This Fact Keeps Me Up At Night

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  • Опубліковано 6 бер 2022
  • 🎓Become a Math Master With My Intro To Proofs Course!
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    Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
    #math #brithemathguy #pi

КОМЕНТАРІ • 394

  • @BriTheMathGuy
    @BriTheMathGuy  2 роки тому +63

    🎓Become a Math Master With My Intro To Proofs Course!
    www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

    • @ingenuity23
      @ingenuity23 2 роки тому

      6 days ago?!

    • @cara-setun
      @cara-setun 2 роки тому

      How did you make this comment before the video was uploaded?

    • @ingenuity23
      @ingenuity23 2 роки тому

      @@cara-setun i guess privating the upload and making it public today

    • @BS-bd4xo
      @BS-bd4xo 2 роки тому

      @@ingenuity23 but why would he do that?

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 роки тому

      @@BS-bd4xo It's common for creators nowadays. He privates it to test the waters first, and if the video needs further editing post upload, then he can make it public.

  • @brucelavoie7333
    @brucelavoie7333 2 роки тому +723

    Engineer's approximation: e = pi = sqrt(10) = 3 +5%/-10% Good enough for estimating purposes.

    • @anic1716
      @anic1716 2 роки тому +37

      = sqrt (g)

    • @dvoid4968
      @dvoid4968 2 роки тому +64

      @@anic1716 g=10

    • @merselfares1123
      @merselfares1123 2 роки тому +4

      dafaq...

    • @SlimThrull
      @SlimThrull 2 роки тому +11

      Metric or imperial? (It matters, ask NASA engineers.)

    • @MrKA1961
      @MrKA1961 2 роки тому +14

      none is a better approx than 355/113 though

  • @FAVYSM
    @FAVYSM 2 роки тому +217

    Nah, gotta stick with 22/7. It's way more accurate.
    To be precise:
    sqrt(2)+sqrt(3) is off by around 0.15%
    22/7 is off by only 0.04%

    • @apuji7555
      @apuji7555 2 роки тому +76

      355/113

    • @ycombinator765
      @ycombinator765 2 роки тому +4

      @@apuji7555 perfection

    • @osaether
      @osaether 2 роки тому +32

      How about 314159/100000 ? I bet you can do even better ;-)

    • @cheems1337
      @cheems1337 2 роки тому +44

      @@osaether Nah thats actualy shit compared to the continued fraction approximations

    • @cara-setun
      @cara-setun 2 роки тому +14

      Or you could just memorize 3.1415926

  • @aweebthatlovesmath4220
    @aweebthatlovesmath4220 2 роки тому +353

    ≈ ≠ =
    (Proximating doesn't equals to equals)
    LOL

    • @damian4601
      @damian4601 2 роки тому +67

      I’m an engineer and I can confirm ≈ = =

    • @beautyofmath6821
      @beautyofmath6821 2 роки тому +47

      Approximately is equal to equal, well approximately..

    • @aweebthatlovesmath4220
      @aweebthatlovesmath4220 2 роки тому +19

      @@damian4601 yeah fundamental theorem of engineering

    • @KrasBadan
      @KrasBadan 2 роки тому +17

      @@damian4601 ≈≈=

    • @damian4601
      @damian4601 2 роки тому +3

      @strawberry pup if it’s close enough then it’s good enough

  • @reibarata
    @reibarata 2 роки тому +32

    damn engineers

  • @TheWingAnthony
    @TheWingAnthony 2 роки тому +231

    this was how the digits of pi was approximated centuries ago. divide the polygons into polygons with even more sides and you get an even better approximation. then they stopped once Newton came around with his discoveries, I believe

    • @navbhatthal1972
      @navbhatthal1972 2 роки тому +21

      u watch veritasium to

    • @LeoStaley
      @LeoStaley 2 роки тому +5

      Who else knew this before Veritasium's history lesson? I sure didn't.

    • @thomasfevre9515
      @thomasfevre9515 2 роки тому +1

      @@LeoStaley I learned that in highschool. That being said, i'm not american so...

    • @LeoStaley
      @LeoStaley 2 роки тому +1

      @@thomasfevre9515 I learned about the polygon method in school but not about newton's method.

    • @thomasfevre9515
      @thomasfevre9515 2 роки тому

      @@LeoStaley I was talking about the polygon method too. We have a good education system but not that good.

  • @GlorifiedTruth
    @GlorifiedTruth 2 роки тому +218

    I love it when I pause the video, do the math, and then see that I did it right without inserting 50 careless errors. To find the side length of the polygon, I used the radius of the circle to create a 30/60/90 triangle within it and went from there, arriving at the same answer.

