graph the lines yal X and Y = RX + a

where R and A are positive real numbers

with r less than 1 these intersect at

the point where the X and the y

coordinate are both a over 1 - R the

vertical distance on the y- AIS between

the two curves is a and that means this

horizontal distance is also a because it

forms an isoceles triangle using the

line yal X as we continue to step up the

staircase diagram we can measure the

vertical distance as the difference

between two curves and the horizontal

distance will always be the same as that

one because it again forms an isoc

triangle with yal x the vertical and

horizontal lengths in the staircase

diagram are all of the form a * R to the

I as I increases and this means that the

widths decrease as the staircase

converges to the intersection point if

we focus only on the horizontal line

segments and start dropping them to the

x-axis we see that a plus a r plus a r 2

plus a R cubed and so on eventually

creates the entire line segment on the

x- axis from 0 to a over 1 - R therefore

the infinite sum of the geometric series

with first term a and common ratio R

must equal a over 1 - r

Infinite Geometric Series Sum

This is a short, animated visual proof demonstrating the infinite geometric series formula for any positive ratio r with r less than 1 and with positive first term a. This series is important for many results in calculus, discrete mathematics, and combinatorics.

If you like this content, you can support my work at https://www.buymeacoffee.com/VisualProofs  Thanks!

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

This animation is based on a proof by The Viewpoints 2000 group from the October 2001 issue of Mathematics Magazine page 320 (https://www.jstor.org/stable/2691106 ).

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

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


par

[Música]

[Música]

non

Visual proof of the theorem Pythagoras! #maths #geometry #pythagoras_theorem #youtubeshorts


here's an interesting way to approximate

Pi on the positive integer grid up to

101 shade the coordinate XY blue if the

greatest common divisor of X and Y is

one and shade the coordinate XY red

otherwise if you do this you get this

image shown here it's a famous result

that the quantity 6 over pi^ 2 is

approximately equal to the ratio of the

number of blue dots to Total dots it

turns out in this picture there are

6,287 blue dots and 10,21 total dots

this means that 6 over pi^ 2 is

approximately 0.

61631 and from this we conclude that Pi

is approximately

3.12 not the best approximation but

definitely a fun and beautiful one

A rough Pi approximation from coprime integers

In this short, plot all the points in the positive integer grid up to 101. If a coordinate (x,y) has numbers x and y that are relatively prime, we shade the dot blue; otherwise we shade the dot red. A famous result states that roughly 6/pi*pi of the dots will be blue. So we can use this image to get an approximation for 6/pi*pi and in turn use that to get an approximation for Pi. We get 3.12, which is not really good, but it’s kind of amazing that such a small grid gets us that close. Using a larger grid will give you a better approximation.

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

For more information about why this works, check out the wikipedia site:

https://en.wikipedia.org/wiki/Coprime_integers

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

#manim #approximation  #relativelyprime #zetafunction #baselproblem #irrational #Pi #mathvideo #math #mtbos  #animation #iteachmath #mathematics #piday

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


in the 12th century Rabbi Abraham Barr

here discovered an amazing visual proof

of a circle's area so imagine a circle

filled with concentric circles whose

radiuses are and circumference is 2 pi r

then cut along one of the radius length

and open the circle into straight line

segments if repeated this process for

infinitely fine line segments the final

shape will be a triangle whose altitude

is r and bass is the circumference of

circle 2 pi r the triangles area is then

half times two pi r times R which

evaluates to be pi r squared therefore

the circle's area is pi times r squared

Unbelievable Visual Demonstration of Circle's Area

Visual Proof of Circle's Area | Area of Circle by Rabbi Abraham bar Ḥiyya| Stay Tuned

In the Mid 12th century, Rabbi Abraham bar Hiyya Hansai discovered an amazing visual proof of a circle's area.

Know About Rabbi Abraham bar Hiyya Hanasi:
https://en.m.wikipedia.org/wiki/Abraham_bar_Hiyya

Area of a Circle by Rabbi Abraham bar Hiyya Hanasi:
https://u.cs.biu.ac.il/~tsaban/Circles.html

About the Video:
Welcome to our mind-blowing showcase of jaw-dropping visuals demonstrating the awe-inspiring concept of a circle's area! In this captivating video, we unravel the secrets behind calculating the area of a circle through an extraordinary visual representation like never seen before. Prepare to be amazed as we unveil mesmerizing geometric patterns and unveil the mathematical wonders hidden within circles. Our experts walk you through each step of the process, providing clear explanations and enlightening insights. Whether you're a math enthusiast or just curious about the beauty of geometry, this video is a must-watch! Join us on this unique journey of exploration and expand your understanding of the remarkable properties possessed by circles.

Stay Tuned brings learning to life through its captivating educational videos.

#AreaOfCircle #Mensuration #AreaOfTriangle #Math #Mathematics #CircleArea #GeometricPatterns #MathematicalWonders #MathEnthusiast #GeometryBeauty #VisualDemonstration #JawDroppingVisuals #CalculateArea #StayTuned

Related Keywords:
circle, area of a circle, geometric patterns, mathematical wonders, math enthusiast, beauty of geometry, visual representation, jaw-dropping visuals, calculating area, remarkable properties, mind-blowing showcase, extraordinary, mesmerizing, captivating video, math concepts, unique journey, proof of area of circle

Schedule: New Video Every Fortnight

To stay updated, subscribe to our YouTube channel: https://youtube.com/channel/UCUHsuNo5zuz_FYu2rwVkIKA


let's check out an interesting infinite

series you may be familiar with this

infinite series the sum where n ranges

from 0 to Infinity of 1 / n factorial

this infinite sum is famously equal to

the value e consequently this infinite

sum where n ranges from 1 to Infinity of

1 n + 1 factorial must equal eus 2 what

if we modify this series slightly and we

consider the sum where n ranges from 1

to Infinity of n / n +1 factorial this

is the infinite sum 1 over 2 factorial

plus 2 3 factorial plus 3 over 4

factorial and so on what's this infinite

sum this is a series of positive terms

and we can show using a geometric

dissection argument that the partial

sums are bounded above moreover the

dissection argument will show us the

actual sum of the

series let's consider a square whose

side lengths are both one we can cut

this Square in half and shade one of

those halves so that we've shaded one

copy of one2 then we can cut the

remaining half into three equal pieces

each of Area 1 six and shade two of them

so that we've shade 2 * 1 over6 or 2 * 1

3 factorial we can cut the remaining 1

over 3 factorial area into four equal

pieces each of Area 1 24 or 1 4

factorial when we shade three of them

we've shaded 3 * 1 4 factorial now cut

the remaining 1 over4 factorial into

five equal pieces and shade four of them

thus shading 4 * 1 over 5 factorial

repeat this process taking the 1 over 5

factorial that's left cut it into six

equal pieces and shade five of them

shading 5 * 1 over 6 factorial let's do

this again cutting the remaining 1 over

6 factorial into seven equal pieces and

shading six of them for a total of 6 * 1

over 7 factorial we can continue this

process indefinitely each time cutting

the remaining 1 n factorial into n + 1

equal pieces and shading n of them

shading a total area of n / n + 1

factorial at each stage

but now we see that the infinite sum of

interest is the limiting value of the

shaded areas and more importantly we see

that in the limit the entire square area

is shaded therefore the infinite sum

that we're interested in must equal the

area of the square which is 1 * 1 so we

have seen by a geometric dissection that

the sum where n ranges from 1 to

Infinity of n / by n + 1 factorial must

be 1 this is truly amazing that it's

such a simple value given that the two

infinite series that this infinite sum

was related to had values of e and e

minus 2 which are both

irrational

A Factorial Series Geometric Dissection (visual proof)

This is animated visual proof demonstrating the sum of the infinite series of of factorials, scaled by an appropriate value each time. This series is closed related to the famous infinite series for Euler's constant e.

