Make Any Number From Four π's!

[@shashankgurav7192]

4:12 this will make the formula a little bit easier to read...... *REALLY*

[@Thesoorajtom]

Did anybody else click the "REALLY" ?

[@vincentdehaldolf1636]

NO IT DOESNT

[@AdhiNarayananYR]

@@Thesoorajtom i did 😂

[@keshavagarwal9167]

@@Thesoorajtom me too 😆

[@civilengineer057]

@@Thesoorajtom Why would you do that

[@ghsdftfm]

ah yes, my favourite mathematical operation, the backspace button

[@drenzine]

Deletion go brrr

[@EHMM]

@@drenzine E

[@wittyusername9544]

I can never distinguish CE and C.

[@Sourja]

😆😆😆😆

[@SuperYoonHo]

yes

[@yurenchu]

Instead of introducing the cosine function, I think it's more elegant to write 2 as log(π)/log(√π) , since we were already using square root and the logarithmic function anyway. So, 7 = log( log(√√√√√√√π)/log(π) ) / log( log(√π)/log(π) )

[@heimdall1973]

I was thinking in the same direction, actually. But more general: first two π-s give 2^n (rather than just 2) and the second 2^m, so when you log them, you get m/n - not just an arbitrary integer but an arbitrary fraction.

[@dormeur20092010]

If you like consistency in formulas, the thumbnail can also be done this way: (-cos(π) - cos(π) - cos(π))! - cos(π)

[@reborn6411]

Also (π/π) + (π/π) = 2

[@generalKnack]

@@reborn6411 it is not valid for any integer of n.

[@tommydecarite4121]

That one video where he doesn't say "Did you figure it out?"

[@RGP_Maths]

I'm glad I didn't try to figure this one out!

[@Juhamakiviita2.0]

The annoying thing is it was the only video where I ACTUALLY figured it out.

[@KasabianFan44]

He hasn’t said it for ages mate

[@wolf-od3zw]

Juha Mäkiviita how the heck did you figure this out

[@treyquattro]

only Paul Dirac-level geniuses need apply

[@char8169]

note: change of base formula does not have to use ln, it works with any base log (ex. log a / log b )

[@NestorAbad]

That formula is pure poetry, I love it!

[@thecoolring6431]

How the hell your comment is 4 days old !?

[@saritavedyadav6221]

@@thecoolring6431 yes it's strange

[@YuvrajSingh-kw6tg]

@@thecoolring6431 yeah bro

[@ades9625]

@ToxicPaad lol he is patreon member and he get the video before us there is no need to report

[@MrDragonorp]

@@thecoolring6431 unlisted video probably, maybe he was a patron or something

[@arttukettunen5757]

"Use pi 4 times to get the number 0" *Easy, it's pi-pi+pi-... * "Then 1, 2, 3..." Aw man

[@rishabhlakhwani2089]

Arttu Kettunen it can be n(pi/pi*pi/pi)

[@vanlr]

@@rishabhlakhwani2089 lol thats 1 not 0

[@rishabhlakhwani2089]

Vaneler thats why i said n*(pi/pi*pi/pi)

[@rishabhlakhwani2089]

Vaneler “n” is the number which has to be the answer

[@vanlr]

@@rishabhlakhwani2089 oh didnt notice that n lol

[@SuperCryptic9]

I just found out today that there is someone who can recite 70,000 digits of pi

[@konomi9205]

ok

[@devanshusharma9768]

Well whats the good name of that Godly man??

[@Felixkeeg]

@@devanshusharma9768 epic math time has a video on him

[@floriannom6691]

"I just found out today that there is someone who can recite 70,000 digits of pi". So can I. Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero ... "70,000 digits of pi" is not difficult. It is "70,000 digits of the mantissa of pi" that is difficult. ... Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero, Zero ...

[@alexting827]

@@floriannom6691 ??

[@nick012000]

You can get up to n=4 by using different combinations of trig functions. sin(pi) = 0 and cos(pi) = -1, so to get 4, you just do -cos(pi)-cos(pi)-cos(pi)-cos(pi), and for each number below that you just replace one cos(pi) with a sin(pi).