    • @Yadobler
      @Yadobler 2 роки тому +4

      ye, you should get (no sides * 2) triangles, with the base being 1/2 of each side, and the height being r
      and b/r = tan(angle)
      so it's (6 * 2) * (r * tan(angle)) where angle is 360 degrees / (6 * 2) = 180/6 = 30deg
      pretty interesting, never really wondered or had to wonder with a perimeter of a polygon circumscribing a circle

    • @mike1024.
      @mike1024. 2 роки тому +2

      That's the right way to do a hexagon, no trigonometry needed!

  • @Fytrzaczek21
    @Fytrzaczek21 2 роки тому +429

    This approximation is pretty good. The error is less than 0.15%. You should place that in the vid, also remove the clickbaity equal sign from thumbnail. Nonetheless, nice one

    • @elidoz7449
      @elidoz7449 2 роки тому +51

      the clickbait got me, I knew it was an approximation because pi isn't algebraic, but it exalted the video

    • @gaurav.raj.mishra
      @gaurav.raj.mishra 2 роки тому +21

      This channel is the infographics show of math channels.

    • @anshumanagrawal346
      @anshumanagrawal346 2 роки тому

      @@elidoz7449 Same lol

    • @alexsere3061
      @alexsere3061 2 роки тому +28

      It's an approximation sign

    • @chaic4211
      @chaic4211 2 роки тому +20

      @@alexsere3061 He may well have changed it after reading this

  • @tumak1
    @tumak1 2 роки тому +113

    Why the equal sign in the title of this video presentation?? It should be a squiggly "equal" sign. Cheers

    • @DubioserKerl
      @DubioserKerl 2 роки тому +24

      yeah, maximum clickbait, sadly

    • @namantenguriya
      @namantenguriya 2 роки тому +2

      Click bait 😅🤣😌👀

    • @jasnoor8-d-155
      @jasnoor8-d-155 2 роки тому +6

      @@namantenguriya dude its a very small mistake have you ever tried to make videos ? Its a very long process and such a small error is common not clickbait and anyways many people don't even know the difference

    • @davidcovington901
      @davidcovington901 2 роки тому +9

      @@jasnoor8-d-155 You sound like a professional apologist. Or someone not aware of math. Go back to your horseshoes and hand grenades.

    • @jasnoor8-d-155
      @jasnoor8-d-155 2 роки тому

      @@davidcovington901 thank you for thinking I am an apologist but I am not an apologist I am very aware of math I can do maths of higher grade then I am I neither have horseshoes or grenades I am a normal guy like you :) I just meant to clarify that some minor mistakes are common and somebody aware of the squiggly equal and equal sign already know its not true but clicks on the video anyway for fun and somebody who is not to know got to know about the approximation sign and equal sign its a win - win please don't point out the minor mistakes of the video see the main concept
      Have a good day !

  • @Anonymationsthecoolanimator
    @Anonymationsthecoolanimator 2 роки тому +23

    Hahaha, that way
    SquareRoot(10) = pi
    pi^2 = g(acceleration due to gravity)

    • @neutronenstern.
      @neutronenstern. 2 роки тому

      actuallyif you define a meter to be the length of a mathematical pendulum with a period of 2s for small angles ( it converges to 2s for α->0) then π^2m/s===g if g is measured in m/s with this exact definition of a meter.

  • @bdharral
    @bdharral 2 роки тому +70

    Not the first to say this, but it's more interesting to me that 22/7 is a better approximation than sqrt(2) + sqrt(3); it's even more interesting to me that 355/113 is ridiculously close to pi.

    • @stevenlubick2689
      @stevenlubick2689 2 роки тому +2

      Please be more exact than 355/113 thank you👍👍

    • @eduardoxenofonte4004
      @eduardoxenofonte4004 2 роки тому +3

      22/7 is just brute forcing pi, this is way more beautiful

    • @Salsmachev
      @Salsmachev 2 роки тому +3

      @@eduardoxenofonte4004 I'll take an accurate value I can easily calculate over a hard to calculate value that is inaccurate but happens to be "beautiful" (whatever the heck that means). Like, I don't have time to use the babylonian method to calculate root3 and root 2 just to get three digits of pi.

    • @eduardoxenofonte4004
      @eduardoxenofonte4004 2 роки тому +1

      @@Salsmachev no, it's not pratical, no i don't expect you to use it, and i won't either. it's just a nice and elegant way to compute pi, but by no means efficient, if you're confused by what i mean by beautiful, watch this video from 3blue1brown: ua-cam.com/video/HEfHFsfGXjs/v-deo.html&ab_channel=3Blue1Brown

    • @bobbun9630
      @bobbun9630 2 роки тому +2

      @@stevenlubick2689 Given a continued fraction representation of pi, you can calculate a sequence of convergents. To generate the sequence, you simply terminate the (infinite, for irrational numbers) continued fraction at successively deeper points and unwind the nested fractions to compute a simple fraction. Ultimately what you're looking for could be described as "the next convergent in the sequence associated with the canonical continued fraction representation of pi." Each fraction in the sequence is progressively closer to the value of pi, but the convergence is not regular. If you have an especially good approximation at some point, the next convergent in the sequence is likely to be only slightly better, even though the numerator and denominator are quite a bit larger--in this case the value you're looking for is 103993/33102, courtesy of Wolfram Alpha.
      This method is actually a fairly good way for computing rational approximations for square roots and sums of square roots like the subject of the video because there's a fairly simple algorithm for finding a continued fraction representation of quadratic roots and they have simple, repeating continued fraction representations. The calculation is less straightforward for other irrationals.

