Two Infinite Series Sums from Regular Polygons (visual proof)
HTML-код
- Опубликовано: 5 окт 2024
- This is a short, animated visual proof computing the sums of two series - one of reciprocals of triangular numbers (i.e., certain binomial coefficients) and the other a classic series that is used to demonstrate telescoping series.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoff.... Thanks!
For some related videos see these videos:
• Infinitely Many Infini...
• Infinite series: summi...
This animation is based on a visual proof by Paul Stephenson from the December 2022 issue of Mathematics Magazine (doi.org/10.108... page 572 ).
#mathshorts #mathvideo #math #infiniteseries #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #triangularnumbers #animation #theorem #proof #iteachmath #series #infinitesums #infiniteseries #pascaltriangle #binomialcoefficients #binomials #binomialtheorem #telescoping
To learn more about animating with manim, check out:
manim.community
I have no words to describe my wonder at this.
👍
I mean, if you wanna get really technical, the limiting shape when you increase sides of a regular polygon isn’t a circle, but an Apeirogon. Though if you could see an Apeirogon in its entirety it would be indistinguishable from a circle, so in the end it doesn’t change the final result
I don't think these 2 are the same. But maybe that's because I don't know what an Apeirogon is.
If you could see infinity, there would be a slight difference
OK, I checked the definition of an Apeirogon and I must say, accprding to its definition, it is not at all related to circles
And @daniel_77 what does "if you could see infinity" mean? And how does it imply that an Apeirogon and a circle are different. They are not similar at all to start with.
A (degenerate) apeirogon is a circle in the context of projective geometry, but in affine geometry, where these types of mathematics are being done, they are completely different shapes.
@@angelmendez-rivera351 WOW! That's totally unexpected! Care to tell us what the definition in both contexts and explain the difference please?
Well, in a sense for the visual proof, we recognized the telescoping series as the one we originally wanted to find
For sure :) still kinda telescoping.
Beautiful! :-D
Thank you! 😊
@@MathVisualProofs what's the proof a circle has 0 sides
Beautiful
Thanks!
That was impressive
😀
Beautiful!!!!
Thank you!!
Is the limiting circle really that much larger? I would not have expected...
In this case, the circle should actually be infinitely large. It would just be a horizontal line
@@duane6386 ah, right. I had in mind an inscribing circle, which this is not.
@@wieneryron
Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*.
(Read and write it in a paper at the same time)
Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon.
Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments.
Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ.
In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides:
𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏.
Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁,
𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2.
It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base.
The area of 𝚫𝐀𝐁₁𝐃₁ is a half of the area of the triangle 𝚫𝐀𝐁₁𝐂₁.
The are of the polygon is equal to 𝒏 multiplied by the are of the triangle 𝚫𝐀𝐁₁𝐂₁.
Therefore, the area of the polygon is equal to
𝒏•2• (1/2)•(𝐚/2)•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] =
𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4.
Determining the limit of the are of the polygon when 𝒏 goes to infinity:
𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] =
𝒏 → ∞
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] =
𝒏 → ∞
𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive.
When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values.
When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore:
𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
𝐚²/4 is always positive. Therefore:
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
Conclusion: the "circle" in as big as you want it to be.
@@wieneryron
Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*.
(Read and write it in a paper at the same time)
Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon.
Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments.
Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ.
In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides:
𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏.
Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁,
𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2.
It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base.
The area of the polygon is equal to 𝒏 multiplied by the area of the triangle 𝚫𝐀𝐁₁𝐂₁.
Therefore, the area of the polygon is equal to
𝒏•(1/2)•𝐚•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] =
𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4.
Determining the limit of the are of the polygon when 𝒏 goes to infinity:
𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] =
𝒏 → ∞
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] =
𝒏 → ∞
𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive.
When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values.
When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore:
𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
𝐚²/4 is always positive. Therefore:
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
Conclusion: the "circle" is as big as you want it to be.
@@wieneryron
Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*.
(Read and write it in a paper at the same time)
Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon.
Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments.
Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ.
In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides:
𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏.
Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁,
𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2.
It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base.
The are of the polygon is equal to 𝒏 multiplied by the are of the triangle 𝚫𝐀𝐁₁𝐂₁.
Therefore, the area of the polygon is equal to
𝒏•(1/2)•𝐚•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] =
𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4.
Determining the limit of the area of the polygon when 𝒏 goes to infinity:
𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] =
𝒏 → ∞
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] =
𝒏 → ∞
𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive.
When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values.
When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore:
𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
𝐚²/4 is always positive. Therefore:
(𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞
𝒏 → ∞
Conclusion: the "circle" is as big as you want it to be.
Fantastico!
😀
Well... the picture was neat
you the MVP fr
😎
What is the name of the program you use to paint?
Cool!
👍
I see y=sqrt(-x)
Hello, can I buy the source code in one of your videos, because I want to learn it, can you, for example, the source code of the intersecting string theorem, I like your videos very much
Could an emergence process of spherical symmetry, represented by 4πr², forming and breaking, shape the characteristics of three-dimensional space by creating statistical entropy with the potential for ever-greater symmetry, similar to cellular life? Photons with energy ∆E=hf continuously transform potential energy into the kinetic energy of matter, Eₖ=½mv², in the form of electrons. Could this process establish a design pattern or template based on spherical geometry that allows for self-organization and the emergence of complexity?
🎖🎖🎖🎖🎖
😀
I'm learning algebra 1 myself by watching videos i do problems
I understand many some not I'd have any maths book shelf guide anything like that i just downloaded a pdf of algebra self study but i can't relate to that I have studied in videos (why is that ) I'm 19
cabbage
Not 100% if this is rigorous as a proof - seeing as there is no limiting shape (as you cannot have an infinitely large circle).
To turn it into a proof, I imagine all you have to do is prove that the limit (of the numerical series) exists
@@SteveThePster The limiting shape is an apeirogon as someone pointed out in another comment.
They go on about this TOSH, and can't even tell what 1/76 scale weight of 126t is.