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

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

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

This animation is based on a visual proof by John Mason  from the November 1995 issue of The College Mathematics Journal (https://www.jstor.org/stable/2687383 - page 381).

 #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #factorials #factorial #series #infiniteseries #fractal #infinitesum  #algebra #eulerconstant #e

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


have you ever wanted to trisect an angle

stick around and I'll show you one

technique to do this

the problem of trisecting an angle

requires us to take a given angle Alpha

and using classical tools find two lines

that trisect the angle into three pieces

all with an angle measure of alpha over

three it turns out that you cannot

trisect a general angle using a straight

edge Encompass but it might be possible

if we add an extra tool

polar curve R equals Theta is called the

Archimedes spiral let's take a look at

the graph of this spiral right here

you cannot construct the Archimedes

spiral using Straight Edge Encompass

however if we add the spiral as one of

our tools we may be able to use this

tool along with the straight edge

Encompass to trisect the angle we won't

actually need the entire Archimedes

spiral to do this just the bit that lies

between 0 and 2 pi

let's imagine that we have the unit

circle drawn like this

and that we have two points one along

the x-axis and the other one creating a

line with an angle of alpha the arc on

the unit circle subtending the angle

Alpha has an arc length of length Alpha

now let's draw the Archimedes spiral R

equals Theta this intersects our Top

Line at a point let's call the point P

because P lies on the line created by

angle Alpha it has an angle of Alpha and

because it lies on the Archimedes spiral

it must create a radius of length Alpha

from the origin we can then use our

straight edge Encompass to divide the

line from the origin to the point P into

three equal pieces like this we use our

Compass to create this Arc of radius

Alpha over 3. it intersects the

archimedian spiral at a point q and

because Q has a radius of alpha over 3

its angle must be Alpha over 3.

similarly we construct this Arc of

radius 2 Alpha over 3 which intersects

the Archimedean spiral at a point R

because R has a radius of 2 Alpha over

three its angle is 2 Alpha over three as

well

and now we can finish off our problem we

connect the origin to point q and the

origin to point R

and sure enough we have divided the

angle Alpha into three separate angles

all of measure Alpha over three

this means that we've used the

Archimedes spiral to trisect any angle

Alpha

Trisect an Angle with Archimedean Spiral (visual proof)

This is a short, animated visual proof demonstrating how to trisect any angle using the Archimedean spiral. #manim #math #mathshorts #mathvideo  #trisect #trisectangle #impossible #geometry #chords #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #inequality

This animation is based on a visual proof from the wonderful book Charming Proofs by Claudi Alsina and Roger Nelsen: (Affiliate link; I may get small commissions) https://amzn.to/3LbcpgG

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


[موسيقى]

n

[موسيقى]

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

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

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


satu motion and leonardo da vinci

devised an ingenious method to find the

area of a circle

partition the circle into slices

split the circle in half and rearrange

corresponding slices to form a

parallelogram-like

shape

if we increase the number of slices and

repeat the entire process

the rearranged shape will approach the

shape of a rectangle

whose width is equal to the radius of

the circle and height is equal to half

the circumference or pi times r

the circle's area

is equal to the rectangle's area

which is r times pi r

and that equals pi r squared

thus the circle's area is pi r squared

incredible

Do you know Leonardo da Vinci's circle area proof?

Happy Pi Day! Sato Moshun and Leonardo da Vinci independently devised an incredible visualization to find the formula for a circle's area. #Shorts

(Apologies there was no early access on Patreon--I finished making the video just in time for Pi Day).

Wikipedia
https://en.wikipedia.org/wiki/Area_of_a_circle#Rearrangement_proof
Cut The Knot area of a circle
https://www.cut-the-knot.org/Curriculum/Geometry/AreaOfCircle.shtml
Area of a circle, formula explained (mathematicsonline)
https://www.youtube.com/watch?v=YokKp3pwVFc
Understanding the Formula for Area of a Circle #shorts (mathematicsonline)
https://youtu.be/0RGz_rvXVpk

Subscribe: https://www.youtube.com/user/MindYourDecisions?sub_confirmation=1

Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

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2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.

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foreign

triangle with sides of length a b and c

draw the inner circle of this triangle

and draw the in radii that are

perpendicular to the three sides call

the length of the in radius R the angle

bisectors meet at the in center of the

triangle and together they create these

three triangular regions which we can

unfold

each of these triangles has a height of

length R so the area of the triangle is

one half times a times R plus one half

times B times R plus one-half times c

times R factoring this out we see that

the area of the triangle is R times the

quantity a plus b plus C all divided by

two

so the area of a triangle is given by

the product of the length of the in

radius and the semi-perimeter

Triangle area formula

This short video shows how to prove that the area of the triangle is given by the product of the inradius and the semiperimeter. Here is a longer, alternate version: https://youtu.be/lRDtdIyFhpE

Thanks to Colin Wright for suggesting this particular visualization.
_______________________________
Music in this video:
Prelude No. 9 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/


[موسيقى]

Visual proof of:Surface area of Sphere and Volume of sphere#maths #mathematics


कई बार चीजों का एक बहुत ही सुंदर विजुअल

सॉल्यूशन भी होता है जैसे इस शेप को कहते

हैं एलर रिंग और इसका एरिया इसके इनर सरकम

फरेंस के कॉड टेंज पर ही निर्भर करता है

यह चीज आप बड़े सर्कल से छोटे सर्कल का

एरिया सबस्टैक करके फिर पाइथागोरस थ्योरम

लगा के भी निकाल सकते हो अब अगर विजुअल

अप्रोच देखें तो मैं अगर इस एनल रिंग को

छोटे-छोटे स्लाइसेज में डिवाइड करूं तो यह

शेप अब आपको एक कैमरा एपर्चर की तरह दिख

रहा होगा और अगर मैं इसको स्क्विज करूं तो

यह कैमरा अपर्ट क्लोज होके लगभग एक सर्कल

का एप्रोक्सीमेशन बन जाएगा

क्योंकि य स्लाइस की साइज बड़ी थी तो यह

सर्कल का केवल एप्रोक्सीमेशन है पर अगर

मैं काफी छोटे छोटे स्लाइस यूज करूं तो

आपको एक परफेक्ट सर्कल मिलेगा जिसका एरिया

होगा पा स्क्वा यानी कि एनर रिंग का एरिया

केवल इनर क टेंज के स्क्वायर पर डिपेंड

करता

Visual Proofs


दोस्तों मैच की एक फेमस इनिक्वालिटी है

तो उनका अर्थमैटिक मिन ए + बी / 2 या तो

बड़ा होगा या बराबर होगा उनके ज्यामितीय

मिन से माने स्क्वायर रूट अब से ऐसा क्यों

होता है विजुअल समझते हैं एक पॉइंट पी के

दोनों तरफ ड्रॉ करते हैं लाइन सेगमेंट्स

लेंथ ए और बीके अब ए + बी को डायमीटर मां

कर और पॉइंट सी को सेंटर मां के ड्रॉ करते

हैं एक सर्कल तो ये सर्कल का रेडियस है और

भी का अर्थमैटिक मैं माने ए + बी =

येलो लाइन से अब अगर पॉइंट पी पे डायमीटर

के परपेंडिकुलर एक और बनाएं और कॉर्ड के

आधे हिस्से को नाम दे ह तो सर्कल की कॉल्ड

इंटरसेक्शन थ्योरम से मिलेगा ह * ह = अब

या ह = स्क्वायर रूट अब माने ह बना ए और

बी का ज्यामितीय मैं तो फाइनली लेफ्ट वाली

येलो लाइन की लेंथ है ए और बी का

ज्यामितीय मैं और सेंटर वाली येलो लाइन की

लेंथ है अर्थमैटिक मैं क्लियर है की ए और

बी किसी भी प्रोपोरेशन में हो सेंटर वाली

लाइन की लेंथ लेफ्ट वाली लाइन की लेंथ से

बड़ी होगी या उसके बराबर होगी मतलब

अर्थमैटिक मैं हमेशा जोमेट्रिक मैं के

बराबर होगा या उससे बड़ा होगा

Visual Proof - AM GM Inequality

Visual Proof - AM GM Inequality


draw three mutually tangent unit circles

like this these three circles enclose an

area as shown here can you determine the

area of the enclosed shape pause the

video and think about it before I show

you an answer one way to answer this

problem is to realize that the centers

of the three circles can be connected to

form an equilateral triangle with a side

length of two this has an area ofare <

TK of three but then this equilateral

triangle can be decomposed into these

three wedge shaped pieces in the area we

want each of the wed shaped pieces has

an area ofk 6 so the area we want is the

< TK 3 - 3 * > 6 which simplifies to the

< of 3 - pi/ 2 let me know in the

comments if you solved it a different

way

What’s the area?