  • @Kokurorokuko
    @Kokurorokuko 2 роки тому +20

    I'll have to dislike this video because you have the equals sign in the thumbnail
    UPD: he changed it. I removed the dislike.

  • @DeJay7
    @DeJay7 2 роки тому +18

    Using irrational numbers to approximate a transcendental number.
    Sure, why not

    • @createyourownfuture3840
      @createyourownfuture3840 2 роки тому

      What is the difference between irrational and transcendental numbers?

    • @andreipopa3163
      @andreipopa3163 2 роки тому +7

      @@createyourownfuture3840 I think a transcendental number is not the solution to any algebraic polynomial equations or something, for example square root of 2 is not transcendental because it is a solution for x^2 - 2 =0

    • @alephcomputer
      @alephcomputer 2 роки тому +2

      me when ln(-1)/i

    • @neutronenstern.
      @neutronenstern. 2 роки тому

      well its not that bad to approx it with irrational numbers, as long as these arent transcendendal

    • @DeJay7
      @DeJay7 2 роки тому

      @@neutronenstern. But why have irrational numbers? 22/7 is also a very good, but rational approximation

  • @adandap
    @adandap 2 роки тому +35

    One minor nitpick: just because a shape is enclosed within another one it doesn't follow that its perimeter is less. In the case of a square inside a circle it's true, but imagine a many pointed star inside the circle - by taking more and more points we can make its perimeter arbitrarily large.

    • @nicolararesfranco9772
      @nicolararesfranco9772 2 роки тому +6

      I am thankfull that someone pointed this out. It works with the area, but not with the perimeter (at least in general; this one was a rather special case, so it works out)

    • @adb012
      @adb012 2 роки тому +28

      It still works if all closed shapes being compared are convex.

    • @nicolararesfranco9772
      @nicolararesfranco9772 2 роки тому +3

      @@adb012 Yeah, that's true. That would be a subcase in which this argument holds

    • @alganpokemon905
      @alganpokemon905 2 роки тому +2

      @adandap @adb012 very interesting point from both of u

    • @ralphy1054
      @ralphy1054 2 роки тому +1

      @@nicolararesfranco9772 I wouldn't call it a _special_ case... as @adb012 said, the only requirement is that the shapes are convex.

  • @chiranjiviupreti
    @chiranjiviupreti 2 роки тому +21

    Interesting. But your video thumbnail is misleading.
    Edit - March 8, 2022: For people wondering, originally the video thumbnail had √2 + √3 = π, which is now changed to √2 + √3 ≈ π.

  • @MUJAHID96414
    @MUJAHID96414 2 роки тому +2

    Your editing is so good

  • @brucelavoie7333
    @brucelavoie7333 2 роки тому +16

    Another estimation: "Rule of 72", take annual rate of return in %. Divide 72 by this number, this yields the approximate number of years it will take to double your money. 12% yield -> 72/12=6 years. works best near 8%. There is one point where it is exact. Left to the reader to figure out. The obvious failure shows up that at 72% yield you should double your money in a year which contradicts the 72% yield.

  • @rubenvela44
    @rubenvela44 2 роки тому

    You cannot think how glad I am to see you again

  • @peterdecupis8296
    @peterdecupis8296 2 роки тому +2

    this video describes intuitively the pi evaluation; in the case of point, it is claimed that the circular perimeter is measurable by a real value that is intermediate between the perimeter radii of an inner tangent polygon and of an outer tangent one. This traditional approach becomes rigorous from the point of view of modern Maths in the following way:
    we build a ricursive procedure for "duplication" of sides both for inner and outer tangent polygons; then we can define the sequences p(n) and P(n) of the perimeters of inner and outer polygons with 2^n × k sides (e.g. if you start from the hexagon, let k=6); the evaluation of sequence terms is performed by means of simple recursive formulas based on Pitagora's Theorem; the numerable Recursion Theorem allows us to span the definition of the sequences to the entire Natural set of indexes n; then it can rigorously proved that the sequence p(n) is crescent and upper bounded, P(n) is decrescent and lower bounded, therefore on the ground of the modern theory of Real Numbers Set (Cauchy completeness, Bolzano Theorem, etc.. ) they are both convergent when you take the limit n to the infinity; as a final point we can prove that p(n)

  • @GameInOne
    @GameInOne 2 роки тому +1

    Me preparing for chemistry exam
    Him :- makes hexagon
    Me :- confused screaming

  • @Squidward1314
    @Squidward1314 2 роки тому

    How does the circle lying in between the two other shapes imply that the circumference is also between the other ones?

