Your new favorite pi approximation.

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  • Опубліковано 12 чер 2024
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КОМЕНТАРІ • 181

  • @jerrysstories711
    @jerrysstories711 2 місяці тому +244

    Nah, mine is 355/113 forever. It has a pleasing 113355 pattern in it, and it's so crazy close that if you used it to calculate the circumference of the Earth, you'd be off by just 11 feet!

    • @jameskislanko8364
      @jameskislanko8364 2 місяці тому +19

      And the rational approximation to pi with error less than 10^-10 and smallest denominator is 312689 / 99532

    • @nicogh1spy
      @nicogh1spy 2 місяці тому +41

      Feet 💀💀💀

    • @mtaur4113
      @mtaur4113 2 місяці тому +6

      How close is the Earth to spherical, even?

    • @mtaur4113
      @mtaur4113 2 місяці тому +4

      Like we know there are mountains and trenches. We could use sea level to get something smoother, but we know the tides change, though I suppose not by 11 feet. But the error is probably on the order of several inches.

    • @mtaur4113
      @mtaur4113 2 місяці тому +1

      I am doing some guesswork here and I don't know where you would learn fully accurate science about it.

  • @tonyennis1787
    @tonyennis1787 2 місяці тому +46

    12:21 This was those rational numbers' moments. The sun was shining down on them in their moment of glory. You took that away from them. You monster.

  • @SgtSupaman
    @SgtSupaman 2 місяці тому +13

    355/113 is one I worked out myself as being the best approximation that isn't too long and difficult to remember, so I really doubt I will ever have a pi approximation that I like more than that one.

    • @theodoresweger4948
      @theodoresweger4948 2 місяці тому +2

      I came to same conclusion long time ago like it better than the old 22/7 Well done.

    • @wyattstevens8574
      @wyattstevens8574 Місяць тому +1

      And as far as I know, it's the only one whose accuracy (in d.p.) trumps the number of digits in its fractional representation!

  • @nagoranerides3150
    @nagoranerides3150 2 місяці тому +8

    pi=3. Good to

    • @forrestgreen9369
      @forrestgreen9369 2 місяці тому +1

      Yep. 3 is what I use 90% of the time. If I need better than that I use the pi key on my calculator.

    • @mytriumph
      @mytriumph 2 місяці тому

      And if you *do* need to be closer, without a calculator (for whatever reason), 3.1 is good to

    • @wyattstevens8574
      @wyattstevens8574 Місяць тому

      ​@@mytriumphOfc 22/7 (n=1 integral) is more accurate, and 355/113 more accurate still!

  • @dkravitz78
    @dkravitz78 2 місяці тому +16

    When you said error

    • @soupisfornoobs4081
      @soupisfornoobs4081 2 місяці тому +7

      I think his maths degree should be revoked

    • @wpbn5613
      @wpbn5613 2 місяці тому

      pi being irrational is not trivial!

    • @Sir_Isaac_Newton_
      @Sir_Isaac_Newton_ 2 місяці тому

      "his maths degree should be revoked"
      (proceeds to forget the +C on some integral)

  • @jimskea224
    @jimskea224 2 місяці тому +18

    Strangely, the number in the thumbnail doesn't appear in the presentation.

    • @ffggddss
      @ffggddss 2 місяці тому +2

      I too was looking for that. He has a whole class of such approximate fractions outlined at the end, but it looks like that thumbnail fraction doesn't occur in any of them, either.
      For the n=1 case, the bracketing fractions are in 4-digit terms, the approximation being just 1- and 2-digit numbers, arrived at by using common denominators on those bounding fractions. For the n=3 case, the common denominator will be pretty big. But does the average of the bounds reduce to what was in the thumbnail? I haven't worked that out.
      To me, the interest is in the ability to generate ever-closer rational approximations to π with increasing n, albeit at enormous computational expense.
      Fred

    • @wyattstevens8574
      @wyattstevens8574 Місяць тому

      Maybe it's the fraction that drops out of the n=2 case?

  • @braydentaylor4639
    @braydentaylor4639 2 місяці тому +12

    I'm just gonna say it. 22/7 is overrated.

  • @lucaswilkins9217
    @lucaswilkins9217 2 місяці тому +5

    When you say half the interval size is the maximum error, that's not right. It would be right if the approximation was exactly half way between the ends, but that's not what you have. Pi could be close to one end of your interval, and the approximation at the other. It means the errors you calculate from your bounds should be twice as big.

