Incredible Formula - Numberphile
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- Опубліковано 21 жов 2024
- Dr James Grime discusses a couple of clever formulas which are pandigital - using all the numbers from 1-9.
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Numberphile. The only channel where a formula is described as "cute". So adorable.
it is quite 'cute',for its not mathematically complicated,yet gives a interesting result.just like a adorable magician preforming a simple trick,you know it's simple,but got amazed anyway.
No clue if this is a compliment or not.
Its not even a formula, it's an expression. So...
*turns minecraft person into real person into candy, then eats it*
nichrun we use that in my class very often
How to remember e to more decimal places than you'll ever need?
It's 2 point 7 followed by birth year of Lev Tolstoj (1828) followed by birth year of Jules Verne (1828) followed by angles of isosceles right triangle (45 90 45)
e = 2.7 1828 1828 45 90 45 ...
Now you can't say I didn't learn anything at mathematics class
whoa
Thanks!
wilkatis
Yeah, it's just basically
2.7 1828 1828 45 90 45
Incredibly easy to remember lol
My maths teacher gave the same kind of grouping for the first ten digits, except that he only used one author born in 1828, namely Henrik Ibsen. Thanks for the triangle mnemonic for the next six digits!
I had a colleague (teaching maths myself) who told me it's "Two-Point" and then the Andrew Jackson sequence: 7th president, elected in 1828, elected in 1828.
Which didn't help me remember e that much, but I know more about Andrew Jackson now than I did before :)
(Oh, yes, not the US here)
For those asking for full steps:
3^(2^85) = 3^[2*(2^84)] = (3^2)^(2^84) = 9^(2^84) = 9^[2^(2*42)] = 9^[(2^2)^42] = 9^(4^42) = 9^[4^(6*7)]
Thanks. My only concern was 3^[2*(2^84)] = (3^2)^(2^84) but after some deep thought, I understand why.
@@rjohnson8ball It's a general rule that a^(b*c) = (a^b)^c = (a^c)^b
Just exlain it briefly
love Dr Grime. ...his enthusiasm is so relatable
To blow your mind:
81= 9^2= 3^4=70+6+5
No. Way.
Also, this is kinda cheating but still...
49^1=07^2=85-36
0≠1≠2≠3≠4≠5≠6≠7≠8≠9 *bam*
badman jones You sir, deserve an applause and a cookie
How about (9^8^7^6^5^4^3^2)^0 = 1?
Can we call approximations of e E-proximations?
Ganaram Inukshuk You must be pun at parties.
PJ Vis You bet he Vis
Russian accent
You aren’t so lim-ted xD
Yes
1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100
123 - 45 - 67 + 89 = 100
12345 - 6 - 7 + 89 = 100
-1 * 2 + 3 + 4 + 5 - 6 + 7 + 89
= 100
(1 - 2 + 3) * ( 4 * (.5 + 6 + 7 + 8 - 9) = 100
OfficialHuMan ummmm no
@@shreccc9326 I see you're too serious to get the joke. Also, 1+2^3456789=100
1:20 I case you were wondering, Dr. James Grime was actually calculating that number on the fly, from the equation, while talking to you.
whattt
lol
0:14
So 'Zero' is basically me during my friends' road trips 😂
I feel you bro
Raja Jinnah at least you have friends. Feelsbadman
Ace Reaction Hahaha 😁
Fun fact, it's also the number of friends I have :D
Me too tanks.
I don't see why everyone has a problem with the 6*7 not just being 42. First, the mathematical curiosity explored in this video is not a formula, but a number. The whole thing can be expresed as one number, which is very close to e. Every part of the number can be expressed differently and it doesn't matter what signs, brackets or symbols are between them. If 4^2 were to show up in these type of expressions, of course you can write it just as 16, but you need the digits 4 and 2, just as you need 6 and 7 in this case. Why would that be considered cheating?
well cuz that wouldn't be pandigital anymore
The point is to make all the digits show up in the formula, so while there are ways to express numbers in other ways, by specifying that 42 is 6*7, he avoids any accusations of cheating by forcing the formula to look a certain way.
Well you could write e like that, but i doubt it just stops, so this way is simplified.
Yeah the formula for e is much nicer.
And far more accurate
SuperWifiBattler The problem is that you are only allowed to use all numbers 1-9 only once...