  • @joshuak.n5553
    @joshuak.n5553 2 роки тому +10

    What's there in this to keep you up at night bro

  • @DaBaSoftware
    @DaBaSoftware 2 роки тому

    Very cool! So many ways to get the value of pi

  • @sal8290
    @sal8290 2 роки тому +10

    I'm sorry, but at 2:19, a < b < c does NOT in general imply that b ≈ (a+c)/2 (usually not the case). Take e.g. a=-100, b=-99 and c=100 (so a < b < c holds) - then (a+c)/2 = 0, but 0 !≈ -99, not even close. Even an example with small values like ours, with 2sqrt(2) < pi < 2sqrt(3), all we can really conclude for sure is that pi is between 2.828 and 3.464, the true value could be skewed towards either side. It's just a coincidence here that the mean of the two is somewhat close to pi.

    • @ruth747
      @ruth747 2 роки тому

      Bro √2+√3 is equal to 3.146 approx so we can use π as √2+√3

    • @duckman62
      @duckman62 2 роки тому

      distance cant be negative

    • @gulgaffel
      @gulgaffel 2 роки тому +2

      They didnt use a

    • @jatloe
      @jatloe 2 роки тому +1

      Ok idk why the replies aren’t saying anything about this. It is indeed just a coincidence, and the best you can say is that it “looks like an arithmetic sequence” from the diagram.

    • @Zxv975
      @Zxv975 2 роки тому +1

      -99 is still an approximation to 0. For the most part it's a *bad* one, but it's still an approximation because there is no hard ruling for what constitutes an approximation, as this is application dependent. When measuring a house, an absolute difference of 2 orders of magnitude might be relevant, but on an astronomy scale it's irrelevant.
      It is a coincidence that a < b < c leading to (a+c)/2 being a *good* approximation for b, I agree, but once you set up the inequalities it is a natural thing to check. Which is exactly what is done in this video. And the intuition comes from visually inspecting the three shapes, so it's not even unmotivated. You can clearly see the circle is roughly in the middle.
      The takeaway is this: applying hard rigour to geometrically and physically motivated approximations is a fool's errand. Not everything has to be derived from first principles. It doesn't even begin to make sense when you're talking about approximations either, which purely exist because first principles is too hard to do for most real world applications anyway.

  • @aioia3885
    @aioia3885 2 роки тому

    2 * side length of the inscribed square + 2 * side length of the inscribed equilateral triangle = 2 * √2 * r + 2 * √3 * r = 2 * r * (√2 + √3) ≈ 2 π r. You can use this to easily approximate a circumference using rule and compass

  • @jahnoun7
    @jahnoun7 2 роки тому +12

    Great video but I assume many people would like you to replace the '=' with a '≈'
    since it makes this video look like a clickbait.

  • @axbs4863
    @axbs4863 2 роки тому +36

    I’m just wondering if you can cover the topic of how (1/2)! = (sqrt(pi))/2
    And great video as always! 😁

    • @BriTheMathGuy
      @BriTheMathGuy  2 роки тому +9

      Great suggestion!

    • @Zxv975
      @Zxv975 2 роки тому +2

      @axbs have you encountered the gamma function before? I believe it's often taught alongside complex analysis, but I think you don't even need that to understand this result. You might need to be familiar with the Gaussian distribution to prove the result directly, but it just relies on a simple substitution and a fairly straightforward derivation after that.

    • @axbs4863
      @axbs4863 2 роки тому +1

      @@Zxv975 I kinda just wanna know why it’s exactly encountered with 1/2 and it’s relationship with e and stuff, I’m not very knowledged on the subject but it’s so interesting and these vids really help me understand it lol