[@harshchaurasia5869]

This formula showed all the beauty, power and simplicity of mathematics. All right.. 'Simplicity' can be understood to be removed from the comment 😅

[@naturelover4148]

Maths is simple if you are simple😊😊😊

[@harshchaurasia5869]

@@naturelover4148 Definitely

[@vaibhav7480]

69th like

[@naturelover4148]

@@vaibhav7480 69 🤣😂🤣 Only legends will understand "69".

[@mochigojo9753]

Harsh Chaurasia it's so simple yet so complex

[@shototodoroki4636]

HAPPY PI DAY! HERE'S VALUE OF π UPTO A HUGE NUMBER OF DIGITS : π=3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229485493038...

[@henrytang2203]

This cheat code function now takes the fun out of many number puzzles.

[@koldovalnya]

Presh: shows a table Me: Oh I CaN UsE ThE BaCkSpAcE BuTtOn tO eRaSe EvErYThInG

[@erikkonstas]

Whoops, he accounted for that and didn't include numbers!

[@konomi9205]

@@erikkonstas well if you backspace all the pis and 'n', then you get nothing = nothing which is technically true

[@erikkonstas]

@@konomi9205 Yeah, Google's calculator (a part of which is shown as the table of operations) doesn't have the "= n" part in the input.

[@shivammishra-vw6eb]

So futuristic 👌🤔😁

[@alaeriia01]

That's like wraparound integration by parts. It always feels like cheating.

[@potentiallyunaffiliated4285]

(before seeing the solution in the video) I can do it in just two pi (Or alternatively, 4 pi can give any rational number by dividing two arbitrary integers, each made from 2 pi): Take pi/pi=1 The % sign on an Android calculator divides the previous answer by 100. I press this button |n| times (for some integer n), resulting in 10^(-2|n|) Take the square root, giving 10^(-|n|) I then hit the log button, which takes log base 10, giving -|n| Now just remember to have placed a negative sign at the start, and you have |n|=n for positive n. (For even n, you can cut half of the %s and skip the square root step to save time. And for negative n, just don't negate the penultimate result.) 22/7=log((pi÷pi)%%%%%%%%%%%)÷log(√((pi÷pi)%%%%%%%)) Edit: my Android calculator apparently considers adjacent %s invalid (i. e. 9%% is not allowed), but it *does* allow percents stacked within parentheses (so (9%)% *is* allowed). So: 22/7=log((((((((((((pi÷pi)%)%)%)%)%)%)%)%)%)%)%)÷log(√((((((((pi÷pi)%)%)%)%)%)%)%)) Edit 2: Wow, I came up with a better solution than the video!

[@смок]

it isn't even a formula and it ain't better

[@potentiallyunaffiliated4285]

@@смок It's as much a formula as the video has. It's "better" in that it takes fewer pi(s) to make an arbitrary integer (2

[@sapien153]

Amazing. Better solution than the video indeed

[@sapien153]

btw You don't even need two pis. Just one is enough. Replace pi/pi with -cos(pi)

[@potentiallyunaffiliated4285]

@@sapien153 Incredible. One integer per pi. Well done, sir.

[@3geek14]

Someone once sent me a fun way to use 1 through 9, in order, to get 100. His method was to start at 123456789 and use only square roots and factorials (and taking the floor before a factorial). It doesn't generalize as cleanly, but I believe you can reach anything with those three operations, as long as you start with 3 or something greater.

[@max-skateboarding2674]

log base pi(pi^n) = ln(pi^n)/ln(pi) = n Uses 2 pi

[@erickperez5930]

Playing with this I found something really interesting. It is kind of an inverse, it is using only integers to find pi. I found that 2 times integral from 0 to 7 of sqrt(x^2+1)-x is a really good approximation of pi. Is this casual?, do someone know of a similar formula for pi?. Maybe adding more terms I can get a better aproximación

[@scoutskylar]

This was my answer: π-π+log(sqrt((π÷π)%%%%%)) with n % symbols. For negative n, change the + symbol before the log to a - symbol. Note that Android's calculator uses log_10 for log.

