[Music]

suppose that a b c and d are all real

numbers then we can build a right

triangle whose leg lengths are given by

the absolute value of B times the

absolute value of c and the absolute

value of a times the absolute value of C

if we assume that c is non-zero then we

can scale this triangle by the value

absolute value of D over the absolute

value of C producing this triangle which

has leg lengths absolute value of B

times absolute value of D and absolute

value of a times absolute value of D

because this new triangle is a scaled

and rotated copy of the original right

triangle when the two triangles are

placed side by side like this they

produce a right angle so that this

picture triangle is a right triangle as

well we can use the Pythagorean theorem

on the original two right triangles to

see that the leg lengths of this new

right triangle are the absolute value of

D times the square root of a squared

plus b squared and the absolute value of

c times the square root of a squared

plus b squared because this new triangle

is also right angled we can use the

Pythagorean theorem to find the length

of the high hotness to be the square

root of a squared plus b squared times

the square root of c squared plus D

Squared but now we know that the

absolute value of AC plus b d is less

than or equal to the absolute value of a

times C plus the absolute value of B

times D from the triangle inequality on

the other hand from this trapezoidal

shape we see that the base length which

is the absolute value of a times C plus

the absolute value of B times D must be

less than or equal to the length of the

hypotenuse of that top triangle which is

the square root of a squared plus b

squared times the square root of c

squared plus D Squared therefore the

absolute value of AC plus b d is less

than or equal to the square root of a

squared plus b squared times the square

root of c squared plus D Squared where a

b c and d are real numbers this is often

referred to as the two variable Koshi

bonyakowski Schwartz inequality

foreign

Cauchy-Bunyakovsky-Schwarz Inequality II (visual proof)

This is a short, animated visual proof of the two-dimensional Cauchy-Schwarz inequality (sometimes called Cauchy–Bunyakovsky–Schwarz inequality) using "Garfield's trapezoid" and scaling of right triangles.

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

To see an alternate proof of this fact, check out
https://youtu.be/mKg_gVagHy8 (wordless)
https://youtube.com/shorts/Xy97t7l6MYg (shorts version)

This animation is based on a visual proof by Claudi Alsina and Roger Nelsen from the April 2015 issue of Mathematics Magazine (https://www.jstor.org/stable/10.4169/math.mag.88.2.144 - page 144-145).

#math #inequality  #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #areas #mathematics #cauchyschwarz #algebraicidentity #mathshorts #mathvideo #mtbos

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
When We Found The Horizon by Late Night Feeler


let's see four iconic proofs without

words in under 60 seconds by using two

triangular arrays of unit squares to

form an N by n +1 rectangle we see that

the sum of the first n positive integers

is equal to n * n + 1 / 2 next by using

four congruent right triangles to

partially fill a square diagram and then

measuring empty space in two different

ways by moving Triangles around we see

that c^2 = A2 + B2 where A and B are the

legs of a right triangle with hypotenuse

C third by rotating the bottom half of

odd Stacks around their central element

we obtain a square showing that the sum

of the first n positive odd integers is

n s and finally by taking a 1x one unit

square and repeatedly shading half of

the unshaded remaining area we get a

formula for the infinite sum of the

powers of 1/2 in particular this

infinite sum is one

Top four visual proofs?

In this short, we show animations of four of the most famous proofs without words: the formula for the sum of the first n integers; the pythagorean theorem using negative space/sliding rectangles/Chou pei suan ching; the formula for the sum of the first n odd integers; and the infinite geometric series of positive powers of 1/2 (using a rectangle/square dissection of a unit area square). We include brief justifications for these visual proofs.

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

For a slightly different version of this short without words, see:
https://youtube.com/shorts/eG72HxNceEs

For longer videos with related animations, see
Summing Integers: https://youtu.be/eHbtc50-qXo
Pythagorean Theorem:  https://youtu.be/o2mbN452ywA (and old version: https://youtu.be/yfGtbNgcrQ8)
Summing Odds: https://youtu.be/I4A2T55k0cg
Geometric Series: https://youtu.be/JteQEN1XPyc

The first proof was known to the ancient greeks (cited by Martin Gardner), the second proof is adapted from the Chou pei suan ching (around 200 BCE according to Roger Nelsen), the third is attributed to Nicomachus of Gerasa by Roger Nelsen, and the final one is attributed to Warren Page (from  the September 1981 issue of Mathematics Magazine, page 201 - https://www.jstor.org/stable/2689632 ). The first three can all be found in Roger Nelsen's first compendium, "Proofs Without Words: Exercises in Visual Thinking: https://bookstore.ams.org/view?ProductCode=CLRM/1 .

#math #manim #pythagoreantheorem #pythagorean #triangle #animation #theorem #pww #proofwithoutwords #visualproof #proof  #mathshorts #mathvideo #geometricseries #infiniteseries #finitesums #infinitesums #oddsumformula #sumsofintegers #sums #shorts

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


we can make sense of the powers of three

by starting with one triangle taking

three copies to form this triangle with

three small triangles taking three

copies of that to represent three

squared and so on each time taking three

copies of the previous diagram to

produce the next power of three if we do

that here's a representation for three

to the N plus one but now let's see

something interesting we can take two

copies of three to the n and stack them

next to each other on top of that take

two copies of three to the N minus one

like this and keep going until we have

two copies of three squared on top of

that we place two copies of three and

finally we place two copies of one if we

then stack one last copy of one on top

it turns out that we produce the exact

same diagram as the diagram for three to

the N plus one so we get equality

between these two formulas from this

equality we see that the sum where I

ranges from zero to n of three to the I

must be three to the N plus one minus

one over two this is a formula for the

geometric sum of the powers of three

Summing Powers of Three

This is a short, animated visual proof demonstrating the sums of finite geometric series of powers of 3.

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 longer, wide format version: Geometric series: sums of powers of 3 (visual proof)
https://youtu.be/9IAm75UY2U8

This animation is based on a visual proof by David B. Sher from the Spring 1997 issue of Mathematics and Computer Education page 190.

#mathshorts #mathvideo #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricsums #series #fractals #finitesums #induction

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


have you ever wondered that how can you

replace sin x by simply X for smaller

values of X well those zoman expansion

might know that you can just ignore the

higher powers of X but here is a visual

proof for it if you plot the function

sinx and the function y = x then you can

see that in this region both of these

curves almost overlap each other that

means that no matter what input you give

in this region the output of both of

these functions is almost equal to each

other and you can do the same thing for

more functions like tangent X sin

inverse X tangent inverse X and even sin

hyperbolic X and tangent hyperbolic X

behaves the same if you're interested in

such science and math video subscribe

for more

Visual proof Of Sin(x) ≈ x

Visualization of sin(x) = x for smaller angles proof using Maclaurin series expansion and graphical method to visualise it.


start with this 2x2 matrix a and

consider the apine linear

Transformations ax and ax + 1 Z these

two Transformations are shown here

applied to the plane now start with any

point in the plane and apply both

Transformations then choose one point at

random and apply both Transformations

again repeat this process over and over

selecting one of the outputs at random

and applying both apine linear

Transformations let's repeat this

process 25,000 times to see what happens

each point is colorcoded yellow or blue

depending on the apine linear

transformation that outputs that point

as we iterate this procedure the points

stay within this enclosed region this

fractal shape is known as the Twin

Dragon and it's the limiting shape or

the attractor of this iterated function

system produced by these two apine

linear transformations

Twin Dragon from Chaos Game

In this video, we show how to use a random process of iteratively applying two (affine) linear transformations in the real plane to generate a 2-rep-tile known as the Twin Dragon as the attractor of the iterated function system. What happens if you do something similar with different affine linear transformations?

Here is a bit longer version (without words): https://youtu.be/fFIyqMiycs4

Check out these videos for related constructions:
https://youtube.com/shorts/dklWNdM9WSg
https://youtu.be/rLxx6b-oJGo
https://youtu.be/NBI_5GdvqUo

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

This animation was inspired by an article written by Lorelei Koss that appeared in the Proceedings of Bridges 2015: https://archive.bridgesmathart.org/2015/bridges2015-423.pdf . For more information, you should also check out the related article "Fractal Tilings in the Plane" by Richard Darst, Judith Palagallo and Thomas Price from the February 1998 Mathematics Magazine: https://www.jstor.org/stable/2691339

#chaos #chaosgame #twindragon #twindragon #mathvideo #math #mtbos #manim #animation #theorem #iteachmath #mathematics #dynamicalsystems  #iteratedfunctionsystem #dynamics #fractals

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


many people draw this diagram when they

want to represent a four set Venn

diagram unfortunately this is not valid

can you figure out what's missing from

this diagram

but we can actually draw a valid four

set Venn diagram here's one possible way

to draw such a diagram we can check that

it's valid by finding all the regions

here's the regions for elements only in

a the ones only in B the ones only in C

and the ones only in D next are the

double intersection regions here's the

region for elements only in A and B this

one for those only in a and C this one

for those in A and D this one for those

in B and C only and in B and D only and

this one for those in C and D only next

we have the triple intersection regions

a b and c only a b and d only a c and d

only and b c and d only here's the

region for elements in all four sets and

here's the region for elements in none

of the four sets we've pictured all 16

possible regions so this is a valid Venn

diagram

Four Set Venn Diagram?