  • @mtaur4113
    @mtaur4113 2 місяці тому +9

    Hmmm, integrating powers of a function that is less than 1 to make the integral small, and using the fixed denominator to guarantee an arctan term in the antiderivative. Moderately slick.

    • @idjles
      @idjles 2 місяці тому +1

      Very slick!

    • @la.zanmal.
      @la.zanmal. 2 місяці тому +1

      Thanks, this makes it clearer why the generalization is possible; since the polynomial in the numerator has an exponent which is a multiple of 4, presumably one can show that the remainder of the long division always works out to 4.

  • @xinpingdonohoe3978
    @xinpingdonohoe3978 2 місяці тому +13

    It's better than 4/1-4/3+4/5-...
    It looks cool on paper, and is easy to derive with integrals, but takes 100 years to converge to something resembling π, I swear.

    • @ffggddss
      @ffggddss 2 місяці тому

      Yes, I like to point out that this, 4x the infinite alternating odd harmonic series, is both the most beautiful and the most computationally useless way to calculate π.
      Fred

    • @jay_sensz
      @jay_sensz 2 місяці тому +1

      The average of subsequent partial sums of this series converges a lot faster though.

    • @benhur2806
      @benhur2806 2 місяці тому +2

      using the related 4arctan(1/5)-arctan(1/239) taylor expansion is quite efficient though... xP

    • @ffggddss
      @ffggddss 2 місяці тому +1

      ​@@jay_sensz Yes, that's a common method of convergence-acceleration. In this case it amounts to changing the final (& only the final) term in each partial sum to half its value.
      This mars its "beauty" but improves its "performance." Even so, it's still pretty slow, compared with some other series.
      One improvement is the Taylor series for 6sin⁻¹(½) = π.
      And of course, even faster is the Machin formula given in @benhur2806's comment, which gives ¼π.
      Then there are ridiculously fast, but more difficult-to-compute series, starting with one given by Ramanujan...

  • @holyshit922
    @holyshit922 2 місяці тому +7

    I used Euler's transformation of arctan(x) series and I have got
    def mysum(n):
    s = 0
    p = 1
    for k in range(n+1):
    s += p
    p *= (k+1)/(2*k+3)
    return 2*s
    It should set one correct bit per iteration

    • @BridgeBum
      @BridgeBum 2 місяці тому

      I love this, so simple but reasonably fast convergence.

  • @Sam_on_YouTube
    @Sam_on_YouTube 2 місяці тому +4

    So if Matt Parker wanted to do this for the next Pi Day and get up to 600 decimal places, how big would he have to make x? And while that is harder for a computer to do, would that be less error prone for a group of humans than the method he did this year, allowing for a faster solve time by the humans?

  • @pablojesusmolinaconcha4504
    @pablojesusmolinaconcha4504 2 місяці тому +3

    One of my favourites is sqrt(g), where g is the gravitational pull of the earth ~9,81

    • @bonzinip
      @bonzinip 2 місяці тому +3

      But did you know that it's not a coincidence? The initial definitions of the meter and second were related so that a pendulum of length 1 meter would have a period of 2 seconds, which gives g=pi^2 m/s^2.

  • @aweebthatlovesmath4220
    @aweebthatlovesmath4220 2 місяці тому +5

    This is so much cooler then using Continued fractions!!

    • @ingiford175
      @ingiford175 2 місяці тому +4

      I was thinking of taking those 2 fractions and seeing how the continued fraction of each looks like

  • @MrMilkdashake
    @MrMilkdashake 2 місяці тому +2

    Top tier approximation of Pi is:
    Pi ~= 31^(1/3)

  • @user-rizzwan
    @user-rizzwan 2 місяці тому +1

    The accuracy of calculation is dependant on how many digits the number you're multiplying with.

  • @jardozouille1677
    @jardozouille1677 2 місяці тому

    I had never heard about these wheels. Thanks.

  • @charleyhoward4594
    @charleyhoward4594 2 місяці тому +2

    all this talk about rational no. caused me to check the lotteries ...

  • @wyattstevens8574
    @wyattstevens8574 Місяць тому

    I guess the n=2 integral is "left... as an exercise for the viewer?" He did jump from n=1 (22/7 w/error < 1/2520) to n=3 (error less than 0.0000000001- yes, that's 10 d.p.).
    But when I did the math for n=2, the approximation was "pi is 9637/3003 with absolute error of magnitude 853/5,250,960 or less."

