here's a fascinating way to approximate

the serinsky triangle with a single

curve in the

plane let's consider a relatively simple

Linden Meer system we start with the

string a and the production rules tell

us that at each stage we replace a by B-

a-b and we replace B With A + B + a and

we will let dashes and pluses stay the

same at each stage let's see the first

few stages of this process start

starting with a the first stage results

in B-

a-b then the second stage results in the

string a plus B+ a dash B- A- B Das a +

B+ a we can continue this process over

and over as long as we like each time

applying the production

rules eventually we can stop at a

particular string and we can use the

string to walk on the plane

if we see a character of A or B we will

step one unit forward in the direction

we're facing if we see a plus sign we

will rotate 60° to the left and if we

see a dash we will rotate 60° to the

right let's see what shape emerges from

this Turtle Graphics process when we

investigate the string that results

after the 10th iteration of the

production rules described by this L

system here are the first few characters

of the process let's use this string to

walk on the plane to some dramatic

[Music]

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music

[Music]

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n

[Music]

if you liked this video check out the

links in the description for other

videos drawing other shapes using Turtle

graphics and L systems

Sierpinski Arrowhead Curve L-system (118098 steps!)

In this video, we describe two how to draw a single amazing curve - the Sierpinski Arrowhead curve - using the simple processes of Turtle graphics and L-systems. We draw 118098 line segments, which is double 3 to the 10th. It is amazing that such a simple process can draw these complex curves.

If you liked this video, consider liking or subscribing. Or you can support me by buying me a coffee at https://www.buymeacoffee.com/VisualProofs
Thanks!

If you want to see two other constructions of like this, see
https://youtu.be/8Q_zuRKKh2g

If you want to see other ways to construct this curve or other ways to build the Sierpinski triangle, check out this video:
https://youtu.be/IZHiBJGcrqI

If you want to know more about L-systems or turtle graphics, check out these wikipedia sites:

https://en.wikipedia.org/wiki/L-system
https://en.wikipedia.org/wiki/Turtle_graphics

To learn more about animating with manim, check out:
https://manim.community

#turtlegraphics #fractal #math #manim #mathvideo #Lsystem #sierpinski #sierpinskitriangle #triangle #arrowheadcurve

___________________________________________
Music in this video:
Majestic by Alex-Productions | https://onsound.eu/
Music promoted by https://www.free-stock-music.com
Creative Commons / Attribution 3.0 Unported License (CC BY 3.0)
https://creativecommons.org/licenses/by/3.0/deed.en_US


I bet you've seen an infinite chocolate

trick before but have you seen one that

allows you to share the chocolate with a

friend start with this chocolate bar and

split it into these pieces as we've done

here there are actually two squares of

chocolate left inside you can split

those apart eat one yourself and share

the other with your favorite friend but

now to make sure that nobody knows that

you've done this you can take the

remaining pieces of the chocolate and

reorganize the triangular pieces like

this on either side and then take both

of the L-shaped pieces and drop them

down into place this way once you've

done this the full original chocolate

bar has been restored and no chocolate

is missing this triangular diagram is

based on the curry triangle which is a

dissection fallacy that allows you to

quote unquote prove that 60 59 and 58

are all equal

The Infinite Chocolate Trick with a Friendly Twist #shorts

In this short, we show a dissection fallacy that allows you to show visually (though incorrectly) that 58=59=60. This is based on the Curry triangle.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

For more on the Curry triangle shown here, check out this longer video that exposes the dissection fallacy: https://youtu.be/iMgFDhpa000

This visual was created by L. Vosburgh Lyons based on work by Paul Curry. See the MathWorld page for more information:
https://mathworld.wolfram.com/CurryTriangle.html

@JamesTantonMath did a real life cake version of this! https://youtu.be/o0_ysMQ9ne0

#math #paradox #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #dissection #area #currytriangle #missingsquare #mathshorts #mathvideo #infinitechocolate #chocolate #sharing #fallacy

Here are some other resources with infinite chocolate games:
https://en.wikipedia.org/wiki/Missing_square_puzzle
https://youtube.com/shorts/wGYrdpKDrb4?feature=share (maxdesignpro)
https://youtube.com/shorts/6aDB9VLnyZQ?feature=share (mindyourdecisions)
https://youtube.com/shorts/rBA0ujypLgI?feature=share (mindyourdecisions)
https://youtube.com/shorts/j8bok-ce-iU?feature=share (kylestook)
https://youtube.com/shorts/SkipjSkqifE?feature=share (_vector_)

To learn more about animating with manim, check out:
https://manim.community


suppose we have a right triangle with

legs of length A and B and a hypotenuse

of C we can draw an A by a square off

the side of length a we can draw a b by

b square off the side of length B and we

can draw a c by C Square Off the

hypotenuse though we leave the C by C

Square unshaped

Pythagorean theorem 📐| Pythagorean theorem visual proof | #fypシ #fyp #shorts  #pythagoras_theorem #ytshorts #pythagoreantheorem #pythagorean #maths #mathstrick #mathematics


[Music]

what do these diagrams have to do with

the arithmetic mean geometric mean

inequality let's find out

[Music]

start with a right triangle with legs of

length A and B

use four copies of the triangle rotated

so that we create this shape which is a

square with a side length of a plus b

inside there's a missing Square reflect

each triangle about its hypotenuse

creating this diagram using eight

triangles notice there's still a missing

Square inside

[Music]

to verify that this missing inside shape

is a square notice that the side length

can be computed as the difference of the

leg lengths of the original triangle so

we see that the square has a side of

length a minus B

now we can find the area of this shape

in two different ways first we note that

it is an A plus b by a plus b rectangle

so the area is a plus b quantity squared

on the other hand we've used eight

triangles plus the inner Square so we

have an area of eight times a half a B

plus the quantity a minus B squared we

can simplify the eight triangles to have

the area four times a times B now we

note that the quantity a minus B squared

is always non-negative so if we leave it

off we obtain the inequality a plus b

squared is greater than or equal to 4

times a times B

as we vary A and B we see that the inner

Square gets larger and smaller depending

on the sizes of A and B but we also note

that the square is always there except

in the one instance when A and B are

exactly the same so that that square

shrinks to zero area

from this we conclude that we have

equality amongst the inequality if and

only if a minus B is zero so that a

equals B

upon taking square roots and dividing

both sides by 2 we obtain the inequality

a plus b over 2 is greater than or equal

to the square root of a times B this is

the arithmetic mean geometric mean

inequality

AM-GM Inequality VI (visual proof)

This is a short, animated visual proof demonstrating the arithmetic mean geometric mean inequality using 8 triangles arranged in a famous shape for visual proofs. #mathshorts #mathvideo #math #amgminequality #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #inequality #arithmeticmean #geometricmean

For other visual proofs of this fact see:

https://youtu.be/Pt8IX_Q5U1A
https://youtu.be/62G9fak1vyk
https://youtu.be/HFGSsPbYc-s
https://youtu.be/SEPkdPWKdO0
https://youtu.be/jNx05gXSblE
https://youtu.be/IJGwvvQdfgg

This animation is based on a visual proof from Ayoub B. Ayoub that appeared in the Spring 1997 issue of Mathematics and Computer Education (vol 31, no 2 p191).

To learn more about animating with manim, check out:
https://manim.community


[Music]

[Music]

keep

going to Le but

Trapezoid Area Formula: Visual Proof with Maths Animation #ssc #maths #ytshorts. In this video, we simplify the Trapezoid Area Formula using engaging visual proof and maths animation. Perfect for SSC exam aspirants and anyone looking to understand geometry concepts more clearly, this short tutorial breaks down the formula step by step. Whether you're struggling with this topic or just want to reinforce your knowledge, our clear and concise explanation will help you grasp the concept quickly. Make sure to watch till the end for a full understanding and share with your friends who might find this useful. Don't forget to like, subscribe, and hit the bell icon for more educational content!
alfaproton
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Hashtags

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Tags
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[Music]

recall that the binomial coefficient n

choose a represents the number of ways

of selecting a unordered objects from a

set of size n

consider the lattice of points in the

plane with non-negative integer

coordinates let's label the origin and

the point a b

how many northeast lattice paths are

there from 0 0 to a b where a northeast

path is one that only allows steps

directly right or up

let's enumerate them for a equals 2 and

b equals 3.

as we build all the paths we realize

that each path has the same number of

steps namely a plus b

moreover we see that each path has

exactly a steps to the right and b steps

up

so we only need to keep track of where

and when the right steps occur

[Music]

we can keep track of these steps by

listing off the order of the steps

for instance

the first path is r r u u u because the

right steps happen first and the last

path is u u u r r because the two right

steps happen at the end

a consequence of this encoding is that

the north east lattice paths to av

correspond to strings of r's and u's

with ars and bu's

thus such a lattice path corresponds to

a selection of a unordered positions

from a plus b spots to place the rs and

the u's go in the other spots

hence northeast lattice paths to a b are

counted by the binomial coefficient a

plus b choose a

returning to our lattice paths

we see that we can group them by their

final step

so in our example

four end with a right step and six end

with an up step

notice that all the paths ending in r

must first pass through the point a

minus one comma b

[Music]

similarly all the paths ending in an up

step u must first pass through the point

a comma b minus one

[Music]

these two collections of paths are

clearly disjoint and make up all of the

possible northeast lattice paths to the

point a b

thus

a plus b choose a

is equal to a minus 1 plus b choose a

minus 1 which is the number of lattice

paths to a minus 1 comma b

plus

the binomial coefficient a plus b minus

1 choose a which is the number of

lattice paths to a comma b minus 1.

replacing a plus b by n

results in n choose a equals n minus 1

choose a minus 1 plus n minus 1 choose a

this is the binomial recurrence

[Music]

you

The Binomial Recurrence from Lattice Paths (visual proof)

This short animated proof demonstrates one combinatorial proof for the recurrence satisfied by the binomial coefficients. #mathshorts #mathvideo #math #mtbos #manim #animation #theorem #visualproof #proof #iteachmath #mathematics #binomialcoefficients #latticepaths #discretemath #combinatorics #combinatorialproof

This animation is based on the well-known interpretation of binomial coefficients as counting lattice paths. You can find more in many textbooks such as Oscar Levin's Discrete Mathematics An Open Introduction (http://discrete.openmathbooks.org/dmoi2/sec_counting-binom.html).