In this video, we show a typical non-example of a proposed four-set Venn diagram. We then show an example of a four-set Venn Diagram by highlighting all sixteen possible regions.

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 longer, wordless video, showing the same thing along with a five set Venn diagram as well : https://youtu.be/WBot4kqc3ro

The image in this video was inspired by the wonderful article "Venn Diagrams and Independent Families of Sets" by Branko Grünbaum  (https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/BrankoGrunbaum.pdf)

#logic #settheory #union #propositions  #intersection #setdifference  #setminus #setconnectives #subsets #venndiagram #visualproof #foursets #math #manim #discretemathematics #fivesets #complement

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


here's a visualization of the

intersecting chords theorem

stick around and we'll see a formal

statement of the theorem along with a

visual proof

start with a circle

and draw two chords

inside assume that these two cords

intersect at a point

that then cuts each cord into two pieces

use the two pieces from each cord

as the side lengths of a rectangle

as pictured here

now let's compute the area of these

rectangles

it turns out that no matter how you draw

the chords as long as they intersect the

areas of these two rectangles will be

exactly the same as pictured here

[Music]

the intersecting chords theorem can be

stated as follows

suppose two chords in a circle intersect

at a point dividing one chord into

lengths of a and b and the other chord

into lengths of c and d

then the product a times b equals the

product c times d

let's check out a proof

first construct two triangles by

connecting the chords like this

now there are two angles one in each

triangle subtended by this arc

there's this angle

and this angle

because these angles are subtended by

the same arc they must be congruent

angles

likewise there's an angle in each

triangle subtended by this arc

this one here

and this one in the other triangle

once again because these two angles are

subtended by the same arc they're

congruent

[Music]

but this means that the two picture

triangles that we created are similar

now we can scale the two picture

triangles the top one we can scale by b

and the bottom one we can scale by c and

connect them like this this construction

creates a parallelogram and therefore

the opposite side lengths a times b and

c times d must be equal just as we need

for the intersecting chords theorem

[Music]

you

Intersecting Chords Theorem (visual proof)

This is a short, animated visual proof demonstrating and proving the intersecting chords theorem from geometry. #manim #math #mathshorts #mathvideo  #intersectingchords #geometry #chords #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #inequality

This animation is based on a visual proof from Hubert Shutrick that appears on the beautiful math website Cut-the-knot (https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/HubertIntersectingChords.shtml)

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


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thank you

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thank you

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foreign

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Geometric series: sums of powers of 8 (visual proof)

This is a short, animated (wordless) visual proof demonstrating the sums of finite geometric series of powers of 8 using cubes. #math #manim #mathvideo #calculus #mtbos #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricsums #series #squares #finitesums #induction #discretemath

This animation is based on a visual proof by Rajib Mukherjee from the December 2021 issue of The Mathematical Intelligencer (https://doi.org/10.1007/s00283-021-10087-5) pages 60-61.

Here are two a related proofs for powers of 3, 4 and 9:
https://youtu.be/9IAm75UY2U8
https://youtu.be/yTpzDEDP090
https://youtu.be/Ch7GFdsc9pQ

____________________________________
Music in the video:
Mountain Trail 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


let X be a positive real number and

first suppose that X is greater than or

equal to 1 start with a right triangle

with legs of length x and 1 take a

rotated copy and then scale that copy by

the value 1 /x creating a right triangle

with side lengths of 1 and 1 /x the

purple triangle that's pictured has an

area of 1/2 * X while the smaller red

triangle has an area of 1/ 12 *

1/x but now notice that the area of

these two triangles encloses the area of

the 1X one square as shown here and

therefore 1 12 * x + 1 12 * 1/x is

greater than or equal to 1 and

multiplying both sides by 2 we get that

x + 1X is greater than or equal to 2 by

interchanging X and 1 /x we see that

this inequality holds for all positive

real numbers X can you see from the

diagram when equality holds

Sum of positive number and its reciprocal from two triangles

This is a short, animated visual proof demonstrating that sum of a positive real number and its reciprocal is always greater than or equal to 2.

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

For a wider format and slightly longer version of this proof, see https://youtu.be/BkQvnriVblY?si=cldTUiQAtPuZLvto

It turns out that this theorem is equivalent to the Arithmetic Mean-Geometric Mean inequality. The equivalence is implied by the fact that his video resembles a similar visual proof for the AM-GM inequality found here: https://youtu.be/62G9fak1vyk

This animation is based on a proof by Roger B. Nelsen from the December 1994 issue of Mathematics Magazine, page 374 (https://doi.org/10.2307/2690999).

#math #manim #visualproof #mathvideo  #geometry #mathshorts  #algebra #mtbos  #animation #theorem #pww #proofwithoutwords #proof #iteachmath #inequality #amgm #maths #mathematics #education

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


let's play the chaos game inside of an

equilateral triangle start with a dot

anywhere inside the triangle now

iteratively add a point one half of the

distance between the current dot's

location and a uniformly randomly chosen

vertex

here are the first few iterations of the

chaos game

but don't just do this a few times let's

do this 30 000 times and see exactly

what happens

[Music]

the limiting shape is the famous fractal

known as the sierpinski triangle

check out the linked video in the

description to find other constructions

of this awesome triangle

Chaos game - Sierpinski triangle

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 play the chaos game using randomness to find order and create the Sierpinski triangle. Can you explain why this works?

#math #manim #fractal #fractal #sierpinski #chaosgame #chaos #mathshort #visualmath

_____________________
Music in this short (a 10 second cut):
When The Lights Go On 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 a right Isles triangle of

area one we can cut this triangle into

eight congruent right isoceles triangles

if we fill in exactly seven of those

triangles like this then we've shaded an

area of 7/8 now we can take that diagram

and scale it and put it in the last 1/8

therefore shading an extra 7/8 of 1/8 or

7 over 8^ squar if we do this again we

shade 7 over 8 cubed and then 7 over 8

4th and so on as we repeat this process

we keep shading 7/ 8 the N of the

triangle as n increases if we do this

indefinitely meaning we consider a

limiting process we eventually shade the

entire triangle therefore the Shaded

area is also one one interpretation of

this equality is that 7 repeating in

base 8 is equal to 1 another

interpretation is that 7 * the infinite

sum of the powers of 1/8 is 1 so that

the infinite sum of the powers of 1/8 is

0.7777… = 1 (in base 8)

This is a short, animated visual proof showing the sum of the infinite geometric series with first term 7/8 and ratio 1/8, which in turn allows us to compute the sum of the series of powers of 1/8 and determine an interesting base 8 representation of 1.

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

For a wide-format wordless version of this geometric series dissection, check out

https://youtu.be/wPgwgoj_aKA?si=W12LT6bUYEhgaFNv

This animation is based on a visual proof from Roger B. Nelsen in the February 2006 issue of Mathematics Magazine (https://doi.org/10.2307/27642904 or https://www.jstor.org/stable/27642904), page 60.

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

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


created ternary branching tree by

starting with a line segment of unit

length scaling the line segment by a

half and rotating by the multiples of 2

pi over three repeat this scaling and

rotating process for each new set of

branches

if we continue this process for as long

as we can and eventually remove all but

the last level of the leaves you should

see an interesting shape appear

to see other amazing ways to construct

the sierpinski triangle check the video

that's Linked In the description

Sierpinski from ternary branching

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 create the Sierpinski triangle using a ternary branching process. Can you explain why this works?

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


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so

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you

The Beauty of Math Constants Visualized

A geometric visualization of 9 major mathematical constants showing the elegance of their derivation from first principles.

Submission for 3Blue1Brown's Summer of Math Exposition, #SoME1. Stay tuned for my submission this year for #SoME4, it'll be on Algebraic Numbers!

Music: "Zero" - Sebastian Kamae x Aylior


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thank you

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foreign foreign

Pythagorean Theorem | Visual Proof | Geometric Proof | Proof Without WORDs

Unveil the elegance of geometry in our illuminating YouTube Shorts video, where we present a mesmerizing visual proof of the Pythagorean theorem: c² = a² + b². Through clever animations and intuitive illustrations, we unravel the mystery behind this timeless theorem, showcasing how the relationship between the sides of a right triangle can be proven with sheer geometric insight. Prepare to be captivated as right angles, squares, and triangles harmoniously come together to unveil the profound connection between areas and side lengths. Whether you're a math aficionado or simply curious about the secrets embedded within shapes, this concise video will leave you with a newfound appreciation for the artistry and logic embedded in the realm of mathematics.

