Tree-house Numbers - Numberphile
ฝัง
- เผยแพร่เมื่อ 18 มิ.ย. 2024
- Matt Parker goes on a mathematician's journey and shows us Heegner Numbers (and the Ramanujan Constant). See part 1 (Caboose Numbers) here: • Caboose Numbers - Numb... --- More links & stuff in full description below ↓↓↓
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Heegner Numbers on the OEIS: oeis.org/A003173
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11:45
"At least I tried"
*brady zooms in on the parker square*
that just feels mean...
I had to laugh out loud. So savage.
Yeah, just the sneaky diss of zooming in made me laugh
🤣
“I hate to be that on brand”
Right in the feels
So, is this a Parker discovery or a Parker proof?
I like when "soon on Numberphile" means "now on Numberphile but you have to click on a link in the description."
That's pretty soon!
Can they doe math on the Bible effects on the mind ?
@@VestalNumbre what?
Loooolll
@@VestalNumbre given the state of your mind, maybe you need a new Bible?
When is Matt gonna write a book called "Things to make and do with terrible Python code"?
async def fork():
while True:
await fork()
await fork()
@@KBRollersomehow I feel that code would just recurse rather than fork bomb
@@TechnoHackerVid Well, yes and no. It'll be a fork bomb, but only until the stack overflows, and then it'll just crash 😂 The proper fork bomb form would be to spin off a thread or process for each of those forks, but that's more code than I wanted to write in a TH-cam comment 😁 (And actually, it may *only* work with multiprocessing; I'm not sure. But Python threads don't work the way most threading does, because of the GIL, so there's a chance a Python thread-based fork bomb wouldn't even work.)
@@KBRoller Just spin off additional python processes with multiprocessing.
@@pierrecurie ...that...is what I said.
e^(π√163) is a Parker integer.
which basically meant it is NOT an integer XD
That classic Parker enthusiasm when he got to e^(π√19). I expected him to say, "It's not an integer, but it's less than 1 away from an integer!"
@@blue_tetris "But aren't all non-integers less than 1 away from an integer?" "Ignoring that and moving on..."
😂
@@blue_tetris I can do you one better: it's less than 0.5 away from an integer!
"The integers are a field"
ah yes, a Parker field
I guess a ring shall be known as a Parker field from now on.
@@mattlm64 an integral domain. I think calling Z/4Z a Parker field would be like calling the square with all 1s a Parker square.
BTW I think he was actually to say “rationals”. I see this state very familiar when he tried to explain what a field is in a simplest way possible off the bat and pausing each word; I find contexts like this blocking my own brain in some way and then I’m derailed and make mistakes; maybe something like that happened. (The joke about a Parker field is still obviously gold unrelated to this attempt at explanation.)
that zoom on the end was hilarious
"The integers are a field"... Such a Parker thing to say
Came here in the comments looking for this comment. Thx!
But... are integers green?
Well, a ring is sort of a Parker field, wouldn't you say?
@@RolandHutchinson actually, i think british mathematicians call fields parker rings
Came here to say this..
8:29 The true journey of mathematics is the stunning joy of discovering something only to realize someone did it before...
And just like Ramanujan was wrong here, he was wrong about the sum of all integers being equal to -1/12
@@threadripper979Actually, Ramanujan never knew about e^(π√163) and the sum of all integers does equal -1/12 in some sense, that is not the usual sense.
@@Nolord_ Sure, if you ignore converging series vs. diverging series rules. LOL
@@threadripper979 Yeah and complex numbers do not exist, LOL.
@@Nolord_ Get real; be rational.
The Caboose numbers fizzled out: I wasn't impressed.
The treehouse numbers give _near_ integers: I wasn't impressed.
Matt finds a connection between the two: I WAS SUPER IMPRESSED.
Matt reveals he didn't find that connection: I wasn't impressed.
Matt gives a little talk on the connection: I was impressed again.
You missed out "Matt writes terrible Python code: I wasn't surprised".
Sorry Matt, I do love you and your terrible Python code.
This is the true armchair mathematician's journey.
What a rollercoaster!
e^π(√-1) gives -1.0000000000000, which is arguably closer to an integer than e^(π√163)
If you follow that through to the corresponding caboose number, n²-n+0 is a prime for all positive integers less than 0. Amazing!
e^π(√-1) = -0.999999999999999999999999.... as many 9s as you could possibly want.
