Two Amazing Machin-like Formulas for Pi/4 (Pi Approximation Day Visual Proofs)
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- čas přidán 20. 07. 2023
- This video describes a relatively new visual proof of two different Machin-like formulas for Pi (or for Pi/4 more accurately), which is a formula that allows for relatively speedy computation of the digits of Pi. These two formulas are called Hutton's formula and Strassnitzky's formula although at least Hutton's was likely known to Machin.
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This animation is based on a proof due to Roger Nelsen from the December 2013 issue of Mathematics Magazine (www.jstor.org/stable/10.4169/... page 350).
Here are three more Pi day gems:
• Machin Formula Visuali...
• Approximating Pi with ...
• e^ Pi vs Pi^e: which i...
To learn more about animating with manim, check out:
manim.community
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nice! the animation makes it simple enough that a child could understand it…and then hopefully start thinking mathematically with various representations and follow their curiosity
Thanks! I always appreciate your comments :)
That's the Fibonacci one. You can continue it.
pi/4=arctan(1/2)+arctan(1/5)+arctan(1/8)
Notice we are missing 1/3 and 1/5 and 1/8 are consecutive. We kill 1/8 and replace it with 1/13 and 1/21
pi=arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/21)
pi=arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/34)+arctan(1/55)
I think there is a way to get the even terms but I don't remember. Maybe alternate sign? This can be applied to all quadratic type recursive numbers, or at the very least the metalic means ratios. It probably has something to do with some sort of derivative or integral. I don't know enough about those yet.
This is a beautiful observation, though of course you meant π/4 (and not π).
The formula you used (proved below as Result 1) is:
arctan(1/F₂ₙ)=arctan(1/F₂ₙ₊₁)+arctan(1/F₂ₙ₊₂), for n≥1,
where Fₙ is the nth term of the Fibonacci sequence (defined by the recursive relation Fₙ=Fₙ₋₁+Fₙ₋₂, where F₁=1 and F₂=1, or, equivalently, F₀=0 and F₁=1, terms 1 to 10 being
1,1,2, 3,5,8, 13,21,34, 55).
So
π/4=arctan(1/1)=arctan(1/F₂)
=arctan(1/F₃)+arctan(1/F₄)
=arctan(1/2)+arctan(1/3)
=arctan(1/2)+(arctan(1/F₅)+arctan(1/F₆))
=arctan(1/2)+arctan(1/5)+arctan(1/8)
(the 2nd formula in the video)
=arctan(1/2)+arctan(1/5)+(arctan(1/13)+arctan(1/21))
etc as you explained, where you express π/4 as the sum of the arctans of the reciprocals of even terms of the Fibonacci sequence from the 2nd up to any point then add the arctan of the reciprocal of the next (odd numbered) term.
An immediate corollary is that we can express π/4 as the sum of an infinite series:
π/4=∑(n=1 to ∞) arctan(1/F₂ₙ)
=arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/34)+...
Result 1
arctan(1/F₂ₙ)=arctan(1/F₂ₙ₊₁)+arctan(1/F₂ₙ₊₂), for n≥1,
where Fₙ is the nth term of the Fibonacci sequence.
Proof
For n≥1, let α=arctan(1/F₂ₙ₊₁), β=arctan(1/F₂ₙ₊₂),
so tan α=1/F₂ₙ₊₁, tan β=1/F₂ₙ₊₂
So tan(α+β)=(tan α+tan β)/(1-tan α tan β)
=(1/F₂ₙ₊₁+1/F₂ₙ₊₂)(1-1/F₂ₙ₊₁×1/F₂ₙ₊₂)
=(F₂ₙ₊₂+F₂ₙ₊₁)/(F₂ₙ₊₁F₂ₙ₊₂-1)
=F₂ₙ₊₃/(F₂ₙ₊₁F₂ₙ₊₂-1)
(using the Fibonacci recursive relation)
Now, using the formula Fₙ₊₁Fₙ₊₂-FₙFₙ₊₃=(-1)ⁿ
(proved below as Result 2)
and replacing n by 2n, we get
F₂ₙ₊₁F₂ₙ₊₂-F₂ₙF₂ₙ₊₃=(-1)²ⁿ=1,
so
F₂ₙ₊₁F₂ₙ₊₂-1=F₂ₙF₂ₙ₊₃
Note that if we were to start with arctan(1/F₂ₙ₊₁) (instead of arctan(1/F₂ₙ)) this step wouldn't work as the (-1)ⁿ term would change sign, leading to a +1 in the previous line.