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

Link to playlist "Scientific WorkPlace Crash Course":
https://www.youtube.com/playlist?list=PLtxDg_npmNDCrZ4Q6wVPz4gk6c7LlYM6v

#shorts #mathtricks #math #mathematics #mathideas #mathanimation #animation #animated #visualization #geometric #geometricpattern  #funfacts #mathfacts #abstractart #abstractmath #mathvisualization #mathart #geometry #shapes #amazingahorts #amazingfacts #factshorts #intution #intutive #funmath #mathideas #mathworld #magicmath #magicmathtricks #mathmagic #mathzone #spirograph #spirography #spirograhart #spirographdesign #pattren #art #creative #creativity #creativeart  #geometric #proof #proofwithoutword #algebric #algebra #algebricequation #algebricexpression #visualproof #geometricproof #amazingmath  #equation #expression #square #proofing #proofof #proofofconcept #amazingahorts #amazingfacts #factshorts #intution #intutive #funmath #mathideas #mathworld #magicmath #magicmathtricks #mathmagic #mathzone #pythagoras_theorem #pythagoreantheorem #theorem


सवेर का मेरे तू सूरज लागे तू मेरा कोई ना

होके भी कुछ लागे तू मेरा कोई ना होके भी

कुछ लागे तू मेरा कोई ना

visual proof of sum of first n positive integers😱📝#viral #trending #dsssb #math #shorts


suppose we have a function f and we draw

the graph y as a function of X like this

now imagine we have an input x equals a

and we want to know what the instant

rate of change of the function at x

equals a is how can we figure this out

one way is to take a separate Point

labeled X and draw the secant line

connecting the two points the slope of

this secant line is the average rate of

change of f between X and A and it's

given by the slope formula f of x minus

F of a over x minus a now here's the

trick we can compute this average rate

of change for any value X except for x

equals a because that would require us

to divide by zero so what we do is we

actually take a limiting value of the

slopes of the secant lines as X

approaches a and this gives us the

instant rate of change of F at x equals

a so we can Define f Prime of a to be

this limit as X approaches a of f of x

minus F of a over x minus a provided the

limit exists we call this the derivative

of F at a and it measures the slope of

the tangent line or the instant rate of

change of F at a

Visual Derivative Definition!

This video animates the idea behind the derivative of a function. We show how to think about the definition of the derivative of a function visually by using a limiting process of slopes of secant lines to obtain the slope of the tangent line (which measures the instant rate of change of a function at a point).

#manim #calculus #derivatives #derivative #tangentline #slope #parabola #mathvideo #mathshorts #math #visualmath #graph #linearapproximation #secantline #instantrateofchange #averagerateofchange


[موسيقى]

Olympiad Right Triangle Inequality (Canada 1969 - visual proof)

This is a short, animated proof without words demonstrating an inequality about the side lengths of a right triangle that was Problem 3 in the 1969 Canadian Mathematical Olympiad. #manim #math #mathvideo #righttriangle  #inequality #mathshorts #geometry  #animation #theorem #pww #proofwithoutwords #visualproof #proof #algebra  #proof #mathematics #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 from Roger B. Nelsen's second proofs without words compendium (page 12).

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Sprite Star from Saidbysed


let's plot the function f of x equals

sine of X on this axis below

we notice that if we plot a point on

this curve we can draw the linear

approximation or the tangent line to the

curve and we can compute the slope of

this tangent line

we can then move the point along the

curve and we note that the slope of the

corresponding tangent line changes as we

move the point we can see that the slope

of the tangent line ranges between

negative one and one in this picture

so let's take this one step further and

let's plot the slopes of the tangent

lines on a separate axis system as we

move the point along the curve f of x

equals sine X we map out the slopes of

the tangent lines on the second axis

below

it turns out and you can see it visually

that the slope function of f of x equals

sine X is the function f Prime of x

equals cosine X to prove this requires

more work but this is a nice visual

representation of the fact that the

derivative of sine is cosine

Visualizing the derivative of sin(x)

A visual of the derivative of f(x)=sin(x). We show how to think about the derivative of a function visually.

#manim #calculus #derivatives #derivative #tangentline #slope #parabola #mathvideo #mathshorts #math #visualmath #graph #sinefunction #trigonometry #linearapproximation


play the chaos game in a regular hexagon

we apply an iterative algorithm with a

random component start with a DOT

anywhere in the interior of the polygon

and then iteratively add a new Point

two-thirds of the distance between the

current dot and a uniformly randomly

chosen vertex on the hexagon you can

think of rolling a fair six-sided die

and then moving two-thirds of the way

from the dot you're currently at to the

vertex corresponding to the die roll

here's the amazing shape that arises

after applying this process 35

000 times no matter the original Point

location or the random outcomes of the

dice rolls along the way the limiting

shape that you see here is always the

limiting shape that results from this

process

Chaos Game in Hexagon (bestagon?)

In this short, we show what happens when iterating the procedure of choosing a hexagon vertex at random and moving wo thirds the distance from the current dot to the chosen vertex.

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

#chaos #chaosgame #hexagon #mathvideo #math #mtbos #manim #animation #theorem #iteachmath #mathematics #dynamicalsystems  #iteratedfunctionsystem #dynamics #fractals

Here's a longer version of this video: https://youtu.be/NBI_5GdvqUo

If you want to know more about the Chaos game, see the following links:
https://en.wikipedia.org/wiki/Chaos_game
https://mathworld.wolfram.com/ChaosGa...
https://math.bu.edu/DYSYS/chaos-game/...
https://youtu.be/kbKtFN71Lfs
https://youtu.be/IGlGvSXkRGI

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


foreign

(a+b)² through Visual Proof. #visualproof #maths #realnumbers #iba #jobs #shorttrick #tricks #jobsea


foreign

ER can be represented by a square array

of squares with a side length of 2K Plus

and the nth triangular number can be

represented by a triangular array of

dots where the bottom row has n dots

when we multiply these two numbers

together the product can be represented

by a triangular array of square arrays

as pictured here the number of small

squares in the base is 2K plus 1 times n

but now each Square array that's at the

end of one of the rows has a triangular

array with base length K in the upper

right corner which we can swing around

and drop down into place like this

resulting in an array with a base of 2K

plus 1 times n plus k

if we look carefully this array is

almost triangular but it's just missing

a base K triangular array from the top

therefore we see that 2K plus 1 squared

multiplied by the nth triangular number

T sub n

is given by the 2K plus 1 times n plus K

triangular number minus the K triangular

number T sub K therefore the product of

[Music]

Products of Odd Squares and Triangular Numbers (visual proof - updated version)

This is a short, animated visual proof demonstrating that the product of an odd square and a triangular number is the difference of two triangular numbers. #mathshorts #mathvideo #math #squares #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #triangularnumbers

This animation is based on a visual proof by Brian Hopkins from the February 2018 issue of Mathematics Magazine (https://doi.org/10.1080/0025570X.2017.1409534).

To learn more about animating with manim, check out:
https://manim.community
__________________________________________________________________
Music in this video:
Reflections by MK2


in an equilateral triangle with area one

first start by tricing the three sides

of the triangle then bict the three

sides of the triangle as well connect

vertices to opposite Tri sectors like

this and connect the remaining Tri

sectors to the midpoints each of these

six shaded regions makes up exactly 17th

of the triangle as you can see by

rotating here therefore we've shaded 67

of the triangle the center triangle is

therefore 17th of the area of the

original we can repeat this process on

the inner 17th shading 67 of 17th so

shading 6 of 7^ squar repeat this

process over and over again each time

shading 67 of the remaining 1 over 7 to

the N if we consider the infinite sum

which corresponds to the limiting

process eventually the entire triangle

will be shaded and that means that the

infinite sum we have is equal to 1 one

interpretation is that 6 repeating in

base 7 is equal to 1 another one is that

6 * the infinite sum of the powers of

17th is 1 so that the infinite sum of

the powers of 17th is 1 16

0.66666… = 1 (in base 7)

This is a short, animated visual proof showing the sum of the infinite geometric series with first term 6/7 and ratio 1/7, which in turn allows us to compute the sum of the series of powers of 1/7 and determine an interesting base 7 representation of 1.

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, wordless version of this animation with two other proofs, see https://youtu.be/L-pxcCk3Xi4?si=hCnLn61vVxSOp2b4

Also, check out my playlist on geometric sums/series: https://youtube.com/playlist?list=PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw

This animation is based on a proof by Stephan Berendonk (2020) from the November 2020 issue of The College Mathematics Journal, (https://doi.org/10.1080/07468342.2020.1830649 p. 385)

#mathshorts #mathvideo #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricsums #series #infinitesums #infiniteseries #geometric #geometricseries #equilateraltriangle

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


[Music]

consider a semi-circle and then along

the diameter of the semicircle draw two

smaller tangent semicircles the Shaded

region shown here lying inside the

larger semicircle and outside the

smaller semicircle is called an arbolos

the term arbolos literally means

shoemaker's knife by varying the sizes

of the inner two semicircles we get

various different arbolo shapes all

lying inside the same outer semicircle

here's an amazing theorem about the area

of the arbolos consider the semi chord

from the tangent point of the circles to

the larger semicircle and consider this

circle that has that semicord as its

diameter as we vary the sizes of the

inner semicircles we see that this new

circle's area changes just like the

arbolos area changes the amazing

principle here is that the area of the

arbolos is actually equal to the area of

that Circle that has the semi chord as

its diameter let's see a visual proof

justifying this fact

first let's complete the semicircle so

that it's now a full circle and then

rotate the semi chord down to this

bottom semicircle we can use this

semi-cord to create this triangle which

happens to be a right angled triangle by

thales triangle theorem next we can

decompose this right triangle into these

two smaller right triangles and split

the diagram into three separate pieces

one for each of the right triangles

for each one of these triangles we'll

draw semicircles off the three sides of

the triangle like this

according to the Pythagorean theorem the

sum of the areas of the semicircles off

the legs of the right triangle must be

equal to the area of the semicircle off

the hypotenuse of the right triangle so

in each of these diagrams the blue area

is equal to the pink area but now we can

glue these three diagrams back together

one at a time remembering that pink

areas in blue areas counteract one

another when we do that we see that the

only area surviving is the pink area of

the arbolos and the Blue Area

corresponding to the circle whose

diameter is the semicord but remember

that the sum of the blue areas was equal

to the sum of the pink areas and

therefore the area of the arbolos is in

fact equal to the area of the pictured

Circle no matter how we drew this

diagram thales triangle theorem would

hold for the semi chord in question and

so the area of the arbolos is always

equal to the area of the related Circle

Arbelos Area (visual proof)

This is a short, animated visual proof proving that the area of an arbelos is equal to the area of the circle with diameter given by the half circle chord obtained from the arbelos construction (from the tangent point of the two interior circles).