    • @Zxv975
      @Zxv975 2 роки тому +5

      @@axbs4863 ok, that's fair. Bri can make a full video on it which explains the details, but I can post a sketch which shows the relationship and where π comes from, because that's probably what you're interested in.
      1) You need to extend the factorial. The factorial only works for whole numbers, but it crucially satisfies the condition (n+1)! = (n+1)n!. This is the key property that we demand that the factorial extension must satisfy.
      2) Turns out there is an extension function, and it's called the gamma function, given the Greek letter for capital G, Γ. This function has the property that Γ(n+1) = (n+1)×Γ(n), and if you stare at this long enough you'll realise that's exactly the factorial relationship above that we wanted to preserve. The Gamma function is defined in terms of an integral involving e^x. However, none of these terms need to be integers, so it turns out the Gamma function works for more than whole numbers: it works for fractional numbers too! So we can basically define the extension x! = Γ(x), where x is any real number. For example, (1/2)! = Γ(1/2). Note that the factorial on the LHS normally can only take integer values. Note, I am sweeping a lot of technical details under the rug for simplicity, and I'll put a footnote at the bottom to give a bit of detail about what I'm ignoring.
      3) Ok, so we have a relationship for (1/2)!, but how do we know what Γ(1/2) is? Well, I said it involves an integral and e^x, right? Turns out this puts it in the form of the Gaussian distribution, which revolves around e^(x^2)). To solve the integral here we use a very clever trick of converting to polar coordinates (the details are given on the wiki page of the Gaussian distribution), and it turns out this is exactly where π comes from! It originates when we convert to polar coordinates to solve the Gaussian integral, which is the form that Γ(1/2) is in.
      Details:
      - so it turns out there are many ways to extend the factorial, but only the Gamma function satisfies a reasonable set of assumptions about how the extension should be done. The wiki page on the gamma distribution has a discussion of the infinite alternatives, but it's enough to know that the Gamma function can quite easily be considered the most natural extension, satisfying the most reasonable properties we'd want.
      - I'm actually using a different definition to Γ compared to what most people probably use. The normal definition is shifted by 1, in that x! = Γ(x+1). Or maybe it's x! = Γ(x-1). I can never remember, so I decided to say "to hell with it" and I use my own definition. It seems silly to me that the most key property of the factorial isn't exactly preserved in the standard definition, so I'll use my own that seems much more sensible.
      - Worth noting that the Gamma function doesn't work for *everything*. It is still undefined whenever the input is a negative number. So we can't use this to evaluate (-1)!.

    • @axbs4863
      @axbs4863 2 роки тому +2

      @@Zxv975 wow thank you! Took the time out of your day to write all that :D

  • @swapnil72
    @swapnil72 2 роки тому +1

    You're UA-cam channel name says your the math guy & you're justifying that name, I Appreciate that

  • @ak_the_gr8
    @ak_the_gr8 2 роки тому +1

    Hey I have a question
    See if we take 3 numbers for example like 1, 5, and 6.
    Then 1 < 5 < 6.
    So 5 ≈ (1+6)/2 or 3.5
    But that's still not much accurate.
    But in the video it was approximately equal. So it doesn't work in all cases
    (I know that it's not "=" and is "≈")

  • @ghilas1216
    @ghilas1216 2 роки тому +3

    We are not engineers here ≈ ≠ = 🤮🤮

  • @l.-.l5451
    @l.-.l5451 2 роки тому

    If we replace the square and hexagon with polygons of higher number of sides...can we get a better approximation of the value of π?
    Suppose the formula is
    lim(2ntan (180/n))>2π
    When n tends to infinity

  • @Laurah847
    @Laurah847 2 роки тому +2

    Mathinception! A square inside a circle that’s inside a hexagon, that’s inside your dream! No wonder it keeps you up at night lol

  • @georgejo7905
    @georgejo7905 2 роки тому

    The beginnings of calculus from ancient times as polygons approach infinity

  • @EdwardCurrent
    @EdwardCurrent 2 роки тому +72

    This was my first visit to this channel, and because the thumbnail tricks viewers by showing a false equation, it will be my last. And, there's no reason why this curiosity should keep someone up at night. It's interesting but it's like saying I'm kept up at night by panda bears not really being bears.

    • @GlorifiedTruth
      @GlorifiedTruth 2 роки тому +8

      I'll admit, if this were my first visit here, I might react the same. But I'm already favorably disposed to Bri, having been a subscriber for a while. (Here, he could've avoided the sin by simply using squiggly equals rather than equals.) As for the "keeps me up at night" thing, it's a very UA-camy title. You know, UA-cam--the place where you yourself had to explicitly state on your channel that your videos were PARODY so that half the people who stumbled across them wouldn't cry and report you. You gotta factor in the stupidity. At least he didn't entitle it "This FACT keeps me UP AT NIGHT: Brian REACTS to MIND-BOGGLING inequality!!!!"

    • @marttielvisto3519
      @marttielvisto3519 2 роки тому +4

      Did he change the thumbnail?

    • @EdwardCurrent
      @EdwardCurrent 2 роки тому +5

      @@GlorifiedTruth I'm glad he changed the thumbnail. I was just really surprised by that from a math channel.

    • @lagomoof
      @lagomoof 2 роки тому +6

      Probably for the best that "pandas aren't bears" isn't keeping you up at night because they're actually bears. The whole "not bears" thing was a combination of science at one time not being sure, despite them looking a lot like bears and because _red_ pandas _aren't_ bears (the red ones are more closely related to racoons). DNA tests proved that the bear-sized, bear-looking, black and white kind of pandas are indeed bears. 🐼

    • @ngiorgos
      @ngiorgos 2 роки тому

      @@GlorifiedTruth it is my first visit, though I came after he changed the sign.
      That said, I don't think I even noticed the approximation symbol anyway, I just noticed "π" and "sqrt(2)+sqrt(3)" and I knew it's supposed to be an approximation.
      I like those quirky approximations of π, so when I saw this, I just though to myself "Ooh, that looks _so_ neat! I've never seen that one before!"