SuperWifiBattler if anyone can do that why was it only ever done recently, and not say the day of e, as a constant, being applied
but it isnt pandigitallllllllllll
would like your comment if it did not have 314 likes
4:15 "I Love e SO much..."
Me too, but I don't go around telling everyone about it!
Anamnesia I do
Anamnesia It's Adam and Eve, not Adam and e!
Anamnesia . Well u just did tell everyone that u love e.
Hah, you should!
e
james is my favourite on numberphile
sing with banana
Wrong
Wow that's an amazing formula. The precision is remarkable.
2 Dr. Grime videos in a row? It's almost like it's Christmas!
Ba dum tss...
@MarioFanGamer
well when I wrote the comment it was still the 23th in my place so..
23th or 23rd lol
I am sorry I had to vote up your comment. You had 227 up votes, and that is a prime number. Quick, please, someone else vote it up to its twin!
479!! prime likes again :D
I smiled when he mentioned how accurate it was
dani gómez likewise
dani gómez me too! I'm still smiling !
Amazing! So simple yet it's very impressive that he was able to do this.
If you're reading this, have an amazing Christmas! 🎄🎄🎄
So now I will see you at Numberphile videos too? I have already seen you at every Mumbo Jumbo video!
CyanGaming | ᴹᶦᶰᵉᶜʳᵃᶠᵗ ⁻ ᴳᵃᵐᵉᴾᶫᵃʸ Thank you, random stranger on youtube, have a great Christmas as well!
Cyan? I didn't take you for the kind of person to write these comments... that's disappointing.
the main celebration was 4 days ago and it was called Winter Solstice!!
David -flamingsword1 did you just assume that those jokes were still funny in 2017?
If you put e / ((1+9^(-4^(6*7)))^3^2^85) on wolfram alpha, the result is "e"... weird
I think what's happening is that it is rounding the (1+9^(-4^(6*7))) to 1, since it can only work with a finite precision. Then 1 raised to any power is 1, so the denominator is calculated as 1 rather than e.
However, it properly calculates ((1+9^(-4^(6*7)))^3^2^85) itself...
It might be, but as Trias00 mentioned, if I don't put it on the denominator (or even on the numerator, which also gives a wrong result) it can calculate to a fair precision.
It doesn't. Raising that to power of 85 would take a lot of time, my guess is that it's hard coded to recognize that number and simply return e, which is why it breaks when the number is in different form, like in OP formula...
Wolfram Mathematica gives me an overflow and underflow error, it's too much for the pc. I tried with Python because it keeps trying even if it's too much data, but it's been going for a while I don't think it's going to finish.
I wonder what that paper feels like...
That is life's biggest question.
Andrew Kovnat If your question may be rephrased as "what is this paper feeling?", the answer is "nothing", because it's paper and it doesn't have a working nervous system.
If it's "what's the sensation of touching this paper?" instead, then I suggest you hand over your life savings to Numberphile and ask for a sheet of their paper in return.
Andrew Kovnat I think its just brown wrapping paper. You should be able to find some pretty easily, if you don't already have some lying around.
Interesting.
I imagine it's like one of those chill giving materials (not pleasant to touch)
That's what I was thinking originally.
Those pandigital formulas are kinda parker squared though tbh...
Not the one on “e”though. That was neat. The others were parker formulas.
This statement is false: l
Thank you for highlighting this little mathematical curiosity. IMO a nicer expression for e is the one seen at 4:04 that uses the digits 1 to 8. It isn't as accurate, but it uses an additional finesse.
1/ln (1+x) = 1/x + 1/2 + o(1)
(1/x + 1/2 + o(1)) ln (1+x) = 1
(N + 1/2 + o(1)) ln (1 + 1/N) = 1
(1 + 1/N)^(N+.5) ~~ e
For a given N, the expression is more accurate with that .5 in the exponent than without. In case you didn't spot that expression, it's
(1 + 2^-76)^(4^38 + .5)
4:16 "I love e so much" - Dr James Grime aka J-Grizzle
My favorite Numberphile ever.
My favorite thing about this formula is that it contains the number 42. I think all formulas for fundamental mathematical constants should.
Wow! That's incredible! You guys make mathematics so interesting! Your fun, lightheaded approach to the subject makes it accessable to the layperson (me) without the feelings of intimidation that math usually conjures.
P.S. I love the paper used in your videos...
This isn't maths though, it's arithmetic.
I actually just audibly gasped at the reveal
I love videos with this guy because he so excited about this. You can see the joy in his eyes
That last formula was such a parker square...