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

For other similar visual proofs check these out:

https://youtu.be/p3_7bOjiWm8
https://youtu.be/-IKyUjRcFDw

This animation is based on a visual proof from Roger B. Nelsen that appeared in the April 2002 issue of Mathematics Magazine (https://www.jstor.org/stable/3219152  p. 144)

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

#math #arbelos #geometry #pythagoreantheorem  #thalestheorem  #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #area #circle #geometricconstruction
_______________________________________
Music in this episode:
Descent by Houses of Heaven


[音楽]

[音楽]

Sum of first N odd natural numbers  - Visual Proof #maths #visualproof #funlearning

Visual proof of showing that the sum of first n odd natural number is n^2.

#maths #visualization #sum #series #learning #educational #funlearning


one way to represent the sum of the

first n positive integers is with a

triangular array of squares where the

first row has one square and the bottom

row has n squares then we can cut this

triangular array with this diagonal

creating one large triangle and N small

one by one triangles therefore the sum

of the first n positive integers is

given by the area of all these triangles

together the large triangle has an area

of one half N squared and each of the

small triangles has an area of one-half

giving a total combined area of one-half

n therefore the sum of the first n

positive integers is one half N squared

plus one half n

Summing positive integers with triangles

This is a clip from a longer video showing a dozen ways to sum the first n integers. This proof uses triangular areas to get the famous sum formula. Full video is here : https://youtu.be/eHbtc50-qXo

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 the following source:

Ian Richards, Proof without Words: sum of integers, Mathematics Magazine (March 1984 page 104): https://www.maa.org/sites/default/files/Richards31975.pdf.

#manim #math #proof #visualproof #mathshorts #proofwithoutwords #sums #integers #sumformula

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


we can represent three finite sets a b

and c by the Venn diagram shown here now

suppose we want to find the cardinality

or the number of elements in the union a

union B Union C to do this we start by

taking the number of elements in a and

adding that to the number of elements in

B and adding that to the number of

elements in C unfortunately this is too

much because we've over counted the

elements in A and B so we remove the

cardinality of a intersect B

likewise we need to remove the

cardinality of a intersect C since we've

counted those elements twice and we need

to subtract the cardinality of B

intersect C since we've counted those

elements twice but now notice that the

elements that are in the intersection of

all three sets have been added three

times and subtracted three times so we

need to re-add the cardinality of a

intersect B intersect C this formula is

known as the three set inclusion

exclusion formula

Visual Inclusion/Exclusion

In this video, we show how to begin to deal with the general Sum principle for Sets when you have more than two sets involved. Can you see how to keep extending this idea for more and more sets?

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

To see a longer (slower) version of this video, without words, check this out: https://youtu.be/vVZwe3TCJT8

If you want to see an alternate video, check this: https://youtu.be/wXdoqPMlVRU

To see more details about this, check this longer video: https://youtu.be/sbIcWGYNA5I

#logic #settheory #inclusion/exclusion #intersection #setdifference  #setminus #setconnectives #subsets #venndiagram #visualproof #math #manim #discretemathematics #sumprinciple

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


[Music]

let's see a geometric way to find this

infinite sum

start with a positive real number X

that's less than one then draw a square

with side length given by the infinite

sum of the powers of X in particular the

side length can be decomposed into a

bunch of line segments of length 1 x x

squared x cubed x to the fourth and so

on

applying the sum formula for an infinite

convergent geometric series we see that

the side length of this Square must be

one over one minus X

now we can use the labeled side lengths

to decompose the square into a bunch of

rectangles first in the lower left

corner we have a one by one square with

area one next to that we have two x by 1

rectangles both containing area X so

together they enclose an area of 2x

on top of those two rectangles we can

shade in a 1 by x squared rectangle and

X by X rectangle and an x squared by one

rectangle

together these three rectangles enclose

an area of 3x squared on top of those

rectangles we can place four rectangles

each with area X cubed and closing an

area of 4X Cubed on top of those

rectangles we have five rectangles each

enclosing an area of x to the fourth for

a total area of five x to the fourth and

then we have these six rectangles

enclosing a total area of 6X to the

fifth

we can continue this process

indefinitely each time we add n

rectangles each enclosing an area of x

to the N minus one but now we can

imagine that this process has been

carried out infinitely often that means

that we're looking at a limiting process

in this infinite limiting process the

entire Square will be shaded by these

rectangles that means that the infinite

sum that we're thinking about must equal

the area of the square but the area of

this Square can be computed just by

taking the product of the two side

lengths which are both one over one

minus X

therefore the infinite sum where n

ranges from 1 to Infinity of n times x

to the N minus 1 must equal the quantity

one over one minus x squared and this is

in fact the formula for the sum of the

differentiated geometric series

General Differentiated Geometric Series | visual proof

This is a short, animated visual proof computing the sum of a differentiated geometric series with ratio given by positive real number x less than 1. The proof uses a dissection proof of a 1/(1-x) by 1/(1-x) square using the formula for the infinite geometric series with ratio x.

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 Roger B. Nelsen from the September 2001 issue of The College Mathematics Journal (https://www.jstor.org/stable/2687557 p. 257).

For related videos, see:
https://youtu.be/2d4CTG1uJs8
https://youtu.be/2WuSNveg-1c

#math #infiniteseries #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #geometricseries #series #mathshorts #mathvideo #arithmeticogeometricseries #differentiatedseries

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

_______________
Music in this video:
Event Departure by Silent Partner


suppose you have a square that has a

side length of one

then the diagonal line that cuts through

the square has a length given by the

square root of 2 by the Pythagorean

theorem

here's a related question what if we

start with a regular pentagon with side

length one what's the length x of the

diagonal cutting through this shape

let's find out using a visual proof

first we draw a second diagonal line

like this which creates this shaded

isosceles triangle when we draw another

diagonal like this we create a second

triangle shaded in red that's a

subtriangle of the blue triangle

now if we consider the three pictured

angles where the top one is labeled

Alpha and we draw the circumscribed

circle we see that Alpha is subtended by

an arc and so are the other two pictured

angles with question marks but all three

of those arcs are exactly one-fifth of

the entire circle and therefore each of

the angles must be Alpha this also means

that the other angle that is in the blue

triangle and the red triangle must be 2

times Alpha

but now the four pictured angles give us

the full measure of the triangle so we

see that 5 times Alpha must be 180

degrees and therefore Alpha must be 36

degrees but this then means that both of

the pictured triangles the blue and the

red have 36 degree angles and 272 degree

angles and both of them are therefore

isosceles triangles but it turns out

that we also see we have a third

triangle shaded in green here this

triangle has two 36 degree angles and

therefore it is isosceles as well

using the fact that these three

triangles are isosceles we can see that

this side length here is one so that the

short side length on the red triangle is

x minus 1.

but now we can use the fact that the two

triangles shown the red and the blue are

similar as they have exactly the same

angle measurements

the fact that the two triangles are

similar means that the ratio of their

side lengths are equal so that x divided

by one must equal 1 divided by x minus

one now when we clear denominators we

get that x times x minus 1 equals one

and this means that x squared minus x

minus 1 is equal to zero this equation

has two roots and only one of them is

positive which is x equals Phi which is

the golden ratio

therefore the length of the diagonal in

this regular pentagon with side length 1

must be Phi the golden ratio

here's a cool way to use this diagram

let's suppose that the golden ratio is

rational that means that it can be

written as a ratio of two integers say A

over B and let's assume that this ratio

is in lowest terms

the fact that Phi equals A over B means

that there is a regular pentagon like

this where the diagonal is length a and

the side length of the Pentagon is B

once again we can swing the side length

down like this to see that the small

subtriangle created inside the larger

triangle pictured here must have length

a minus B but once again these two

triangles are similar and therefore A

over B must be equal to B over a minus B

but Phi equals A over B so that means

Phi equals B over a minus B and this is

a contradiction because B is less than a

and a minus B is less than b so this

contradicts that we had Phi in lowest

terms when we wrote it as A over B but

that means our original assumption was

wrong so that the golden ratio fee

cannot be written as the ratio of two

integer values and so the golden ratio

is irrational

Golden Ratio is Irrational from a Regular Pentagon (visual proof)

In this video, we show how to prove that the golden ratio is the length of a diagonal in a regular pentagon with a side length of 1. We do this using similar triangles and angle analysis using the circumscribed circle. As a bonus, we use nearly the same diagram (just scaled appropriately if needed) to prove that the golden ratio is irrational, that is, that it cannot be written as the ratio of two integers.

