An amazing approximation of pi!
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- Опубліковано 11 вер 2024
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Did UA-cam screw this up? I got a 3:53 minute video that starts at the tail end of a discussion on the factorial of the derivative operator that cuts midway into the tail end of pi^3 being about 31.
Yes, UA-cam did something weird and spliced two videos together -- it was ok at first.... I am chatting with them right now to try and figure this out.
Seems like we got a glitch in the matrix 😅
It was a matrix in the glitch.
I thought that we slipped from Christmas into April Fools.
My favorite approximation of pi is still 355/113. Correct to 7 decimal places and easy to remember: just write the first 3 odds twice each and then split into 2 numbers: 113|355. :-D
This approximation is way better than the 31. And easier to use, just a division. What use is the cube root approx, when is it useful?
@@MarcusCactus 31 and 3 have 3 digits between them, and the approximation is accurate to 4 digits. 335 and 113 both have 6 digits between them, and the approximation is accurate to 7 digits. An approximation that is accurate to roughly as many significant digits as the total number of digits in all numbers involved is completely unremarkable. Give me a formula with 6 digits total that approximates pi to 12 significant digits - that would be remarkable.
FYI and for those who are interested, these are the approximations of pi in the similar form mentioned in this video:
3^(1/1), 10^(1/2), 31^(1/3), 97^(1/4), 306^(1/5), 961^(1/6), 3020^(1/7), 9489^(1/8), 29809^(1/9), 93648^(1/10), ...
Those integers are in the sequence numbered A002160 in OEIS.
Why there is clip of other video ( about d/dx!) in the middle of this video. Was it intended?
I thought is was to distract you from the indexing error of using 1 instead of 0, but when the correct content resumed it still showed 1 and he corrected it later.
No, but these videos already have pretty poor editing so something like this was bound to happen eventually
PI ~= (2143/22)^0.25
is a pretty good approximation due to Ramanujan. The error is
-1.00714711325660620363E-9
The more complex the formula is, the more precise the approximation is, not surprising.
By the way, you may have a ≈ sign on your phone keypad. Well, unless you are typing on a computer, then depending on the operating system you have different ways to find a given character, for example from a list of characters on the Internet.
Wish you all the best ♥️
10:47 to 12:11 is a section from a completely unrelated problem.
Yeah 🤣 now he gonna become same like other youtubers 😭
This video is (also) sponsored by the video "Finding the Factorial of the Derivative!" 😉
Thanks to pointing this out. I didn’t edit this video but i did input a fix using UA-cam’s editor, hopefully it’ll be corrected soon. Just waiting on UA-cam to process it.
Sounds like those math teacher jokes everyone makes.
"So we just logically derived how 1+1=2, now solve the Riemann Hypothesis"
@@NotoriousSRG Hey, the repost is wrong too! Now the video START with the section from the derivative video.
306^(1/5) is better
28658146^(1÷15) is just 0.0000000002 bigger than π
306.0196848^(1/5) even better.
It uses more numbers, so it is less efficient, that is not better just more precise. This time less is more.
∛31 - π = 0.0002120012 (4 digits accuracy)
22/7 - π = 0.00126448927 (3 digits)
355/113 - π = 0.00000026676 (7 digits)
ln(640320³ + 744) / √163 - π =0.000000000000000000000000000000223 (30 digits)
“Indiana Pi Bill” - π = 0.0584 (2 digits)
more cube roots minus pi (from the end):
32 -> 0.0332 (2 digits)
32-1 -> 0.000212 (4 digits as noted already)
32-1-5/27 -> -0.00648 (3 digits and a sign change)
32-1-5/27+32/125 -> 0.00218 (still only 3 digits)
It seems like there’s something missing here as 31 looks like a sweet spot that remains unexplained.
Than Ln over SQRT(163) is soooo Ramanujan...
@@TimMaddux ...not really, because he separated the first term (32/27) in an integer and a rational part (1+5/27). If you consider the series term by term (without splitting the first one) it gets better and better (converges) as you would expect.
You made it make sense; thank you!
My personal favorite approximation of pi is 355 / 113, which is rememberable as "1 1 3 3 5 5".
kingbeauregard
Yes; mine too.