[@yurenchu]

That formula actually gives negative n. For _positive_ n, the + symbol before log must be replaced with a - symbol. In case "log" doesn't represent the base 10 logarithm, your formula can be easily amended into log(sqrt((π÷π)%%...%%%)) / log(sqrt((π÷π)%)) or, even better, log((π÷π)%%...%%%) / log((π÷π)%)

[@johningham1880]

It looked like it was going to be a monochromatic 3brown1blue video when the four pi creatures lined up at the beginning

[@jocabulous]

3blue1brown*

[@johningham1880]

jocab henke Of course. Thank you

[@alexblack1572]

Actually, that formula is written with using n (have a look: sqrt(sqrt...(sqrtpi)...) n times is just pi^(1/(2^n)) ). So as far as this formula consider n, it isn’t magic. It’s like pi+pi-pi-pi+n=n. The real problem is to find out equations with using only pi (or any other numbers) and calculator’s operations, which doesn’t consider n anyway

[@yurenchu]

- "It’s like pi+pi-pi-pi+n=n." No, it's not. The question asks, for a _given fixed (integer) value_ of n , to find an expression that equals n, using only four pi's and the given mathematical operations. For example, for n = 7, Presh's solution would result in an expression that uses the square root operator √ seven times, which is allowed. However, your solution would result in an expression that uses the digit "7", which isn't allowed.

[@alexblack1572]

@@yurenchu well, as I said, solution would still consider n, just because all of that roots, strictly speaking, are some degree of the number pi (pi^(1/2^n)). That strange writing of n roots is a veiled use of given n. And I don't give my solution for 7, for example. I just say that this solution doesn`t use n directly, but it does. By the way, this kind of problem is really famous: do the same thing but using only +-:*() with 123456789. And that solutions don't have the same idea, they really don't consider n

[@yurenchu]

@Alex Black , - "solution would still consider n," Of course every solution for a given fixed value of n would somehow consider n, since every expression needs to somehow take into account the value that we're trying to get as the outcome. For example, 3 = (pi + pi + pi)/pi ... but wait a minute, the numerator (pi + pi + pi) = 3*pi, so we took the number 3 into account! Well, duh! 0 = ( sin(pi) + sin(pi) + sin(pi) ) / pi 1 = ( sin(pi) + sin(pi) + pi ) / pi 2 = ( sin(pi) + pi + pi ) / pi 3 = ( pi + pi + pi ) / pi Formulaic description of this solution: n = (3 - n) times sin(pi) plus n times pi , all over pi Works for n = 0, 1, 2, 3 . - " I just say that this solution doesn`t use n directly, but it does. " What does that sentence even mean? - "By the way, this kind of problem is really famous: do the same thing but using only +-:*() with 123456789." Not sure what your puzzle exactly is (you don't say clearly which numbers and how many we can use, and you don't say what each expression should be equal to), but it seems to me that this type of puzzle has already been done too many times. Furthermore, using only binary operators makes the space of possibilities really limited, we can just brute-force it and make an inventory of all possible outcomes. Really boring and not much of a problem at all.

[@alexblack1572]

@@yurenchu sorry for my english, not a native speaker, anyway I mentioned this video kzbin.info/www/bejne/Y6PYdGh3bqqtsKc and I think it shows my thoughts. So the solution of this kind of problem doesn't consider n. Talking about pi-problem, here we don't press n on a calculator, however, if we write this equation mathematically clearly, we have to use n (no one write sqrt(sqrt...sqrtpi))...) n time; it is equal to 1/2^n)

[@sam_la_chevre]

tfw I thought I was smart with (-cos(π)-cos(π)-cos(π))! - cos(π) = 7

[@pandu2749]

Mine was π+π-π-π =0

[@onepunchman9991]

I thought n(cosπ × cosπ × cosπ × cosπ)😂😂😂

[@shiro2738]

@@onepunchman9991 lol it was so easy

[@huseyinerguven958]

if that is allowed, this should also be allowed: -(cos(pi)+cos(pi)+...n-times)=n. If you have extra pi, you can add them as sin(pi).

[@yurenchu]

But how would that work (for example) for n = 7? You have to use exactly four pi's. You can't use seven instances of pi. However, you can use seven instances of the square root operator √ .

[@anandk9220]

I was busy hoping to find different solutions like those in 6s puzzle. But after having watched the general formula in video solution, I must say - This is CRAZY and BEAUTIFUL !!!!!!!!!! 😙😙😙😙😙😙😙😙😙😙

[@maxjacobs6625]

Happy Pi-day!