This visual proof was inspired by the work of Steven Klee with his undergraduates:

https://cs.uwaterloo.ca/journals/JIS/VOL21/Klee/klee2.pdf

and

https://cs.uwaterloo.ca/journals/JIS/VOL23/Klee/klee4.pdf .

To learn more about animating with manim, check out:
https://manim.community


[Music]

[Music]

an odd square number can be visualized

as a square array with side length two n

plus one

this array has a triangular array

with base n sitting in the upper left

corner

if we swing the triangular array out and

down it forms the beginning of a larger

triangular array

with side length 3n plus 1. to complete

this larger array

we need a base n triangular array at the

top

therefore two n plus one squared is

equal to the three n plus first

triangular number

minus the nth triangular number that is

each odd square is a difference of two

triangular numbers

[Music]

you

Odd Squares as Difference of Triangular Numbers (visual proof)

This is a short, animated visual proof demonstrating how to visualize odd squares as the difference of two triangular numbers. #mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #triangular

This animation is based on a visual proof by Roger B. Nelsen from the March 1996 issue of The College Mathematics Journal (https://www.jstor.org/stable/2687399 - page 118).

Note that this technique is also utilized in the related result proved in one of my previous videos : https://youtu.be/sDbPEjOtp50

To learn more about animating with manim, check out:
https://manim.community
__________________________________________________________________
Background music, used according to CC BY-SA 3.0, is "The Road" from Esteban Orlando:

Esteban Orlando:
https://soundcloud.com/orlando-esteba...
https://www.instagram.com/estebanorla...

https://creativecommons.org/licenses/...
Music provided by RFM: https://youtu.be/b4KCM1bIq50


ขอ

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ดูหนังโป๊หน่อย

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ม.ค

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Cauchy Bunyakovsky Schwarz Inequality I (visual proof)

This is a short, animated visual proof of the two-dimensional Cauchy-Schwarz inequality (sometimes called Cauchy–Bunyakovsky–Schwarz inequality) using the Side-angle-side formula for the area of a parallelogram. #math #inequality  #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #areas #mathematics #cauchyschwarz #algebraicidentity #mathshorts #mathvideo #mtbos

If you like this video, please click "like" and consider subscribing and checking out my other videos.

This animation is based on a visual proof by Sidney H. Kung from the February 2008 issue of Mathematics Magazine (https://www.jstor.org/stable/27643084 - page 69).

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Cyberpunk Gaming Electro | STARDUST by Alex-Productions | https://www.youtube.com/channel/UCx0_M61F81Nfb-BRXE-SeVA
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


start with a right triangle that has

legs of length a

and b

and a hypotenuse of length c

rotate and shift a copy of this triangle

so that they align like this and notice

that the two connected angles are

complementary by construction

we can add this line to make a trapezoid

and the interior triangle here is right

angled by construction

let's compute the area of the resulting

trapezoid called garfield's trapezoid in

two different ways

first of all the area of the trapezoid

is the product of the average of the two

bases with the height so the area is a

plus b over two times a plus b which

expands to a squared plus two times a b

plus b squared over two

alternatively

the trapezoid is built from three

triangles so its area is the sum of the

areas of those three triangles

[Music]

the first two triangles in this sum are

congruent and they have the same area of

one-half times a times b

or a times b over two

the last triangle has area one half

times c squared so the area of the

trapezoid is two times a times b plus c

squared over two

but this means that a squared plus two a

b plus b squared over two

is equal to two times a times b

plus c squared over two

multiplying this equality by 2 and

subtracting 2 times a times b results in

a squared plus b squared is equal to c

squared this is the pythagorean theorem

[Music]

you

Pythagorean Theorem IV (visual proof; Garfield's trapezoid)

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using the trapezoid that is now attributed to President James Garfield. This theorem states the square of the hypotenuse of a right triangle is equal to the sum of squares of the two other side lengths. #mathshorts #mathvideo #math #pythagoreantheorem #pythagorean #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof

This animation is based on a proof due to James A. Garfield. For a static version of this proof, see Roger Nelsen's first "Proof Without Words: Exercises in Visual Thinking" compendium (page 7) : https://bookstore.ams.org/clrm-1. You can also check out Howard Eves' "Great Moments in Mathematics (Before 1650)" page 34-36.

For other proofs of this same fact check out:
https://youtu.be/q28H5NY5UVU
https://youtu.be/yfGtbNgcrQ8
https://youtu.be/JRjPsuJ7S-g

To learn more about animating with manim, check out:
https://manim.community
________________________________________
Music in this video:
Harmony by Vlad Gluschenko | https://soundcloud.com/vgl9
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


which is larger e to the pi or pi to the

E to figure this out graph the function

y equals the natural log of x over X

this function has a derivative equal to

1 minus the natural log of x over x

squared and this derivative is zero if

and only if x is equal to E

this means that the function attains a

maximum value at x equals e and in

particular that maximum value is 1 over

e

now because Pi is not equal to E when we

plug Pi into the function we get the

value natural log of pi over pi and we

see that this value is less than one

over e

therefore we conclude that the natural

log of pi over Pi is less than one over

e multiply both sides by E and Pi then

apply one natural log rule followed by

another natural log rule at this point

we found our winner calculus has proved

to us that pi to the E is less than e to

the pi

[Music]

Pi to e versus e to pi: which is larger? (Pi day short)

This is a short, animated visual proof demonstrating which of the values e^pi or pi^e is larger (where pi is the circle constant and e is Euler's constant) . #math #inequality #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #mathshorts #mathvideo #mtbos #euler #pi #piday

This animation is based on a visual proof by Fouad Nakhli from the June 1987 issue of Mathematics Magazine (https://www.jstor.org/stable/2689563).

To see the longer, wordless video with this result, check out https://youtu.be/JE_YeVlSqLY

For two visual proofs of a more general version of this inequality, see:
https://youtu.be/SaYs856QYlI
and
https://youtu.be/DOcsgckdvxU

To learn more about animating with manim, check out:
https://manim.community


what's one over one plus one over two

plus one over three so on forever you

keep adding positive values so maybe

it's infinity but the numbers are

getting smaller and smaller so maybe

this sum is finite

here's the one represented as an area

here's the one half as an area the

one-third and so on forever

and here's the function one over x

graphed from one to infinity

notice how the rectangles line up nicely

and more importantly notice that the

area of the rectangles is larger than

the area under the curve of the function

you can use technology or calculus to

find the area under this curve and

realize it's infinite

yet the area of the rectangles is bigger

than that

since the area of the rectangles

represent our sum

we have our answer

so now what if i ask you this question

Every Student Should See This

This has always been one of my absolute favorite visual proofs!

This sum 1 + 1/2 +1/3 +... is call the harmonic series. This is a great math proof showing why the harmonic series diverges! (We basically use the integral test for series)

►BECOME A CHANNEL MEMBER
https://www.youtube.com/channel/UChVUSXFzV8QCOKNWGfE56YQ/join

►WEBSITE
https://www.brithemathguy.com

►MY COURSE
Prove It Like A Mathematician! (Intro To Math Proofs)
https://www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #shorts


[Music]

foreign

with any triangle such as the one

pictured here

then place a square on each side of the

triangle so that we have three squares

as pictured

these three squares allow us to create

three new triangles connecting the

adjacent vertices of adjacent squares as

shown here's an interesting question

about the four triangles that we've

created which of them has the greatest

area remember that I'm asking this

question in general not for the specific

triangle that I drew so I want to know

the answer no matter which triangle we

start with is it possible that the

answer doesn't depend on the starting

triangle or will the answer change as

the initial triangle changes

pause the video and see if you can

figure this out on your own

[Music]

it turns out that this is a bit of a

trick question if we erase the three

squares and rotate each of the three

created triangles 90 degrees we get this

shape pictured here in this shape each

of the three created triangles has a

side length that extends one of the side

lengths of the original triangle and

matches that length now when we draw

three altitudes in the original triangle

one perpendicular to each side we see

something interesting not only do the

newly created triangles share a base

with the original triangle they also

share a perpendicular height with the

original triangle this means that the

areas of each of the three created

triangles is equal to the area of the

original triangle but that means that

all four triangles have exactly the same

area

this argument holds No Matter What

triangle we started with it turns out

that whatever triangle you start with

you create three new triangles and when

you rotate them 90 degrees you will

always have one side length that matches

the side length of the original triangle

and you will always get this nicely

glued together three triangle shape so

that when you draw altitudes for the

original triangle those altitudes become

altitudes of one of the three created

triangles

Four Triangles: Which is Largest? (visual proof)

This is a short, animated visual proof demonstrating a surprising relationship between three triangles created in a special way from a given fourth triangle.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

For similar visual proofs see:

https://youtu.be/zhriP_4QpJQ
https://youtu.be/_emvEr4IB_Y
https://youtu.be/PNQxS6antBY
https://youtube.com/shorts/NA8UsiZ3mMI

This animation is based on a visual proof from Steven Snover that appeared in Roger B. Nelsen's second Proofs without Words Compendium (page 15).