The best approximation, easy to remember & execute is "355/113". Write 113355 (the first three odd numbers in doubles), draw a line in the middle, then put the first half as denominator & the latter half as numerator of the fraction that we seek. The error is 2.7X[10exp(-7)], about a quarter millionth.
A quick Desmos-based investigation seems to suggest that the Fourier series for x(pi-x) should start at n = 0, not 1. Excellent video as always :D
He fixes that later.
I've always been intrigued by how crazy good 355/113 is. Using the old method of approximating the circle as a polygon, it's easy to show that 3/1 < pi < 22/7. Then with lots of paper computation the same method can show that the series (3+22n)/(1+7n) gets closer and closer to pi from below all the way to n=15, (333/106). But then at n=16, (355/113) is such a good approximation that you can't use the polygons method anymore. It'd be years of paper computation to even prove if it's < or > pi.
That's why until Newton changed the game, our best proven bounds were 333/106 < pi < 22/7.
But 355/113 is so close that if you used it to draw a circle the size of the Earth, your equator would be off by just 12 feet!
@jerrysstories711 Greetings Jerry. Aren't you that one who helped me prove an inequality ?
@@zeggwaghismail827 Maybe. I do things like that sometimes. 🙂
@@jerrysstories711 your old account is jerrymouse, isn't it !
@@zeggwaghismail827 Nope, that's not me. Some other math-lovin' Jerry must have helped you.
I'd love to see the details of that calculation! You or Michael should make a video on that!
I'd love to see a video on other orthogonal bassis functions, like the Legendre or Laguerre polynomials!
That come up a lot in physics, so I was wanting some more mathematical flavor.
13:12 Note that the next term left out is 32/343 or by as similar crude approximation, about 1/10, i.e., we'd next end up with 31 - 1/20 instead of 31 + 1/20. So 31 is quite in the centre of the oscilation of the alternating series, so is pretty close to the limit
Nothing is more amazing than 355/113 approximation.
There are better answers, Westerner.
@@Sh4dowbannedWesterner????????
I'm sure there's probably a trivial reason for this but I was playing around w this and it seems like the expression:
(3×10^n)^(1/(2n+1))
Always seems to give a halfway decent estimation for pi. Of course, if you tweek the 3×10^n part you can get it closer. Eg for n=2
300^(1/5)=3.12913...
Whereas
306^(1/5) = 3.14155...
But (3×10^n)^(1/(2n+1)) seems to get you in the ballpark.
Great, but the hard bit is to guess what function to use as the basis of the Fourier expansion in the first place. Presumably this result was originally noticed by accident when someone expanded x(pi - x) as a Fourier series. Is there a systematic way of figuring out what function to use in situations like this?
It clearly wasn't a good place to _start_
Merry Christmas and Happy New Year, Michael and all readers of this comment.
got some nice ones too;
38153603,19^(1/15,25) = 3,14159265358979
and the real pi = 3,14159265358979 (excel approximation)
so at least about 10^-14 accuracy.
or
55666641,05^(1/15,58) = 3,1415926535897
which is to 10^-13 accurracy
or
67625500,9^(1/15,75) = 3,14159265358969
which is to 10^-13 accuracy aswel
I prefer 2*arccos(0), but this is really nice, too.
PI ~ Acos (-1) is pretty good and avoids any multiplication.
@@bernhardbauer5301 Basically the same thing as mine, just without multiplication.
@@SyRose901 it's more stable numerically, when computing in a program
Those are not approximations, though.
@@whycantiremainanonymous8091 Ah, yeah. Pretty sure this is just what pi is supposed to be.
At 10:48, there is a sudden change of subject... At 12:10, the original subject is resumed.
I'm only seeing a 3 minute 53 second video that starts with matrices?
@@prk55 Yeah, they reposted, but the repost is wrong too.
113355
355/113 is a very close approximation of pi.
yes, it was a good place to stop, but it wasn't a good place to start.
😂😂
One way to think of it is informationally: pi minus the cube root of 31 is about 2*10^-4; is that very remarkable? Well, we have to specify two digits of an integer, and one digit of an exponent, and the operation "yth root of x" out of all the simple operations we could possibly do, and that seems not so far off from encoding 4+ digits of information, so I guess that's about in line with the "goodness" of approximation we could expect to get this way.