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 an argument in Roger Nelsen’s book Nuggets of Number Theory (I may receive a commission from Amazon, at no cost to you, if you click the link below):

https://amzn.to/3RfbRf2

Here are other videos related to the Golden ratio:
https://www.youtube.com/watch?v=yeHDXdv5KH4
https://www.youtube.com/watch?v=eMMErKcw8yM
https://www.youtube.com/watch?v=1ctRFgWwg5E&t=4s

#goldenratio #irrationalnumbers #realnumbers #manim #math #mtbos #animation #theorem #visualproof #proof #iteachmath #mathematics #irrational #pentagon #proofbycontradiction #algebra #infinitedescent #similartriangles #triangle #trianglesimilarity #similarity

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


Welcome to our Channel today we are

going to discuss important question from

CBC class 10 mathematics specifically

trometry let us consider a question if 3

c a = 4 check whether 1 - tan a / 1 + t²

a = cos² a - sin

a to solve this let consider a right

angle triangle ABC with the right angle

at B give given that t a given that 3 C

= 4 Rite the equation c a = 4

by3 we know the

relation tan a = 4 by3 since t a equal

to opposite side ided by Adent side we

have tan a = to BC / by AB now let's now

let's substitute the value BC = to 3K

units and a = 4K units where K is a real

positive number we need to calculate

hypotenuse AC using Pythagorean theorem

AC

s a² = to a² + BC squ after substituting

the value AC = to 5K

unit now we have hypotenuse value so we

can calculate sin a and cos a sin AAL to

opposite side divid hypot which is equal

to BC / AC with 3x 4 similarly cos a

equal to Adent side/ hypotenuse which is

equal to a / AC with value 4 by 5 to

check the given equation let consider

the left hand side LHS 1 - tan S A / 1 +

t s

a substituting the value of T A we got

got 1 - 3x 4² 1 + 3x

4² you simplified 7 by 25 now let us

consider the right hand side rhs cos² A-

sin S A substituting the value of cos a

and sin a we got 4x 3² - 3x b² which is

Al which is simplified to 7 by

25 since LHS equal to rhs

we have proved that given equation is

correct this is the one of the easiest

way to solve this problem

If 3 cot A = 4, check whether (1 – tan2A)/(1 + tan2A) = cos2 A – sin2 A or not.

rigonometry Challenge: Prove the Identity!

In this video, we'll tackle an intriguing trigonometry problem:

Given: 3 cot A = 4

Prove: (1 - tan²A) / (1 + tan²A) = cos²A - sin²A

We'll break down the problem step-by-step, using fundamental trigonometric identities and algebraic manipulations. You'll learn how to simplify complex expressions and arrive at the desired result.

Why Watch This Video?

Clear and concise explanations
Step-by-step solutions
Visual aids for better understanding
Tips and tricks to solve similar problems
#class10maths #trigonometry #cotA #tanA #cosA #sinA #trigonometricidentities #mathstutorial #education #studytips #exampreparation #cbse #ncert #trigonometryproofs #mathproblemsolving

class 10 maths, trigonometry, cot A, tan A, cos A, sin A, trigonometric identities, math tutorial, education, study tips, exam preparation, cbse, ncert, trigonometry proofs, math problem solving, prove trigonometric identity, class 10 trigonometry problems, trigonometry practice problems, how to solve trigonometry problems, trigonometry tricks, trigonometry formulas
CBSE Class 10 Maths
Class 10 Maths
10th Maths
Maths Tutorial
Math Tricks
CBSE Board
NCERT Solutions
Exam Preparation
Study Tips
Specific Topic Keywords:

Trigonometry
Algebra
Geometry
Arithmetic
Statistics
Probability
Real Numbers
Polynomials
Pair of Linear Equations in Two Variables
Quadratic Equations
Arithmetic Progressions
Coordinate Geometry
Introduction to Trigonometry
Trigonometric Identities
Heights and Distances
Circles
Constructions
Areas Related to Circles
Surface Areas and Volumes
Statistics
Probability
Long-Tail Keywords:

CBSE Class 10 Maths Chapter-wise Solutions
CBSE Class 10 Maths NCERT Solutions
CBSE Class 10 Maths Important Questions
CBSE Class 10 Maths Previous Year Papers
CBSE Class 10 Maths Sample Papers
How to score good marks in CBSE Class 10 Maths
Best tips for CBSE Class 10 Maths
Easy way to learn CBSE Class 10 Maths
Fun way to learn CBSE Class 10 Maths


let a and b be two positive real numbers

and consider line segments of length a

and b

use these two line segments to create a

circle of diameter a

and a circle of diameter b

then move these two circles together so

that they are tangent

the lengths of the radii of the two

circles are a over two and b over two so

the length of the line segment

connecting the centers of the two

circles is given by the sum a plus b

over two the horizontal line coming from

the center of the smaller circle and

intersecting the diameter of the larger

circle creates a line segment of length

absolute value of a minus b over 2. to

find the length of the line segment that

we just drew that represents the

distance between the x-coordinates of

the centers of the circle we can use the

pythagorean theorem since it forms one

leg of a right triangle

the square of a plus b over two minus

the square of the absolute value of a

minus b over two is equal to the

quantity a times b so the desired length

is the square root of a times b

but then from this diagram we can see

that the value a plus b over 2 which is

the hypotenuse of a right triangle is

greater than or equal to the square root

of a times b which is the length of a

leg of the same right triangle no matter

how we change the sizes of a and b we

see that a plus b over 2 is always the

hypotenuse and square root of a b is

always the leg of a right triangle

they only become equal if a and b are

exactly the same so we see that a plus b

over 2 is equal to the square root of a

b if and only if a equals b the

inequality that we've just proved using

this circle diagram is called the

arithmetic mean geometric mean

inequality

you

AM-GM Inequality IV (visual proof)

This is a short, animated visual proof demonstrating the arithmetic mean geometric mean inequality using two tangent circles. #mathshorts #mathvideo #math #amgminequality #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #inequality

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

This animation is based on a visual proof from Roland H. Eddy that appeared in the June 1985 issue of the College Mathematics Journal (https://www.jstor.org/stable/2686572)

To learn more about animating with manim, check out:
https://manim.community
_______________________________________
Music in this episode:
Procession -- Geographer


[Music]

start with a right triangle with legs of

length a and b and hypotenuse of length

c

draw the circle containing the triangle

with radius of length c and given by the

hypotenuse

extend the leg labeled a to a full

diameter by adding the radius of length

c and a segment of length c minus a

connecting the lines from the ends of

this diameter to the top of the triangle

like this yields a right angled triangle

according to thaley's triangle theorem

consequently these two triangles sharing

an edge of length b are similar

this implies that c plus a over b

is equal to the quantity b over c minus

a

in turn

this implies that c plus a

times c minus a is equal to b squared

and expanding the product we get c

squared minus a squared is equal to b

squared so that c squared is equal to a

squared plus b squared this is the

pythagorean theorem

[Music]

you

Pythagorean Theorem III (visual proof)

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using the semicircle and Thales triangle theorem. 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 Michael Hardy from the November, 1986 issue of The College Math Journal (https://doi.org/10.2307/2686255).

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

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


consider the set of positive real

numbers and draw the graph of the

function f ofx = 1 /x on this domain the

derivative of this function is frime of

x = -1 /x^2 if we consider the input x =

1 we see that the point on the graph is

1 1 and the derivative frime of 1 is 1

this means that the tangent line to the

function 1 /x at x = 1 has a slope of -

1 and therefore the equation of the

tangent line is Y = 2 - x because FX =

1X concave up we see that every point on

the tangent line lies below the

corresponding point for the same x value

on the curve 1/x this means that in

general 2 - x is less than or equal to 1

/x by adding X to both sides we see that

for any positive real number x 2 is less

than or equal to 1x + x can you see from

the diagram when equality holds

Sum of a Positive Number and its Reciprocal from Calculus

This is a short, animated visual proof demonstrating that sum of a positive real number and its reciprocal is always greater than or equal to 2.

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

It turns out that this theorem is equivalent to the Arithmetic Mean-Geometric Mean inequality. The equivalence is implied by the following two proofs: https://youtu.be/62G9fak1vyk
https://youtu.be/IghOHBl0Do8

And here are two alternate ways to prove this fact via the Pythagorean Theorem and triangle areas:
https://youtu.be/zBUK8C6wSqs
https://youtu.be/BkQvnriVblY

This animation is based on a proof by Roger B. Nelsen from the December 1994 issue of Mathematics Magazine, page 374 (https://doi.org/10.2307/2690999). You can find this here too: https://www.maa.org/sites/default/files/Nelsen99959975.pdf

#math #manim #visualproof #mathvideo  #geometry #mathshorts  #algebra #mtbos  #animation #theorem #pww #proofwithoutwords #proof #iteachmath #inequality #calculus #derivative #tangentline #slope #amgm

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


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start with a square of side length y

covering an area of y squared

then remove a square from the lower left

corner with side length z so that the

resulting area is y squared minus z

squared

we can cut the resulting nomen in half

along the diagonal creating two

trapezoids

move one of the trapezoids down

rotate it

and then flip it to connect it to the

other trapezoid resulting in a rectangle

the rectangle has side length y minus z

and the other side length is y plus z

therefore the original area of y squared

minus z squared is the same as the

product of y plus z

and y minus z

this formula is known as the difference

of squares formula

we can plug in x plus a over 2 for y and

a over 2 for z so that y plus z is x

plus a and y minus z is x

then

the resulting equation becomes x plus a

over 2 squared minus a over 2 squared is

equal to x plus a times

x

but the latter quantity is just x

squared plus a times x

therefore the quantity x plus a over 2

squared minus a over 2 squared is equal

to x squared plus ax

this equation is known as completing the

square

and therefore we have used the

difference of squares formula to

complete the square

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you

Completing The Square via the Difference of Squares (visual proof)

This is a short, animated visual proof showing how to derive the formula for "completing the square." This proof first uncovers the formula for the difference of squares. The resulting algebraic formula can be used to write a polynomial of the form x^2 +ax as a difference of two squares. #mathshorts #mathvideo #math #completingthesquare #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #completingthesquare

This animation is based on a visual proof by Munir Mahmood from the January 2014 issue of The College Mathematics Journal  (https://doi.org/10.4169/college.math.j.45.1.021) page 21.