So
F₂ₙ₊₃/(F₂ₙ₊₁F₂ₙ₊₂-1)
=F₂ₙ₊₃/(F₂ₙF₂ₙ₊₃)
=1/F₂ₙ
As 0
@@MichaelRothwell1 oh yeah, I forgot about tangent addition ((v+w)/(1-v*w)), this works because of that. Which is weird cause I'm playing with that a lot right now. I'm looking at tension and E^2. So... I think I might know how to continue this. The experssion sec(arctan(i*v/c))+i*v/c is the quadratic formula.
sqrt(-v^2/c^2+1)+iv/c
sqrt(c^2-v^2)/c+iv/c
I don't know how to express this the way you do. But c=2a, v=b, and a factor of c in the sqrt is the other c.
(sqrt(4*a*c-b^2)+i*b)/(2a)
(sqrt(-1(-4*a*c+b^2))+i*b)/(2a)
(isqrt(-4*a*c+b^2)+i*b)/(2a)
So, I know this works on pell numbers, but maybe it's because of that a being equal to c. The pell numbers are probably hidden because the sequence converges.
Sorry about the bad math. I only have grade 12.
wow, very cool
@@imperfectclark Michael Penn addressed it. It's about the Cassini Identity. It can be generalized.
this channel is answering the real needs !!!! +1 supporter!
Thanks! Happy for your support
Very intuitive video!
Could you do a video to prove the one that goes
pi/4=4arctan(1/5)-arctan(1/239)?
I’ve seen it in matt parker’s video and he mentioned that it was used in the 1800s to calculate pi to over 500 digits. But i was wondering where the formula came from.
Yes! Here it is from this years pi day : czcams.com/video/enQdwZwoyyo/video.html 😀 hope this helps.
@ Ninja: it is possible to prove the formula using complex numbers.I m trying to figure out exactly how to write down a proof.
This is cool....and smart !
WoW
I'll bet there are several other similar ratios of sides that would fit this pi/4. And then there must be similar number of squares to prove other fractions with pi as the numerator.
Here are some more : czcams.com/users/shortsXJ-GaUrY-LE?feature=share 😀
@@MathVisualProofs Very cool!
Can we get an animation about the following series:
pi ≈ (4/1) - (4/3) + (4/5) - (4/7) + (4/9) ...
It's elegant, though I imagine there are pi approximation algorithms that converge more rapidly.
I know of a source for it, but I have had a tough time figuring out the best way to animate the decomposition because the pieces get rather small fairly quickly. I'll brainstorm a bit more to see if I can make it work :)
that's the machinlike formula pi/4 = arctan 1
To a layman like me those "arctan" might as well be "arcane" - it's a whole blackbox of magic
Can these be further decomposed to explain arctan as something more primitive?
In a right triangle, the tangent of one of its acute angle is the ratio of the length of the leg opposite that angle to the length of the leg adjacent to that angle. If we know those two lengths, then the arctangent of that ratio gives us the angle.
Bro what is your qualification in math I want to do the same
I am a math professor. I have a phd and have taught/done research at a small college for the past 15 years. This channel arose out of my love for proofs without words and my desire to be able to deliver better digital content to my students (as required by the pandemic). This channel is a way for me to keep practicing and improving my animation and coding abilities.
Why not draw a single right angle triangle with angles π/3,π/6,π/2 and then you get tan(π/3)=sqrt(3)
Therefore π/3 =atan(sqrt(3)) ,finally
π=3•atan(sqrt(3)).
This is much simpler and you need only one triangle.
Yes. But root 3 not as easy to work with when computing approximations of arctan using Taylor series.
It is even possible to use a right isosceles triangle and then you get that
π=4•arctan(1) which is even simpler and doesn't contain any irrational numbers at all.
Here is curious fact:
π/4=∑(n=1 to ∞) arctan(1/F₂ₙ)
=arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/34)+...
where Fₙ is the nth term of the Fibonacci sequence.
For a proof, see my (first) reply to the comment by @thomasolson7447, which generalises the 2nd formula,
π/4=arctan(1/2)+arctan(1/5)+arctan(1/8),
which is the sum of the arctans of the reciprocals of certain Fibonacci numbers.
the main question is: "So what?" How can we use this?