  • @MrKA1961
    @MrKA1961 2 місяці тому +1

    22/7-pi is approx 0.012, which is about 1/84, so, the error is by far more than 1/2520

  • @eliasrodriues6614
    @eliasrodriues6614 2 місяці тому +4

    A moment of rationality of Mr. PENN

  • @dneary
    @dneary 2 місяці тому +1

    This identity is wonderful! What is its origin, and how was it derived?

  • @PawelS_77
    @PawelS_77 2 місяці тому +24

    My favorite approximation of pi is sqrt(2) + sqrt(3).

    • @jimskea224
      @jimskea224 2 місяці тому +3

      sqrt(10) isn't far off either

    • @di-riso
      @di-riso 2 місяці тому +6

      Ahhh yes replace the irrational number with a sum of two other irrational numbers,might as well use π/2+π/2 and get maximum precision shall we??

    • @billcook4768
      @billcook4768 2 місяці тому +1

      @@di-risoWhat about tau/2?

    • @dcterr1
      @dcterr1 2 місяці тому +2

      Good one! Not as crazy as the near inequality pi^4 + pi^5 = e^6 though!

  • @byronwatkins2565
    @byronwatkins2565 2 місяці тому +4

    At 11:30, you are mistaking "maximum error" with "round-off error." Any epsilon>0 from either endpoint leaves effectively the entire interval on the other side. The probable error is about 1/2e of this interval.

    • @robertveith6383
      @robertveith6383 2 місяці тому +1

      If epsilon is meant to be in the denominator, then this needs grouping symbols: 1/(2e).

    • @byronwatkins2565
      @byronwatkins2565 2 місяці тому

      @@robertveith6383 Not epsilon; e=2.718...

    • @byronwatkins2565
      @byronwatkins2565 2 місяці тому

      @@robertveith6383 Parentheses are implied; otherwise it would be e/2. Also, it is e=2.718... not epsilon.

  • @mtaur4113
    @mtaur4113 2 місяці тому +2

    The error is less than half the width if you average the upper and lower estimates, so I think you should have written out the average for good measure.

  • @la.zanmal.
    @la.zanmal. 2 місяці тому +2

    This would be better if it explained at the end why the generalization is possible. It's clear that values of n > 1 produce tighter bounds, but not why those bounds still include pi.

    • @Yoshinoyo1
      @Yoshinoyo1 2 місяці тому +1

      I guess, when you perform the Euclidean division of the numerator by (x^2+1), you end up with some 1/(x^2+1), which integrate to arctan, hence pi

    • @ZekeRaiden
      @ZekeRaiden 2 місяці тому

      The form of the polynomial long division always produces 4/(x^2+1) as the remainder, because all you're doing is multiplying all of the components of the equation by x^4 except the x^0 component.

  • @maxhagenauer24
    @maxhagenauer24 2 місяці тому +1

    Lim n -> oo [ 2^n * sqrt( 2 - sqrt( 2 + sqrt( 2 + sqrt ( 2 + ...)))) with n many total roots. Use a large natural number for n, and that should give a good approximation.

  • @dg8620
    @dg8620 2 місяці тому

    The best thing about the 22/7 approximation is that the final decimal point matches pi.

  • @NLGeebee
    @NLGeebee 2 місяці тому +1

    0:26 is the integral a good approximation of pi because of of just because it gives a good enough value?

  • @chrisrichardson111
    @chrisrichardson111 2 місяці тому +3

    The bit I don't get is the equation around 6:44. If it actually equals 22/7 - pi then the given integral doesn't approximate pi it approximates zero.

    • @yuvrajguglani821
      @yuvrajguglani821 2 місяці тому

      yes, the integeral would be _exactly_ zero had the value of pi been 22/7 the answer to the given equation basically gives us the value of pi. the vid tries to approximate the answer to 22/7 to find a good approximation for pi. you can of course increase the values of n like shown in the vid and try to find the another equation and approx pi (left as an exercise to you, if you need help try re-watching the vid)

    • @chrisrichardson111
      @chrisrichardson111 2 місяці тому

      ​@@yuvrajguglani821Point missed, I think. Your claim that "the answer to the given equation basically gives us the value of pi" is incorrect. At the timestamp given it approximates zero.