Here is another visual proof of the same fact:
https://youtu.be/T0HyWIFbsHQ

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


T

[音楽]

Area of Circle, visual proof

Watch as we visually demonstrate the fascinating formula for calculating the area of a circle! Using simple geometric shapes and intuitive explanations, this short video breaks down how πr² comes together. Perfect for anyone curious about the magic behind circles and their areas. Like, subscribe, and share for more engaging math insights!"
#youtube #viral #math#areaofcircle


have you ever wondered about the

infinite sum shown here 1 4 plus 1 4

squared plus one-fourth cubed and so on

one way to make sense of this sum is to

start with an equilateral triangle and

decompose it into four equal pieces then

shade the middle triangle which

represents one-fourth of the area now do

the same thing with the upper triangle

and repeat over and over again each time

shading one-fourth of the remaining

upper triangle so that we're shading

powers of one-fourth

notice that the total shaded area in

this equilateral triangle can be

represented by the infinite sum of the

powers of one-fourth

but now here's the amazing trick this

process has divided the original

equilateral triangle into three pieces

which are all stacks of triangles at

each level the triangles are the same

size so we've divided the equilateral

triangle into three equal pieces

therefore the sum of the powers of

one-fourth is equal to one-third

A visual infinite sum like you’ve never seen!

This is a short, animated visual proof demonstrating the sum of the infinite series of the powers of 1/4. #shorts #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricseries #series #infiniteseries #fractal

To see the original wordless version of this video, check out: https://youtu.be/8i1xj5ORwUw

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

This animation is based on a visual proof by Rick Mabry from the February 1999 issue of Mathematics Magazine (https://www.jstor.org/stable/2691318 - page 63).

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

_______________________________
Music in this:
Prelude No. 6 by Chris Zabriskie is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/

Source: http://chriszabriskie.com/preludes/

Artist: http://chriszabriskie.com/


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oh

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#trending #mathematics Visual Proof: Factors of Quadratic Equation Polynomials

#trendingshorts  #mathematics Visual Proof: Factors of Quadratic Equation Polynomials

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


here's a classic visual infinite sum

start with a unit square cut it into

four pieces and shade three of them

shading an area of 3/4s cut the

remaining 1/4 into four pieces and shade

three of them so that we've now shaded

an additional 3 over4 squar continue

repeating this process cutting the top

one quarter into four equal pieces and

shading three of them at each stage this

means we've shaded 3 over 4 to the end

of the total area of the square if we

continue this indefinitely Or imagine

the limiting process we see that in the

limit the entire Square diagram will be

shaded and therefore we've shaded an

area of one this means that 3 * the

infinite sum of the powers of 1/4 is one

one interpretation is that 0.3 repeating

in base 4 is 1 and another

interpretation is that the sum of the

infinite poers of 1/4 is

0.3333… = 1 (in base 4)

This is a short, animated visual proof demonstrating the infinite sum of the powers of 1/4.

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

To see two other versions of this sum, check out
https://youtu.be/8i1xj5ORwUw?si=5SlPwm_0ym2NyGWl
https://youtu.be/UePC62RciUc?si=gdtUgL7gPSgiImYC

To see the related base 10 fact, check out:
https://youtube.com/shorts/XJtjU_Ipjno?feature=share

And to see the related binary fact, check out:
https://youtube.com/shorts/RoovN01q83Y?feature=share

To see related base-3 fact, check out:
https://youtu.be/vfEDDI3vfHU
or
https://youtube.com/shorts/GwIVlNRV-M0

To see a related base-5 fact, check out
https://youtube.com/shorts/ltKZJ8wumUI

This animation is based on a proof by Sunday A. Ajose from the June 1994 issue of Mathematics Magazine (https://www.jstor.org/stable/2690616 page 230).

#mathshorts #mathvideo #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #geometricseries #series #infinitesums #infiniteseries #shorts

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


[Music]

GL loneliness

Why Area of Circle = πr^2 | Visual Proof | Math fun | Chetan Setia | Best Coaching Institute

ਮਾਸਟਰ ਕੈਡਰ ਦੀ ਤਿਆਰੀ | ਚੇਤਨ ਸੇਤਿਆ
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About Chetan Setia: He is the founder and Director of Chetan Setia Education.
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#areaofcircle #visualeffects #visualization


here's the graph of the function f

defined by f of x equals 1 over X along

with one label at the input x equals one

now consider another input labeled X

that can vary and draw a line from the

x-axis up to the point on the curve as

we move x to the right we can measure

the area under the curve and above the

x-axis between 1 and the input X the

area is 0 when x equals one and

increases as X increases we can do more

let's actually plot the area on a

separate set of axes like this as X

varies we end up with the curve of a new

function we denote this new function by

capital f of x equals the integral from

1 to X of 1 over T DT and we call this

an area accumulation function of the

function f this particular area

accumulation function is actually one

way to define the natural log notice

that we can let the x value vary less

than one as well resulting in a negative

orientation for the area so this defines

the natural log of x for any X greater

than zero

Defining the natural logarithm

The definition of the natural log as an area accumulation function. #shorts #math #manim #logarithms #calculus #integral


here's a non-standard way to find the

using areas the sum of the first n odd

positive integers 1 + 3 + 5 and so on up

to 2 N minus 1 can be represented by an

array of 1x one squares like this where

the first row has one square the second

row has three squares and so on all the

way down until the very last row which

has 2 N - one squares this array has a

height of size n and a base of size 2 N

minus one like this

to find the area of this array we can

cut it into pieces using this diagonal

line of slope - one2 starting at the

bottom right corner of the top

square we can use that line to break the

diagram up into a variety of rectangular

and triangular pieces as shown here

first on the left side we see a 1xn

rectangle the triangle in the middle is

2 N - 2 by nus1 and the nus1 triangles

on the right are all 2 2x1 triangles the

rectangle has an area of n * 1 or n the

large triangle has an area of 12 N - 1 *

2 N - 2 which is n -1 * n -1 or n^2 - 2

n + 1 and finally the N -1 triangles on

the right all have an area of 1/ 12 * 1

* 2 or 1 combining these areas together

we get that the sum of the first n odd

positive integers is n + n^ 2 - 2 n + 1

+ N - 1 but this simplifies to n^2 and

therefore the sum of the first n odd

positive integers is

n^2 to write this more compactly we have

that the sum where K ranges from 1 to n

of 2 K minus 1 must equal n^ 2 this is a

classic formula but we proved it using

areas of rectangles and

triangles

Adding Odds using Areas (Sum of odds visual proof IV)

This is a short, animated visual proof showing one method to find the sum of the first n odd positive integers decomposing an array of unit squares into rectangular and triangular pieces and using area formulas for those pieces.

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 other methods for finding the sum odd integers visually, check out the following videos:
https://youtu.be/jq5AYCgkciE?si=U6bxLBmWjyBD_HQZ
https://youtu.be/I4A2T55k0cg
https://youtu.be/x3qfFBNRRDg
https://youtu.be/ZeEOgbLo0Rg
https://youtu.be/6H6nCpGW1nU

For other sums related to odd integers, check out these visuals:
https://youtu.be/WKxoi2i0EaE (summing odd squares)
https://studio.youtube.com/video/9BC1klE8jMk/edit (alternating odd sums)
https://youtu.be/MmOTqrtbtFQ (odd sum ratios)
https://youtu.be/fTBvVeURb3Q (odd sum ratios II)

This animation is based on a visual proof due to Samuel G. Moreno that appeared in the October 2017 issue of Mathematics Magazine (https://doi.org/10.4169/math.mag.90.4.298 page 298).

You can find other diagrams in Roger Nelsen's book: (I may receive a small commission at no cost to you for this affiliate link): https://amzn.to/47kfG7w

#manim #math #mathvideo #mathshorts #sequences #triangles #animation #theorem #pww #proofwithoutwords #visualproof #proof #squarenumbers #squares #oddnumbers #odds #sums #pww #proofwithoutwords  #proof #algebra #finitesums #mathematics #mathvideo #mtbos

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


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suppose the fraction a over b is less

than or equal to the fraction c

over d both of these fractions can be

represented by the slopes of hypotenuses

of right triangles and the inequality

means that the slope of the hypotenuse

of the triangle with legs a and b

is less than or equal to the slope of

the hypotenuse of the triangle with legs

c and d

aligning the two triangles this way

allows us to draw them inside of a third

right triangle

this new right triangle has legs of

length b plus d and a plus c

now comparing the slopes of the

hypotenuses of these three triangles

reveals that a over b is less than or

equal to the slope of the big triangle

which is a plus c over b plus d and that

in turn is less than or equal to the

slope of the top right triangle

which is c over d the middle quantity

a plus c over b plus d is known as the

median

of the two fractions a over b and c over

d

therefore this is known as the mediant

inequality

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you

The Mediant Inequality I (visual proof)

This is a short, animated visual proof demonstrating what one might called the mediant inequality. #mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #fractions #mediant #inequality #mathematics #fareyfractions #farey

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).