I was going to leave the same comment, but I saw yours~ 👍
@@softy8088 it's weird and wonderful how that statement of yours is true. No counterexamples exist to make it false, so it is not false. So it's true. Or maybe independent. But who's going to make an axiom stating "n²-n+0 is prime for all positive integers less than 0"?
This is related to Г(n²-n)
Brady seems to have forgotten that he's already made a video about these numbers 12 years ago, it was titled '163 and Ramanujan Constant'.
a numberphile classic. back when they used to just do one number and talk about it.
Funny cause I just watched that video lol
I was convinced this could not be the first time Heegner numbers had been mentioned.
I knew this sounded familiar!
Thanks for mentioning that! I've found the video.
Recommendation to the new viewer: They say at the start of the video that you don't *need* to see the first video, and that is true, but the conclusion of this video is deeply pleasing if you watch the other video first.
You need the foreplay to appreciate the climax
"The integers are a field." - Matt Parker, 2024
A Parker field.
Maths God to Euler and Ramanujan: "Why is it always the two of you if something mathemagically strange happens?"
In this case, the answer is: because it's a meme. Ramanujan didn't discover this number and he didn't lie to people about it being an integer. Martin Gardner did as an April Fool's joke and just claimed it was Ramanujan who said it.
Them: "idk why'd you put it there"
Usually it"s Euler and Gauss
Don't forget Gauss!
@@ArawnOfAnnwnYeah Gauss is just crazy, the greatest imo
I think that Matt might have misspoken at 10:20.
The integers do not constitute a field as they aren't closed under taking multiplicative inverses (dividing two integers generally does not result in an integer). Matt was probably thinking of the rationals or the reals
Wouldn't they be a field if restricted to the addition and subtraction operators?
@@dojelnotmyrealname4018 Not as such. Typically in algebra, we don't think of subtraction as a binary operation, but instead as a shorthand for expressions of the form a+(-b).
We can instead equip the integers with the operations of addition and multiplication, which turns the integers into what we call a ring. But the integers still only contain inverses with respect to addition.
In order to obtain a field, we need to extend to the rational numbers (or something bigger).
Or to something smaller! (Modular arithmetic)
it's a parker field
I'm almost certain that he meant the integers modulo prime numbers constitute fields. He definitely misspoke, but I think that's what he was going for.
"At least you tried."
Zooms in on the Parker Square. savage.
"Suspiciously close to an integer" is my new favorite maths term of all-time 😂
I mean, just the pure, firm rigor of the concept 😂😂
Reminds me of a college student's answer to an indefinite integral he could not solve, so he wrote as the answer: 0 + C. When asked why in the world he provided that answer when he knew that it would obviously be wrong, he stated that this is in fact the correct solution to the problem and any indefinite integral problem. His rationalization of this was that give that the constant is arbitrary by definition, we can make said constant take on whatever value we want to make the solution correct for any integral. This proves once again the old adage that a false hypothesis can always lead to a vacuously true conclusion.
They definitely fall within the set of interesting numbers
They could have just defined "Treehouseness" of a number n as the relative deviation from an integer it gives in that the exp(pi*sqrt(n)).
Wait, this wasn't found by Ramanujan! Wolfram Mathworld says: Although Ramanujan (1913-1914) gave few rather spectacular examples of almost integers (such e^(pisqrt(58))), he did not actually mention the particular near-identity given above. In fact, Hermite (1859) observed this property of 163 long before Ramanujan's work. The name "Ramanujan's constant" was coined by Simon Plouffe and derives from an April Fool's joke played by Martin Gardner (Apr. 1975) on the readers of Scientific American. In his column, Gardner claimed that e^(pisqrt(163)) was exactly an integer, and that Ramanujan had conjectured this in his 1914 paper. Gardner admitted his hoax a few months later (Gardner, July 1975).
how do people discover things like that before computers. it blows my mind :D
@@vsm1456 Apparently by not having computers and therefore nothing but time on their hands to compute stuff like this by hand :D
@@vsm1456 Hermite didn't discover it by computing lots of numbers blindly and hoping to find a pattern. There's a special function in mathematics called the j-invariant. On the one hand, there is some theory that tells us that j((1 + sqrt(-d))/2) is an integer if a certain number system (the "ring of integers of Q(sqrt(-d))") has unique factorization. In particular, the ring of integers of Q(sqrt(-163)) has unique factorization. On the other hand, there is an infinite series expression for j that tells us that j((1 + sqrt(-163))/2) is very close to e^(pi*sqrt(163)). If you know this underlying theory then you can deduce that e^(pi*sqrt(163)) is very close to an integer without having to explicitly calculate e^(pi*sqrt(163)).