This is a short, animated visual proof finding the area bounded between three mutually tangent unit circles.

Have a different solution? Share it in the comments!

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

Check out these related videos:
https://youtu.be/5VaQVXECaVM
https://youtu.be/6DnnJlzbCC4
https://youtu.be/PNQxS6antBY

This animation is based on an exercise from Roger Nelsen's and Claudi Alsina's wonderful book "Icons of Mathematics". You can find the book and other information about it at the links below

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

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

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

 #manim #math #mathvideo #mathshorts #geometry #hexagon #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #root3 #pww #proofwithoutwords  #proof #algebra #areas #mathematics #mathvideo #mtbos

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


visual proof of area of circle split the

circle into four parts and arrange them

like this do it again with eight

partitions do it again with 16

partitions repeat this pattern infinite

times

and you will see that entire circle fits

in a rectangle the dimensions of the

rectangle will be pi r and r

which gives us the area effectively pi r

square

[Music]

Visual proof that Area of Circle is equal to Area of Rectangle #VisualRepresentation #MathAnimation

Visual proof of Area of Circle.

As we know the area of a rectangle is equal to “base x height”.

The height of our rectangle will be the same as the radius.

To find the base we will need the circumference of the circle.

When we compared the base and the circumference we will get the base is equal to one-half the circumference.

Since the circumference is equal to 2πr, the base has to be πr.

Now “base x height” becomes “πr x r”.

That gives us πr² which is equal to the area of the rectangle which is equal to the area of the circle.

For more such educational reels follow @championsacademy.in

Tag your friend
Share the knowledge!

#ProofOfAreaOfCircle #AreaOfCircle #Radius #Circumference #Diameter #Circle #Algebra #Rectangle #Base #Height #Geometry #Mathematics #Diagrams #VisualRepresentation #StudyFromHome

#ChampionsAcademy #AcademicSystem #Coaching #Learning #Online #JEE #NEET #Physics #Chemistry #Biology #CBSE #ICSE #EasyLearning #LearningMadeEasy #OnlineLearning


[సంగీతం]

[ప్రశంస]

[సంగీతం]

[ప్రశంస]

[సంగీతం]

[ప్రశంస]

ఆ

[సంగీతం]

Happy Tau Day!

In this short video, we show one way to visualize the constant tau, which is twice pi, as the area between the curves 1/(1+x^2) and its negation over the entire real line.

Happy Tau Day if you celebrate!

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

#mathshorts #mathvideo #math #short #math #manim #radian #circle #radius #tauday #taurus
#calculus

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


[音楽]

あ

Math visual proofs, geometry of binomial theorem, visual proof of binomial theorem, #viral #shorts

geometry of binomial theorem,
Math visual proofs,proof without words, math, Mathematics visualization, mathematics, Mathematical visual proofs, maths, Easy math, maths for kids, Math basic concepts, basic maths for beginners, maths basic concepts,
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[Music]

consider an

x by x square a joined

with an a by x rectangle

together these shapes enclose a total

area

of x squared plus a times

x now separate the a by x rectangle

cut it in half along the side with

length a and attach the two resulting

smaller rectangles to adjacent sides of

the square

like this these three rectangles now

almost fill a larger square with side

length x

plus a over two however

there is a smaller a over two by a over

two square

missing from the lower right corner

therefore

we see the original enclosed area is

quantity x plus a over two squared

minus the quantity a over two squared

converting x squared plus ax

to the form x plus a over 2 squared

minus a over 2 squared

is called completing the square

Completing the Square (visual proof)

This is a short, animated visual proof showing what is meant when we "complete" the square algebraically and justifying why the formula holds for positive real numbers. #mathshorts #mathvideo #math #completingthesquare #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra

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Background music, used according to CC BY-SA 3.0, is "The Road" from Esteban Orlando:

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Music provided by RFM: https://youtu.be/b4KCM1bIq50


[Music]

consider a right triangle with legs of

length a

b and hypotenuse of length c

the in circle of this triangle is the

largest circle contained in the triangle

it is tangent to each of the three sides

the radius r of this circle is called

the in radius of the triangle

the inradius can be used to decompose

the triangle into four triangles and a

square as demonstrated

if we glue a second rotated copy of this

triangle to the original along the

hypotenuse we end up with a rectangle

decomposed into eight triangles and two

squares

we can then arrange the pieces of this

rectangle dissection to form a different

rectangle

[Music]

from the decomposition we can see that

this new rectangle has length given by a

plus b

plus c

and the width of the rectangle is given

by the in radius r

because the areas of these two

rectangles must be equal

we can equate the areas to see that a

times b

must be equal

to r times

a plus b plus c

but then we can use this equality

to find that r

is equal

to a times b divided by a plus b plus c

giving us a formula for the n radius of

a right triangle in terms of its side

lengths

but that's not all

we can do more the four triangles from

the original decomposition

tell us extra information

the leg of one triangle is a minus r and

the leg of the other is b minus r

but these two legs side by side create

the hypotenuse of the triangle so that c

is equal to a minus r

plus

b minus r

we can again use this equation to solve

for r

and we'll see that r is equal to a plus

b minus c all divided by two

we have found two formulas for the n

radius of the right triangle in terms of

the side lengths

[Music]

[Music]

you

Inradius of a Right Triangle (visual proof)

This is a short, animated visual proof of two different formulas for the inradius of a right triangle in terms of its side lengths. #manim #math #mathvideo #mathshorts #geometry #circle #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof #incircle #inradius #righttriangle

This animation is based on a proof due to Liu Hui. You can read more in the following sources:

Article by Philip D. Straffin Jr. in the June 1998 Mathematics Magazine (https://doi.org/10.2307/2691200) pages 163-181

Mathematical Morsels by Ross Honsberger (MAA, 1978)

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

To learn more about animating with manim, check out:
https://manim.community
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Spirit Of The Dream by Keys of Moon | https://soundcloud.com/keysofmoon
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start with a right triangle with lengths

a b and c create four congruent copies

of such triangle arrange the congruent

triangles to form a square the empty

yellow square has a side length of C

arrange the congruent triangles so that

they look exactly like the yellow Square

the small red square has a side length

of B minus a the area of the four

triangles is 12 a b * 4 plus the area of

the small red square which is the square

of B minus a

these are all equal to the area of the

yellow Square which is c² 12 of ab * 4

becomes 2ab plus the expansion of the

square of B minus a which is square of a

- 2 a + b² = c² 2 a and -2ab is 0 so the

resulting equation is square of a + b² =

c² this is the Pythagorean theorem

Visual + Algebraic Proof of Pythagorean Theorem #mathviral #maths #algebra #mathtrick  #mathisfun


[موسيقى]

Triangle angle sum property visual proof #academicdiary #maths #triangle

a simple visual proof of angle sum property of a triangle.


