an amazing and mysterious approximation for pi!
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- Опубліковано 6 лют 2025
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(sinx)*(cosx) is just (sin2x)/2, the Taylor Series is easy to find 5:30
3:25 Engineers
I'm an engineer, but I'm more of a 3.14159 kind of engineer.
@@major__kong I went to engineering school between slide rules and TI-30's. Memorized 8 digits for my four banger: 3.1415927. (Yes, the last digit is a round-up.) Still remember the friendly warm glow of the red LED seven segment display! And don't forget to bring a spare 9 volt battery to the test, just in case!
e=pi=3
@@Maths_3.1415 pi-e = 3/7
A physicist is a mathematician who knows how to approximate. An engineer is a physicist who knows when to approximate.
Just a small detail, In the Newton final fórmula should be 3sqrt(2) in the denominator.
Great video, as always
20:49
This is kinda cool but seems pretty useless when you can get simple rational approximations that are closer. For example, 312689/99532 already gives you a better approximation than the one at the end (and if you were calculating it you still have the issue of needing to know sqrt(3) accurately, which I guess is much easier than pi but still).
You can get even better approximations by searching "digits of Pi" on Google. I think the actual point of the video is to teach maths.
Those rational approximations come from convergents of pi's continued fraction representation, I think it's cool to see approximations coming from entirely different representations even though not as accurate
Yeah, and 355/113 is only better than 1 part in 1 million. Both approximations are good enough for most terrestrial applications.
Its cool therefore it's good.
It's probably _easiest_ simply to learn a decimal approximation of pi to the required number of digits. I mean I can recite 3.14159265358979323846 from a well-known mnemonic.
But the easiest way is often the least fun.
x= pi/4 for n=1 has relative error of about 2.27x10^-3, and x= pi/6 (still 1) gives 12*(2sqrt(2)-1)/7, which has relative error of about 4.31x10^-4.
The thumbnail (n=3, x= pi/6) has relative error of 4.13x10^-11 + O(10^-13).
The guy's not named Snellins, but Snellius. It is the latinised version of the name Snel (meaning quick).
yesterday I saw a cheese going for 3.14. I bought it.
Did you make a cheese pi with it?
it would be interesting to see this for padé approximations of the basis functions. but that would be even more hard to follow on a black board. 🙂
nice but I do not like the sqrt(3) in the final answer. it is an irrational. would be nicer if it were just rationals in the approximation. You still have to calculate sqrt(3)...
rational approximations to sqrt(3) can be generated very fast with pell equations
the primitive solution to x^2 - 3y^2 = 1 is (2, 1)
then for the point (a, b), the next point is (2a + 3b, a + 2b)
(2, 1) -> (7, 4) -> (26, 15) -> (97, 56) -> (362, 209)
sqrt(3) = 1.7320508...
362 / 209 = 1.7320574...
Yesterday I was reading Dan Pedoe’s Geometry in which he proved the equivalence of Snell’s Law and Fermat’s Principle of Least Time using Ptolemy’s Theorem. I thought "This is something Michael might turn into a video." Today I turned on UA-cam and you mentioned Snell’s Law. Though relatively simple Pedoe’s proof is still interesting, at least to me. As is Snell's Window.
Calculus of variations would be a nice topic generally for the channel I think. It's clearly very similar to normal calculus, but used in a very different way.
Do you have any content on hypergeometric series?
The UA-cam thumbnail is (currently) wrong saying "80405" instead of "80504"
Thank you I did the calculation from the thumbnail and I was like all that for 3.13? Not even 2 decimal places?
I did the same.
22/7 is a better approximation.
So,does this system of equations has a solution always?
I know that we can use the rank theorem and say that rank of the matrix with and without the (1000...) cell are the same which implies that the solution exists,but it is not much obvious i wish...
Wow this is interesting❤. How about (355/113)-Tan(355/113), where do you think it comes from?
That is really close. Error = 6.32792512 ∙ 10⁻²¹
But calculating that symbolically is almost impossible.