To learn more about animating with manim, check out:
https://manim.community

#mathshorts #mathvideo #math #triangle #area #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometry #algebra #inequality

_______________________________
Music in this video
Til Death Parts Us by Aakash Gandhi


start with a unit square with an area of

one connect each vertex to an opposite

Edge midpoint like this the region

bounded between the four Drawn Lines

forms a square itself can you find the

area of this Square pause the video and

think about it before I reveal an

answer we can rotate the four small

triangles pictured here to form four

squares congruent to the original shaded

square but that means the Shaded area

must be

A SquAREA within a Square?

This is a short, animated visual proof of a classic proof about the area of the square obtained by connecting the vertices of a square to the midpoints of the opposite sides of the square.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

This animation is based on a proof that appears in the following sources:

Mathematical Morsels by Ross Honsberger (MAA, 1978)

Proofs without Words II by Roger B. Nelsen (MAA, 2000) (https://bookstore.ams.org/clrm-14/)

And can be traced to an article by an author known as Nev R. Mind.

To learn more about animating with manim, check out:
https://manim.community

#manim #math #mathvideo #mathshorts #geometry #square #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #onefifth


so you need to divide a circular Pizza

seven equal ways and all you have is a

straight edge Encompass no worries I've

got you covered first take the diameter

of the pizza and cut it into seven equal

pieces which you can do with a straight

edge Encompass now use your compass to

draw these seven semi-circles you should

have one semi-circle of diameter ND for

each value of n from one up to seven

like this

now here's the trick you want to do the

exact same construction on the bottom

half of the pizza except this time from

right to left

these are the resulting seven pieces

that you have and they're all equal in

area if you want to know more about why

this works check out the link in the

description to the longer video

describing all the details

Divide pizza into 7 using only straightedge and compass

In this short, we show how to decompose a circle into seven pieces each with the same area. This process can be done using ruler and compass constructions, unlike the process of constructing the regular heptagon. #manim #mathshorts #mathvideo #math #geometry #mtbos #animation #theorem #visualproof #proof #iteachmath #circle #mathematics #rulerandcompass #construction #pizza

Here is a longer video filling the details that the seven regions are actually equal in area: https://youtu.be/KhfZK5IIK9E

Here is a video showing how to divide a line into seven equal lengths: https://youtu.be/afZiAGaKEY4

This animation is based on a visual proof from Roger Nelsen's and Claudi Alsina's wonderful book "Icons of Mathematics" (Chapter 14.1 on page 76). You can find the book and other information about it at the links below

(I may receive commissions from this paid link) : Icons of Mathematics from Amazon https://amzn.to/3L5U44S

https://www.maa.org/press/maa-reviews/icons-of-mathematics-an-exploration-of-twenty-key-images

https://bookstore.ams.org/dol-56/

To learn more about animating with manim, check out:
https://manim.community


[Music]

a

oh

#mathematics #calculus Indeterminant Form (0xinfinity): L'Hospital rule

#mathematics #calculus Indeterminant Form (0xinfinity)

Mathematics Shorts: https://www.youtube.com/playlist?list=PLN1Cp7TpZOL7LH2zGFtSejZ18P4swqnDA

Other Videos
Fibonacci Sequence, Golden Ratio, How to draw Golden Spiral:
https://youtu.be/yHBa3EvDPco
Radius and Tangent Vectors:
https://youtube.com/shorts/KHSafRET3Sk
De Morgan's laws: Proof by Venn Diagrams:
https://youtube.com/shorts/pR-REo4aBYk?feature=share
Visual Proof of Roots of Quadratic Equation
https://youtube.com/shorts/FdE7rHDZRd8
Visual Proof: Factors of Quadratic Equation PART-1 x^2+3x+2
https://youtube.com/shorts/lIqv5zKAKKE
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2-3x+2
https://youtube.com/shorts/5pRYbKFWjUM
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2+2x-3
https://youtube.com/shorts/AzAsz5itOfQ
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2+2x-3
https://youtube.com/shorts/AzAsz5itOfQ
Visual Proof: Trigonometric Identity sin^2 (x) +cos^2 (x)=1
https://youtube.com/shorts/bPAm1nRUJWw


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let A and B be two positive real numbers

first draw a square with side length a

and adjacent to that draw a square with

side length B in each Square we can draw

the diagonals and use the intersection

to draw a line from the center of the

square to the midpoint of the base of

the square this process has created

isosceles right triangles in either

square one of them with a side length of

a over two and one of them with a side

length of B over 2. now we can connect

the center points of the two squares to

create a right triangle pictured here

the legs of this right triangle are a

divided by the square root of 2 and B

divided by the square root of 2 since

both of those are hypotenuses of the

previously described right triangles but

then we can use the Pythagorean theorem

to find that the hypotenuse of this

triangle is the square root of a squared

over 2 plus b squared over two the two

isosceles right triangles and the Shaded

purple triangle create a trapezoid from

this trapezoid we see that the one side

length given by a over 2 plus b over 2

is less than or equal to the longer side

length which is given by the square root

of a squared over 2 plus b squared over

two no matter this sizes of A and B we

see that we can always create this

diagram with an a by a square and a b by

b square and a triangle connecting their

Center points this triangle is part of a

trapezoid and the trapezoid has one long

side and one short side so the

inequality is always strict except in

the one situation where the side length

a equals the side length B therefore a

over 2 plus b over 2 is less than or

equal to the square root of a squared

over 2 plus b squared over 2 and

equality occurs if and only if a equals

B A over two plus b over 2 is the

arithmetic mean of A and B and the

square root of a squared over 2 plus b

squared over 2 is the root mean square

of A and B so this is known as the

arithmetic mean root mean Square

inequality

Arithmetic Mean-Root Mean Square Inequality (visual proof)

This is a short, animated visual proof demonstrating the arithmetic mean - root mean square inequality using two adjacent squares.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

For other visual proofs of inequalities like this see:

https://youtu.be/jNx05gXSblE
https://youtu.be/Pt8IX_Q5U1A
https://youtu.be/62G9fak1vyk
https://youtu.be/HFGSsPbYc-s
https://youtu.be/SEPkdPWKdO0

or this playlist: https://www.youtube.com/playlist?list=PLZh9gzIvXQUuKcFHdtjFkwAVELDxu9nH4

This animation is based on a visual proof from Juan-Bosco Romero Márquez that appeared in the March 2001 issue of the College Mathematics Journal (https://www.jstor.org/stable/2687116  p. 118)

To learn more about animating with manim, check out:
https://manim.community

#mathshorts #mathvideo #math #amrmsinequality #arithmeticmean #rootmeansquare #rootmean #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #inequality
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Music in this episode:
Procession -- Geographer


here's one way to make sense of the

infinite sum of the powers of one-half

start with an isosceles triangle of Area

1 and cut it into two congruent

isosceles triangles shading one of them

so that we've shaded an area of a half

then repeat on the unshaded isosceles

triangle cutting it in half and shading

one so that we've shaded a quarter we

can continue this process indefinitely

each time we take the unshaded isosceles

triangle cut it in half and shade one

piece the area of this piece will be one

half the area of the previously shaded

triangle but as we do this infinitely we

end up with the Shaded region being the

sum of the powers of one-half on the

other hand as we do this forever we end

up filling the entire isosceles triangle

which has area one therefore the sum of

the powers of one-half is equal to one

foreign

Adding powers of 1/2

This is a short, animated visual proof demonstrating the infinite sum of the powers of 1/2 using an isosceles triangle. #mathshorts #mathvideo #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricseries #series #infinitesums #infiniteseries

Here is a bit longer and slower version (with dramatic music and no words): https://youtu.be/saJJGNlsfn8

This animation is based on a proof by Roger B. Nelsen from the February 2006 issue of Mathematics Magazine page 60. (https://doi.org/10.2307/27642904).

To learn more about animating with manim, check out:
https://manim.community


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##mathematics #calculus Visual : Generalized  Parabola y=a(x-h)^2+k

Other Videos
Fibonacci Sequence, Golden Ratio, How to draw Golden Spiral:
https://youtu.be/yHBa3EvDPco
Radius and Tangent Vectors:
https://youtube.com/shorts/KHSafRET3Sk
De Morgan's laws: Proof by Venn Diagrams:
https://youtube.com/shorts/pR-REo4aBYk?feature=share
Visual Proof of Roots of Quadratic Equation
https://youtube.com/shorts/FdE7rHDZRd8
Visual Proof: Factors of Quadratic Equation PART-1 x^2+3x+2
https://youtube.com/shorts/lIqv5zKAKKE
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2-3x+2
https://youtube.com/shorts/5pRYbKFWjUM
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2+2x-3
https://youtube.com/shorts/AzAsz5itOfQ
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2+2x-3
https://youtube.com/shorts/AzAsz5itOfQ
Visual Proof: Trigonometric Identity sin^2 (x) +cos^2 (x)=1
https://youtube.com/shorts/bPAm1nRUJWw


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a positive integer is said to be perfect

if it is equal to the sum of its proper

divisors

for example here's the number six with

its proper divisors two three and one

and here's the number 28 with its proper

divisors 4 14 2 7 and 1. we can see that

6 is equal to 1 plus 2 plus 3 and 28 is

equal to one plus two plus four plus

seven plus fourteen therefore both 6 and

28 are perfect numbers

separated by about 2000 years Euclid and

Euler proved the two halves of the

theorem that states that n is an even

perfect number if and only if n can be

written as the form 2 raised to the

quantity P minus 1 multiplied by 2 to

the P minus 1 where 2 to the P minus 1

is prime according to this theorem if n

is an even perfect number we can arrange

n dots in a rectangle with Dimensions 2

raised to the P minus 1 and 2 to the P

minus one but then the side of length 2

to the P minus 1 can be split into two

pieces one of length two raised to the P

minus 1 and the other with a length of 2

raised to the P minus 1 minus one now

consider the Triangular array with base

of length 2 raised to the P minus 1

minus 1 sitting in the upper right of

this rectangle we can swing this array

down forming a larger triangular array

in this case the base of the new

triangular array is 2 to the P minus 1

plus 2 to the P minus 1 minus one which

simplifies to 2 to the P minus 1 but

this means that any even perfect number

can fit into a triangular array with a

base of length 2 to the P minus 1. so

that n is an even perfect number if and

only if n is the 2 to the P minus 1

triangular number where 2 to the P minus

1 is prime it turns out that there

aren't many primes of the form 2 to the

P minus 1 that we know these primes are

called mercen primes and it's an open

problem to figure out if there are

infinitely many such primes equivalently

it's an open problem to figure out if

there are infinitely many even perfect

numbers

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Perfect Numbers (even ones) are Triangular (visual proof) -- with bonus open problem!