[Music]

start with a right triangle with legs of

length a

and b

and a hypotenuse of length c

the square drawn using the side of

length a has an area of a squared

similarly

the square drawn using the side of

length b has an area of b squared

and finally the square drawn using the

hypotenuse of length c has an area of c

squared

[Music]

then add two more copies of the right

triangle one reflected about the angle

between the legs and one rotated and

shifted to the bottom of the square off

of the hypotenuse

next we draw two lines of symmetry

separating two hexagons inside of the

diagram like this

[Music]

we use the new regions as follows

this top quadrilateral is congruent to

its reflection along the first line of

symmetry

the resulting quadrilateral is congruent

to its rotation of 90 degrees clockwise

about the left triangle vertex and this

quadrilateral is congruent by 180 degree

rotation to the other half of the second

hexagon cut by the second line of

symmetry

therefore the two overlapping hexagons

in the diagram are both built from two

copies of the same quadrilateral

that means that these two hexagons

doesn't close the exact same area

but notice that both of these hexagons

use two copies of the original right

triangle

so removing those triangles from both

diagrams shows that the large square

encloses the exact same area as the two

smaller squares which means that c

squared is the same as a squared plus

b squared

[Music]

remembering that a and b

are the lengths of two legs of a right

triangle with hypotenuse of length c

we have shown that for such a right

triangle c squared is equal to a squared

plus

b squared

this is the pythagorean theorem

as proved by da vinci

[Music]

you

Pythagorean Theorem V (visual proof; Leonardo da Vinci)

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using the a diagram that is now attributed to Leonardo da Vinci. The proof uses reflection and rotation symmetry arguments. 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 #leonardodavinci #davinci

01:30 I said something incorrect so made a small cut. I meant to say the two hexagons are made from congruent parts.

If you liked this video, please consider subscribing to the channel and checking out my other videos.

This animation is based on a proof due to Leonardo da Vinci. For a static version of this proof, see Roger Nelsen's first "Proof Without Words II: More Exercises in Visual Thinking" compendium (page 5) : https://bookstore.ams.org/clrm-14/. You can also check out Howard Eves' "Great Moments in Mathematics (Before 1650)" page 29--31 and Roger Nelsen's book "Math Made Visual: Creating Images for Understanding Mathematics" : https://bookstore.ams.org/clrm-28.

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

To learn more about animating with manim, check out:
https://manim.community
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Tranquil Fields (Tense) by Alexander Nakarada | https://www.serpentsoundstudios.com
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[Music]

[Music]

[Music]

[Music]

[Music]

[Music]

hmm

[Music]

[Music]

[Music]

so

[Music]

[Music]

hmm

[Music]

you

Pell Number Lattice Path Enumeration and visual recurrence formula (synthwave; visual proof)

This synthwave enumeration shows all of the lattice paths with steps (1,1), (-1,1) and (2,0) from the origin (0,0) to the line y = n-1. The number of such lattice paths is counted by the Pell numbers, and we use the visual enumeration to show how to produce the (linear) recurrence formula for the Pell numbers. To see other lattice path enumerations, check out https://youtu.be/9YUl0k2FYzc, https://youtu.be/KYpOoA2D63w, and https://youtu.be/__w_MDQeUYc.

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

#pellnumbers #pell #mathvideo #math #mtbos #manim #animation #theorem #visualproof #proof #iteachmath #mathematics #linearrecurrence #linear #latticepaths #discretemath #combinatorics #enumeration #synthwave #recurrence

To learn more about animating with manim, check out:
https://manim.community
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here's an intriguing three-dimensional

way to sum

squares start with 6 1X 1 by 1 cuboids

as shown here each of these cuboids

contains a volume of 1^ 2 * 1 so the

total volume of these six is 6 * 1

squared we can glue these six together

to form this cuboid now we can enclose

this shape using six 2x 2x1 cuboids two

on top and two on bottom and one in the

front and the back each each of these

cuboids contains a volume of 2^ 2 * 1 so

when we glue them together to form a

shell we add a volume of 6 * 2^ 2 now

let's create a new shell for this shape

by using six 3x3 by1 cuboids two on the

left and the right each and one each in

the front and the back each of these

cuboids contains a volume of 3^ 2 * 1 so

the full shell created here encloses a

volume of 6 * 3^ 2 we can continue this

process as long as we want at each step

we add six copies of a k by K by1 cuboid

alternating between putting two on top

and two on bottom and two on the left

and two on the right when we finally

stop our last stage consists of adding

six copies of an N byn by one cuboid

each of these cuboids contains a volume

of n squ * 1 so the final shell shown

here encloses a volume of 6 * n^ 2 this

process of adding shells at each stage

has built a filled in cuboid whose

volume is given by this finite sum shown

here 6 * 1 2 + 6 * 2^ 2 added all the

way up to 6 * n^ 2 on the other hand

this cuboid has Dimensions n by n + 1 by

2 n + 1 because the final cuboids we

added were n by n by 1 but this means

that the volume of the cuboid is n * n +

1 * 2 n + 1 from this we see that the

finite sum where I ranges from 1 to n of

i^ 2 is n * n + 1 * 2 n + 1 all divided

to recap we've just proved the famous

sum of squares formula by stacking

cuboidal arrays inside of cuboidal

shells as shown here the resulting

formula says that the sum of the first n

squares is equal to n * n + 1 * 2 n + 1

all divided

An Amazing way to Sum Squares (visual proof V using cuboids)

This is a short, animated visual proof for the sum of squares formula using cuboids and cuboidal shells. This is the fifth in a series showing various ways to find the famous sum of squares formula.

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

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

https://youtu.be/a8j3sBrXchg
https://youtu.be/WidzHiUFWNA
https://youtu.be/-tJhH_k2LaM
https://youtu.be/UqVmocdLFGc
https://youtu.be/Q-frL0Ot2m4

This animation is based on a visual proof by Sasho Kalajdzievski from the June 2000 issue of The Mathematical Intelligencer (https://doi.org/10.1007/BF03025282 pages 47--49 ).

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

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


foreign

[Music]

Visual Proof of Pythagoras' Theorem

Made using Pivot Animator 5. As the triangles move, the grey area doesn't change, so we can equate the two. #shorts


using the classical tools of a straight

edge and compass for which n can we

divide a circle into n parts of equal

area

if we start playing around with this

problem at first it doesn't seem too

interesting we can construct two

antipodal points on a circle and that

divides the circle into two pieces of

equal area

we can construct an equilateral triangle

on the circle and that will allow us to

divide the circle into three pieces of

equal area

proceeding in a similar manner we can

construct the regular square

on the boundary of the circle dividing

the circle into four pieces of equal

area

and then we can do the same thing with

the regular pentagon and the regular

hexagon dividing the circle into five

pieces of equal area and six pieces of

equal area respectively

we might think we can continue this

pattern and just construct the regular

heptagon on the circle which then

divides the circle into seven equal

pieces

but there's a problem here it turns out

the regular heptagon is not

constructable using the classical tools

of a straight edge and compass

so that leaves us with a question can we

divide a circle into seven equal parts

using just a straight edge and compass

let's find out

start with a line segment that will

serve as the diameter of a semi-circle

like this

then divide the diameter of the

semicircle into seven congruent pieces

which can be done with a straight edging

compass let's call the length of these

pieces d

after bisecting segments when necessary

use the compass to construct these

nested semicircles for a total of seven

pictured semicircles

let's compute the area of the

semicircles first the first semi-circle

has area pi times d squared over eight

the second one is four pi d squared over

eight then nine pi d squared over 8 and

so on each time we have the nth square

times pi d squared over 8.

if we call the area pi d squared over 8

the value x

then the areas of the circles are x 4x

9x 16x and so on each time we have the

next square times x

we can use these values to determine the

areas in the circle the first region has

an area of x then 3x then 5x 7x and so

on each time we're taking the difference

of successive squares to get the next

odd number

and now we apply our trick

this time we construct the exact same

regions in the bottom semicircle but

this time from right to left

once we adjoin the regions like this we

see that each of the seven created

regions has an area of 14 times x

that means that each of these regions

has exactly the same area

consequently this means that we have

divided the circle into seven pieces of

equal area using a straight edge and

compass

you

Dividing a Circle into Seven Equal Areas (visual proof)

