just a fun exercise I tried out, crazy to think people can find exact numerical solutions to these kinds of things.
Glad to have watched this in the last 5 minutes to 2022!
the 1 dimensional version is my favorite. r/2!
I love that this is correct even when viewed as factorial
@@phenax1144i didn’t even consider that, amazing. 5:17
I'd wager any problem unsolvable, should be considered a big problem.
This is amazing. I could see myself working out something like this when I get into college.
An interesting question me and my friends came up with when we were in school, which looks very similar to this, was how far apart do the centres of two circles (of radius 1 wlog) need to be so that the three areas are all equal, named the venn diagram problem because of the shape. You can also get pretty far with just school-level maths!
of course generalise to your heart's content
Y'know that's actually a kind of interesting question. Best of all, once you solve it, you could use it to draw the perfect Venn diagram of equal areas haha! Thanks for sharing. Edit: after trying it for a bit, unlike in this video haha, I was unable to find an exact answer. But I could approximate it: unless I made a mistake, the two centres should be placed approximately 0.8079455066 units apart (if we are dealing with circles of radius 1).
Interesting. So as James said : As the dimension N of the problem increases, the problem gets easier to solve and eventually, as N goes to infinity, the length of the rope goes to √2. But dimension 1 is also very easy to solve : imagine a line of length 2 (so "radius" 1), and the the goat tied to one end of the line. Then for the goat to graze half of the line, its rope must be length 1. Now the question is : is the length of the rope always bound between 1 and √2 ? Also, as we saw, in dimension 2 it's approximately equal to 1.16, in dimension 3 to 1.23, in dimension 4 to 1.27, and in dimension 5 to 1.29. Is the length of the rope always strictly increasing with N ?
This is a non sequitur. The solution tending to the square root of two does not make the problem easier to solve unless you are asking for the limit itself.
Yes, the length of the rope is always strictly increasing with dimension n. Proof sketch: Assume the n-dimensional case is solved by length r. Try using the same length of rope for n+1 dimensions. Consider an n-dimensional slice that contains the center of the (n+1)-ball and the tether point. By assumption, the goat can reach exactly half of the grass in the slice. In any other n-slice parallel to this slice the goat can reach strictly less than half of the grass. In total, the goat can reach strictly less than half of the grass. Hence the rope needs to be longer in n+1 dimensions. Q.E.D. by induction. By "n-slice" I mean intersection with an n-dimensional hyperplane.