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


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#trending #mathematics #maths #calculus Visual Proof: Trigonometric Identity sin^2 (x) +cos^2 (x)=1

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


here's how to find the total number of

gifts in the 12 Days of Christmas using

a geometric argument on the first day of

Christmas you get one partridge in a

pear tree on the second day you get the

Partridge in the pear tree and the two

turtle doves on day three you get three

French hens and the gifts from day two

on day four you get four calling Birds

and the gifts from day three on day five

you get five golden rings and the gifts

from day four on day six you get six

geese and all the gifts from day five on

day seven you get seven swans and all

the gifts from day six on day eight you

get eight maids and all of the previous

gifts on day nine you get nine ladies

dancing in all the previous gifts on day

10 you get 10 Lords and all of the

previous gifts on day 11 you get 11

Pipers piping and all of the previous

gifts and finally on day 12 you get 12

drummers and the previous gifts so how

many total gifts have you obtained over

the 12 days to figure this out we employ

a trick we imagine three people getting

all of these gifts so that we actually

have three copies of the gifts we take

the gifts from the second person and

glue them together like this with the

gifts from the first person we then form

this stack finally we can use the stack

of gifts from the third person to lay

them down in order like this as we fill

in this diagram we see that the final

shape will be a perfect rectangle by

construction the length of this

rectangle is two more than the total

number of days so it's 12 + 2 which is

14 on the other hand the width of this

rectangle is equal to the total number

of new gifts each day so it's equal to

the sum of the first 12 positive

integers but this sum is the total

number of gifts obtained on day 12 which

is half of this 13x 12 rectangle

pictured here so the sum of the first 12

positive integers is 13 * 12 / 2

therefore this construction shows us

that 3 * the total number of gifts

received is equal to the area of this

rectangle which is 14 * 13 * 12 12 ID 2

but this means the total number of gifts

is 14 * 13 * 12 / 6 which is

364 in the song The 12 Days of Christmas

they've given 364

gifts

Geometry of the Twelve Days of X-mas (visual proof)

You might know that in the song "The Twelve Days of Christmas" there are a total of 364 gifts given over the 12-day span. You might even know how to prove this mathematically (there are great videos about this - see links below). But did you know that you can demonstrate this fact geometrically? In this video, we show one way to think about the sum of the first 12 positive integers. Not that this technique also can work in general; see this earlier (more dramatic, and wordless) video showing the general formula and general visual proof:

https://youtu.be/N0ETyJ5K6j0

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 Monte J. Zerger from the December 1990 issue of Mathematics Magazine (https://www.jstor.org/stable/2690899- page 314).

To see other amazing videos discussing the math behind the Twelve Days of Christmas, check out:

https://youtu.be/fC8W4s6N9HQ?si=5r2gRH-r1o9V6LIq by @TippingPointMath
https://youtu.be/cNl7GypEH_A?si=v4ZDSYrmwxDapMpo by @MindYourDecisions
https://youtu.be/7adXOcUweaY?si=t-5QxwlypzdQtu5W by @polymathematic

 #manim #math #mathvideo #mathshorts #sequences #triangles #animation #theorem #pww #proofwithoutwords #visualproof #proof #triangularnumbers #sums #pww #proofwithoutwords  #proof #algebra #finitesums #mathematics #mathvideo #mtbos

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

Thumbnail for video includes part of an image from Xavier Romero-Frias, entitled English: The Twelve Days of Christmas song poster. This image was used according to Creative Commons Attribution-Share Alike 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en ).


[موسيقى]

Visual Proof of the Pythagorean Theorem | 1

In this short the most famous and intuitive visual proof of the Pythagorean theorem. #maths #visualization #geometry #algebra #python #proof #visualproof


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do

Visual proof of difference of squares of two numbers || visualize math || FS bukhari official #maths #mathematics_facts #mathlogic #mathematics


foreign

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foreign

Factor a Sum of two Squares (visual proof)

This is a short, animated visual proof of showing how to factor a sum of two squares. #manim #math #mathvideo #arithmeticprogression #squares #mathshorts #geometry  #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #algebra #pww  #proof #areas #mathematics #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 from Roger B. Nelsen's second proofs without words compendium.

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Friends Forever by FSM TEAM
Music by https://www.free-stock-music.com


foreign

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suppose we want to walk from the point 0

0 to the point n comma n Only traveling

north and east and not passing below

this diagonal Y equals X

this problem consists of counting what

are called the Northeast integer lattice

paths and they look like this

in a previous video I enumerated lots of

these and showed how to find a recursive

formula for counting the number of such

lattice paths

right here we see that the first view

values of this sequence are one one two

five Fourteen and here we see the 42

lattice paths traveling from 0 0 to 5 5.

as I said we know already how to find a

recursion for this sequence but if we

let C sub n be the number of such

lattice pass to n comma n can we now

find a closed formula for this integer

sequence

here's one way to provide a closed

formula for the number of such paths

here we let N equal 5 represent the

general situation instead of counting

the restricted lattice paths we can

first enumerate all the lattice paths

traveling Northeast they go from 0 0 to

n n in a previous video I showed that

this is counted by a binomial

coefficient but we want to avoid the

paths that pass below Y equals X those

must be the ones that intersect this

line y equals x minus 1.

for each such path that intersects the

red line we can reflect the remaining

part of the path after the first

intersection across that line

here we do it with all of the lines that

we don't want to include in our final

count

[Music]

as we reflect and remove the bad lattice

paths we notice something special about

all of these Reflections look at what

happens to the end point of these

lattice paths they all seem to be

hitting on the same point so what we're

getting from this collection of

reflected lattice paths seems to be a

collection of lattice paths that still

starts at the origin zero zero but this

time ends at a different point

in fact all of the reflected lattice

paths end at the point n plus 1 comma n

minus 1 which is the reflection of the

point n n over the red dotted line but

now if we enumerate all the Northeast

lattice paths from 0 0 to the point n

plus 1 comma n minus one we see that

each of them crosses the red dotted line

and therefore must have been one of the

reflected paths because the reflection

process is invertible we see there's a

one-to-one correspondence between the

bad paths traveling to n comma n and all

of the paths traveling to n plus one

comma n minus one but now we have our

strategy to count the good lattice paths

the ones from 0 0 to NN traveling

Northeast and not passing below the

diagonal in order to do this we must

first have all the paths that travel

Northeast from 0 0 to NN as I mentioned

before this is counted by the binomial

coefficient 2N choose n then we're going

to subtract all of the bad paths but we

know that those correspond to the

Northeast lattice paths traveling from 0

0 to n plus one comma n minus 1 which is

also counted by the binomial coefficient

2N choose n plus one but then we can use

binomial coefficient identities and

formulas to see that c sub n is equal to

1 divided by n plus 1 times 2N choose n

this is a closed formula for the

sequence c n which are known as the

Catalan numbers

you can find this result in any standard

text describing the Catalan numbers

Walking city streets: Catalan Closed Form (visual proof from lattice paths)

In this video, we show how to provide a closed form for the number of northeast lattice paths to the point (n,n) that don't pass below the line y=x. The number of such lattice paths is counted by the famous Catalan numbers.

For other videos discussing lattice paths (and those mentioned in this video), check out
1) https://youtu.be/9YUl0k2FYzc
2) https://youtu.be/KYpOoA2D63w
3) https://youtu.be/__w_MDQeUYc

If you like this video, check out my others and consider subscribing. Thanks!

#catalannumbers #catalan #manim #math #mtbos  #animation #theorem #visualproof #proof #iteachmath #mathematics #binomialcoefficients #latticepaths #discretemath #combinatorics #enumeration #closedform

This animation is based on the well-known interpretation of Catalan numbers as counting restricted lattice paths. I recommend the Wikipedia site or the books Enumerative Combinatorics I and II by Richard Stanley for someone wanting to know more about Catalan numbers.