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

Perhaps this should be titled "The Best Way to Cut Pizza into 7 pieces" based on the comments in this video :)

There were also a number of comments asking about how to divide a line into seven equal lengths, so I have created a short, wordless animation showing that procedure if you're interested: https://youtu.be/afZiAGaKEY4

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

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

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

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

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

Source: http://chriszabriskie.com/divider/

Artist: http://chriszabriskie.com/


[Music]

you cannot square the circle using

classical tools but you can if you use

other techniques

if we let b be greater than zero then

the polar curve r equals b times theta

is called the archimedean spiral and it

looks like this

let's use the archimedean spiral to

square the circle

start with a radius a greater than zero

then

consider the circle of radius a centered

at the origin like this

[Music]

this circle

has an area given by pi times the radius

squared

so the area of the circle is equal to pi

times a squared

extend the radius to the other side

creating a line segment of length two a

and then draw the archimedean spiral a

times theta

this spiral intersects the y axis at the

y coordinate a pi over two

so we can draw a rectangle using these

two line segments and the rectangle has

lengths 2a and a pi over 2.

therefore we see that the area of the

rectangle

must be a pi over 2 times 2a which is pi

a squared or the area of the circle

now let's square the rectangle

to do this we can draw the circle

with radius given by xy which is the

short edge of the rectangle and we can

extend

line segment xz out to this circle to

create a point called e

now use e and z

to find the midpoint

which we'll call m

[Music]

next

draw the circle centered at m

with radius m z

like this

[Music]

then extend

the line segment y x

with a half chord down to the circle

creating an intersection point s

we can use the line segment x s

to draw a square like this

[Music]

according to the intersecting chords

theorem we see that x s multiplied by x

s

is going to be equal to xz multiplied by

xe

but because xe equals xy

that means that the area of the square

and the area of the rectangle must be

equal

putting these three facts together

we see that the area of the square is

equal to the area of the rectangle which

in turn is equal to the area of the

original circle

but this means

that we have used the archimedean spiral

to square the original circle

[Music]

you

Squaring the Circle with the Archimedean Spiral (animated visual proof)

This is a short, animated visual proof that we can square the circle IF we use the Archimedean spiral. Unfortunately, this is not a solution to the squaring the circle problem from antiquity because that requires it to be done with only a straightedge and compass. #mathshorts #mathvideo #math ##geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #circle #mathematics #thales #squaringthecircle #squaringtherectangle

This animation is based on a visual proof from Roger Nelsen's and Claudi Alsina's wonderul book "Charming Proofs: A Journey into Elegant Mathematics" (Theorem 9.5 from page 145). You can find the book and other information about it at the links below

https://bookstore.ams.org/dol-42/

https://www.maa.org/press/maa-reviews/charming-proofs-a-journey-into-elegant-mathematics

https://www.amazon.com/Charming-Proofs-Mathematics-Mathematical-Expositions/dp/0883853485

Here is another video that squares the circle with a nonclassical technique: https://youtu.be/0D3KiCmum90

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


ही तकदीर तो क्या टूटी शमशीर तो क्या टूटी

शमशीर से ही

हो कर हर मैदान पते

Sum of odd no.(Visual proof)😳📝#viral #trending #dsssb #math #shorts


here's a visual take on a math fact that

often creates controversy what is the

number 0.9 repeating start with a circle

of Area 1 and take a scaled copy that

has an area of 1 10. then the

surrounding annulus has an area of nine

tenths now we can repeat this process on

the inner circle this time we take one

tenth of one-tenth for the smaller

Circle which leaves us with nine tenths

of one-tenths for the annulus region

which is nine over a hundred imagine

we've repeated this process infinitely

so each time we have a small circle we

take a smaller sub circle of Area 1 10

and shade the annular region of nine

tenths of the previous small circle the

total area shaded is nine tens plus nine

one hundredths plus nine one thousandths

and so on but because the sequence of

radii of the inner circles converges to

zero this means every single point in

the circle except the center point is

contained in one of the annular regions

by the Archimedean principle so the

total shaded area is the area of the

circle which is one but this infinite

sum is precisely 0.9 repeating which

means 0.9 repeating is one

0.999999… = 1

In this short, we use a geometric representation and the Archimedean principle to show why 0.99999... equals 1. This fact tends to be a bit controversial in the mathematics classroom, but the only way to make sense of an infinite sum is to treat it as a limit. Thus, we can use the term equality when discussing infinite sums like 0.999... and there is no number this sum could equal besides 1.

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

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

To see a longer video with a similar fact (and perhaps more justification at the end), see
https://youtu.be/BGsJjHbjcnA

Here is a playlist with other geometric sums dissection proofs:
https://youtube.com/playlist?list=PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw

#infiniteseries #paradox #manim #math #mathshorts #visualproof #proofwithoutwords

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


[Music]

[Applause]

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[Applause]

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[Applause]

[Music]

[Applause]

n

(a+b)^2 formula, visual proof|a plus b whole square formula |#square #maths #classviii

(a+b) ^2 formula visual proof
step by step explanation
class 8 formula proof
subscribe for watching more like this


approved Pythagoras Theorem really

quickly I've drawn a square here with

another Square inside it and this gives

me four congruent right angled triangles

which have these side lengths A and B

all equal to each other if we calculate

the area of the large Square we can do

that in two different ways I can just

think of it as a large square that has

side length a plus b so its area is a

plus b squared and if you multiply out a

plus b squared you get a squared plus b

squared plus 2 a b or I can think of it

as these four congruent right angle

triangles their areas are half base

times height so a half a times B so I

get a half a times B times four because

there are four of them and then I also

need to include this Square in the

middle which is a square of side length

C so it has area c squared so four times

a half is two that's two a B plus c

squared but these two expressions

represent the same area so they must be

equal to one another but they both

contain two a b so I can subtract that

from both sides of the equation so that

leaves us with Pythagoras theorem in the

familiar form a squared plus b squared

equals c squared and so that's how

Easiest Proof of Pythagoras' Theorem

Proof of Pythagoras' Theorem using just basic areas and algebra in less than a minute!
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let's consider a right triangle with

perpendicular sides are a and b units

and the hypotenuse is c units

then according to pythagoras theorem a

square plus b square is equal to c

square

that is sum of squares of perpendicular

sides is equal to the square of

hypotenuse but my question is why

why is this true for any right triangle

for the detailed explanation see the

comment box and for more videos please

visit and subscribe the channel

Pythagoras theorem - proof with visual explanation | Hypotenuse of Right triangle | Mathematics

Link for explanation : https://youtu.be/0SorlZmPNOo

Any queries regarding the subject or videos are invited.
Gmail ID : problemanalysis99@gmail.com
Instagram ID : problem_analysis

#shorts #shortvideos #maths #mathematics #pythagorastheorem #righttriangle #hypotenuse #geometry


[موسيقى]

h

[موسيقى]

#trending #mathematics Visual Proof of Roots of Quadratic Equation

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


start with a circle where the radius is

length R here's an interesting question

how many arcs of length R can we fit

around the circle

certainly one is not enough and we can

see pretty quickly that two Arc lengths

with a length of R is not enough how

about three that looks like it could

almost cover half

we can check that 4 won't do the job

and that 5 won't do the job either as

there's still plenty of circle left over

however when we try to cover this circle

with six arcs of length R That Just

About Does it and only leaves a little

bit if we try with seven different arcs

of radius R it covers the entire circle

and there's overlap so we can fit six

arcs of length R around the circle each

such Arc of length R subtends an angle

of one radian so this means that we can

fit more than six radians in a circle in

fact the total number two Pi of radians

it fits in a circle is approximately

How many radii fit around a circle? #tauday #tau

In this short, we investigate how many radii of a circle would fit if you lay them in arcs around a circle. The method of laying a single radius around the circle creates an angle subtended by that arc - that angle is known as 1 radian. So in this video we see that there are more than 6 radians and less than 7 radians in a circle. The full measurement is actually around 6.28 and is the constant 2*Pi.