Hey, I logged on to youtube to relax and take a break from politics, and I see an LGM commenter and a Bernie shirt. (Love to see both.)
I was always partial to sqrt(g), the gravitational acceleration on the surface of the Earth. Because it makes the π and the g cancel out in the formula for period of a pendulum.
Also e^π-π=20, of XKCD fame (number 207).
UA-cam Editor did the exact opposite of what i was trying to do. Will fix asap.
not your fault, youtube has been eating shit lately
If someone want to calculate it with paper and pencil I give you some hints
(10a+b)^3 = 1000a^3+300a^2b+30ab^2+b^3
(10a+b)^3 - 1000a^3 = 300a^2b+30ab^2+b^3
(10a+b)^3 - 1000a^3 = (300a^2 + 30ab + b^2)b
(10a+b)^3 - 1000a^3 = ((300a^2+b^2)+30ab)b
1.
At first you divide number to three digits groups starting from decimal point in both directions
2.
Then choose digit such that differece between leftmost three digits group and cube of this digit is smallest nonnegative number
It will be first digit of actual approximation
3.
To the difference we append next three digits group
On the side we write triple square of actual approximation and append to it square of last digit of next approximation
To this number we add triple product of actual approximation and last digit of next approximation shifted one position to the left
(this triple product of actual approximation and last digit of next approximation is shifted one position to the left)
This number we multiply by last digit of next approximation then we subtract it from the actual rest
4. Repeat step 3. until rest is equal zero or we have got enough digits
No it wasn't an error. The Matrix just glitched. Look out Morpheus.
Best i know is still the easiest to remember:
355/113
True. The sequence of digits 113355 from bottom left to top right makes it my favourite.
Cool video but the index should start at zero for 2n+1.
*edit* Dangit! Should have waited before commenting!
When reindexing the sum at 8:45, you accidentally removed the n=1 term. The sum should either start at n=0 or the 2n+1 should be 2n-1 instead, either fixes the issue.
Edit: He corrects it later in the video
Wait, what do you mean at 8:45 because this video is less than 4 minutes long!
Oh, are you talking about his other video about (d/dx)!, and if so, then why not comment on that video instead?
@@megauser8512 I think like half the video cut off or smthm
Editing mistake at ~ 10:48
The video is 3 minutes 53 seconds long tho…
What are you talking about?
@@fullfungo this is why you dont do drugs lol
Pi most aproximate
is 28658146^(1/15)
the divergence of
π and 28658146^(1/15) is
about (-2.23) x 10^(-10)
5:13 Nice Chrono Trigger reference
Approximation is off by about 0.000212
I guess there's a sequence such that the nth root of the nth term produces an approximation for pi:
a_n ={3, 10, 31, 97, 306, 961, 3020, 9489 ... }
Surprisingly, it appears as though each successive approximation is not necessarily better than the previous one (I have not proven this. Just used excel to explore the pattern).
I'm curious to know if there are other higher degree polynomial equations whose solutions approximate pi. Some of the linear equations include:
x - 3 = 0 (the solution to this equation is the first term in the above sequence)
7x - 22 = 0
106x - 333 = 0
113x - 355 = 0
x^2 - 10 = 0 is a quadratic equation whose solution is the second term in the above sequence. Is there another quadratic equation whose solution approximates pi better than sqrt(10)?
More generally, are there equations of degree n whose solution better approximates pi than that of x^n -a_n = 0?
Look up "integer relation algorithm" for the main tool used to find these polynomials
What does the factorial of the derivative have to do with this?
Edit mistake
Can someone please resolve this? I can't wrap my head around it.
e^(πi) = -1
also
e^(3πi) = -1
It must then follow that
e^(πi) = e^(3πi)
But if we take ln on both sides we get
ln(e^(πi)) = ln(e^(3πi))
ln and e's cancel and we're left with
πi = 3πi
HUH?
Just like asking: "Since cos(π)=cos(3π), so π = 3π ?"
No, because arccos is not univalued. (cos⁻¹ for you Americans who confuse inverse and reciprocal).