This is a short, animated visual proof showing that all even perfect numbers are triangular numbers (and thus all currently known perfect numbers are triangular numbers). #math #euler #euclid #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #mersenne #openproblem #unsolvedproblem

This animation is based on a visual proof by Roger Nelsen from the May 2016 issue of The College Mathematics Journal (https://doi.org/10.4169/college.math.j.47.3.171 ).

To learn more about animating with manim, check out:
https://manim.community
__________________________________________________________________
Music in this video:
Til Death Parts Us by Aakash Gandhi


let's say you have to multiply two

binomials for example x + 3 and x - 2

you multiply the first terms of each

binomial the outer terms the inner terms

and then the last terms you lastly

combine any like terms to get the final

result this is known as the foil method

and here's a visualization showing how

it works say we have a rectangle where

its area is equal to its width time its

height if the width and height are each

broken into two variables A and B and C

and D we would have to multiply these

binomials to find the area of the re

rectangle an alternative approach would

be to break the rectangle down into four

smaller rectangles it's easier to

visualize their area and we can find the

area of the total rectangle by adding up

all the smaller areas by doing so we

find the total area is equal to AC plus

a d plus BC plus BD which yields the

exact same result as if you were to

follow the foil method the same approach

can be used to visualize multiplying

larger polinomial the order in which you

add the terms doesn't matter as long as

you're multiplying every possible pair

between left and right comment your

thoughts below and subscribe for more

stem content

FOIL method visual proof #maths #animation #education

Here’s a potentially useful visualization of binomial multiplication. Comment below for what content you’d like to see form this channel!
Music: Inhale - Recall


the halleyuda pascal triangle is a

triangular array of numbers built as

follows

every row begins with a one and ends

with a one

and every other entry in the triangle is

obtained by adding the two entries

directly above the entry in question we

can fill in the rest of the triangle

using this pattern

here we've only pictured up to row seven

in the halle utah pascal triangle

now let's go through this triangle and

shade only the odd entries

we get a picture that looks like this

but if we zoom out to show 256 rows of

pascal's triangle and shade only the odd

entries something amazing happens we

produce the sierpinski triangle

you

Sierpinski from Pascal

This is a recreation of a short clip from a long form video showing six different ways to construct the Sierpinski triangle: https://youtu.be/IZHiBJGcrqI

In this short, we shade odd entries of the Halayuda/Pascal triangle to obtain the Sierpinski triangle. Can you explain why this works?

#math #manim #fractal #fractal #sierpinski #pascaltriangle #pascal #mathshort #visualmath


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#trending #mathematics Visual Proof: Algebraic Identity (a+b)^3=a^3+3a^2b+3ab^2+b^3

#trending #mathematics Visual Proof: Factors of Quadratic Equation Polynomial PART-4 x^2-2x-3

Other Videos
Fibonacci Sequence, Golden Ratio, How to draw Golden Spiral:
https://youtu.be/yHBa3EvDPco
Radius and Tangent Vectors:
https://youtube.com/shorts/KHSafRET3Sk
De Morgan's laws: Proof by Venn Diagrams:
https://youtube.com/shorts/pR-REo4aBYk?feature=share
Visual Proof of Roots of Quadratic Equation
https://youtube.com/shorts/FdE7rHDZRd8
Visual Proof: Factors of Quadratic Equation PART-1 x^2+3x+2
https://youtube.com/shorts/lIqv5zKAKKE
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2-3x+2
https://youtube.com/shorts/5pRYbKFWjUM
Visual Proof: Factors of Quadratic Polynomial PART-2 x^2+2x-3
https://youtube.com/shorts/AzAsz5itOfQ


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Visual Proof of PT 😇 #shorts #youtube #trending #viral #youtubeshorts #maths #mathstricks #pt #fun


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Geometric Series: sum of powers of 1/6 (visual proof)

This is a short, animated visual proof showing the sum of the infinite geometric series of powers of 1/6 (starting with 1/6) is 1/5 using self similar trapezoids in a regular pentagon. #manim #math #mathshorts #mathvideo  #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #series #geometricseries #infiniteseries  #pentagon #trapezoids

This animation is based on a proof by James Tanton from the March 2008 issue of The College Mathematics Journal page 106. (https://www.jstor.org/stable/27646594).

To learn more about animating with manim, check out:
https://manim.community
___________________________________
Music:
Tale Forest by Vlad Gluschenko | https://soundcloud.com/vgl9
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


foreign

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Cubes from double sums (visual proof without words)

In this (wordless) video, we demonstrate how each integer cube is a special double sum.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

Here are a few other related visual proofs:
https://youtu.be/OECW5qqZN9I
https://youtu.be/936eKowyvR4
https://youtu.be/WidzHiUFWNA
https://youtu.be/BBrtHbRcpXs
https://youtu.be/YQLicI8R4Gs

This animation is based on a visual proof by Roger B. Nelsen from the March 2002 issue of the College Mathematics Journal  (https://www.jstor.org/stable/1559005 page 171 ).

To learn more about animating with manim, check out:
https://manim.community

#math #manim #mathvideo #cubes #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #finitesums #discretemath #induction #doublesum

___________________________________________
Music in this video:
Dance Party | JUMP by Alex-Productions | https://onsound.eu/
Music promoted by https://www.free-stock-music.com
Creative Commons / Attribution 3.0 Unported License (CC BY 3.0)
https://creativecommons.org/licenses/by/3.0/deed.en_US


draw the circle of radius 1 at the

origin rotate an angle Theta the point

on the unit circle for Theta has an x

coordinate of cosine Theta and a

y-coordinate of sin Theta if we draw the

tangent line to the circle at the point

and connect to the x-axis by triangle

similarity the distance labeled D here

satisfies d/ 1 is equal to sin thet over

cosine thet which is the tangent of thet

also by triangle similarity the

hypotenuse of the picture triangle is 1

over the cosine of theta so it's the

secant of theta if we extend the tangent

line to the Y AIS like this then the two

picture triangles are similar so that

the labeled distance D satisfies D 1al

cosine Theta over sin Theta which is the

cent of theta and the hypotenuse of the

green triangle is 1/ sin Theta which is

cosecant Theta these definitions of the

six standard trig functions can be

extended to the other quadrants as well

and this animation demonstrates their

range restrictions the prefix Co stands

for complimentary because the two

picture triangles have complimentary

angles Theta and pi/ 2 minus Theta so

the cosine is the sign of the

complimentary angle

Six Trigs in 60 Seconds! #math #trigonometry

In this video, we show a single diagram consisting of various triangles that connects the six primary trig functions (sine, cosine, tangent, secant, cosecant, and cotangent) to lengths of line segments created from the unit circle (circle of radius 1).  We also briefly discuss the fact that the "co" on three of the functions refer to the "complementary angle."

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

For a longer version of this video with more details and information, see
https://youtu.be/dUkCgTOOpQ0

This animation is based on a classic diagram for describing the six main trig functions. If you want to know more trig identities, check out my playlist:
https://youtube.com/playlist?list=PLZh9gzIvXQUvc6SufrIpkVrVLdhcGthSy&si=2IXs0Jx5j30_-spb

or check out this Wikipedia article:
https://en.wikipedia.org/wiki/List_of_trigonometric_identities

Here is an interactive version from Tien Chih using Desmos:

https://www.desmos.com/calculator/26ftp9is44

Here are alternate videos with the same diagram:

https://youtu.be/Dsf6ADwJ66E?si=1N8GEyBe3abyoan1

https://youtu.be/68Pi4RJ1Rnw?si=aTCeeGa73-R3eApe

#math #mathvideo #math #trigonometry #trigidentities #identity  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #pythagoreantheorem #obtuseangle #acuteangle #angle #sine #cosine #tangent #secant #cosecant #cotangent #learnmaths

To learn more about animating with manim, check out:
https://manim.community


the Fibonacci sequence starts with fub1

equal to 1 so f1^ squar is the area of a

1x1 Square FS2 is also equal to 1 so FS2

^2 is the area of a 1x one square as

well when n is greater than two the nth

Fibonacci number is the sum of the

previous two Fibonacci numbers notice we

can keep attaching squares to this

diagram in such a way that the side

length is the sum of the previous two

Fibonacci numbers and is therefore for

the next Fibonacci number so as we add

squares to this diagram the I Square

will have an area given by this square

of the I Fibonacci number we can

continue this process as long as we like

if we eventually stop say at the number

F subn then we have a rectangular

diagram enclosing an area given by the

sum of the squares of the first n

Fibonacci numbers on the other hand the

dimensions of this rectangular diagram

are the N Fibonacci number by the N plus

1 Fibonacci number so the enclosed area

is the product of these two numbers this

means that the sum of the squares of the

first n Fibonacci numbers is the product

of two consecutive Fibonacci numbers

Sums of Fibonacci Squares

This is a short, animated visual proof demonstrating the sum of the squares of the first n consecutive Fibonacci numbers.