  • @jimmykitty
    @jimmykitty 2 роки тому

    OMG!!!! How's this possible!!!!
    I liked the formula of a side length of circumscribed polygon ❤❤🌿🌿🌿

  • @ekszentrik
    @ekszentrik 2 роки тому +1

    Hmm, intuitively I would never have arrived at it (naturally), but before seeing the explanation, my intuition is that "oh yes, that makes sense", as to describe a circle, you need just two shapes: a triangle and square, and these can be encapsulated via their roots. Purely conceptually, not necessarily mathematically, of course.

  • @GEMSofGOD_com
    @GEMSofGOD_com 2 роки тому

    Now what about a hexagon and an octagon? Or elsewhat. This is actually super epic, I totally luv it!

    • @ngiorgos
      @ngiorgos 2 роки тому

      Archimedes started with hexagons and kept dividing them until he reached 96-gons. That gave him that
      223/71 < π < 22/7
      Edit: I corrected the approximations

  • @derwolf7810
    @derwolf7810 2 роки тому +2

    I prefer the (also much better) approximation 355/113, which is in also easy to remember (the first three odd natural numbers 113355).

  • @pinedelgado4743
    @pinedelgado4743 2 роки тому +3

    This is great stuff!!! Thanks lots for posting!! Your videos are a gift from Heaven and The Heavens since The Heavens encompass all of nature, the consequences of which define the truth of mathematics.

  • @charlo7308
    @charlo7308 2 роки тому +2

    A few days ago I found that e^cbrt(3/2) was an even better approximation for pi. I don't know if there's an explication for this like the one you made

    • @neutronenstern.
      @neutronenstern. 2 роки тому +1

      well then approx e at first

    • @neutronenstern.
      @neutronenstern. 2 роки тому

      and what is way way bether is 355/113

    • @charlo7308
      @charlo7308 2 роки тому

      @@neutronenstern. Indeed! I knew about that one but I didn't know that it was that good!

    • @Elrog3
      @Elrog3 2 роки тому

      @@neutronenstern. Why memorize 355/113 when I already have the first 10 digits of pi memorized from just seeing them all the time?

  • @rattina7784
    @rattina7784 Рік тому

    this is like some alien language
    to me

  • @theimmux3034
    @theimmux3034 2 роки тому

    oh so thaaats's why it's eerily close

  • @albertwood8836
    @albertwood8836 2 роки тому +3

    1:50 - this logic doesn't follow. Just because one shape is contained in another, doesn't mean the perimeters satisfy an inequality (think of a spiky star inside a circle). It is true for convex shapes, however, which is suitable for this application - though the proof takes some thought...

  • @firasnizam
    @firasnizam 2 роки тому

    very fine, this is first time I see this idea, i like it

  • @TikeMyson69
    @TikeMyson69 2 роки тому

    Is there a subset of natural numbers such that the sum of their square root equals pi? If there is, is there only one such subset?

    • @davidellis1929
      @davidellis1929 2 роки тому +1

      No, because pi is transcendental, meaning it is not the root of any polynomial equation and cannot be expressed as the sum of square roots of natural numbers.

  • @hussainfawzer
    @hussainfawzer Рік тому

    The thing I don’t get is that, when we look at the figure drawn.. the perimeters are so different, in fact we could never say (at least according to me) the perimeters approximately equal. But the final answer is a good approximation of pi…

  • @pmxi
    @pmxi 2 роки тому +2

    This is known as Archimedes method for computing π

    • @ngiorgos
      @ngiorgos 2 роки тому

      It's similar to Archimedes' method, but not exactly.
      The biggest difference is that this is a one-time approximation, while Arcimedes' method produced a series of increasingly precice appriximations

  • @horizon210
    @horizon210 2 роки тому +1

    speaking of pi, have you noticed what you get when you take the cube root of 31?

  • @qwertyasdfzxcb
    @qwertyasdfzxcb 2 роки тому

    The title was quite funny

  • @jamcdonald120
    @jamcdonald120 2 роки тому

    "this is a very good aproximation"
    [shows an approximation worse than 22/7]
    me: is it though? might as well just use 3.14 than that approximation

  • @arostheautistic1045
    @arostheautistic1045 2 роки тому

    You can prove the lim(n->inf)[ntan(pi/n)]=pi which I guess could be a way to approximate pi

  • @lumi2030
    @lumi2030 2 роки тому

    why did i think that the 1st second is a 3b1b intro xD

  • @masoncamera273
    @masoncamera273 2 роки тому

    The thumbnail made me question my sanity

  • @MarioDSLife
    @MarioDSLife 7 місяців тому

    I would say “good approximation” is subjective, and it depends on what you want to use the approximation for.