To learn more about animating with manim, check out:
https://manim.community
_________________________________________
Music in this video:
Cylinder Two by Chris Zabriskie is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/

Source: http://chriszabriskie.com/cylinders/

Artist: http://chriszabriskie.com/


foreign

[Music]

here's an intriguing number pattern one

plus two is three

four plus five plus six is seven plus

eight

nine plus ten plus eleven plus twelve

is 13 plus 14 plus 15 and 16 plus 17

plus 18 plus 19 plus 20 is 21 plus 22

plus 23 plus 24 and so on this pattern

continues as long as we take it out the

question I have for you is can you

determine the general pattern and the

general formula and if so can you

provide a proof that this formula always

holds pause the video and give it a try

[Music]

if you need a hint the first term and

every sum on the left is of the form N

squared for an integer n

in the first term in every sum on the

right is of the form N squared plus n

plus 1 for the same value of n let's

investigate A visual representation of

these sums in order to uncover the

formula Take N squared times n plus 1

cubes and place them in a cuboid with

Dimensions n by n plus 1 by n on top of

this cuboid place a triangular array of

cubes with a base of length n this adds

an extra n times n plus 1 divided by two

cubes on top of the cuboid foreign

[Music]

utilizing this structure in two

different ways will reveal the formula

that we want if we first separate the

structure into horizontal arrays like

this we see that all of the green

rectangular arrays are n by n

so there is a total number of N squared

plus N squared plus 1 up to N squared

plus n cubes on the left on the other

hand if we separate the structure

vertically we end up with rectangular

arrays of the size n plus 1 by n with

cubes on top so there's a total of N

squared plus n plus one up to N squared

plus n plus n total cubes on the right

side

therefore the formula that we're looking

for says that the sum of the numbers

from N squared to N squared plus n is

equal to the sum of the numbers from N

squared plus n plus 1 up to N squared

plus 2N

Unbelievable Equal Consecutive Sums (3D visual proof)

This is a short, animated visual proof showing how two related sums of consecutive integers are equal using a 3D cuboid with an extra triangular array on top. #manim #math #mathvideo #mathshorts #triangularnumbers  #numbertheory  #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof

This animation is is based on a proof without words by Roger B. Nelsen from the February 1990 issue of Mathematics Magazine (https://doi.org/10.2307/2691505) page 25.

Here is a recent related video turned out by @Mathologer : https://youtu.be/XR2u7izwZ04

To learn more about animating with manim, check out:
https://manim.community
____________________________________________________
Background music:
Run Until Your Wings Grow
Late Night Feeler


here's four proofs that the area of the

regular do deagon in a unit circle is

three the higher the level the better

the

[Music]

proof

yeah yeah sing in the street the sun is

in the East

yeah yeah calling out the Beast the

tiger underneath

yeah yeah I'm starting to believe the

world is at my feet

yeah I'm ready for the feast it's all

within my reach it's all within my

reach I'm going all the way I'm going

yeah my blood is Rebel I'm going all the

way I'm going to another level

[Applause]

[Applause]

if I'm on the loose running after you

yeah yeah there's nothing you can do

I'll catch you when it's through

yeah yeah hle for the juice yeah 120

poof yeah yeah yeah yeah I'm blowing up

the roof blowing up the roof got nothing

left to

Pro I'm going all the way I'm going yeah

my blood is reel going all the way I'm

going to another

[Applause]

level let

go

got to get my hands

dirty chain breaker Earth Shaker rain

maker yeah another level chain breaker

Earth Shaker rain

maker yeah another level chain breaker

Earth Shaker rain

maker yeah another level chain breaker

Earth Shaker rain maker oh to another

level I'm going all the way I'm going

yeah my blood is re I'm going all the

way I'm going to another

[Applause]

level another

Lev let's go let's go let's

go to another

level

you let's get

it let me know in the comments which of

the four is your favorite thanks for

watching

Area of a Regular Dodecagon | Proofs without Words Levels 1 to 4

This is a compilation of four animated visual proofs demonstrating that the area of a regular dodecagon inscribed in the unit circle has an area of exactly 3.

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

Level 1 is a standard construction using trigonometry that you can find in many sources.

Level 2 is based on a dissection from the book The Penguin Book of Curious and Interesting Puzzles by David Wells. https://amzn.to/3O4CnUF : I may receive commissions from this link

Thanks to Matthew Scroggs for pointing me to the source.

Level 3 is based on a proof by Roger B. Nelsen from the January 2015 issue of The College Mathematics Journal page 10 (https://doi.org/10.4169/college.math.j.46.1.10 ).

Level 4 is based on a dissection by J. Kürschák 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/)

See these videos for standalone versions of these dissections:
https://youtu.be/LbCghvu6G5w
https://youtu.be/aSdo2ug4EyE
https://youtu.be/7lCJDdD_fHs
https://youtube.com/shorts/uxL7K2IZBf8
https://youtube.com/shorts/f_y-V75O93c
https://youtube.com/shorts/AgdgwQbWFvM

#math #manim #visualproof #mathvideo  #geometry #mathshorts  #geometry #mtbos  #animation #theorem #pww #proofwithoutwords #proof #iteachmath #dodecagon #area #dissection

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


[Music]

start with the real number r between 0

and 1.

draw an infinite vertical line connected

to a horizontal line of length given by

the sum of the positive powers of r

which is a geometric series and thus

converges to a finite value

from the end of the horizontal line

build a staircase type diagram where the

stair heights are all one and the stair

depths are the positive powers of r so

that each stair platform gets narrower

this object has been called gabriel's

staircase it has an infinite perimeter

but encloses a finite area

let's compute the enclosed area in two

ways

first we decompose the staircase into

columns of rectangles the first one has

height 1 and width r so it contains an

area of 1 times r

the next one has height 2 and width r

squared so it contains an area of 2

times r squared

the third one has height three

and width r cubed so it encloses an area

of three times r cubed and the next one

encloses an area of four times r to the

fourth we continue this process forever

with narrower and narrower rectangles

giving an infinite sum for the area

we can also decompose the staircase into

rows of rectangles

the first such rectangle has height 1

and the width is the sum of the

geometric series with ratio r and first

term r which converges to r over 1 minus

r so the area enclosed is r over 1 minus

r

the second rectangle also has height 1

and this time the width is the geometric

series with ratio r and first term r

squared which converges to r squared

over 1 minus r so that the area is r

squared over 1 minus r

the next two rectangles are also of

height 1 and widths of geometric series

and thus they enclose areas of r cubed

over one minus r

and r to the fourth over one minus r

we again continue this process forever

with smaller and smaller rectangles

giving a different infinite sum for the

area

these two infinite sums can be shown to

both converge to a finite value because

r is between 0 and 1. the sums enclose

the same area so they must be equal thus

we have that the sum from one to

infinity of k r to the k

is equal to the sum from one to infinity

of r to the k over one minus r

but we can factor one minus r out of the

ladder sum leaving us with the geometric

series with ratio r and first term r

again which we know converges to r over

one minus r thus the sum from one to

infinity of k

times r to the k is equal to r over the

quantity one minus r squared provided r

is between zero and one

[Music]

[Music]

you

Arithmetico-Geometric Series Visualized (visual proof using Gabriel's Staircase)

This is a short, animated visual proof computing the sum of the differentiated geometric series with terms of the form k times r^k where r is between 0 and 1. The proof uses a fascinating object called Gabriel's staircase, which has infinite perimeter but encloses a finite area. #math #infiniteseries #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #geometricseries #series #mathshorts #mathvideo

If you like this video, consider subscribing to the channel. Thanks!

This animation is based on a visual proof by Stuart G. Swain from the June 1994 issue of Mathematics Magazine (https://doi.org/10.2307/2690612 p. 209).

To learn more about animating with manim, check out:
https://manim.community
__________________________________________________________________
Music in this video:
Autumn Day by Kevin MacLeod is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/

Source: http://incompetech.com/music/royalty-free/index.html?isrc=USUAN1100765

Artist: http://incompetech.com/


here's a problem from a 1976 Journal of

recreational mathematics issue that

Roger Nelson calls the Christmas tree

problem start with a circle and inside

the circle inscribe an isoceles triangle

along the diameter like this inside of

the isoceles triangle let's inscribe a

circle the bottom radius shown bcts the

diameter we can use the two quadrants

created to inscribe two more circles

like

this the the problem in the journal asks

you to prove that these three small

inscribed circles are all congruent

pause the video and think about this

before I reveal one possible

answer are you ready to see a solution

it turns out that we can prove this fact

visually by changing our point of view

we take a second copy of the Outer

Circle and shift it up by a distance

equal to the radius then rotate 45° or <

over 4 radians from this perspective we

see that the top circle inscribed in the

triangle is actually one of the pair of

the two circles inscribed in the two

quadrants of this new Circle therefore

these two circles are congruent to the

original two in the bottom quadrant of

the original Circle and this means that

all three of the original inscribed

small circles are congruent to one

another thanks for watching if you like

this video and others on my channel you

might enjoy Roger Nelson's books proofs

without words there are three different

compendia Linked In the

description

Recreational Math Christmas Tree Problem (with visual proof)

Like this video? You might like Roger Nelsen's Proofs without Words books (I may receive a small commission at no cost to you for these affiliate links):
https://amzn.to/41Cci6j
https://amzn.to/41BtR6n
https://amzn.to/48mjeX1

In this short video, we pose a fun problem from the 1976 Journal of Recreational Mathematics and give a visual proof that is featured in Roger Nelsen's Third Proof without Words Compendium.

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 geometric proofs like this check out:

For a static version of this proof, see Roger Nelsen's third "Proof Without Words: Exercises in Visual Thinking" compendium (page 39) :
(I may receive a small commission at no cost to you for these affiliate links):
https://amzn.to/41Cci6j

#math #circle #pythagorean #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof  #mathshorts #mathvideo #incsribed #recreationalmath

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


e o peitinho primeiro é

[Música]

Visual Proof for 1/3 + 1/9 + 1/27+ 1/81 + ... = 1/2 #eduwow #reels #shorts

Visual Proof for 1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2

#shorts #math