To buy me a coffee, head over to https://www.buymeacoffee.com/VisualProofs
Thanks!

To see a slower, wider high definition version of this visual identity (without words), check out this video: https://youtu.be/2EsoiBdnIAM?si=3bW8M_zxxx-ueAhC

Also, check out my playlist on other finite sums:
https://youtube.com/playlist?list=PLZh9gzIvXQUubr38YfIlul9j7_54hXZy_

This animation is based on a proof by Alfred Brouusseau from "A Primer for the Fibonacci Numbers by M. Bicknell and V.E. Hoggatt Jr (eds.). You can also find this proof in Roger Nelsen's first Proof Without Words compendium (page 83).

Get Nelsen's Amazing Book (Affiliate link so I may receive a commission): https://amzn.to/3QF6Zvz

#fibonacci #math #manim #mathvideo #calculus #mtbos #mathshorts #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #series #finitesums #finiteseries #geometric #squares

To learn more about animating with manim, check out:
https://manim.community


you can use a 1x one square to make

sense of the infinite sum 1/2 plus 1/4 +

1/8 and so on first cut the square into

two parts and shade 1/2 then shade 1/2

of the remaining half shading a quarter

then shade 1/2 of the remaining 1/2

shading an eighth continue this process

over and over again each time shading

1/2 of the remaining 1 over 2 to the N

if we continue shading indefinitely that

is if we consider the limiting process

eventually in the limit the entire

Square will be shaded by the sum of

powers of 12 but the square has an area

of 1 so that the infinite sum where K

ranges from 1 to Infinity of 1 2 the K

is equal to 1 another interpretation of

this infinite sum is that 0.1 repeating

in base 2 is equal to 1

0.1111… = 1 (in base 2)

This is a short, animated visual proof showing the sum of the infinite geometric series with first term 1/2 and ratio 1/2 using a 1 by 1 square. This infinite sum allows us to determine an interesting base 2 (or binary) representation of 1.

To buy me a coffee, head over to https://www.buymeacoffee.com/VisualProofs
Thanks!

For an alternate wide-format wordless geometric series dissection of the same fact, check out

https://youtu.be/saJJGNlsfn8?si=7eCh6og1dkB9GrWB

This animation is based on a visual proof by Warren Page from the September 1981 issue of Mathematics Magazine, (page 201 - https://www.jstor.org/stable/2689632 ).

 #manim #math #mathshorts #mathvideo  #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #series #geometricseries #infiniteseries  #base2 #binary #zeno #paradox

To learn more about animating with manim, check out:
https://manim.community


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#VISUAL #PROOF #OF #DIFFERENCE #OF #SQUARES #x² — y²

Dear viewers,
Welcome to    MATH BRIDGES  FUN WITH MATH
To enhance the learning process  and  make the math concepts alive  and fun loving through online medium for all..


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Sum of First n Odd Integers | Visual Proof | Geometric Proof | Proof Without WORDs

Thanks for watching! If you found these tips helpful, please like and subscribe to our channel. Share your reviews in the comments below!

#shorts #mathtricks #math #mathematics #mathideas #mathanimation #animation #animated #visualization #geometric #proof #proofwithoutword #algebric #algebra #algebricequation #algebricexpression #visualproof #geometricproof #amazingmath #funfacts #mathfacts #abstractart #abstractmath #mathvisualization #mathart #geometry #shapes #equation #expression #square #proofing #proofof #proofofconcept #amazingahorts #amazingfacts #factshorts #intution #intutive #funmath #mathideas #mathworld #magicmath #magicmathtricks #mathmagic #mathzone #abstractview #proofs


here are two methods to find the area of

a circle using the method of exhaustion

in the top image we start peeling off

circumferences one at a time and lay

them flat as lines because the

circumference varies linearly with

respect to the radius we see that the

shape will eventually form a triangle in

the bottom diagram we split the circle

into an even number of wedge-shaped

pieces as we create smaller and smaller

wedges we can split the circle apart

into those wedge shaped pieces and glue

the wedges together to form shapes that

look more and more like a rectangle it

turns out that in the top shape we end

up with a triangle that has a height of

R and a base of length 2 pi r and in the

bottom shape we end up with a rectangle

where the length is given by half the

circumference which is pi * R and a

height given by the radius of R so in

the top diagram we get an area of 1 12 *

2i R * R which is p piun r^ 2 and in the

bottom diagram we get an area of piun *

r^ 2 let me know in the comments which

visualization you like better

Circle area by exhaustion (two visual area techniques)

In this short, we show two fascinating methods of determining the area of a circle using the "method of exhaustion." Which one do you like better?

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

The top animation unrolls successive circumferences of nested circular shells. In this manner, the circle area gets transformed into the area of a triangle with base 2pi times r and height r.

Here are other versions from me:
https://youtu.be/GjmtmCT5N8Q
https://youtube.com/shorts/3L_GorNnk4Q

To see an alternate (and quite viral) video showing this in stop motion, check out this one from  @MinutePhysics  : https://youtu.be/whYqhpc6S6g

For more information about this construction, see
https://personal.math.ubc.ca/~cass/courses/m446-03/exhaustion.pdf

or check out this nice survey article by David Richeson from the May 2015 issue of The College Math Journal: https://doi.org/10.4169/college.math.j.46.3.162 .

This bottom animation of a classic visual proof showing how to find the area of a circle by using more and more wedges and arranging them in a rectangle.

This proof can be traced to both Satō Moshun and Leonardo da Vinci (see Smith, David Eugene; Mikami, Yoshio (1914), A history of Japanese mathematics, https://archive.org/details/historyofjapanes00smituoft, page 130-132 and Beckmann, Petr (1976), A History of Pi, St. Martin's Griffin, page 19).

Here are other versions from me:
https://youtu.be/-UBLutigNMM
https://youtube.com/shorts/X-kTZgFWUFQ

You can also read more about this in a great NYT article by Steven Strogatz: https://archive.nytimes.com/opinionator.blogs.nytimes.com/2010/04/04/take-it-to-the-limit/

#math #manim #visualproof #proofwithoutwords #circle #circlearea #archimedes #radius #area #areaofcircle #pi #piday #shorts #circle #archimedes #infinite #methodofexhaustion

To learn more about animating with manim, check out:
https://manim.community


start with a triangular array of

triangles with n triangles on the base

imagine placing one token in the first

triangle for one squared tokens

then place two tokens in each of the

triangles in the second diagonal adding

two squared tokens

place three tokens on each of the three

triangles in the third diagonal adding

three squared total tokens and continue

this process

so that the n minus 1 tokens in each of

the n minus 1 triangles in the next to

last diagonal add n minus 1 squared

tokens

and the n tokens in each of the n

triangles in the last diagonal add n

squared tokens

so the total number of tokens is the sum

of the first n squares

now create a copy of the diagram

and rotate it 120 degrees

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then

create a copy from this one and rotate

it another 120 degrees resulting in

three total copies

notice that at any position if you add

the number of tokens at that position

over the three copies you always get two

n plus one

thus when we stack all three copies

together we have a triangular array of

stacks of tokens with two n plus one

tokens so the total number of tokens is

two n plus one

times

the nth triangular number

but this is equal to two n plus one

times n times n plus one over two

consequently

three

multiplied by the sum of the first n

squares

is equal to two n plus one times n

times n plus one over two

dividing both sides by three

gives the formula for the sum of the

first n squares to be equal to two n

plus one times n times n plus one over

six

[Music]

you

Sum of Squares II (visual proof)

This is a short, animated visual proof of the formula that computes that sum of the first n squares using 3 copies of a triangular array containing stacks of tokens equaling the sum of the first n squares . #mathshorts #mathvideo #math #sumofsquares #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #finitesums #discretemath

This animation is based on a visual proof by Sidney H. Kung from the May 1989 issue of The College Mathematics Journal  (https://doi.org/10.2307/2686767) page 205.

Here is another visual proof of the same fact:
https://youtu.be/-tJhH_k2LaM

To learn more about animating with manim, check out:
https://manim.community


let d and n both be positive integers

this means that we can arrange d dots in

a line

and we can arrange n dots in a line

the theorem we're interested in says

there exists unique integers q and r

with 0 less than or equal to r less than

d satisfying the equality n equals q

times d plus r

to prove the existence of q and r we

start subtracting copies of d from n

as we do this we create a list of

decreasing positive integers

by the least integer principle

of positive integers we know that this

set must have a least element as we see

here where we can no longer subtract off

d

this is going to be the value of r

as we see in the diagram we have

arranged the n dots into an a d by q

array with r dots left over and a line

on top

this means that n equals

q

times d

plus a remainder of r

to prove uniqueness of the pair q and r

let's imagine that we have another pair

q prime and r prime satisfying the same

conditions because these two are equal

we can subtract to see that q prime

times d minus q times d is equal to r

minus r prime

but now because r and r prime are both

between 0 and d their difference is

between negative d and d

this means that q prime times d minus q

times d is a multiple of d that lives

between minus d and d

from this we see that q prime times d

minus q times d must be zero and

consequently that means r minus r prime

must be zero

dividing the second equality by d which

is non-zero shows us that q prime minus

q is zero and r minus r prime is zero

therefore r must be equal to r prime and

q must be equal to q prime this means

that r and q are unique

this concludes our proof of the theorem

known as the division theorem or the

division algorithm

[Music]

you

Division Theorem/Division Algorithm

This is a proof, with a visualization, of the classic number-theoretic proof of the division theorem/division algorithm. The video includes a "visual demonstration" of the existence part of the theorem (where we also implicitly show how it is an algorithm). We also investigate the well-ordering principle (or least integer principle) and discuss existence and uniqueness style proofs. #math #proof #manim #numbertheory #visualmath #mtbos #visualproof #division #introductiontoproofs #divisiontheorem #divisionalgorithm #existence #uniqueness #subtraction

If you liked this video, please consider liking and subscribing and checking out a few of my other videos.