  • @adrianolubini5079
    @adrianolubini5079 2 роки тому

    And the number e is approximately the sum of the cubic roots of 2 and 3

  • @Xayuap
    @Xayuap 2 роки тому

    it doesn't have to follow your rules of perimeters
    that is true but for areas

  • @N-methyl1phenylpropan-2-amine
    @N-methyl1phenylpropan-2-amine 2 роки тому +1

    Isn't this how people used to approximate pi before newton? I mean they just kept adding more and more sides to the polygon

    • @fly7188
      @fly7188 2 роки тому +1

      this is how Archimedes did it in fact 200 bc

    • @N-methyl1phenylpropan-2-amine
      @N-methyl1phenylpropan-2-amine 2 роки тому

      @@fly7188 Yeah he started it. And then people kept adding more and more side (even like millions), until Newton came. Forgot the exact process there's this Veritasium video on this

  • @prantosaha9201
    @prantosaha9201 2 роки тому

    Whooaa, amused

  • @rogerkearns8094
    @rogerkearns8094 2 роки тому +1

    Here, get some sleep at last: 22/7 is closer.

  • @rainjar
    @rainjar 2 роки тому

    Interesting. If you replaced the inscribed square with an inscribed regular pentagon, then the perimeter would be 10 x sin of 36 degrees (approx 5.877852) and the average with that of the circumscribed hexagon (approx 6.928203) would be 6.403027 which gives us π approximating 3.20... .... less accurate. But if we also replace the circumscribed polygon with an octogon, its perimeter would be approximately 6.627408, so the average would be even less accurate.

  • @anshumanagrawal346
    @anshumanagrawal346 2 роки тому +3

    Lmao you changed the thumbnail

  • @Yadobler
    @Yadobler 2 роки тому +3

    Everyone complaining about clickbait, but the dude already changed his thumbnail from equal to approx

    • @cara-setun
      @cara-setun 2 роки тому

      I made my complaint before then

  • @spase667
    @spase667 2 роки тому

    Geez, when does this guy sleep.

  • @thomaskaldahl196
    @thomaskaldahl196 2 роки тому +1

    Now explain why e^e^e^(-2) is an even better approximation for pi lol

  • @noahbrimhall3370
    @noahbrimhall3370 2 роки тому

    This video is great! It’s like 3blue1brown but I actually understand what’s happening

  • @bobbun9630
    @bobbun9630 2 роки тому

    The cube root of 31 is quite a bit better. Perhaps a bit tougher to use in practice, but if you're worried about "in practice", it's easy to memorize a half dozen digits of pi and use that as your approximation.

    • @richardbloemenkamp8532
      @richardbloemenkamp8532 2 роки тому

      Why not use 4*arctan(1) directly. It is exact and also works very well in most programming languages when you can't find pi.

    • @bobbun9630
      @bobbun9630 2 роки тому

      @@richardbloemenkamp8532 That's an option, of course. Depending on circumstances. You might be using a handheld calculator that doesn't have inverse tangent available.
      It's probably worth mentioning that the exact value of pi isn't numerically expressible on a computer at all--numeric values on computers are all rational numbers. The calculation you give does not, therefore, give an exact value, just a better approximation. You could represent pi symbolically without ever converting it to a number, of course, in much the way you might represent i if writing an implementation of the complex numbers.

  • @Kaslasiyo
    @Kaslasiyo 27 днів тому

    Damn

  • @robertthe19th
    @robertthe19th 2 роки тому

    My question is, why is it approximately pi and not exactly pi?

  • @soudyg9080
    @soudyg9080 2 роки тому +1

    It is not necessary that the perimeter of the inscribed should be lesser than the circumscribed

    • @DeJay7
      @DeJay7 2 роки тому

      Acthally, the way they were drawn, yes it is

    • @titiyop7286
      @titiyop7286 2 роки тому

      Yeah but the explanation he gives isn't enough.

  • @itamarshiponi8758
    @itamarshiponi8758 2 роки тому

    sqrt(0.02003) + sqrt(9) so I have a lot of reasons not to sleep

  • @tahaelaminekassabi5720
    @tahaelaminekassabi5720 2 роки тому

    sqrt(2)*sqrt(3)*4rt(e) is even closer

  • @Inspirator_AG112
    @Inspirator_AG112 2 роки тому +1

    I know how to obtain tan(67.5°) with an octagon.

  • @Tetraverse
    @Tetraverse 6 місяців тому +1

    I always liked 355/113

  • @SuperYoonHo
    @SuperYoonHo Рік тому +1

    wow!

  • @pedrosso0
    @pedrosso0 2 роки тому +1

    Bruh. I saw an equal sign here, I expected an equal.
    ≈ ≠ =

  • @ultrakatiz
    @ultrakatiz 2 роки тому

    yeah 3.14 is closer to pi and simple to remember, so i'll just keep using that
    nice method though

  • @erenkyojin7735
    @erenkyojin7735 2 роки тому

    I also discovered why sin(pi/4) = sqrt2/2, Nice

  • @KingGisInDaHouse
    @KingGisInDaHouse 2 роки тому

    Cube root of 31 or 22/7 is closer and easier

  • @user-xz7sj7vm2r
    @user-xz7sj7vm2r 2 роки тому

    NASA uses 40 digits after the decimal point when using Pi.
    do you think is it a good approximation?