Similar result for 22/7, with an error of 6.73945∙ 10⁻¹⁰
That comes from the initial number being (pi + eps). Tan(pi + eps) will be approx. (eps + a*(eps^3)), so subtracting it will cancel the linear error, and leave you with only a cubic error term.
I was wondering if you are currently thinking about that Hank Hill sum problem or whatever it was called. It looked to me like there was an underlying theme to the recent videos on the channel and was curious if you were occupied with that intriguing sum. It looks preposterous, that its convergence is unknown, but the more I thought about it the more I saw, and understood that it was reasonable that it is a difficult problem. It still seems weird that someone hasn't shown that the reciprocal cube term totally dominates the reciprocal squares of sines at a given magnitude of n, even though that latter term does become arbitrarily large. But, I guess that's the interest of the problem, we need a proof of that if it's true.
I can't seem to find anything by typing "Hank Hill sum problem" on Google, what does the problem want us to solve?
I'd be impressed if the number of digits that matched Pi was greater than the number of digits required to express the formula (19).
I'm wondering if there's any method to pop out 355/113 ~ pi, or if someone just noticed it was only off by less than one part in a million?
One way for that one is to use the continued fraction for pi. (Which of course requires you to have a good value for pi to begin with.) The fraction starts out pi = 3+1/(7+1/(15+1/(1+1/(192+...
Since 192 is large, 1/(192+.. is small, so if you approximate 1/(192+... as 0 you get pi=3+1/(7+1/16) which works out to 355/113
9:17 that's not right. That fraction gives me an answer of 3.940603027...
The pi approximation of Ramanujan would be interesting. But it is probably out of reach(?)
@sarcastic_math343 maybe some glimpse towards that topic would be nice to see.
@@thomashoffmann8857 Mathologer had a more accessible approximation
ua-cam.com/video/ypxKzWi-Bwg/v-deo.htmlsi=sYr7hmusH9xhd2xU
I believe it's "Snellius", not "Snellins" 😅
Calculus was still in its infancy at the time of Newton. Would he have had access to Taylor expansions like that? Or this presentation in modern terms, when Newton would have thought of it very differently?
I'm not really in a position to say something about this with certainty, but I have followed a course on the history of mathematics. It seemed to me that people like Newton were already aware of polynomial approximations of sin(x) and the likes, but I'm not sure if they would have had a proof of their correctness as we have today. In either case, to me it seems very likely that the method described in the video uses Newton's ideas in some capacity.
The Madhava series for sine and cosine have been known since the mid 1300's, a good 300 years before Newton was born.
@@tracyh5751 Known to someone, but not necessarily known to Newton? I have no idea what the state of maths communication between Western Europe and India was like during the Renaissance.
There must be a way to invert the matrix of coefficients in general, right?!
Makes sense that the error at the end (10^-10) would be as good as twice the number of digits in the approximation (5 digits)
Sin(x)*cos(x) = 0.5*sin(2x).. It seems not to match the product of the two aporoximation... Why?
22/7 quite good
Isn’t pi just the relationship between coordinate systems? Doesn’t the fact that it’s not resolvable prove that the systems are not extensions of each other? So exploring approaches to pi gives insights into the shortcomings of mathematical processes. Or am I missing something?
No, no, no, yes, respectively 🙂
@@WackyAmoebatrons Okay, I deserved that. But my point concerning the nature of π is that it emerges using an orthogonal coordinate system to sum the space bounded by a circle, called ‘Area’. Since π is irrational, it follows that the area of a circle is not ‘on the same line’ as the area of a triangle, or any polygon . It is fundamentally different because the orthogonal coordinate system cannot resolve the boundary of a circle properly. So long as ‘area’ remains tied to axes this will be true. You can’t get round it! Newtonian methods of summing increments of change (ie integration and differentiation) stem from trying to solve this issue.
Taylor's expansion for the sinx should be x^(2n+1) not just x^n
Please for the love of god stop messaging my niece on furaffinity. This will be the last time I ask
But don't forget that for most real life applications 22/7 is an *acceptable* approximation.
22/7 is a horrible approximation, especially when something as simple as 355/113 (note how the odd numbers line up) is much better
Is this how those projects for calculating 1 million digits of pi work?