@@unvergebeneid Funny you say that. The term "computer" was originally used to refer to people who did all of the heavy calculations involved in creating mathematical tables for trigonometric functions, that were widely used until the mid-20th Century in a number of fields (engineering and navigation).
@@vsm1456 Hermite didn't discover it by computing lots of numbers blindly and hoping to find a pattern. There's a special function in mathematics called the j-invariant. On the one hand, there is some theory that tells us that j((1 + sqrt(-d))/2) is an integer if a certain number system (the "ring of integers of Q(sqrt(-d))") has unique factorization. On the other hand, there is an infinite series expression for j that tells us that j((1 + sqrt(-d))/2) is very close to e^(pi*sqrt(163)). If you know this underlying theory then you can deduce that e^(pi*sqrt(163)) is very close to an integer without having to explicitly calculate e^(pi*sqrt(163)).
“Just off the top of my head” … camera points to the top of Matt’s head
Big respect to Matt for sharing all the near misses that happen when doing math.
Like missing the correct definition of a field, by including integers as an example.
This epiphany happened for me in discovering a recurrence relation for multiplicities of eigenvalues in the stochastic matrices generated by the move-to-front rule for the linear search problem. I still remember the moment and it was over 30 years ago. Mathematics is the best.
"If? we'd like to watch more videos with Matt Parker?" Of course we want to watch more videos with Matt Parker. We've aready watched all those.
well, we'd nearly like to watch more videos. We'd almost like to watch more videos. We're suspiciously close to wanting to watch more videos. But sadly, in the end, we're just slightly off wanting to watch more videos.
652 and 1467 after 163 are just 163*4 and 163*9, which means the numbers you obtain afterwards are just the square and the cube of the original Ramanujan's constant.
But that makes me think... If we write Ramanujan's constant as (N-eps), then, we get that (N-eps)^2=N^2-2*N*eps+eps^2. Now, eps is small, but 2*N*eps seems to be much larger than 1, which means that, in principle, it (and by extension the final result, eps^2 is clearly negligible) has no reason to be also close to an integer. And yet, (N-eps)^2 is still very close to an integer? Same with -3*N^2*eps for the cube, and also +3*N*eps^2, which is probably less than 1 but still about 100 times the offset of the cube from the integer, which means that that term also plays a role in correcting the number, if I did the computations right.
I know that there is a lot of complicated maths behind these numbers, but is there a chance we might hear more about this?
look up the j-invariant
A mathematical proof that e^(π√163)=integer was reported in the Mathematical Games column in Scientific American magazine in April 1975. Years later, I realized that it was an April Fools hoax. The same column announced other amazing discoveries, one of which had to do with the Four Color Map Conjecture.
Interestingly, the numbers that Matt found, 652 and 1,467, are just 2^2 and 3^2 times the Heegner number 163. (noticed this because this is such a fundamental mystery in modern number theory that we would've heard of the numbers 651 and 1,467 if they weren't so directly related to 163)
If exp(pi*sqrt(n)) ≈ integer
Then exp(pi*sqrt(x^2*n)) = (exp(pi*sqrt(n))^x ≈ integer^x
which is integer if x is integer
@@yonimaor1005 I don't think that explains it, though. exp(pi*sqrt(163)) is about 10^17, and is about 10^-12 away from an integer. At that scale, the error is more than enough to change the square by 1, which means if you round exp(pi*sqrt(163)) to an integer and then square it, the answer is more than one away from exp(pi*sqrt(163))^2. So there must be a different reason why exp(pi*sqrt(163))^2 and exp(pi*sqrt(163))^3 are almost integers.
@@animaniacsfan2 You are right. I was wrong.
This is an example of the symbol ≈ being mal-defined and should not be used.
Brady is king of naming numbers!
Parker is the best exemple of not being afraid to give It a try. Wish I was like that.
It's learnable, to some extent. Start small. Doesn't really matter what, so long as you a) Actually do try, and; b) Remember that failure is a great teacher, and use that to your advantage.
Can take years and years, but it's well worth the effort.
In honour of Brady, I will forever call Heegner numbers Treehouse numbers instead.
underrated
This side of the pond we spell it “honor” so get on the program soldier.