[@uoyspoob]

|= sideways pi

[@BurningShipFractal]

My solution: first I realized that I can use the percentage sign for n times to π and that gives π%%… = π(1/100)(1/100)… = π((1/100)^n) Because x% = x*(1/100). And to get rid of (1/100)^ , I used the log. (Logarithm base 10.) log(π%%…) = log(π((1/100)^n)) = logπ + log((1/100)^n) = logπ + nlog((1/100)) = logπ - 2n I also need to get rid of logπ, so I used one more π. logπ - log(π%%…) = logπ - (logπ - 2n) = 2n Finally I need to get rid of 2. Let f(n) = logπ - log(π%%…) = 2n then f(n)/f(1) = 2n/2 = n f(n)/f(1) uses exactly four πs. So it is the solution. f(n)/f(1) = (logπ - log(π%%…))/(logπ - log(π%)) = n

[@kryptokly1435]

This Formula is so amazing and impressiv. A great example for the beauty of math.

[@fingerdash683]

That also means you can create any integer from exactly 6 real numbers (and complex if I'm not wrong) by using the same method and to make the base 2 logarithm : R/R + R/R

[@raileite5994]

R = 0, R = 1, R = Negative🤔🤔 Problem with Log(Ln) appears

[@fingerdash683]

@@raileite5994 R for any real number (except 0)

[@yurenchu]

2 = log(R)/log(√R) , for any real or complex R , except R=0 and R=1 . (where log(a)/log(b) means "the base b logarithm of a".)

[@fingerdash683]

@@yurenchu with exactly 4 complex or real number then

[@yurenchu]

@Fingerdash68 , By the way, 2 = (R+R)/R for any non-zero R (and you can replace R with R! when R = 0 )

[@pepehimovic3135]

Calculation taken quickly...

[@mrpythagoras7079]

LLORENTEEEE

[@aak5810]

Origi

[@RocketRaju2212]

Oriiiiiigi

[@siripiripiri1]

This was more complicated than i thought it will be

[@cubing7276]

I love "how to make a number given the conditions" but this is on a whole nother level

[@livinglifeform7974]

If you have n in the formula you might as well just have it as (л^2/л^2)n

[@squirey]

"One simple formula." I was gonna say that but you beat me to it

[@SidneySilvaCarnavaleney]

Prezado nobre amigo, com todo meu respeito a todos neste singelo canal, venho a público afirmar que a Hipótese de Riemann, existe alguns pontos negativos, foi em cima de grandes pesquisas, várias investigações e alguns estudos, que com uma simples PA(Progressão Aritmética) dentro de uma lousa com um simples giz relato a seguir: Prezado nobres amigos(as) Professores(as), alunos(as), conhecidos(as)deste singelo canal qual o impacto que causaria no Universo da Matemática, ao afirmar que alguns números citados não são primos? e os Primos Gêmeos não existe? 2; 19; 41; 59; 61; 79; 101; 139; 179; 181; 199; 239; 241; 281; 359; 401; 419; 421; 439; 461; 479; 499; 521; 541; 599; 601; 619; 641; 659; 661; 701; 719; 739; 761; 821; 839; 859; 881; 919; 941; 1019; 1021; 1039; 1061; 1181; 1201; 1259; 1279; 1301; 1319; 1321; 1361; 1381; 1399; 1439; 1459; 1481; 1499; 1559; 1579; 1601; 1619; 1621; 1699; 1721; 1741; 1759; 1801; 1861; 1879; 1901; 1979; agora a Hipótese de Riemann cai por terra abaixo, o autor Sr Sidney Silva, portanto sou o detentor do maior número primo já encontrado em todos os tempos..... Segue minha obra para sua apreciação quem sabe talvez comprar está única e majestosa obra sobre o Pi ser Racional e Irreversível. o autor Sr Sidney Silva. www.portaldoslivreiros.com.br/livro.asp?codigo=4410831&titulo=A+Ousadia+do+Pi++Ser+Racional : www.estantevirtual.com.br/mod_perl/info.cgi?livro=2655536635 aeditora.com.br/produto/a-ousadia-do-%cf%80-ser-racional/ www.amazon.com.br/dp/655861281X?ref=myi_title_dp aeditora.com.br/produto/a-ousadia-do-%cf%80-ser-racional/ Onde encontrar minha obra, segue os links acima em epígrafe, e adquirir uma obra onde sua leitura é bem simples e objetiva provando a Racionalidade de Pi, para saber se é Racional ou não, compre minha obra, gratidão sempre o autor Sr Sidney Silva.