L4Vo5 This is the most apt use of that term.
I'm a 27 graphic designer, and I don't know that much in mathematics. I can't explain why I've been so much fascinated by your videos @numberphile, but this has become a serious addiction. Loving your stuff!
And yes, I also think there are too much "points" in pi formula.
he gets so happy about numbers, its adorable. lol
I know.. right? Dr. Grimes has an incredible mind.
It's not fair really. I like numbers too but they vex me more often than not, so they don't make me happy sometimes.
About noombahs, you mean.
i love Dr Grime on numberphile
Wow, that e one.
That was so beautiful.
I love these guys' enthusiasm for math!
If you enter 3^(2^85) in wolfram alpha it tells you that it has 18 457 734 525 360 901 453 873 570 digits, exactly the same number of digits that the video says the formula gives accurately for e.
woah, that is amazingly cool. It was not obvious to me he was just writing an elaborate version of the limit formula until he showed his work, heh
I agree, that pandigital piquation (hey! if the other was a pandigital equation, we can do this now) was not all that impressive. just 10 places? Heck, the well known fraction 355/113 is accurate to 6 places.
Have you ever tried (9^2+(19^2)/22)^(1/4)? Pretty accurate...
I think Ramanujan found the approximation.
That is a nice one, @Geoffrey_Wu . It is accurate to 8 places. I'm not sure I'll remember it the way I remember 355/113, though.
alysdexia well, obviously. I even had to look up _nescient_. 🙄
I got a pandigital formula for 2... 10 there you go
Well, first you have to specify that it's in binary, but it does work. How about 1+(23456789^0)?
Ianwubby Smart and tricky in the same way because you used the 0 as well, but ironically to be the power of all the digits other than 1 to make it a 1 itself, of course summed with the 1 before the brackets to make it a 2
Berniksus dude the answer is 1
Why? We're talking about approximation here. 2 = 10 with an error of 8.
Richard Weiss 2 number system
I LOVE THIS GUY , HE IS GREAT GUY !
HE'S ALSO A SINGINGBANANA. THAT IS HIS UA-cam USERNAME. CHECK IT OUT.
BUT IS HE AMAZIN GUY?
Geebz YEEEE
TomatoBreadOrgasm tnx
NO HE IS PURE EVIL. Don't let that charming smile charm you. His agenda is global domination.
What makes the e formula so much more impressive than pi is that it links back to the definition of the number
Is there any mentionable reason for the "6*7" in the formula and why it's not just 42?
+N3KLAZ so it's pandigital
N3KLAZ to make it pandigital.. other wise the digit would repeat and that wouldnt be cool
The same reason as for the formula itself. Otherwise you can just come up with any huge number and place it in the formula (1+1/N)^N
oh them trolls :) nice one
Because 42 is the answer to the meaning of life, so... I don't know. Help me out.
Actually this was the first numberphile video I saw. I found it by chance on an app called Curiosity and since them I've been in love with this channel!!
Looks incredible
its great to see james again
4:15 James loves his pills
This joke's been done a trillion trillion times hasn't it?
I find the one for Pi, at the end, to be much MORE impressive than the video's main one for E. After all, the one for E is just piggybacking on the large-x limit of (1 + 1/x)^x and only relied on finding a way of divvying up {2,3,4,5,6,7,8,9} to write a power-tower number in two different ways. (Still a bit clever!)
What's a mathematician's favorite dessert?
Pi-e
ARE YOU FOR REAL
No. Numberphile told me that it's cak(e).
0.42331
1+2+3+4+5+6+7...=?
agarRoyale 2002 infinity
These kinds of videos are why I love math.
Holy Cow! My mind has literal-e been blown!
LeiosOS Sorry! Thought this was Google
I think you mean /lɪtɹ̩əli/
LeiosOS
*stares dissaprovingly*
leios Os here what u was doing
Julius The Reformer no he meant /lɪɾɚɹəli/
The way our mind works with numbers and logic is simply beautiful
I find these videos weirdly relaxing even though I have absolutely no idea what's going on. XD
I was just expecting a pandigital formula to equal a pandigital number from the beginning of the video. But this was much better!
Damn, Grime throwing some shade on the Pi formula!
Just as I was about to begin to study German, an INCREDIBLE VIDEO APPEARS!
Bro...I hadn't even watched the video but I will certainly not scrap my claim.
keine problem, die duetsch warten auf du!