    • @wyattstevens8574
      @wyattstevens8574 Місяць тому

      The question was actually essentially "how close to 0 is the integral?" The closer it was, the closer the Q in question would be to pi- in fact, said error would be about abs(22/7-pi).

    • @chrisrichardson111
      @chrisrichardson111 Місяць тому

      @@wyattstevens8574 That is the point.
      "So what I'd like to do today is work through all of the details evaluating this integral and seeing why it's a good approximation of pi". It isn't.

  • @lapaget1
    @lapaget1 2 місяці тому

    From this family of integrals, it’s a way to demonstrate that pi is irrational as these integrals are strictly positive. Thank you, Michael, for the idea.

    • @Happy_Abe
      @Happy_Abe 2 місяці тому

      How do you get that it’s irrational?
      What if it’s some really large numerator over some really large denominator that this method is approaching?

  • @wernerviehhauser94
    @wernerviehhauser94 2 місяці тому +10

    My favourite approximation is 3, if 3 is not precise enough, I'm usually going to need at least 7 digits....

    • @peterkron3861
      @peterkron3861 2 місяці тому +3

      And this was the value legislated by Indiana so good enough for me

    • @trucy1337
      @trucy1337 2 місяці тому +1

      3.000000

    • @wernerviehhauser94
      @wernerviehhauser94 2 місяці тому +1

      @@peterkron3861 with the caviat that setting pi to exactly 3 is simply wrong and using 3 as an approximation means being aware of the error. Some legislators only demonstrate that they are completely unaware of any of their errors....

    • @SubTroppo
      @SubTroppo 2 місяці тому

      @@peterkron3861I shall forever now call three the Locally-Yokelly number and you have reminded me of that song: 'Indiana Wants Me' (obviously 'Me' is probably not a mathematician.)ua-cam.com/video/2p3OfHP5Hmo/v-deo.htmlsi=3I4uiHQK7vN0LcbO

  • @dcterr1
    @dcterr1 2 місяці тому

    Here's a good philosophical question for you guys. Could another universe exist in which pi had a different value?

  • @dcterr1
    @dcterr1 2 місяці тому

    It's crazy to me how precisely pi is known, to something like 100 trillion digits, when there is no known practical reason to know more than 40 or so, other than testing software and looking for patterns in its digits or continued fraction, which don't seem to exist.

  • @StevenSiew2
    @StevenSiew2 2 місяці тому

    The best approximation is 355/113 because all you have to remember is 113355

  • @masterbaraman9372
    @masterbaraman9372 2 місяці тому

    Isn't the whole idea of 22/7 that it's easier to do mentally and by hand?

  • @frobeniusfg
    @frobeniusfg 2 місяці тому

    I guess, using the same logic of integrating p(x)/(1+x) it's possible to find rational approximations of ln(2)

  • @jarikosonen4079
    @jarikosonen4079 2 місяці тому

    10:33 22/7-π~1/791 and most of these rational approximations seem give one more decimal in the approximated value of pi than there is decimals in nominator denominator together. So there is no point to rationalise it if the number of decimals on the ratio is not much lower than in rationalised pi...
    Why does that integral give values near to the pi ?
    Why is the '0' the digit that appears last in π at 33rd place (after others [1...9])?
    Program can give easily values such as 312689/99532.
    Or integrating Taylor series of 4*sqrt(1-x^2) from 0 to 1...

  • @dcterr1
    @dcterr1 2 місяці тому

    I seem to recall seeing a proof of the irrationality of pi involving integrals like these. Can you prove that pi is irrational in this way?

  • @JxH
    @JxH 2 місяці тому

    @12:48 ""Error < 10^-10" Wouldn't that be expected given that the four integers all have at least 10 digits?
    In other words, it'd be a sad result if the error was (for example) "...< 10^-6" given all the ten digit numbers.

  • @gesucristo0
    @gesucristo0 2 місяці тому +3

    Does this mean that as n goes to infinity the error is zero?

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 2 місяці тому +5

      The integral is from 0 to 1. At the endpoints, x^(4n)(1-x)^(4n) will equal 0.
      In between, for 0

    • @mtaur4113
      @mtaur4113 2 місяці тому +1

      I believe this is all correct. The long division and antiderivative should grow in complexity, and to be 100% sure that we get an estimate for pi, we need to prove that the C/(1+x^2) term is nonzero for all n, or at least infinitely many n, but I suspect all n.