[@premsai2050]

We can do by (pi/pi) +(pi/pi) =2

[@shikharsrivastava5020]

No mistakes. You can publish this

[@kristijanross977]

Loool

[@tejaskulkarni4579]

Shikhar Srivastava How did u comment 2 hours ago when the video was posted 25-30 mins back?!?!

[@matthewjames7513]

I suspect he's one of the people that get private access to the video early so that they can spot for mistakes. Once Mind Your Decisions gets the all clear, then he makes the video public

[@Blink5002]

Matthew James ya

[@shikharsrivastava5020]

@@matthewjames7513 Yeah. So basically, if you go to this channel and in the community tab, you'll see a post about this video. When he made the post, the video was unlisted. He usually does this to ensure that there's no mistakes. I am assuming that the post will only be visible to subscribers. So yeah.

[@GioFaber]

My son: "So, if i eat 4 pies i'll know all the numbers" 😂

[@pe3akpe3et99]

XD

[@pkeric2626]

r/thathappened

[@TestAccount-nv2jq]

@@pkeric2626 r/ihavereddit

[@pkeric2626]

Radicality r/whocares

[@dinhanhx]

r/AngryUpvote

[@pachapatate848]

Him saying "Pi pi pi pi = n" My brain: **start singing coffin dance song with pi sound**

[@v0idm1nds75]

I dont know how but you improved my maths significantly

[@evank3718]

I’m so glad this was released on pi day

[@Mark-kc2ke]

I got this and it works. Pi pi pi pi Pi * n / Pi + (Pi - Pi) = n Works for any integer

[@SledgerFromTDS.]

So this Number is gonna be based off The Knuths Up Arrow Notation as it is: - Level 0: Counting (+1) - Level 1: Addition (+) - Level 2: Multiplication (×) - Level 3: Exponential (^) - Level 4: Tetration (^^) - Level 5: Pentation (^^^) - Level 6: Hexation (^^^^) So let's start off with 2^5 = 2 × 2 × 2 × 2 × 2 = 32, 2^^5 = 2^(2^2^2^2) = 2^(65536), 2^^^5 = 2^^(2^^(2^^2^^2)) = 2^^(2^^(65536)) 2^^^^5 = 2^^^(2^^^(2^^^2^^^2)) = 2^^^(2^^^(2^^(65536))) HBUN(1) = 2^^^^5 HBUN(2) = 2(HBUN(1)^)5 HBUN(3) = 2(HBUN(2)^)5 Tip: Keep going on and on until you reach HBUN(100) = 2(HBUN(99)^)5 Note: HBUN means Hector's Binary Upper Number Function: HBUN(C) = 2(HBUN(C - 1)^)5 Size: This number is way bigger than G(64)

[@AlejandroGarcia-co8uk]

If that's the case, then HBUN(-2) = 2^5 = 32 meaning HBUN(-(PI)) has to get rounded to HBUN(-3), So HBUN(-(PI)) + ((PI + PI) ÷ PI) = 12.

[@SledgerFromTDS.]

@@AlejandroGarcia-co8uk Well heres some extra Calculations based on HBUN Series, This is unofficial because I made it up. I was quite impressed with how it turned out, It's really just Based on the G Series from Graham's Number.

[@AlejandroGarcia-co8uk]

@@SledgerFromTDS. Well at least we get a new and creative way for 12.

[@SledgerFromTDS.]

@@AlejandroGarcia-co8uk What do you Mean by "For 12"

[@AlejandroGarcia-co8uk]

@@SledgerFromTDS. HBUN(-3) = 10

[@pranaysingh4957]

Hey Presh I just realized we can solve this by converting it into something we already know - cos (π) = 1 and we can use π 4 times so this is just making any number with 4 1's - something already done

[@GourangaPL]

i figured it out cuz i knew the "four fours riddlle" :)

[@nrpbrown]

I learn something neat every video, man!

[@legendskys8188]

"So we have 1 simple formula, using pin4 times" Yes, simple

[@amritbahl1223]

Hey presh if it weren't for only using common mathematical operations, can we Euler's formula:- e^(iπ) =-1 , then multiply both sides by n and manipulate??

