an unsurprising appearance of pi
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- เผยแพร่เมื่อ 20 ธ.ค. 2024
- A classic limit for pi.
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I LOVE to see a video about this. Some years ago I deduced this formula for pi (which logically converges very quicly, as the number of sides grows exponentially). I wanted to find a limit expression for pi that didn't use sin or cos. I started with a square inscribed in a circumference (side = √2) and built a recursive sequence that tells you the side of the next 2ⁿ-gon, only by using the Pythagorean theorem.
After not very complicated math I came up with the expression, and i was so happy, but i could not prove it "backwards", and finally i see this "backwards" proof ,starting from the expression and getting pi as the answer! Thank you so much!
7:46 those triangles in the overlay… they should all meet in the middle…
I think the nested radical construction can be built through repeated applications of Pythagoras to the geometrical construction that Michael shows, then 'scaled up' with a 2^n factor to give the perimeter of a 2^n-gon. Nice video, Michael (and Stephanie) - thank you!
I remember this from my student time. It was something like Archimedes way to estimate the circumference of a circle by inscribing regular polygons. I tried to compute pi with machines with 12 digits and found that very soon the computation becomes unstable due to the fact that the square root of the sum becomes too close to 2 and the difference becomes less than the computer (or calculator) precision. It is interesting is this an efficient way to compute pi by using long arithmetic? (I guess no, but it is efficient as compaed to, say, the way based on Machin or Machin-like formula.)
Van Ceulen did it for 35 digits and used Archimedes' method
Oh, I recognized that limit right away! I've worked out pi with polygons before, to understand how the ancients did it!
This was really slick)) The relationship between the limit and diameter of circle was really cool))
related to this, do nested radicals with alternating sign have interesting properties?
ex: √2 + (√2 - (√2 + (√2 - (√...))))
I just tried calculating 2sin(pi/2^k) to see if it resulted in anything like that because the half-angle formula of sin(x/2) gives you a minus sign where cos(x/2) would have a plus sign, but because sin(x/2) also just reflects back to cos(x), you only get one single minus sign and only plus signs after that - and then I noticed that this is exactly the expression that this video started with. So going through this route only gives you a single minus sign at the start, and not infinitely alternating signs.
Using desmos I noticed the expression √(2 + √(2 - √(2 + √(2 - √(...)))) appears to be converging to the golden ratio ϕ which is about 1.618033... This is surprising to me because the golden ratio has another well known nested square root representation which is only √(1 + √(1 + √(1 +...))). Also negating all plus and minus signs in the expression gives the reciprocal of ϕ.
Edit:
The nesting can be simplified by its recursive nature. Asume x = √(2 + √(2 - √(2 + √(2 - √(...)))) We can see that it contains itself: x = √(2 + √(2- x)).
By simplifying we get the following equation: x^4 - 4x^2 + x + 2 = 0.
Considering the following 2 properties of ϕ (derived from the defining equation of ϕ namely 0 = ϕ^2 - ϕ - 1):
1. ϕ^2 = ϕ + 1
2. ϕ - 1 = 1/ϕ
We can see that ϕ = √(2 + √(2 - ϕ)) satisfies the equation.
By squaring both sides:
ϕ^2 = 2 + √(2 - ϕ)
By property 1:
ϕ + 1 = 2 + √(2 - ϕ)
ϕ - 1 = √(2 - ϕ)
By the 2nd property:
1/ϕ = √(2 - ϕ)
1/ϕ^2 = 2 - ϕ
Again by the first property:
1/(ϕ + 1) = 2 - ϕ
1 = 2(ϕ + 1) - ϕ2(ϕ + 1)
0 = -ϕ^2 + ϕ + 1
Arriving at the defining equation of ϕ form which follows that it must be a solution to the original equation.
q.e.d.
All of the solutions to the original equation are x1 = -2, x2 = -1/ϕ, x3 = 1 and x4 = ϕ.
Interestingly the other solutions to the equation don't seem to show up by cutting off the iterative evaluation of the nested square root at some point.
Well i can tell it's a root of this polynomial:
let x=sqrt(2+sqrt(2-sqrt(....))...)
Then x=sqrt(2+sqrt(2-x))
So (x^2-2)^2=2-x
Or x^4-4x^2+x-2=0
@Josua no its not equal to it since if it was then it has to be a root of x^4-4x^2+x-2=0(see my reply) which i used wolfram alpha and none of it's roots were golden ratio...
@@aweebthatlovesmath4220 huh... maybe there is a way to invert this
given a 4th order polynomial, we can write down the root in terms of alternating nested radicals
I came up with this when I was in high school! I used it to try and approximate pi.