    • @cara-setun
      @cara-setun 2 роки тому

      As far as I can tell, NASA only uses 15 digits of pi
      With 40 digits, you could calculate the circumference of the galaxy to within a nanometer, and with 62 digits, you can calculate the circumference of the observable universe to less than a Planck length

  • @dungewar
    @dungewar 2 роки тому +2

    Bri: sqrt(2)+sqrt(3) is a good estimation
    Meanwhile 22/7: [laughs in simpler]
    I don't know how to make memes. I just wanted to point out the 22/7 is way simpler than sqrt(2)+sqrt(3) to pi

    • @cara-setun
      @cara-setun 2 роки тому +1

      It’s also about 4x more accurate than sqrt2 + sqrt3

    • @eduardoxenofonte4004
      @eduardoxenofonte4004 2 роки тому

      @@cara-setun 22/7 is just brute forcing pi, try the first 100 digits of pi over a googol, if you want the most precision

    • @cara-setun
      @cara-setun 2 роки тому +1

      @@eduardoxenofonte4004 that’s more digits of pi than you’d need for… literally anything in this universe

    • @eduardoxenofonte4004
      @eduardoxenofonte4004 2 роки тому

      @@cara-setun welp, it's very precise

  • @Mohanchous
    @Mohanchous 2 роки тому

    How about pi^4 + pi^5 = e^6?

  • @holctomaz2562
    @holctomaz2562 2 роки тому +2

    I mean, the video is quite interesting, but putting an incorrect equation in the thumbnail is kinda misleading. (Its been fixed :D )

  • @data50090
    @data50090 2 роки тому

    very cool

  • @manucitomx
    @manucitomx 2 роки тому

    Fantastic, well presented problem.

  • @somasahu1234
    @somasahu1234 2 роки тому +2

    Amazing

  • @4thalt
    @4thalt 10 місяців тому

    Meanwhile engineers: π = 10

  • @Alkis05
    @Alkis05 2 роки тому +1

    The weird stuff is that if you do that with two hexagons, one inside the other outside, you would expect to get a better approximation, right? WRONG!
    Basically, you get √3+1.5 instead of √3+1.4142. Doing it with 2 hexagons just gives you a smaller range off possible pi value. The fact that √3 and √2 are equally distant from pi is just a concidence.

  • @benheideveld4617
    @benheideveld4617 2 роки тому

    Take a calculator, compute SQRT(2)+SQRT(3) and then square it. You’ll get a smile…

  • @Sci0927
    @Sci0927 2 роки тому

    1.56^0.78 is about √ 2

  • @natashok4346
    @natashok4346 2 роки тому

    Approximation is 03.14 and every year🐶

  • @Crash-yp7ll
    @Crash-yp7ll 2 роки тому

    ... and cubrt(2) + cubrt(3) ~ e

  • @suddeneevee9441
    @suddeneevee9441 2 роки тому

    "This is actually a very good approximation"
    Literally only got 3 decimals right, 4th not even close. More of a funny coincidence than a approximation.

  • @georgeeaster
    @georgeeaster 2 роки тому

    How about the square root of 9.869606

  • @markdenversorino4136
    @markdenversorino4136 2 роки тому

    Interesting...

  • @_lightless
    @_lightless 2 роки тому +1

    Will you ever do one on the sum from n=0 to infinity of n, and its interesting solution?

    • @BriTheMathGuy
      @BriTheMathGuy  2 роки тому +1

      Yep! I'm in the process of planning it.

  • @Blaqjaqshellaq
    @Blaqjaqshellaq 2 роки тому

    While we're at it, pi is pretty close to the square root of 10...

  • @airtonoliveira1732
    @airtonoliveira1732 2 роки тому

    Uhhhh that’s nice 👍🏻

  • @mr.daredevil2181
    @mr.daredevil2181 2 роки тому

    still.....if we add your approximation upto three decimal we will get the value of pie

  • @GAMEDATA1010
    @GAMEDATA1010 2 роки тому

    A better approximation is
    355/113
    And it's only 3 digits and accurate to the a great degree

  • @jimschneider799
    @jimschneider799 2 роки тому +1

    But what about the (real) cube root of 31? cbrt(31) =~ pi to 1 part in 14,817.

    • @SRahul22
      @SRahul22 2 роки тому

      Bro
      What do you want to say ?

    • @jimschneider799
      @jimschneider799 2 роки тому

      @@SRahul22, Bri explained why sqrt(2)+sqrt(3) is an approximation of pi. Is there some reason that the cube root of 31 is so close?