[@patrickboner]

Here is an easier way n+pi-pi+pi-pi You never said I couldn't use other numbers

[@gregoryfenn1462]

patrick boner yeah someone else pointed out you could just do n*pi/pi+pi-pi. You could also do n^(pi/pi * pi/pi) now that I think about it, not sure if powers are allowed on standard android calculators thoughz

[@Thankmel8r]

Get n by using pi exactly four times. And O(n) standard operations. Brilliant! Next candidate, please.

[@ultimatedude5686]

I solved this one a bit differently. You included the % symbol, which I will assume divides the argument by 100. I will also assume that "log" denotes a logarithm base 10. ± log(((π / π)%...)%)/(cos π + cos π)

[@younes-47]

Today 14th March is the Pi day. Albert Einstein birthday and Steven Hawking death day. It's really an interesting day 😄

[@TomRiddle-yo3pt]

@HollyT i know till 25 decimal places 3.1415926535897932384626433...

[@TomRiddle-yo3pt]

Also Galileo's birthday too

[@Seb135-e1i]

@@TomRiddle-yo3pt 25 decimals. Interesting. _They said would you like some pi, I said yes I would~_

[@purrplaysLE]

Amit Bashan I know 46 decimal places 3.1415926535897932384626433832795928841971693993...

[@MrOligi3003]

@@purrplaysLE You accidentaly wrote 9 instead of 0 as the 32nd digit after the decimal point.

[@thisispk...]

wow!!! thanks! I cannot stop myself from sharing this to my friends!!!

[@jacobcollins9821]

if you're allowed to have n on the LHS why not just do n+(π+π-π-π) = n

[@gustavocaetano2166]

If you wanted to make (for e.g.) n=7, you would have 7 in the expression: 7+(π+π-π-π), whereas in his solution, you would have 7 roots, but not an explicit number 7.

[@kshitij7050]

sin π + sin π + sin π - ncos π = n

[@dragonskywalker5507]

Alternate title: How high can you get with four pies

[@jayfredrickson8632]

Maybe if they're pot pies....

[@TheGrapeApe22]

(pi^pi^pi)/(pi%%%%%%%%%%%%...) with infinite percents, that gets you to infinity

[@yatijumari1361]

With this you can make any number from 9 Cirno Wants Know Your Location

[@overlordprincekhan]

4:35 When you get that the -(cos pi) was written twice such that he can get an extra pi

[@lucykitsune4619]

@EliteDz # Chanelle Yea but if it was only once you'd have the log with base 1 which returns infinity/an error for everything except 1 (at least it should because 1^n = 1 (n = ANY real number))

[@lennyvlaminov9480]

Never seen it, amazing solution that is understandable. Thanks for this one!

[@weylinpiegorsch9253]

wouldn't it be far simpler as π - π + N(π/π)?

[@gdtargetvn2418]

No?

[@Firefly256]

You can’t use an n on the left side

[@weylinpiegorsch9253]

@@Firefly256 Presh did....

[@Firefly256]

@@weylinpiegorsch9253 no, That's just the number of square roots

[@weylinpiegorsch9253]

@@Firefly256 There's a couple different ways to explore that. If you assume that the "n" in Presh's example is an instruction to whomever uses the equation, I'll counter that in the "2" example, it was immeidately converted to a formal equation, with "n" formally part of the equation. In Preshs example, n used as an instruction is an informal way of writing it. But given what he did with the "2" example, it's easy to convert his format into a more formal formula. In that context, my formula is using no rules or interpretations that Presh isn't himself using.

[@magaplex6476]

Well, you are using 'n' in the equation, which makes it not just 4 Pi's. But this is still pretty interesting.

[@Maggieeification]

Correct me if i'm wrong, but isn't this the easiest solution: n×π÷π+π−π=n works for any n... sure it's not as beautiful, but hey....:D

[@PaulMutser]

You're introducing an n though

[@prajaswadekar3157]

Lol yup that works😂😂

[@Maggieeification]

@@PaulMutser true, but is it illegal? In the solution of the video you also need n to write the equasion..

[@sapien153]

There's no n or any numbers in the calculator Presh showed

[@Maggieeification]

@@sapien153 you're right, but you need n to know how to write ln√(...√(π)) to get to your solution n.