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 2 місяці тому +1

      @@mtaur4113 x^(4n)(1-x)^(4n), as a real polynomial, will be able to be expressed as a product of linear terms and irreducible quadratics. It's already factored, into a product of just linear terms, so no irreducible quadratic will divide it. That includes the irreducible quadratic 1+x².

    • @mtaur4113
      @mtaur4113 2 місяці тому +1

      @@xinpingdonohoe3978 This could still produce Ax/(1+x^2), unless I missed something.

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 2 місяці тому +1

      ​@@mtaur4113 I guess, look at the example he did. The minimum power for the thing we're dividing is x^(4n). And since we've only got 1 and x^2, then the division chain at the end will go x^(4n), x^(4n-2), x^(4n-4), etc. because each one is what is needed to multiply x^2 with to get the above one, as the long division asks. So you won't see odd powers of x below x^(4n+1), so there won't be an x^1 in the remainder since it won't appear at all in the division.

  • @klausolekristiansen2960
    @klausolekristiansen2960 2 місяці тому +1

    Where did halving the max error come from?

    • @charleyhoward4594
      @charleyhoward4594 2 місяці тому +1

      he stated that the max no. on the integral was 1(set x=1); thus 1/(1+ x^2) = 1/(1 +1) = 1/2

    • @alquinn8576
      @alquinn8576 2 місяці тому +2

      put another way, he bounded the value of pi to within a range of size X. if you find the midpoint of the range defined by the inequalities and set that as your approximation of pi, then the farthest the true value can be is one half of X

    • @klausolekristiansen2960
      @klausolekristiansen2960 2 місяці тому

      @@alquinn8576 thanks

  • @thegoldengood4725
    @thegoldengood4725 2 місяці тому

    22/7 - π is a very good approximation of zero

  • @jaimeduncan6167
    @jaimeduncan6167 2 місяці тому

    Suggestion for everybody, Veritasium video on the calculation of PI : ua-cam.com/video/gMlf1ELvRzc/v-deo.html The Discovery That Transformed Pi

  • @lincolnyu6768
    @lincolnyu6768 2 місяці тому

    I don't understand why you can half the error to 1/2520... Someone help plz

    • @alquinn8576
      @alquinn8576 2 місяці тому +1

      he bounded the value of pi to within a range of size X. if you find the midpoint of the range defined by the inequalities and set that as your approximation of pi, then the farthest the true value can be is one half of X (+X/2 or -X/2 from the midpoint)

  • @Nyth63
    @Nyth63 2 місяці тому

    355/113

  • @JimmyDoyel-by2cp
    @JimmyDoyel-by2cp 2 місяці тому

    Might as well just memorized pi lol.

  • @MikeB3542
    @MikeB3542 2 місяці тому +3

    I'm on the "pi = 22/7" team...as an aside, 19/7 is an excellent approximation of e.
    Could be worse...the engineer in me is tempted to use 3 for both 😅

    • @Qermaq
      @Qermaq 2 місяці тому +1

      3 is a reasonable value. It all depends on your margin of error.

    • @ZekeRaiden
      @ZekeRaiden 2 місяці тому +2

      I'm still partial to 355/113. It's not meaningfully slower than 22/7, but instead of being only accurate to two decimal places, it's accurate to six--and it takes much, much larger approximations to get even small increases. For anything but aerospace work and quantum physics papers, 355/113 is genuinely indistinguishable from the real thing

  • @AnatoArchives
    @AnatoArchives 2 місяці тому +1

    wao pi

  • @BobChess
    @BobChess 2 місяці тому

    π ≈ π-0.00000000000000000000000000000000000000001 🥶

  • @ericfielding668
    @ericfielding668 2 місяці тому +1

    That'll do in a pinch, but I find it easier to simply remember the first 32 decimal digits of pi (I do know these first 32 off by heart.)

  • @Ladka13-rx7eq
    @Ladka13-rx7eq 2 місяці тому

    [Request]: What is the last non-zero digit of $(\dots((2018\underset{! \text{occurs}1009\text{times}}{\underbrace{!)!)!\dots)!}}?$

  • @Qermaq
    @Qermaq 2 місяці тому +1

    Why is pi so easily estimated by roots of integers? 31^1/3 is accurate to 3 places. 306^1/5 is to 4. 29809^1/9 is to 5. 93648^1/10 is to 6. 294204^1/11 is to 7. Is there a reason there are so many cases of this? I mean, there's got to be a pattern.