To learn more about animating with manim, check out:
https://manim.community

_________________________________
Music in this video:
日本庭園 (Japanese Garden) by Alex-Productions | https://www.youtube.com/channel/UCx0_M61F81Nfb-BRXE-SeVA
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


[Applause]

[Music]

[Music]

oh

Best sum visual proof?

In this video, we show an animation of perhaps the best and most iconic visual proof of the formula for the sum of the first n positive integers.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

To see more about this proof and a generalization, check out this video:
https://youtu.be/eHbtc50-qXo?si=ztvGFf1G-dLzr-m3

This proof is attributed to the "ancient Greeks" by Roger Nelsen (referencing Martin Gardner) and can be found in Roger Nelsen's first compendium, "Proofs Without Words: Exercises in Visual Thinking:

(I may receive a small commission at no cost to you for this affiliate link): https://amzn.to/40cQL3k

#math #manim #sumformula #naturalnumbers  #triangle #animation #theorem #pww #proofwithoutwords #visualproof #proof  #mathshorts #mathvideo #finitesums  #sums #shorts #triangularnumbers

To learn more about animating with manim, check out:
https://manim.community


here's my favorite diagram for finding

the sum of the first n squares imagine a

stack of squares is pictured here we see

one squared plus 2 squared plus 3

squared and so on up to N squared where

in this case we've pictured n equals 4.

now we create a tower using the L-shaped

gnomons we have one row with two n minus

one squares two rows with two n minus

three squares and so on all the way down

until we have n rows of one square then

we can use one stack of the sum of the

first n squares on the left side like

this and we can use another stack of the

sum of the first n squares on the right

side of the tower like this to create a

rectangular diagram we've used three

copies of the sum of the squares to

create this rectangular diagram which

has a width of two n plus one and a

length of the sum of the first n

positive integers so three times the sum

of the first n positive squares is equal

to the sum of the first n positive

integers times two n plus one we apply

the formula for the sum of the first n

positive integers to see that the sum of

the first n squares is n times n plus 1

times 2N plus one divided by six

Summing Squares in a Rectangle

This is a short, animated visual proof of the formula that computes that sum of the first n squares using 3 copies of the sum of squares to build a rectangle .

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

Here is a slower version of this video: https://youtu.be/-tJhH_k2LaM

Here are three other visual proofs about the sum of squares formulas:

https://youtu.be/WidzHiUFWNA (or https://youtube.com/shorts/PVH5uxHFIGQ)
https://youtu.be/UqVmocdLFGc
https://youtu.be/Q-frL0Ot2m4

This animation is based on (independently discovered) visual proofs by Dan Kalman from the March 1991 issue of The College Mathematics Journal (https://www.jstor.org/stable/2686447) and Martin Gardner from the October 1973 Scientific American (https://www.jstor.org/stable/24923225).

To learn more about animating with manim, check out:
https://manim.community

#mathshorts #mathvideo #math #sumofsquares #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath


here's a quick proof of the Pythagorean

theorem first take four copies of your

right triangle flip them over as such

then arrange them to form a big Square

that's blue and green notice that these

angles are complimentary so the red

lines form a square Now find the area of

the big blue and green square by taking

the side

lengths next find the area alternatively

by multiplying by four times the

triangles and 1 * the Red Square now now

we have the following equation for the

area which we simplify and finally we

get Pythagoras's beautiful identity

thank you for watching

A Visual Proof of the Pythagorean Theorem

Manim animation of Pythagoras


foreign

[Music]

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Pythagoras visual proof 🧐 #studyhub #interestingfacts #factshorts #pythagoras_theorem

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foreign

[Music]

thank you

[Music]

foreign

[Music]

foreign

[Music]

[Applause]

Geometric Series: sum of powers of 1/9 (visual proof)

This is a short, animated visual proof showing the sum of the infinite geometric series of powers of 1/9 (starting with 1/9) is 1/8 using self-similar trapezoidal shapes in arrays of squares. #manim #math #mathshorts #mathvideo  #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #series #geometricseries #infiniteseries  #pentagon #trapezoids

To learn more about animating with manim, check out:
https://manim.community
___________________________________
Music:
What Words Can’t Describe by Vlad Gluschenko | https://soundcloud.com/vgl9
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


start with rational numbers A over B

less than or equal to C over D we can

represent the rational number A over B

as the slope of the hypotenuse of a

right triangle with side lengths A and B

and we can represent C over d as the

slope of the hypotenuse of a right

triangle with legs of length C and D by

assumption the latter triangle has a

steeper hypotenuse if we shift the

triangle so their vertices connect like

this we can fit them inside of a larger

right triangle with sides of length B

plus d and a plus c by investigating the

slopes of the hypotenuses in this

diagram we see that A over B is less

than or equal to the quantity a + C / B

+ D and that's less than or equal to C

over D the fraction in the middle of

this inequality a plus c over B + D is

known as the mediant of the two

fractions so we can define an operation

where we combine A over B and C over D

to produce a plus c over B plus d and

this is called the median operation due

to this terminology the inequality we

proved is sometimes called the mediant

inequality

Mediant Inequality I

This is a short, animated visual proof demonstrating what one might called the mediant inequality.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

This animation is based on a visual proof by Richard A. Gibbs from the June 1990 issue of Mathematics Magazine (https://doi.org/10.2307/2691137  - page 172).

#mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #fractions #mediant #inequality #mathematics #fareyfractions #farey

To learn more about animating with manim, check out:
https://manim.community


[Music]

consider a triangle with sides of length

a

b

and c

and angle theta opposite to c

draw the circle containing the triangle

using the side length of a as the radius

extend the radius to a full diameter

and extend the side length c to a

diameter by connecting a radius of

length a and a segment of length a minus

c

extending the side length b to the

circle

and connecting that line to the other

end of the horizontal diameter

results in a right triangle according to

thaley's triangle theorem

this right triangle has a hypotenuse of

length 2 times a

this means that the leg adjacent to the

angle theta has length 2a cosine of

theta

therefore the line segment extending the

length

b to the circle has length two a cosine

theta minus b

now we have two intersecting chords in

the circle

[Music]

so by the intersecting chords theorem

we get that the product of two a cosine

theta minus b

times b

equals

the product a minus c

times

a plus c

expanding both products produces the

equality

2 a times b cosine theta minus b squared

equals a squared minus c squared

adding b squared plus c squared to both

sides results in two a times b cosine

theta plus c squared equals a squared

plus b squared

this is the law of cosines

[Music]

you

Law Of Cosines I (visual proof)

This is a short, animated visual proof of the Law of Cosines using Thales triangle theorem and the intersecting chord theorem. This theorem relates the side of a triangle with the other two sides and the angle between those two sides. #mathshorts #mathvideo #math #trigonometry #lawofcosines  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #thalestheorem #intersecting chords

This animation is based on a proof due to Sidney H. Kung from the December 1990 issue of Mathematics Magazine (https://doi.org/10.2307/2690911) page 342.

To learn more about animating with manim, check out:
https://manim.community
____________________________________________________
Background music:
The Temperature of the Air on the Bow of the Kaleetan by Chris Zabriskie is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/

Source: http://chriszabriskie.com/uvp/

Artist: http://chriszabriskie.com/


suppose we start with a regular pentagon

and let's assume the side length is one

the diagonal length that cuts across the

Pentagon here is well known to have a

length given by the Golden Ratio

fee another well-known fact about the

regular pentagon is that this diagonal

shown here along with the original

diagonal drawn creates an isoceles

triangle as pictured here which has

sides of length one and a base of length

1 divided by the golden ratio to see a

proof of this fact check the video

Linked In the description the purple

shaded triangle shown in this diagram is

known as the Golden Triangle the outer

containing triangle is often referred to

as the golden

nomon now we scale the Golden Triangle

by the value one over the Golden Ratio

this allows us to fit a new triangle

right next to the original with sides of

length 1 over the golden ratio and base

given by 1 over the golden ratio squar

this is actually just another Golden

Triangle as it's similar to the original

but now we can repeat this process

scaling the new triangle by one over the

golden ratio and rotating it and fitting

it in the empty space and continuing

again scaling by one over the golden

ratio and fitting the new triangle in

the empty space we can continue this

process

indefinitely each stage we take the

previously scaled copy of the Golden

Triangle and scale Again by the value 1

divid by the Golden Ratio rotate and fit

it in the empty space if we imagine that

we let this process go on forever or

infinitely often the limiting shape is

the entire golden Noman and that means

that we've filled in the entire golden

Noman with these scaled golden triangles

but then we can compute the length of

the long side of the golden Noman

triangle two different ways we know that

on the one hand it's given by the golden

ratio on the other hand it's given by

the sum of the the lengths of the purple

Golden Triangle sitting inside but each

of those purple triangles has a length

given by 1 over an even power of the

golden ratio this means that the golden

ratio is equal to the infinite sum of

the reciprocals of the even powers of

the golden ratio as shown here but we

can also get a related fact by looking

at the pink shaded Golden Triangle

sitting inside this golden nomon on the

one hand the side of the golden nomon

has a length one on the other hand it's

also given by the sum of the lengths of

the pink shaded triangles sitting inside

but each of the pink shaded triangles

has a length given by one over an odd

power of the golden ratio therefore one

is equal to the infinite sum of the

reciprocal of the odd powers of the

golden ratio notice that these two facts

are actually one and the same as we can

obtain the first fact from the second

fact simply by multiplying the entire

equality by the Golden Ratio Fe either

of these two infinite series could be

called the Golden series as they deal

with the infinite powers of the

reciprocal of the golden ratio can you

see what happens if you try to add up

all of the powers of the reciprocal of

the golden ratio

A Golden Ratio Infinite Series Dissection (visual proof)

This is animated visual proof demonstrating the sum of the infinite series of the reciprocals of even powers of the golden ratio is the golden ratio. We also get a related fact showing the sum of the reciprocals of the reciprocals of the odd powers of the golden ratio. This dissection proof utilizes the so-called "golden gnomon" and the "golden triangle".