[@like31000]

(4pi times 0) + n = n

[@hsujui-peng8175]

Ceiling(pi + pi) + (pi - pi) does it work? I have this in mind before watching the video.

[@KingBaller06gaming]

That would equal 2pi

[@total_dk6517]

@@KingBaller06gaming I think you missed the ceiling(). It's equal to 6.

[@prunodagen]

E(pi + pi) + pi/pi

[@KingBaller06gaming]

@@total_dk6517 Because of Order of Operations both parentheses have to be solved first. The first set (pi+pi) is equal to 2pi and the second set (pi-pi) is equal to zero because when you subtract the same term that's just a given. So then you're left with 2pi+0 which then equals 2pi.

[@pavankumarr9163]

That formula awakened the put off spark of mathematics inside me.

[@xiaoshen194]

That song in the end😍... what is that tone Mr.Presh?

[@divyanshjain4962]

Xiao Shen..I too want to know...I asked him EARLIER but.... NO REPLY.

[@GutReconIkaros]

@@divyanshjain4962 yeah same

[@shivammishra-vw6eb]

If you find out Let me know 🙏

[@TheMazinka]

Finally some straightforward rules!

[@jeanr5742]

I don't know why I'm keep watching this, but it's incredible

[@harshith_takkala]

Bro you are awesome Happy pie day for all...😀

[@victorrizkallah6014]

Dude that was the greatest video I've seen in a while.

[@jingchen2188]

EXACTLY what I did before this video came out. I tried making infinity with only 3 digits without using factorials. I came up with 9/sqrt(sqrt(sqrt(...sqrt(2)...)))+1. But then I talked to myself, “Hello? What about logarithms?” And I came up with this, lol.

[@Jay_Scheme]

Why don’t you just do sin(pi) * sin(pi) * sin(pi) + or minus “n” cos(pi) where n is the number

[@draganjonceski2639]

Ye thats what i also thought and it seems alot simpler

[@prestonraab6209]

You can’t use any number except for pi, so n is off limits

[@professor_zilio]

Anyone knows the software Presh uses for his videos? Due to the quarantine in my country, ourcollege has asked us (teachers) to elaborate online videos for the students. So that they don't lose any content.

[@abhisheksabnis5065]

If only this video would be 3:14 mins long, it would be perfect

[@rednassie1101]

Pi-pi+pi-pi+n Yeah, this is big brain time

[@ashishkingdom]

Can we make any number from 4 zeros?

[@yurenchu]

for any positive integer n , n = log( log(√√√√...√√√π)/log(π) ) / log( log(√π)/log(π) ) Replace each of the four π's with cos(0!) (or with sin(0!) , or with tan(0!) ).

[@ashishkingdom]

@@yurenchu Oh wow!

[@shivanshsingh5819]

1.π+π-π-π=0 2.π*π/(π*π)=1 3.π/π+π/π=2 4.(π+π+π)/π=3 Keep it simple

[@Archanfel]

-Cos(π)-Cos(π)-Cos(π)-Cos(π)=4

[@Zeabox]

How is there a comment from 4 days ago when the video is posted 8 minutes ago

[@sudheerthunga2155]

Patreon

[@InfernalBanana]

So for the “Four 4s” puzzle, could you replace the denominator of this equation with ln(sqrt(sqrt(4*4))) and just replace “pi” with “4” everywhere else in the equation? If you need to reduce the equation into log base 2, that should work, right?

[@BenDRobinson]

I'm pretty sure that would work - for the top of the fraction, the two pi's could be two of any positive number and the value will still be 2^n

[@BenDRobinson]

In fact for "Four Xs" replace the denominator with ln( ln( X ) / ln( sqrt( X ) ) ) - using exactly the same approach as the numerator to get our 2. So we've now solved the four anythings puzzle.

[@cristianolaru1249]

"One simple formula" indeed!

[@billyjuarez1579]

Could you explain 2:55? I don't understand why you say both of this will cancell up.

[@academicunboxing1402]

Nobody: Not even a soul: Mathematicians: pi pi pi pi

[@YamiSuzume]

This meme wasnt funny to beginn with. but now its also very old

[@heimdall1973]

I got an arbitrary fraction by accident. I also started with square roots and log. Then I asked myself what to do with the other two π-s. Here it is: m/n = ln [ln (√√...√π) / ln (π)] / ln [ln (√√...√π) / ln (π)] where the first thingy has m nested √ and the second thingy has n nested √.