In the complex plane, exponential is not a real valued function, but actually , fromC to C. Idem for log. hence in the complex, there is no reciprocal from an apparently real result, which is just a particular of a complex value: e(0,π)=(-1,0).
Or equivalently, log is a multi-valued function. Here, both sides give the same answer : (2k+1)πi, k ∈ Z. Just like in the arccos example.
@@MarcusCactus I'm British mate. But that makes sense, the analogy with the trig and inverse trig functions did the trick. Cheers.
e ~ 3 root (20) , phi ~ 3root (31)
I was following along fine when the problem switch to matrix stuff, and I was like, professor, where the heck did that come from?
Did you drop a term when you reindexed? Need to sum from zero so you get the first term (first thing you stick in sin is 3, not 1).
Edit. I see you caught it.
Just in case you see this Michael... I really miss the backflips.
Thank you for the nice content as always.
PI^3 = 31+ 0, 006 = 31 + 1/200 and not 31 + 1/20, am I wrong ?
1:33 Why are we able to get away with only using sines here? I know that on [0, 2pi] we use both sines and cosines because you can’t approximate sin(x) with just cosines (because integrating any cosine times sin(x) over that range gives 0, whereas integrating sin(x) times sin(x) over that range is nonzero). But here, you’re in essence allowing half-period sines, and saying that that is good enough.
I know that Stone-Weierstrass says you can approximate any continuous function uniformly arbitrarily well (and hence any L2 function arbitrarily well in L2) if your space (1) is an algebra, (2) separates points, and (3) contains a nonzero constant. None of these are true, since sin^2(x) is not a linear combination of sines, and f(0)=f(pi)=0 for all f in the space.
The other method I’ve seen to prove density is the Hilbert-Schmidt theorem, which says that if you have a self-adjoint compact operator on a Hilbert space, then its eigenfunctions form a basis. I don’t know what operator you’d use, because I’m not familiar with the standard techniques of this field of functional analysis, but I know you can’t uniformly approximate every function (exactly because of this f(0)=f(pi)=0 problem), so you have to do an argument with either this function specifically (because its value at the endpoints is 0) or with L2 in general.
I hope you’ll explain this issue, either in text or in a video someday.
The function x(π-x) is even about π/2 on the interval (0,π) whereas cos (x) is odd on that interval so when you calculate the Fourier coefficients for the cos terms you get zero.
@@georgedoran9299 Sure, for cos(x), but what about cos(2x)?
Got to search half range of fourier series. U can extend the interval from ( 0, pi ) to (-pi, pi) as odd or even function. For example, f(x)=x(pi+x) for (-pi, 0) and f(x)=x(pi-x) for (0, pi) with period=2pi.
@@khoozu7802 Yeah, that last bit was what threw me. As you say, it's the Fourier series for that piecewise function, which equals our function on the appropriate interval.
@@khoozu7802 This is an awesome and correct answer. To flesh it out, we look at the odd extension to [-pi,pi], whence all cosines are eliminated, and all sines just double their coefficients. That’s why it’s okay to ignore cosines - because you’re secretly integrating against a function where the cosines are killed, even though integrating against the current function they don’t.
It seems that the editting has gone awry with some portion of (d/dx)! coming in front of the (31)^(1/3) as an approximation to pi. There may be a constructivist argument for QA somewhere in here.
The 5th root of 306 is even closer.
13:30 Good Place To Stop
5:16 Michael reveals his true origins
Why is it 2n+1 instead of 2n-1?
*edit:* looks like he caught it later, he had a note on the screen that it should start at 0
At 8:47? I was wondering that too, I think it's just a transcription error
Indexing over all odd numbers, it should indeed either be 2n-1, or the index should start at 0 instead of 1. Note the change to zero indexing during the camera cut at 11:12, where the error was corrected but the edit removed mention of the correction.