#short #math #manim #radian #circle #radius #tauday #tau


start with a right triangle with sides

of length A and B and a hypotenuse of

length C

then draw an A by a square and next to

that draw a b by b square we can glue

these two squares together to form this

shape right here

inside of this shape we can fit one copy

of the triangle vertically like this and

we can fit a second copy horizontally

like this now we can take the two

triangular regions and rotate them as

pictured here the resulting shape is a c

by C Square this means that if a and b

are the legs of a right triangle and C

is the hypotenuse we can take an a by a

square and a b by b square and decompose

it to fit in a c by c squared that means

a squared plus b squared equals c

squared When A and B are the legs of a

right triangle with hypotenuse C this is

the Pythagorean theorem

Pythagorean Theorem IX (hinged visual proof)

This is a short visual proof of the Pythagorean theorem using swinging hinges.

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 playlist with all my videos that prove or feature the Pythagorean theorem: https://www.youtube.com/playlist?list=PLZh9gzIvXQUvlRlnYryejWGHYQquM92rB

This animation is based on Proof 2 in Bogomolny's site cut the knot: https://www.cut-the-knot.org/pythagoras/. Check that link for more about the history of this proof.

#math #manim #shorts #pythagoreantheorem #proofwithoutwords #visualproof #theorem #proof


start with a right triangle with legs of

length a b and a hypotenuse of length C

first take a copy of this triangle and

scale up by the value B

next take another copy of the triangle

and scale up by the value a

because the leg lengths a b and b a are

equal we see that the triangles fit

together this way and because the two

non-right angles in the original

triangle are complementary this larger

triangle we formed is a right triangle

now scale the original triangle up one

more time by the value C we now have two

right triangles with equal leg lengths

this means that the two triangles are

congruent and so their hypotenuse

lengths are equal

therefore a squared plus b squared must

be equal to c squared where A and B are

the legs of a right triangle and C is

the hypotenuse this is the Pythagorean

theorem

Pythagorean theorem proof by scaling

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using the hypotenuses of scaled triangles. 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 visual proof by Frank Burk from the November 1996 issue of The College Mathematics Journal (https://doi.org/10.2307/2687335 page 409).

Here is my original animation of this proof: https://youtu.be/q28H5NY5UVU

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


[Music]

[Music]

you

Heron's formula visual proof | mathocube |

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here's one way to make sense of the

alternating sums of the powers of

one-half start with this nomon shaped

object here and assume that it has area

one we can divide the gnomon into four

smaller gnomons and remove two of them

so we removed half and then we can add

one of those back adding one-fourth back

now we can repeat this process on the

added back nomon removing half and then

adding one half of that back we can

continue this process indefinitely each

time we re-add one of the gnomons we

divide it into four nomons remove two

and then add one back we can continue

this process forever or indefinitely and

it turns out that we've shaded the

infinite sum of the alternating powers

of one-half on the other hand if we look

at the picture we see that we've shaded

exactly two-thirds of the diagram so the

sum of the powers of negative one-half

must be equal to two-thirds

A visual alternating sum!

This is a short, animated visual proof demonstrating the sum of the infinite geometric series with ratio -1/2.

For a longer version of this animation (with dramatic music only), check out : https://youtu.be/wLPsEULfPnk

This animation is based on a visual proof by Roger B. Nelsen from the November 2012 issue of The College Math Journal page 370 (https://doi.org/10.4169/college.math.j.43.5.370).

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

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


let's play the chaos game in a square

start with a point anywhere in the

interior of the square now we're going

to roll an eight-sided die each side

corresponds to a Vertex or a midpoint of

the square we then add a new DOT 2/3 of

the way from the current dot that we're

at and the corresponding Vertex or

midpoint from the dice roll let's

iterate this process 30,000 times to see

what

happens amazingly this is the diagram

that arises as the limiting shape of the

process

this particular object is known as the

serinsky

carpet

Chaos Game in a Square

In this short, we show what happens when iterating the procedure of choosing a square vertex or edge midpoint at random and moving two thirds the distance from the current dot to the chosen point.

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

#chaos #chaosgame #square #sierpinski #sierpinskicarpet #mathvideo #math #mtbos #manim #animation #theorem #iteachmath #mathematics #dynamicalsystems  #iteratedfunctionsystem #dynamics #fractals

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


[Music]

start with a circle of radius r

and then draw an inscribed regular two

end gone like this

let's figure out the area of this

regular two n gone

to do that we can decompose the polygon

into triangles

the radius of the circle is r and there

are two n copies of this triangle that

can be used to build the regular two end

gone

so the area of the regular two n gone is

2n times the area of one of these

triangles

now let's draw the inscribed regular

n-gon like this

and suppose that this n-gon has a side

length of s

then the triangle that we're interested

in has a height of length s over two and

a base of length r

therefore the area of the regular two n

gone is two n times one half times r

times s over two but this simplifies to

r over 2 times n s and n s is the

in the circle

but this means that we have shown that

the area of the regular 2-n-gon

inscribed in a circle of radius r is

equal to r over 2 times the perimeter of

the regular n-gon inscribed in the same

circle

but that's not all we can say

because both the two n-gone and the

n-gon approach the circle as we let n go

to infinity we get the result that the

area of the circle is equal to r over 2

multiplied by the circumference of the

circle if we remember that the

circumference of the circle is 2 pi r

then we see that we've found a formula

for the area of a circle of radius r to

be pi times r squared

Areas and Perimeters of Regular Polygons (visual proof)

This is a short, animated visual proof demonstrating that the area of the regular 2n-gon inscribed in a circle of radius r is equal to one-half the product of r and the perimeter of the regular n-gon inscribed in the same circle. We finish with a bonus proof that the area of a circle is one-half the product of its radius and its circumference. #math #geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #polygons #regularpolygon #circlearea #area #perimeter #mathshorts #mathvideo

This animation is based on a visual proof by Roger B. Nelsen from his second compendium of visual proofs:

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

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Only Memories Remain by Hayden Folker | https://soundcloud.com/hayden-folker
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


[музыка]

[музыка]

A

[музыка]

Visual Proof: Natural Numbers Formula#maths

Sum of first n natural number formula visually.#maths #mindsphere

AlgebraIsAwesome #CalculusCrunch #GeometryGuru #ShapesOfMath #StatsSaturday #ProbabilityPuzzle #PrimePower #NumberNerd #ArithmeticAdventures #TrigTuesdays #MathMagic #MathematicsMatters #GeometricGenius #CalculusConcepts #AlgebraicExpressions #GeometryGoals #StatisticalInsights #ProbabilityPlays #PrimeNumbersOnly #NumberTheory
#AlgebraIsAwesome #CalculusCrunch #GeometryGuru #ShapesOfMath #StatsSaturday #ProbabilityPuzzle #PrimePower #NumberNerd #ArithmeticAdventures #TrigTuesdays #MathMagic #MathematicsMatters #GeometricGenius #CalculusConcepts #AlgebraicExpressions #GeometryGoals #StatisticalInsights #ProbabilityPlays #PrimeNumbersOnly #NumberTheory #ArithmeticArt #TrigonometryTricks #MathematicalMind #MathMania #AlgebraAllDay #CalculusCraze #GeometryGenius #StatisticalStudies #ProbabilityPatterns #PrimeNumberPuzzles #NumberNinja #ArithmeticAddict #TrigonometryTales #MathematicalMaster #AlgebraicAdventures #CalculusChallenges #GeometryGreatness #StatisticalSolutions #ProbabilityProblems #PrimePowerhouse #NumberNerdiness #ArithmeticAficionado #TrigonometryTips #MathematicalMusings #AlgebraAcademy #CalculusConundrums #GeometryGalore #StatisticalAnalysis


here's an interesting chaos game variant

start with these two matrices A and B

and consider the apine linear

Transformations ax and BX

-5.5 these two Transformations are

applied to the plane and shown here

start with a point in the plane and

apply both Transformations choose one of

the two output points at random Say by

flipping a coin and apply both

Transformations again then repeat this

process each time select one of the

output points at random and apply both

Transformations let's do this 25,000

times to see what the resulting shape is

no matter the starting point or the

outcome of the random coin flips along

the way the limiting shape will always

look like this the resulting fractal

shape is called the levy C curve or the

levy Dragon it can be generated in other

ways but this is how we generate it

using an iterated function system

Levy Curve from the Chaos Game

In this video, we show how to use a random process of iteratively applying two linear transformations in the real plane to generate fractal curve known as the Lévy Dragon. What happens if you do something similar with different affine linear transformations?