    • @IoT_
      @IoT_ 2 місяці тому +2

      There's nothing special about pi. Given the density of a^(1/n) you can approximate any real number with any precision.

    • @robertveith6383
      @robertveith6383 2 місяці тому +1

      All of your fractional exponents must be written inside grouping symbols!

    • @Qermaq
      @Qermaq 2 місяці тому

      @@robertveith6383Well yes, if I was presenting this rigorously I would use superscript notation. (1) This is a UA-cam post. (2) You figured it out. (3) Satisfied.

  • @versusstatusquo
    @versusstatusquo 2 місяці тому

    311/99

  • @user-qb8fp8oj1p
    @user-qb8fp8oj1p 2 місяці тому

    Smart❤😂👍🏆💪

  • @manyokivagyok
    @manyokivagyok 2 місяці тому

    nah, mine is π/1

  • @BajrangJat-mr1gu
    @BajrangJat-mr1gu 2 місяці тому +3

    Thanks sir for help we I am from """indiaa"""
    """Shreeniwash ramanujan """
    Country

    • @vestieee5098
      @vestieee5098 2 місяці тому +1

      Srinivasa*

    • @BajrangJat-mr1gu
      @BajrangJat-mr1gu 2 місяці тому +1

      Yes

    • @BajrangJat-mr1gu
      @BajrangJat-mr1gu 2 місяці тому +1

      Ramanujan
      Aaryabattt
      Aacharay madhav who discover calculas and infinite series

    • @Miyamoto_345
      @Miyamoto_345 2 місяці тому

      Bro you are suffering from crippling inferiority complex. Get some love from your parents.

  • @TurdFerguson43
    @TurdFerguson43 2 місяці тому +3

    My favorite approximation is actually just pi.

    • @Qermaq
      @Qermaq 2 місяці тому +1

      Heh. But that's the exact value, it's not an approximation. It's like if the abbreviation for May was "May".

  • @jhawk2402
    @jhawk2402 2 місяці тому +10

    pi^4+pi^5=e^6

    • @woody442
      @woody442 2 місяці тому +3

      Pithagorean Eorem

    • @aweebthatlovesmath4220
      @aweebthatlovesmath4220 2 місяці тому +8

      Tell me you are an engineer without telling me you are an engineer:

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 2 місяці тому +1

      If that were true, what would the consequences be? Could we no longer conclude that one of π+e and πe was transcendental, for example?

    • @mtaur4113
      @mtaur4113 2 місяці тому +1

      ​@@xinpingdonohoe3978 hard to say. They aren't equal, but if they simultaneously are equal, subtract to get 0.000000034=0 or whatever, then multiply both sides by whatever and then every number is equal to zero. Contradictions cut deep.
      Maybe there is some other mathematical structure where something else plays the roles of e and pi. You would have to go outside of looking at Euclidean geometry and metrics or something, and I don't know anything off the top of my head. I saw something trending about computing "pi" for metrics where an exponent different from 2 is used in the Pythagorean theorem. pi is the minimum possible, and I don't remember if the e^6 is the bigger or smaller number. If it's bigger, then you could choose a p, but if smaller, you can't. But this probably isn't the only weird thing you could do with it.

  • @finmat95
    @finmat95 2 місяці тому +1

    My favourite is pi=3.

  • @billcook4768
    @billcook4768 2 місяці тому

    3/1 works for me. But I’m a simple guy.

  • @Poepwoes
    @Poepwoes 2 місяці тому

    No. Pi is an "irrational" number and cannot be described/represented by a string of numbers (incl ratios) in ANY numerical base. Did not watch the vid

  • @RJiiFin
    @RJiiFin 2 місяці тому

    Nah, I'll just use 3 and call it a day

  • @parthhooda3713
    @parthhooda3713 2 місяці тому +1

    Nah mine favourite approximation is π≈π

  • @ludolfceulen
    @ludolfceulen 2 місяці тому

    en.wikipedia.org/wiki/Continued_fraction#Continued_fraction_expansion_of_%CF%80_and_its_convergents

  • @2kchallengewith4video
    @2kchallengewith4video 2 місяці тому +3

    Nice thumbnail