(I am sure that is bad grammar though)
Genau. Ich habe deutsch nur 100 Stunde gelernt, aber...I still instantly saw at least 2 mistakes (deutsch and word order: auf should be the last word) :D
wait but then it would be die deutsch warten dir auf?
the German waits on you?
Is it bad that I am in German 6 AP and do not know?
@Joseph Florentine: It would have been correctly: "Kein Problem, die Deutschen warten auf dich!" Just my two cents.
My favorite is the golden ratio.
(1 + 5^(4/8))/2 + 7(9 - 6 - 3)
oh shi...
to be fair, the golden ratio is an algebraic number, making its pandigital formula rather simple...
agreed with rohan. since the golden ration is exactly equal to (1+sqrt(5))/2, you can pretty easily create a pandigital formula equal to that. here's a few more just off the top of my head:
(1+5^(4/8))/2+(63/9)-7
(1+5^(3/6))/(8/4)+9-7-2
seh
(1+5^(4/8))/2 + 7(9-6-3)
(1+sqrt(5))/2 + 7(0)
(1+sqrt(5))/2
You missed a pair of parentheses. Lol. Its technically not exactly the golden ratio anymore.
I first thought the formula would equal a new pandigital number... that was even cooler! :)
Brady, We love you, We love your videos. It's Christmas time. Stop editing videos and spend it with your wife and dogs.
I love this guy a thorough amount
Yo James you should make videos for your own channel!
Another awesome fact about e: the line that is tangent to log_b (x) (log of x of base b) that also passes through the origin (0,0), for ANY positive b different than 1, touches the curve at x=e. That's a nice relation between the inverse function of e^x and the number e.
((1+9^-4)^6*7)^3^2^85 = e, therefore Half-Life e confirmed.
Still follows the rule that Valve can't count to 3. :)
2011 called
@@stumbling we have HL alyx , waiting for 3 :)
I love James Grime videos!
I identify as pandigital.
Is counting your digits considered sexist?
I am trans-cendental.
Go back to Numblr.
Ugh, sick of your non-binary BS. There are only two digits, 1 and 0.
Support the BDDP community
Binary duodecimal decimal pandigital
This guy must be a quite happy/lucky chap doing what he so much loves (talks maths so passionately) with such a great success (having >2M subscriptions)!
In (fairly) simple terms, how did Sabey decide on that value for *N*?
its a big number and it uses the remaining digits from 1-10, the bigger the number is, the closer it will approximate e regardless of any properties the number has
He needed a formula, which is the reciprocal of another formula that uses the digits he didn't already use. (excluding 1 of course)
this just blew my mind, thank you Dr. Grime
But how did he figure to use 3^(2^85) to start?
Justin Marinelli It could have been any number as long as he used an equivalent number with the other digits he needed to make it a pandigital equation. I'm sure he never actually calculated the number, and he probably played around with other combinations before finding that one.
The formula for e calls for taking n to infinity, aka a really big number.
When he was making the pandigital formula for e, he thought "I have to create a formula that puts in a 'really big number' for n, however that number must also satisfy the pandigitalness of the formula".
So, he chose 3^(2^85) because:
1.) It is really big
2.) It has the characters he needs to make the formula pandigital.
Thanks
I'm so glad james is back
This is really ingenious stuff, I wonder if the other participants used the properties of powers to achieve pan-digital formulas in their submissions. Such an insightful approach to the challenge 👏👏👏
One of them did, for a variant of the problem. To express e with 8 instances of the digit 8, Maksymilian Piskorowski came up with (8/8+8^-8^8)^8^8^8, which is correct to 15,151,335 decimal places. But Richard Sabey was the only one who used it for a pandigital formula, where it is a lot harder to arrange the numbers in a way that works.
I should say that the puzzle website has recently been updated with my own solution, which was inspired by the Numberphile video: (1+.2^9^(6×7))^5^3^84, with is correct to 8368428989068425943817590916445001887164 decimal places -- 14 orders of magnitude better than Sabey's. His record stood for 20 years though. And you can write it as (1+0.2^9^(6×7))^5^3^84 if you want to include the digit 0.
I hate math. I hate numbers. But your videos are some of the most interesting and inspiring on UA-cam.
Your genuine enthusiasm is infectious and it makes learning fun.