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

To see the proof that the golden triangle has the dimensions stated, see this video: https://youtu.be/lL9VN8ELH1M

For more infinite series dissections, check out this playlist:
https://www.youtube.com/playlist?list=PLZh9gzIvXQUv6JqhlqQDlnqQw1NczW5Ro

Here is a longer form video with lots of infinite geometric sums: https://youtu.be/JteQEN1XPyc

This animation is based on a visual proof by Steven Edwards from the March 2014 issue of The College Mathematics Journal (https://doi.org/10.4169/college.math.j.45.2.120 - page 120).

#shorts #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricseries #series #infiniteseries #fractal #infinitesum  #goldenratio #pentagon #algebra #similartriangles #triangle #trianglesimilarity #similarity #goldentriangle #goldengnomon

To learn more about animating with manim, check out:
https://manim.community


foreign

[Music]

the sum of the first K positive integers

is called the kth Triangular number is

denoted T sub K and can be visually

represented by a triangular array of

dots where there's K dots in the last

row and one dot in the first row in this

video we'll investigate divisibility of

triangular numbers by three

because we're dividing by three there

are three cases to consider suppose

first that we're looking at a triangular

array with 3K plus one dots in the base

so that we're looking at the 3K plus

first triangular a number then we can

focus on a trapezoidal array at the

bottom that has 2K plus 1 Dots and a

height of length K we can then take

three copies of this trapezoidal array

rotate it and move it around the diagram

so that it almost fills up the entire

diagram there's only one dot left over

this means that the 3K plus first

triangular number must be equal to one

plus three copies of the number of dots

in the trapezoidal array but the

trapezoidal array dots can be

represented as the 2K plus first

triangular number minus the K plus first

triangular number as it's really the

difference of two triangular arrays so

in this case the number of dots is

congruent to one mod 3. for our next

case let's suppose that the number of

dots in the base of the Triangular array

is divisible by three so that it's three

times k for some integer k then we're

looking at the array representing the

Triangular number T sub 3K this time we

can form a trapezoidal array in the

bottom left with two K dots in the

bottom and the height of size k then we

can rotate and shift three copies of

this trapezoidal array to perfectly fill

in the entire diagram this time this

means that t sub 3K is equal to 3 times

the number of dots in the trapezoidal

array which is given by the difference

of the 2K triangular number and the kth

Triangular number so in this case we see

that t sub 3K is congruent to 0 mod 3.

for our final case we imagine that our

triangular array has 3K plus two dots in

the base so that we're looking at T sub

three K plus 2. once again we form a

trapezoidal array in the bottom left

with a base consisting of two k plus 1

Dots and a height consisting of K plus 1

dots once again we can use three rotated

copies of this trapezoidal array of dots

to fill in the entire triangular array

that we're interested in this means that

t sub 3 K plus 2 is equal to 3 times the

number of dots in the trapezoidal array

which is given by the difference of the

2K plus first triangular number and the

kth Triangular number as shown in this

case we see that t sub 3 K plus 2 is

congruent to 0 mod 3. using these three

cases we see that t sub n is congruent

to 1 or 0 mod 3 and it's congruent to 1

if and only if n is congruent to 1 mod

3. check out the linked video for the

analogous result for square numbers

Triangular Numbers Modulo 3 (visual proof)

We present an animation of a visual proof demonstrating how to visualize the the congruence classes of triangular numbers modulo 3.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

If you want to see a related video (about congruence classes of squares), check out : https://youtu.be/HiDe2Ngcq2c

This animation is based on a visual proof by Roger B. Nelsen from his first Proofs Without Words Compendium: https://bookstore.ams.org/view?ProductCode=CLRM/1

To learn more about animating with manim, check out:
https://manim.community

#mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #modulararithmetic #numbertheory
__________________________________________
Music in this video:
Twinkle in the Night by Aakash Gandhi


let's say we have a line comprised of

segments A and B the total length of

this line would be a plus b how would we

find a square of this let's visualize by

rotating the line by 90° this would form

a square made out of one red square one

blue square and two purple rectangles

the area of the red square is a * a or a

squ so we add a squ to the equation next

we have a purple rectangle of lengths A

and B its area would be AB same goes for

the other purple rectangle this

simplifies into positive 2ab lastly the

area of the blue squared is a b * B

simply b^ 2 therefore we have proven

that a + b 2 is equal to a 2 + 2 a + b s

subscribe to Mika Pine 91

[Music]

Visual proof of (a+b)^2

Thank you @MikaPie91 for helping out!

PS. first math yt video. please give feedback!
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#maths  #mathshorts #visualproof #algebra #matheducation #mathtutorial
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#youtubeshorts #mathproblems


foreign

[Music]

suppose we have a triangle with sides of

length a b and c and suppose that the

angle opposite of C is called Theta the

law of cosines provides a relationship

between these four values to derive the

law of cosines we drop a perpendicular

from the top of the triangle to the base

and that creates two sub right triangles

the hypotenuse of the yellow triangle is

B which means that the height of that

yellow triangle is B times the sine of

theta while the base of that triangle is

B times the cosine of theta this means

that the base of the blue right triangle

is a minus B times the cosine of theta

but now that we have a right triangle

pictured we can draw squares off the

legs of the right triangle and a square

off the hypotenuse of the right triangle

like this these squares have area B

times the sine of theta squared

quantity a minus B cosine Theta squared

and c squared by the Pythagorean theorem

we know that c squared equals the

quantity a minus B times cosine Theta

squared plus the quantity B times sine

of theta squared but now we can apply

algebraic rules along with the

Pythagorean trigonometric identity to

see that c squared equals a squared plus

b squared minus 2 times a times B times

cosine Theta this is the law of cosines

and this means that we've used the

Pythagorean theorem to prove the law of

cosines we can also plug theta equals pi

over 2 into the law of cosines to prove

the Pythagorean theorem and that means

that the two theorems are equivalent

you may be wondering if the diagram

works when the angle Theta is obtuse

instead here is such a triangular

diagram with sides of length a b and c

and an obtuse angle Theta across from C

in this instance we can drop an altitude

and then draw a horizontal line to build

this right triangle inside this blue

right triangle we see a yellow right

triangle with an angle Pi minus Theta so

its height is B times the sine of Pi

minus Theta while its base is B times

the cosine of pi minus Theta however we

know that b times the sine of Pi minus

Theta is actually equal to B times the

sine of theta and B times the cosine of

pi minus Theta is negative B times the

cosine of theta this means that the

pictured blue right triangle has a

height of B times sine of theta and a

base of length a minus B times a cosine

of theta so we can apply the Pythagorean

theorem again to get this exact same

identity that c squared equals the

quantity a minus B cosine Theta squared

plus b times sine Theta squared once

again algebraic manipulation will

provide the law of cosines in this case

Law Of Cosines II (visual proof)

This is a short, animated visual proof of the Law of Cosines using the Pythagorean theorem. This theorem relates the side of a triangle with the other two sides and the angle between those two sides.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

This animation is based on a proof from Timothy Sipka in the October 1988 issue of Mathematics Magazine (https://www.jstor.org/stable/2689363 ) page 259.

For another visual proof of this same fact check out:
https://youtu.be/vNCe7ciFwVc

#mathshorts #mathvideo #math #trigonometry #lawofcosines  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #pythagoreantheorem #obtuseangle #acuteangle #angle

To learn more about animating with manim, check out:
https://manim.community
_________________________________
Music in this video:
Elegy by Asher Fulero


here's an interesting Pi conundrum start

with a semicircle with radius one so

that it has a circumference of Pi now we

can draw two semicircles inside each has

a radius of a half so has a

circumference of pi/ 2 for a combined

length of Pi again now draw two

semicircles in each of those semicircles

so that each has a radius of 1/4 and a

circumference of pi over 4 for a

combined total length of Pi again we can

continue this process over and over

again each time we double the number of

semicircles but we have the radius this

means that the total combined lengths of

their circumferences will always remain

pi as we keep doing this we see that the

semicircles squash down and flattened to

almost match the diameter of the

original semicircle does that mean that

the length of the diameter of the

original semicircle is actually equal to

Pi when we consider the limiting process

is pi equal to 2 of course Pi isn't

equal to two but can you see where this

argument fails

Does Pi Equal 2?? (Spoiler: no)

In this short, we show a thought experiment that arises by doubling the number of semicircles drawn in a semicircle but halving their radii. The process produces an infinite collection of smaller and smaller chains of more and more semicircles that always have combined circumference length equal to pi. If we think about the limiting process, it seems like maybe this technique shows that pi = 2. But Pi can definitely be shown to be greater than 3 - so where does this argument fail?