@@ZedaZ80 yeah, I had to go so I didn't get to finish the video. My bad
@@TJStellmach I had to go, so I didn't get to finish the video. My bad
i'd love to see a video on this D I column integration method. i assume most viewers of the channel are familiar with integration by parts, but i personally have never been taught this generalized strategy, i would usually just do it by parts at each step, which is tideous. the method is pretty self-explanatory, but i think it would still be nice to see a general proof with some examples that showcase the integral structure. might be better suited for the second channel, though
You can check out bprp video about DI method for integration, it's really just integration by parts in disguise so if you're familiar with that you'll have no problem in using the DI method. It's also useful when you have to do successive IBPs because you are more likely to avoid mistakes
Twelfth root of 924,269.1815 is even better 😃
Here is another "easy" approximation: pi = 294204 ^ (1/11). Gives 8 decimal precision.
So a circle of diameter pi has an area of 31 units^2, to a good approximation. Why wasn't I told this before?
Correctly, a circle with radius=π has area=31
@@panPetr0ff Ha! You're right of course, but I was actually thinking of the area of a sphere
Uhm, but how matrix factorial is connected to pi^3? There should be an easter egg or something?
It looks like this video needs to be redone and reuploaded, it's somehow two different videos mashed together.
306^1/5 is closer. 29809^1/9 is even closer. 294204^1/11 is accurate to 7 decimal places.
Taken further, (floor(PI^n))^1/n yields better and better approximations as n increases. Not sure this is any better than just finding better and better a/b approximations of PI.
This might be an interesting question: does floor(pi x pi x pi … infinity) = pi x pi x pi … infinity? If so then pi x pi x pi … infinity is an integer.
Here is another amazing approximation I discovered:
tan (x) ≈ 3x/[3 - (x^2)].
Michael, you might want to try again
Damn. 31^(1/3)+1/4717 is very close.
Indeed, it is only off by approximately 2.046 * 10^-9.
Here, the function seemed to be of the discontinuous sort.
3.14138. 5 digits in, and...
sqrt(2) + sqrt(3) is also pretty cool, though not very accurate
nice bernie shirt!
0 is a very nice approximation too!2
Yeah this is a little off. The start of it seems to be the wrong video
Thank you
What about pi! ?
✓10 is only 0.66% too high
Not sure what the beginning was about?
I think the video was put out with a section of another video inserted in between. In an effort to correct it with ended up with part of the wrong section and part of the end.
I’m sure it will get corrected.
π = 355/113
or
π = 180×(sin p / p) ; limit p ➡ 0
Nice Try & Try Again, Mr Penn 😉 🖋
How are there only 12 comments on this video!?!
10:47 amogus
De javú at 1047
pi~e^(e^(e^-2)))
Chaotic much? #FeedTheAlgorithm
pi ≈ 0
Not much better than (420/1337)*10
@@thychairman5782 nope, I just think it's a funny approximation of pi given the numbers involved
Yeah sorry about that 😅
Pi*e-e
first minute is from other video
What’s with the weird editing in the first minute?
Editing is very off here
Being the nitpicking nerds that we are, I can't resist to mention, in a totally constructively meant way, that you're overcorrecting your pronunciation of the name "Fourier". The stress is on the first syllable. In French, German, English, ... FOOr-ee-ay. What you're saying is more like "Fouriée". 😉
Thank you for mentioning this. You've saved me from unmasking myself as a fellow nitpicker.
French has a lot of regional variation in pronunciation. No-one would look at you funny if you were a Quebecois trying to talk like a Parisian and said that. Lol
Another approximation of Pi
Pi~= (2856/565) - 0.8*(√1.2+√1.68)
Correct to 8 digits
3.141592653
Not better than 22/7...
15 03
Great math. Great shirt.
You're butchering the pronunciation of Fourier.
You ok man? You seem a bit agitated. Beautiful math as always though.
Insert funny comment here
Do something with your voice. Sorry
355/113 = 3.14159292
I know some very nice approximations that I found a long time ago
(sqrt(5)-1)/2*exp(exp(2)) = 1000.088
If x is the root of the equation cos(x) =x,
then x ≈ 2 - qbrt(2)
If x! = gamma_function(x+1),
then 9*(pi!)! = 66425.000018
28658146^(1÷15) is just 0,0000000002 bigger than π
pi ~ ln(262537412640768744)/sqrt(163) ?
Did he try 22/7?
22/7 is approximately 3.14285714285714 but pi is approximately 3.141592653589793
Did he try22/7?