Assuming the geometric property, would it suffice as a proof for the identity?
Conversely, assuming the identity, would it suffice as a derivation of the circumference of a circle? I’m trying to find holes due to circular reasoning (like how we’re not typically allowed to use L’Hôpital’s Rule on sinx/x), since the values of sin and cosine depend on a circle right?
Assuming the geometric property, I feel like it does suffice as proof, because think about it, if geometrically the expression represents a number, algebraically it also has to, logical.
Now, as a derivation of the circumference of a circle? I wouldn't say so, π is defined as the circumference divided by the diameter of a circle, and it's used in the identities of sine and cosine, though I am not qualified to answer that confidently.
Where did the lemma come from? I've never seen this before.
RJ
Concerning nested radicals, do we need to prove they converge?
Would the last part of the video (using geometry) be enough as a proof for this identity?
Thanks Mike❕ That video was so cool.
How do we know the radius of the circle is one half ?
By construction
Does it come from Archimedes' method for pi starting from square
(Two variants of Archimedes' method - one with perimeter of regular polygons (inscribed and circumsribed), second with area,
but original version was with perimeter)
In each step we double number of sides and bisect the central angle
We will get pi as limit of created sequence
This method of calculating pi looks... exhausting.🥁
It’s actually really fun using a calculator, even more so when if doesn’t have pi as its own key.
th-cam.com/video/TdPUnftdpHA/w-d-xo.html
The infinite sqrt(2 - sqrt(2 + sqrt(2 - sqrt(2 + ... )))) tends to (sqrt(5) - 1)/2. 2pi/(sqrt(5) - 1) (log 2) is appx. 2.345728. That's what I thought we were gonna do.
I know that the term in the square root is supposed to converge but after the first 2 you only wrote only one '-'. Shouldn't each nested square root alternate between "+" and "-" ?
No.
@@tomkerruish2982 Yes, I see that now. I thought about it, and decided to test my assumption. And wouldn't you know it does converge!!
@richardcuddy6166 Apologies for the extremely terse reply. I'm often not in a position to be able to type even this much.
Good teaching
looks like something ramanujan did
:)
Archimedes used a similar method, but the approach is much older than that.
totally man
It was Vieta who introduced this series.
6:25 You can't use L'Hopital rule to solve that limit 😭 It's circular logic
i tried for some reason i got ln2 * pi
Sure you can, normally we define sin by power series. Unit cercle definition is useless in undergraduate+ math.
@@SimsHacks Nope, you can't, because you need derivatives (of sine) to define power series in the first place, and that limit is used to calculate the derivate of sine. Circular logic.
And the "unit circle definition" is THE definition, definetely not useless
@@davidcroft95Well, math isn't always constructive.
You can say
Let f(x)=x - x³/3! + x⁵/5! - …
This series converges absolutely for all real x, we define this to be the sine function.
Then you prove using series theory, that it has all the properties as the "unit cercle old definition" had, and furthemore you prove that there exists only one unique function that verifies the properties. Therefore, new definition is fine and you can use it for example to show that the derivative is cos(x) instantly.
Pick a real analysis book and see :)
The unit cercle definition becomes useless and bad beyond a basic calculus class. Defining a function by some length on a circle, not really useful for further work with the function.
Alternatively, you can define sin(x) as the imaginary part of exp(ix), where exp is defined by power series.
@@SimsHacks I have read a bunch of real analysis books (and even took a course in complex analysis, if you're interested) and in neither of those sine is defined in that way and in all of those using L'Hopital rules in that limit is circular logic.
Suck it up.
Great video, but that l'hopital at 6:38 gave me a heart attack lmao
I'd do u=πx and get a sinu/u there
this is fucking awesome
Oh, oh, oh that gnarly, pesky π shows its head again.
Thank you, professor.
Fuck man , i thought i discovered this lemma on my own by experimentation LmAo.
Thanks for ruining my day.
Woah, woah, woah
I would argue using:
L'Hopitals rule on the following limit:
as x approaches zero from above of the function sin(pi*x)/x
IS CIRCULAR REASONING
Your presupposing the derivative of the function sin(pi*x) when the limit is essentially the limit definition of finding the derivative of the same function @ x=0
I think that's not a reasonable objection in this case. We're not proving things from first principles here; we're operating at a level of abstraction where we can assume that both the derivative of sin and L'Hôpital's rule have been proven, so of course we can use them together. The fact that the thing we're proving with those tools closely resembles something that was already proven at some lower level of abstraction when finding the derivative of sin is irrelevant.
Here’s an explanation with an area interpretation.
th-cam.com/video/KqA54IP21dE/w-d-xo.html