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

Here is a related false proof that shows root 2 is equal to 2:
https://youtube.com/shorts/owFwvyozcn8

This false argument was suggested to me by Jeff Stuart as a nice alternative to the more classic Pi = 4 argument.

#math #manim #visualproof #proofwithoutwords #circle #circumference  #radius #pi #piday #shorts #infinite #falseproof

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let's see an intriguing diagram that

gives us five formulas for

pi start with a rectangle with

Dimensions 2 by3 inside the rectangle

draw this triangle as shown here the two

highlighted pink triangles are both 1X

two right triangles so they're congruent

because they fit together at

complimentary angles that means that the

blue shaded triangle is a right triangle

the hypotenuses on the pink triangles

are equal and that means that the blue

triangle has two sides of equal length

therefore the blue triangle is a right

isoceles triangle so the other two

angles are both Pi over4 the pi over4

angle on the left of this diagram can be

obtained as the sum of two angles from

these pictured triangles one of them has

an angle of arctangent 1/3 and the other

one has an angle of arc tangent

1/2 the upper right Pi over4 angle can

be obtained as the difference of two

angles one in this triangle which has an

angle measurement of arctangent 3 and

one in this triangle which has an angle

measurement of AR tangent of a half

another way to get pi over 4 in this

case is as the difference of two angles

in these triangles one with an angle

measurement of arct tangent of two and

the other one with an angle measurement

of the arct tangent of 1/3 the angle in

the upper right corner of the rectangle

is pi/ 2 and this can be obtained as the

sum of three angles where one of them is

the angle measurement of the blue

triangle which is arct tangent one the

next is the arct tangent of 1/2 in this

yellow triangle and the last is the arc

tangent of 1/3 in this yellow

triangle finally the angle on the left

side of the diagram pictured here is pi

and it's the sum of three angles in

three triangles the blue triangle gives

us an angle measurement of arc tangent 1

the bottom yellow triangle here gives us

an angle measurement of arct tangent 2

and the top yellow triangle here gives

us an angle measurement arct tangent

3 this amazing diagram has given us five

formulas for pi let me know in the

comments which of these five you like

the best and check out my Pi playlist

for other videos and animations related

to the circle

constant

Five Pi Formulas in "Pi/2 minutes" (Pi Day visual proofs)

In this short video, we give five different formulas for pi (or related constants) using only a single diagram consisting of a 2x3 rectangle and a triangle inside. Using the fact that the tangent of an angle is the ratio of the opposite and adjacent sides of an angle (and you can get the angle using the arctangent of the appropriate ratio), we decompose angles (of measure pi/4, pi/2, and pi) using certain triangles with visually computable arctangents using the diagram.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

This visual proof is due to Rex Wu and comes from the March 2003 issue of The College Mathematics Journal (https://www.jstor.org/stable/3595783 ) see pages 115 and 138.

For more Pi-related videos, check out my playlist:
https://youtube.com/playlist?list=PLZh9gzIvXQUsGRDvzvXc02lDIKHaVeFwq

#manim #irrational #Pi #mathvideo #math #mtbos  #animation #iteachmath #mathematics #piday #shorts #trigonometry #arctangent #tangent #triangle #rectangle #identities

To learn more about animating with manim, check out:
https://manim.community


in

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this them c i your energy

baby look at you crazy got to cut the

lights ons cut

[Music]

the

got the birds the I don't want to see I

want to see when the be drop theop got

to shake the Sheck

3²+4²=5² Visual Proof, Desmos 3D #Desmos3D #DesmosArt #MathArt #pythagorastheorem

This video was created by:

https://www.twitter.com/tiago_hands

For more mathematics content, check out:

https://www.threads.net/@mathematics.proofs

https://www.instagram.com/mathematics.proofs

Thanks for watching this video and make sure you have subscribed if you want to see more videos like this!


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Visual Proof of (a+b)^2 #math


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yeah

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he

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oh

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[Music]

Infinite Sum of Powers of -1/7 in a Square (visual proof without words)

This is a short, animated (wordless) visual proof demonstrating the sum of the infinite geometric series with ratio -1/7.

If you like this video, consider subscribing to the channel or consider buying me a coffee: https://www.buymeacoffee.com/VisualProofs. Thanks!

Here are two alternating series dissections for powers of -1/2: https://youtu.be/NKFrHfBGDNE
https://youtu.be/wLPsEULfPnk

And here is an alternating series dissection for powers of -1/3:
https://youtu.be/MyCn7gIPxNg

And here is an alternating series dissection for powers of -1/4:
https://youtu.be/SeKGZ2DLAwo

#math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricseries #infiniteseries #fractals #sum #induction #mathshorts #mathvideo

To learn more about animating with manim, check out:
https://manim.community
_______________________________
Music in this video:
Doing My Best by Alex-Productions | https://onsound.eu/
Music promoted by https://www.free-stock-music.com
Creative Commons / Attribution 3.0 Unported License (CC BY 3.0)
https://creativecommons.org/licenses/by/3.0/deed.en_US


[Music]

start with two real numbers theta less

than phi both less than pi over two

on the domain from 0 to pi over 2

graph the functions y equals the tangent

of x which is concave up and increasing

and the function y equals sine of x

which is concave down and increasing on

the domain

secant line connecting the origin to the

point theta sine of theta

has the equation y equals the sine of

theta over theta times x and this secant

line lies above sine of x for x larger

than theta similarly the secant line

connecting the origin to theta tangent

of theta has the equation y equals the

tangent of theta over theta times x and

this secant line lies below the tangent

function

if we plug phi into the sine function we

see that we have sine of phi

and this is less than what happens when

we plug phi into the secant line

which is the sine of theta over theta

times phi

likewise if we plug fee into the secant

line of the tangent function

we get a value of tangent theta over

theta times phi

and that must be less than what happens

when we plug phi into the tangent

function

which gives us tangent of phi

[Music]

we can divide the first inequality by

the positive value sine of theta

resulting in sine of phi over sine of

theta is less than v over theta and we

can divide the second inequality by

tangent of theta

resulting in phi over theta is less than

tangent of phi over tangent of theta

but then we can string these two

inequalities together

and we see that sine of phi over sine of

theta is less than phi over theta which

is less than tangent to phi over tangent

of theta

this is known as aristarchus's

inequalities

[Music]

Aristarchus's Inequality (visual proof)

This is a short, animated visual proofs of the Aristarchus's inequality using the concavity of the sine and tangent functions. #mathshorts #mathvideo #math #aristarchusinequality #aristarchus #inequality #logarithm #tangentfunction #sinefunction #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof

This animation is based on a proof due to Roger B. Nelsen from the February, 1993 issue of Mathematics Magazine page 65 (https://doi.org/10.2307/2690480).

To learn more about animating with manim, check out:
https://manim.community
________________________________________________
Music in this video:
The Voyage by Audionautix is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/

Artist: http://audionautix.com/


πr^2 is de formule voor de oppervlakte van een cirkel

Maar vanwaar komt deze formule eigenlijk?

Ten eerste

Tekenen we een cirkel, en kleuren we de oppervlakte groen.

Daarna verdelen we de cirkel in grote, gelijke delen

En we schuiven we ze in elkaar

Zoals je kan zien benadert dit nog niet echt een rechthoek

Dus daarom gaan we de cirkel in kleinere deeltjes verdelen

En op dezelfde manier schuiven we deze in elkaar

Dit lijkt er al meer op!

Als we de cirkel steeds in kleinere deeltjes blijven verdelen

benaderen de in elkaar geschoven deeltjes steeds beter een rechthoek

Maar hoe klein moeten die deeltjes nu precies zijn

Om een perfecte rechthoek te vormen?

Wel, we blijven de cirkel gewoon verdelen in steeds kleinere en kleinere stukjes

Tot het kleinste stukje dat je kan maken

Eigenlijk zouden we oneindig moeten blijven verdelen

Totdat we de lijntjes niet meer van elkaar kunnen onderscheiden

En pas dan

bekomen we een perfecte rechthoek

Dus de oppervlakte van de cirkel is gelijk aan de oppervlakte van de rechthoek en deze is gelijk aan

basis maal hoogte

De hoogte van de rechthoek

is zoals je kunt zien even lang als de straal van de cirkel

Dus de lengte van de hoogte is gelijk aan de straal

Om nu de basis te vinden

Moeten we eens kijken naar de omtrek van de cirkel

Wanneer we de basis hebben vergeleken met de omtrek van de cirkel, zien we dat de basis

gelijk is aan

de helft van de omtrek van de cirkel

Herinner je dat de omtrek van een cirkel gelijk is aan 2πr

En wanneer we dit combineren met één tweede

dan kunnen we de tweetjes schrappen

en de basis is dus gelijk aan πr

Dus de formule basis maal hoogte wordt dan πr  maal r

Als we de r's bij elkaar zetten, krijgen we

Wat gelijk is aan de oppervlakte van de rechthoek

en dit is dan weer gelijk aan de oppervlakte van de cirkel

Dus de formule voor de oppervlakte van elke mogelijke cirkel is πr^2 of πrr

Area of a circle, formula explained

https://mathematicsonline.etsy.com

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πr(squared) gives you the area of a circle, but where does it come from? Here is a simple explanation. The area of a circle is the space occupied by the circle in a two-dimensional plane. The space occupied within the boundary/circumference of a circle is called the area of the circle. The formula for the area of a circle is A = πr2, where r is the radius of the circle.