So...this prompts a question: what is the biggest number you could write using just the ten digits? I'm ashamed to say this is not immediately obvious to me. Might it be 91 raised to the 80, raised to the 7, 6, 5, 4, 3, 2 (or the reverse?)? I'd like to know -- and I'd like to know whether one can give a simple explanation of a "proof." Maybe I'm just tired, but the answer isn't obvious to me right now...
GetMeThere1 I think it is the reverse
GetMeThere1 Maybe 90^81^72^63^54
Would using factorials be considered cheating? Because that's a way of generating huge numbers...
54^63^72^81^90
2^3^4^5^6^8^91
You want as many exponents as possible and you order them from smallest to biggest (2^100 is much bigger than 100^2 for example)
Thanks for this awesome christmas present!
Why do english speakers outside of America call both parenthesis and brackets "brackets?" Doesn't that get confusing when you end up with equations containing both? Is there a slight inflection difference I'm missing?
Nah, there's no inflection difference. It's just colloquial laziness.
It could potentially be confusing. But, well, if there's a chance of ambiguity then you should just use the correct terminology (or the variants of "square brackets" and "curly brackets" to be explicit about their shapes).
It's not that the terminology is actually different. It's just a colloquial thing not to be too bothered by it.
Ah okay, in case you couldn't tell I'm much more of a linguist than a mathematician, this had always been something I've wondered about. Thanks!
That is such a nice, satisfying formula I'm so happy.
1:20 Aaaannnd I need a new pair of underwear.
I need a new pair of pants as well.
And then he lets all the air out of it by explaining the pathetic little trick but you've already gone to the trouble of shitting your pants.
😉 TIHS TNSAW TAHT
@@Kalumbatsch but how you'll start to something that can take all the digits from 1-9 in all the transformations. It's still something one has to figure out by his brains.
How wonderfully creative.
e is the best constant.
Jason93609 Nope I'm a π person myself
I will be i, let us join together and be one. Shoutout to all who get that joke.
Harnoor Lal nice meme
Harnoor Lal Did you mean; get one and be zero? ;)
Minus one. And all odd integers can join in.
Please make a video of all the greek (or non-english) alphabets or symbols used in maths and their uses, please
"It's really cute." "...It's approximately e."
You are amazing guys! Cheers from Uruguay!
6 hours ago better say something...
golden ratio = (4-3+5^(1/2))/(7+9-6-8)
^ now that's truly beautiful ;)
Wow, thanks for replying. Of course, it took 5 minutes but it was an "early" comment anyway.
I love this guy for smiling 24/7.
What paper do they use ?
Brown paper.
Unbleached paper
TheReel Burke le brown paper
Non GMO paper
Jeff paper
never seen such an enthusiast in ages..maths much more than a fantsy.
Thanks for showing us this. This has got to be one of the most impressive things in mathematics, at least amongst the set of results in maths that's not too difficult for me to understand, ha. First of all just to approximate e to 10^25 decimal places is pretty cool, but to do it with a pandigital formula--quite clever indeed! Great idea for a numberphile video.
This is pretty cool! Numberphile never disappoints me.
Include 0 by adding +0 to any pandigital formula
How about expressing Phi using only one of the ten digits? It is my favorite. But this pan digital formula for e is out of this world. Amazing stuff. Wow.
How about: .5(5/5 + 5^.5)
It uses five 5s, how nice.
I identify as a pandigital number.
Hilarious
Not hilarious
Downstream01 Now it's an official sexual orientation
Downstream01 What would be the gender pronoun of a pandigital number?
The pronoun is 123456789.
Thanks to another of your videos, I recognised e before you said it! You have taught me more than 5 years of 1980s comprehensive school... thank you :)
Who else besides me loves this channel? :)
we all love it!
Armand Stefan you and 2248653 other individuals
On one hand, I love it. On the other hand, I so badly want to know if that's the most accurate pandigital formula for e, or if there's another pandigital formula that's even more accurate than that.
Oh yes.
This really tickles my fancy.
i love it when numberphile does a little bit of calculus
how many digits N has?
orochimarujes Can it get the exact number of digits of π or divide by 0?
Berniksus only in the paid version.
The sum of each pair of digits in the formula from left to right is 10, in the way that 10 = 1+9 = 4+6 = 7+3 = 2+8 = 5+5, where 5 is an exception due to the formulas lack of repeating digits, but added to itself is 10.
This is really e-erie!
I love Dr James so much
When I heard 18 trillion trillion digits my eyes opened up so much😂 I love maths
18 Septillion
The most beautiful formula after e^iπ=-1