Mathematicians already did Heegner dirty, and now you are trying to bring that back. Poor Heegner
@@Slowphoton Neither of the people featured in this video are from your side of the pond bruh
Sadly, Heegner died without his proof of the Stark-Heegner theorem being accepted. Sometimes considered an amateur mathematician, his proof was ignored by many for years.
I came across the Stark-Heegner theorem in the context of studying factorisation and I was very taken aback that these 9 numbers are the only ones that provide for quadratic imaginary number fields whose rings are PIDs. The Wikipedia page on Heegner numbers provides a rabbit hole of possibilities.
That zoom in was iconic, thank you Brady.
Caboose Numbers (Part 1): th-cam.com/video/gM5uNcgn2NQ/w-d-xo.html
9:06 I feel off the chair laughing... the face he makes while saying that deserves a painting
I propose they be called “Parker integers”
They form a Parker field.
( pi + pi + e ) / 3 is also incredibly close to an integer
(3 + 3 + 3) / 3 ≈ suspiciously close to engineer
Try smaller numbers, like -1
Holy frick
I've checked 12 trillion decimal places. All 0s so far. I know there are infinitely many 0s. At least 40% of the decimals are 0s. But I won't stop until I've seen if all of them are 0s.
Here I am writing a Rust program, in like 15 minutes I've checked 8 million numbers for caboose numbers, only to be bested by Matt proving there aren't any!
"Try something, and it doesn't always work." That deserves to be a poster or t-shirt with the Parker Square.
While using -1 as a treehouse number does produce a true integer, adding 1 to it and dividing by 4 does not produce a valid caboose number. It seems we've got a Heisenberg situation with this relationship.
(-1+1)/4 =0, and 0 is vacuosly a carboose number as there are no natural numbers less than zero.
These numbers are also connected to the proof of Fermat's Last Theorem via modular forms. I have no idea in what way but I found that interesting while we're on the topic of unexpected connections between seemingly unrelated mathematical facts or even fields.
0:41 Transcendental numbers, like e and π are like those MVPs, who know they’re super important: ”We don’t need to behave well, because we know we’re super important, for Mathematics, as a whole. We can’t be replaced. You, folks, need us!”. 😅
A process that generates random normal reals should get within 1/10^12 of an integer about 2/10^12 of the time just by chance, so it's not at all obvious that there shouldn't have been an infinite number of these.
I think you might get quite close to an integer if you use the treehouse number -1
;)
Labrador unimpressed - dreaming of its next meal.
I'm getting major Parker Square vibes about this
glad I watched this one before I started writing a script to enumarate more Caboose numbers :D
Alternatively: The Journey is the Reward
The zoom in on the Parker Square though, that was perfect. 🤣
"We need a name for these numbers..."
Kurt Heegner: "What am I? Chopped liver?!"
e^(pi*sqrt(-1)) is an integer.
2:36 In fact, that is the example how you get your name after something while you have nothing to do with it at all. The number was discovered in 1859 (28 years before Ramanujan born) by the mathematician Charles Hermite. And "Mathematical Games" columnist Martin Gardner made the hoax claim that "the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it".
I love how Matt always uses the term “terrible Python code” all the time. It’s really very validating for those who like to code just for fun and to get results and explore stuff. Which honestly outside of the realm of software development really should be what people are using coding for, fun and exploration.
10:20 Integers in fact are NOT a field. He probably though about rational numbers.
For 101×4-1 (which is 403). e^(π√403) isn't close to an integer, but what it is close to is an integer+(3/8)
So this is the Caboose video?
This is the first class carriage
This is the Parker video
This is post-caboose.
This is the caboose to the caboose video
No, this is Patrick.
This is the first Numberphile video in a while that has me completely gobsmacked.
I thought i was about to walk into another "e^π is approximately 20+π" but I'm happy to be surprised
At least that result becomes nicer when you see where it's from, and it also uses π≈22/7 to get e^π≈20+π.
2:49 Parker integer
"Zeroes have same effect as 9s for closeness to an integer"
😂
Yeah thanks mate ❤
That's correct though?
It's a Parker integer!
❤ What a cool coalation. I love when the same pattern appears in two seemingly unrelated places.
Matt already has his square, let Ramanuyan have his integer!
The ending was so perfectly done. 🤗
11:38 "It's good to publish our null results." This was discussed by Richard Feynman in his "Cargo Cult Science" essay/speech, easily found online.