[@pe3akpe3et99]

a tip:always ask you math teacher to ππ

[@detectivl5811]

Wait if you can use n in the calculation, why don't you just use: cos(pi) * cos(pi) * cos(pi) * cos(pi) * n?

[@twy_]

n is not used at all actually in any particular equation, it is just used to derive the general formula

[@detectivl5811]

@@twy_ It's not? So what's the excact equation for n=8 then?

[@twy_]

​@@detectivl5811 ln(ln π / ln √√√√√√√√π)/ln(-cos π - cos π)

[@detectivl5811]

@@twy_ Yeah but you need excactly 8 times the square root! That's cheating... Then I could as well just take 8 times +1

[@twy_]

Since they allow the unlimited use of the provided mathematical operations in their rules... (Therefore, if the operation '%' is allowed, there is also a solution -log(√(1%%%...%%)), where 1 = -cos(π), and the other 3 'π's get eliminated using any method.)

[@lifeofphyraprun7601]

Man!I don't have words to express the beauty of this.

[@sujatasharma8560]

This is improved challenge of a famous puzzle of creating all numbers including fractions, irrationals and complex numbers by only 4 fours. A mathematician even successfully tried for infinity. Do check out the video by searching *power of 4 fours Numberphile* and you will get it....

[@MrAnglow1]

0:03 Mandarin Chinese Tones? pāi pái pǎi pài 🤔

[@gerritrakker9609]

When using log, sqrt, ln you introduce the numers 10, 1/2 and e. I had rather seen using only basic arithmatic.

[@alexandertownsend3291]

That would make sense for the four 4s problem, but you have to go beyond arithmetic when incorporating pi into these formulas because pi is a transcendental number.

[@alexandrucalitescu5822]

The problem stated making any number from only 4 pi-es. You can use whatever you want besides those 4 pi-es.

[@yurenchu]

The "log" as used in this formula doesn't really introduce the number 10, because log(x)/log(a) can be seen as a plaintext-typing-friendly way to express "the base a logarithm of x" or, in other words, logₐ(x) without using subscripts.

[@technoultimategaming2999]

So basically. If you didn't know the formula, you're screwed

[@jordanberson2945]

“Now let’s solve our problem with pi” I never have a problem with pie :D

[@randompython4211]

Happy national π day to everyone !!

[@DRposko]

„one easy formula“ ... yeah, right.

[@legendskys8188]

Me: Doesn't know what a logarithm is Also me: Watches this video the whole way through

[@kxxlzr]

LegendSkys are you okay man. You commented twice

[@pavelrepin4326]

The task was: pi pi pi pi = n. The solution is shown for: pi pi pi pi n = n (n is used in the formula, and in case of n = 7, 7 will be used in formula?). Then with in the formula, we could use: n(pi - pi + pi/pi) = n which is much easier?

[@MartinPoulter]

Look again. n isn't used in the formula. You just repeat the square root n number of times; you don't actually write n.

[@Branthony]

I don't like this solution simply because it uses √. It's inadvertently implying a root of 2. If we have to specify the base for the log function, why shouldn't we specify the base of the root function?

[@LifeIsBeautiful-ki9ky]

I would give a challenge to u , can u explain cubic or quartic equations.

[@VENOM-ol6pv]

Stephen hawkings died this day...😢

[@gregoryfenn1462]

So did about 0.27% of all humans ever, about 270 Million people.

[@VENOM-ol6pv]

Yes bro.

[@datkeks]

1:11 is the same like: 1+1-1-1 = 0 or5 +5-5-5 = 0, 10+10-10-10 = 0. for example: 1+1- (1-1) = 0

[@alphabeta1337]

Read the video description guys. "Why are there comments before the video is published? Get early access and support the channel on Patreon..."

[@maanasagarwal7518]

This is one of the best ones yet💯

[@deni580]

Happy 🥧 day!!!

[@yagi4u8]

pi-pi-pi+pi=0 pi/pi/pi*pi=1 pi/pi+pi/pi=2 -cos(pi)-cos(pi)-cos(pi)+sin(pi)=3 -cos(pi)-cos(pi)-cos(pi)-cos(pi)=4