By FAR, the best Numberphile video EVER!!!
I love you Matt and Numberphile ❤
Oh damn, I suspected when seeing the previous video that there may be a link between Caboose numbers and Heegner numbers, and at the beginning I suspected that the last Caboose number would be 163, but then as I saw that the two lists didn't match I dropped the idea. Seems like there is a link between the two, and I just didn't see the pattern. Nice to know that my intuition was correct though
We almost need a part 3...
i thought for sure he was gonna write the square root of -1...
Numberphile rarely goes near the complex zone
Love this new "off the top of my head" bit Matt used in these.
In his Mathematical Games column in the April 1975 issue of Scientific American, Martin Gardner claims that e^(pi√163) is an integer (he also displayed a map that he claimed needed five colors). The column was an April Fool's joke.
Very funny. I wrote code to find all the complex numbers that conjugate to primes and found this same relation. Nice to see it explained so well here.
I love it when Brady gives a list of numbers a name. Treehouse Numbers! I love it.
Matt Parker is such a legend. I love his videos 😝❤️
This guy always do things almost spectacular. Such a parker square...
For those interested, if the large integer at the beginning represented the circumference of the earth, then the level of precision given by e^pi*sqrt(163) would give the circumference with an error of about 40 micrometers. That's about as big as one of your skin cells can get. Incidentally it's also the lower limit of human vision, so if those two numbers were wrapped around the entirety of the earth, you would not be able to see the difference.
What's truly remarkable is that every number is within 0.5 of an integer.
If we're being pedantic, I don't think that necessarily applies to non-real numbers.
@@billberg1264 I stand corrected.
Good on you Matt, for giving it a go!
I've seen other Numberphile videos, as well as some on Brady's channel, Stand-up Maths, and I've never seen that dog before (the one resting on the couch). Indeed, I never knew Matt or Brady even HAD a dog.
3:14 if you think about it, euler’s identity fits this pattern since i is sqrt(-1), but it actually does give a whole number, -1.
I like to think that many viewers are like the dog on Matt's couch, taking an afternoon or evening nap falling asleep to interesting maths facts
The zoom-in on the Parker Square is both savage and hilarious. 😆
Love how Skylab is relaxing in the background :D
Matt said something that really made me love undergrad physics actually. It's when you can't solve something, or get an experimental conclusion that you expect or looking for BUT YOU KNOW WHY you can't get it.
Like being able to explain why to me makes all the difference. having something mess up and you're in the dark about it sucks haha.
Great video!
It's a Parker discovery.
I was about to say that there seems to be a theme going on here...
Wow, a video to replace the “Parker Square”. ❤ You’re the best, Matt.
All numbers below 2,000,000 which give 6 nines: 478233, 881967, 1053883, 1341615
...which give 6 zeros: 2608, 880111
...which give 7+ nines: 163, 1467, 1844122
...which give 7+ zeros: 652
7:47 Matt heard Brady is cataloguing all the Brown Paper and thought "time to make things difficult!"
It intuitively makes sense that 652 and 1467 work almost as well as 163, since they are 4*163 and 9*163. Apparently, 16*163 also works.
I've probably known Matt Parker sans hair as long as I ever knew Matt Parker with hair, but whenever I see a video has Matt Parker in it, I am always mildly surprised that he no longer has hair.
Discovering something that is already known is such a parker thing to do.
Thanks for sharing, it was new to me.
From now on, zeros are known as Parker nines
The "revelation" actually gave me goosebumps 🤩👍
Nicely done.
It's wild that Ramanujan calculated that to such precision before the age of computers.
absolutely brutal zoom-in on the Parker Square at the end
This was a Parker Maths Journey.
Another very interesting relationship with these numbers is that the corresponding field can be used to show that Fermat's Last Theorem is not true for a non-trivial set of numbers (I think for something up to all integers less than 100, though it's been a while since I did an algebraic number theory course)
No way we already got Parker caboose and Parker tree house numbers
Matt: *points at expression*
"These are just, growing exponentially..."
WHAT?
A THING WITH e IS GROWING EXPONENTIALLY!?
A NUMBER WITH AN EXPONENT GROWING IS GROWING EXPONENTIALLY???
I was about to comment: "Bravo!", until the reveal 🙂 Surely, a number theorist's journey should start at (or involve early on) the OEIS! 🤣