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

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

This animation was inspired by an article written by Larry Riddle that appeared in the the November 2016 issue of Math Horizons:
https://doi.org/10.4169/mathhorizons.24.2.18

See more here (https://larryriddle.agnesscott.org/symmetricfractals.pdf) and on his website: https://larryriddle.agnesscott.org/ifs/levy/levy.htm

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

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


imagine this diagram with two squares 1

1 4 the size of the other encloses a

total area of one now cut both squares

into four equal pieces and remove one

this removes a total of one-fourth of

the area of the diagram on the removed

area cut that into four pieces and

re-add one piece thus adding one-fourth

of one-fourth back we can imagine

repeating this process indefinitely each

time removing one quarter of the area

and then re-adding one quarter of that

removed area back if we do this

infinitely then the resulting limiting

diagram has a shaded area given by the

infinite sum of the powers of negative

one-fourth and now here's the trick we

can shift the Shaded part of the small

square up to the larger Square thus

filling in all of the larger square but

by construction the larger Square was

four times bigger than the small square

and therefore we've shaded exactly

four-fifths of the original area this

means the sum of the powers of negative

one fourth is four-fifths

Alternating 1/4 Sum

This is a short, animated visual proof demonstrating the sum of the infinite geometric series with ratio -1/4.

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, high-definition version of this animation: https://youtu.be/SeKGZ2DLAwo

Here is an alternating series dissection for powers of -1/2: https://youtu.be/NKFrHfBGDNE

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

This proof is based on a visual proof by Ivica Martinjak & Ana Mimica from the May 2021 issue of the College Mathematics Journal (https://doi.org/10.1080/07468342.2021.1909963 page 204).

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

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


foreign

[Music]

start with four positive real numbers a

b x and y and assume a is less than or

equal to B and X is less than y create a

right triangle with legs of length A and

B and let's assume this angle in the

upper left is Alpha

first create a scaled copy of the right

triangle where the scale value is y

then create a rotated and scaled copy of

the triangle where the scale value is X

we see that the angle Alpha appears in

these two locations in the scaled

triangles connecting the other two

vertices of the triangle together like

this creates a right triangle due to the

Pythagorean theorem the legs of this

right triangle are x times the square

root of a squared plus b squared and Y

times the square root of a squared plus

b squared and let's label the angle in

the upper left beta finally we can

create a rectangular diagram from this

trapezoid by adding this right triangle

with legs of length b y minus ax and a y

plus b x the bottom angle in this

triangle must be Alpha plus beta by

construction because it's complementary

to the other acute angle which is

complementary to Alpha plus beta

now we can get an interesting formula by

Computing Alpha plus beta in two ways

first of all it's the arc tangent of a y

plus BX divided by b y minus ax

according to the upper right triangle on

the other hand Alpha is the arctangent

of a y divided by b y according to the

left triangle and beta is the arc

tangent of x times the square root of a

squared plus b squared divided by y

times the square root of a squared plus

b squared according to the inner

triangle notice that we can cancel the

Y's and cancel the square root of a

squared plus b squareds and we end up

with the formula that arc tangent of A

over B plus the arc tangent of X over Y

is equal to the arctangent of a y plus

BX divided by b y minus ax setting both

A over B and X over y to one-fifth in

this equation yields that 2 times the

arctangent of a fifth is equal to the

arctangent of 5 12. setting both A over

B and X over y to 5 12 reveals that 2

times the arctangent of 5 12 is equal to

the arc tangent of 120 over 119 and

finally setting A over B equal to one

over one and X over y equal to 1 over

239 we get the at pi over 4 plus the arc

tangent of 1 over 239 is equal to the

arc tangent of 120 divided by 119. now

let's string these three equations

together use the second equation to

replace arctangent 120 over 119 by 2

times arctangent 5 12. then replace

arctangent 5 12 by 2 times the

arctangent of one-fifth using the first

equation if we rearrange the resulting

formula we get that pi over 4 is equal

to 4 times the arctangent of one-fifth

minus the arc tangent of 1 over 239 this

is called the matron formula and is

attributed to John Machin who discovered

it in 1706 when he used it to compute pi

to 100 decimal places

here are the first 100 decimal places

that Machin would have found

in 1873 William Shanks used the same

formula but this time got 707 digits of

pi as listed here

he was so close but in 1946 an error was

found in his work in the 528th digit

shown here

how many digits of pi do you think

modern computers can compute using this

formula leave a comment let me know

Machin Formula Visualization (Pi day special)

This video describes a relatively new visual proof of the famous Machin formula, which is a formula that allows for relatively speedy computation of the digits of Pi.  As a Pi day bonus, we see Pi to 100 digits, as computed by Machin as well as to 707 digits as computed by Shanks. How many digits do you think it can be used to compute using modern technological tools? #math #pi #piday #geometry #inscribed #circumscribed  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof  #similartriangles #mathvideo #machinformula #machintheorem #digitsofpi #trapezoid

If you enjoyed this video, please consider liking and subscribing. Thanks for watching!

This animation is based on a proof due to Roger Nelsen from the November 2021 issue of The College Mathematics Journal  (https://doi.org/10.1080/07468342.2021.1969860 ) page 355.

Here are two more Pi day gems:

https://youtu.be/-LVpjnaqtwU
https://youtu.be/JE_YeVlSqLY

To learn more about animating with manim, check out:
https://manim.community
____________________________________________________
Background music:
Acoustic Meditation 2 by Audionautix is licensed under a Creative Commons Attribution 4.0 license. https://creativecommons.org/licenses/by/4.0/
Artist: http://audionautix.com/


foreign

let's let T sub K represent the sum of

the first K positive integers the sum

can be represented by a triangular array

of Cubes like the one pictured here

where there's K cubes in the bottom row

and one cube in the top row

so let's consider T sub n plus 1 which

of cubes with n plus 1 cubes on the

bottom

now construct n different copies of this

triangular array and place them side by

side like this this staircase diagram

contains a total of n times T sub n plus

1 with cubes

let's remove the N by n plus 1 rectangle

in the back of this staircase this

rectangle can be decomposed into two

triangular arrays each of these

triangular arrays has n cubes in the

bottom so represents the number T sub n

we can take these two arrays and place

them on either side of the remaining

staircase we see that the staircase

diagram now has a total of n plus two

copies of the Triangular array

representing the number T sub n

therefore the total number of Cubes is n

plus 2 times T sub n but we can use this

formula to see that t sub n plus 1 is

equal to n plus 2 over n multiplied by T

sub n the numbers given by T sub K are

called the Triangular numbers so this is

a recurrence for the Triangular numbers

A Multiplicative Triangular Number Recurrence (visual proof)

This is a short, animated visual proof demonstrating how to multiplicatively obtain the (n+1)st triangle number from the nth triangular number. #manim #math #mathvideo #mathshorts #triangularnumbers  #numbertheory  #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof

This animation is included in the following sources:

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

To learn more about animating with manim, check out:
https://manim.community
____________________________________________________
Background music:
Timeless by Lauren Duski


suppose we are given a circle with

radius r so that its area is a equals pi

r squared

let's paint half the circumference blue

and then roll the circle one half turn

so that we are left with a line of

length pi times r

extend the line by the length of the

radius r and we get a semicircle of

diameter pi r plus r

then create the half chord from the

bottom of the original circle to the

larger semicircle and call its length x

connecting the ends of the diameter to

the end of this half chord results in a

right triangle according to thaley's

triangle theorem

consequently the two subtriangles with

side length x are similar

this implies that pi r over x

equals

x

over r

therefore

x squared equals pi r squared

we can also use the half chord to create

a square with side length x so that its

area

is a equals x squared

but x squared equals pi r squared and

thus we have constructed a square with

the same area as the given circle

that is to say we have squared the

circle

[Music]

you

Squaring the Circle by Rolling (animated visual proof)

This is a short, animated visual proof that we can square the circle IF we allow the circle to roll. Unfortunately, this is not a solution to the squaring the circle problem from antiquity because that requires it to be done with only a straightedge and compass. #mathshorts #mathvideo #math ##geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #circle #mathematics #thales #squaringthecircle

This animation is based on a visual proof by Thomas Elsner from the May 1977 issue of Mathematics Magazine (https://doi.org/10.2307/2689507 - page 162).

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


suppose the golden ratio were a rational

number that means we could write it as A

over B where A and B are positive

integers and this is in lowest terms the

existence of two such integers means

that we can draw this regular pentagon

with a diagonal of length a and a side

length of B the pictured green triangle

is known as a Golden Triangle when we

swing this side length down into the

Golden Triangle we create a second

triangle pictured in red that's also a

Golden Triangle with corresponding side

lengths B and A minus B but because

these two triangles are similar that

means that A over B is equal to B / a

minus B this means that the golden ratio

Fe is equal to B over a minus B but

that's a contradiction since B is less

than a and a minus B is less than b so

FAL A over B couldn't have been in

lowest terms this proof means that the

golden ratio fee must be irrational

Golden IrRATIOnal

In this video, we use the fact that that the golden ratio is the ratio of a diagonal in a regular pentagon to the side length 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!

If you want to see a longer version with justification about the similarity of pictured triangles, see
https://youtu.be/lL9VN8ELH1M

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


start with one Circle representing the

number one take seven copies of this

circle by placing one in the center and

six around the outer ring like this to

represent the number seven repeat this

process as often as you want taking

seven copies of the previous shape by

placing one in the center and six around

the outer ring this technique produces

visual representations for the powers of

seven to create the visual

representation for 7 to the N we can do

this one more time using seven copies of

the visual representation for 7 to the N

minus one but an alternate way to

produce this same diagram is to take the

sum of the powers of 7 from 1 up to 7

the n minus1 and create six different

copies living around the central circle

once we add the central circle we end up

with the same total number of circles in

the diagram as the one from 7 to the N

therefore 6 * the sum of the powers of 7

from 1 up to 7 the N minus 1 + 1 is

equal to 7 the N this gives us a formula

for the finite geometric sum with ratio

Geometric sum of powers of 7

This is a short, animated visual proof demonstrating the finite geometric sum formula for any integer n with n greater than 3 (explicitly showing the case n=7 with k=3). This series (and its infinite analog when x less than 1) is important for many results in calculus, discrete mathematics, and combinatorics.

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

For a longer version of this video with other geometric sums (to see how this generalizes), check out: https://youtu.be/1wIdJxSfUz4

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

This animation is based on a proof by Mingjang Chen from the December 2004 issue of Mathematics Magazine page 373 (https://doi.org/10.2307/3219202).

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

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


[Music]

let's build a centered stack of squares

consisting of one square in the top row

three squares in the second row five

squares in the third row and so on

adding the odd numbers in each

consecutive row

the resulting stack is a representation

of the sum of the first n odd numbers

but we can rotate this stack creating

four total copies and when we do that

the resulting shape is a perfect square

[Music]

the area of the resulting 2n by 2n

square is just 2n times 2n which

simplifies to 4 times n squared

[Music]

therefore

four times the sum of the first n odd

numbers is equal to four times n squared

if we divide both sides by four we see

that the sum of the first n odd integers

is equal to n squared

you

Sum of the first n odd numbers (visual proof)

This is a short, animated visual proof demonstrating how to find the sum of the first n positive odd integers. #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #oddintegers #numbertheory #sum #mathshorts #mathvideo

This animation is based on a visual proof by Roger B. Nelsen from his first compendium of visual proofs:

Proofs without Words I by Roger B. Nelsen (MAA, 1993) (https://bookstore.ams.org/clrm-1)

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Dramatic Interlude by Alexander Nakarada | https://www.serpentsoundstudios.com
Music promoted by https://www.free-stock-music.com
Attribution 4.0 International (CC BY 4.0)
https://creativecommons.org/licenses/by/4.0/


let's find a formula for the sum a plus

ax plus ax squared up to ax to the K

minus 1 where X is greater than 1 a is a

positive number and K is a positive

integer we can make sense of this sum by

considering a line with lengths marked

at 1 x x squared x cubed and so on up to

x to the k then we can draw a

perpendicular line and start to build a

sequence of rectangles of width a over x

minus one the first rectangle has area a

the next rectangle has area ax the next

rectangle has an area of ax squared and

so on all the way until the last

rectangle which has an area of a x to

the K minus 1. so the total shaded area

in this rectangle is the finite sum

we're interested in but we can also just

compute the area of the rectangle in

terms of its endpoints we see the

rectangle has a length of x to the K

minus 1 so that the area enclosed is a

times x to the K minus a divided by x

minus 1. by Computing the area of this

rectangle in two separate ways we've

produced a formula for a finite

geometric sum

Finite Geometric Sum Formula

This is a short, animated visual proof demonstrating the finite geometric for any ratio x with x greater than 1. This series (and its infinite analog when x is less than 1) is important for many results in calculus, discrete mathematics, and combinatorics.

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

For a wide format, slower, wordless version of this video check out https://youtu.be/ROovDaRQVl0

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

This animation is based on a proof by both Carlos G. Spaht and Craig M. Johnson from the March 2001 issue of The College Math Journal page 109 (https://doi.org/10.2307/2687114).

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

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


으

저희

예 5

Pythagoras’ theorem, an animated explanation!

(a^2)+(b^2) does indeed equal (c^2) !!!
A visual proof!

Technical info:
Computer Generated motion graphics, created in Adobe After effects.

Credit:
Sound effects sourced from:
pacdv.com
freewavesamples.com
soundbible.com


[موسيقى]

Visual Proof of the Pythagorean Theorem | 2

Here's a famous visual proof of the Pythagorean theorem.
#maths #visualization #geometry #algebra #python #proof #visualproof

Visualization Tool: Manim library, Python


let's see two trigonometric facts proved

visually in 2

minutes let's draw a right triangle

where one of the angles is labeled by

the Greek letter Theta assume the

hypotenuse has a length of one and the

opposite side then has a length of s of

theta the complimentary angle is Pi / 2

minus Theta we can take a second copy of

this triangle and reflect it creating a

larger isos triangle as shown here this

triangle has an angle 2 Theta and the

opposite side length is then 2 * the S

of theta we can use this triangle in two

different ways first of all from the law

of ss we see that s of 2 thet / 2 sin of

thet must equal the S ofun / 2 - thet /

1 this means that sin of 2 thet = 2 sin

of thet * the sin ofk 2us thet but the S

ofk 2 - thet is equal to the cosine of

thet therefore we see that sin of 2 thet

= 2 * * the S of thet * the cosine of

theta now let's apply the law of cosin

to the triangle pictured from the law of

cosin we see that 2 * the S of th^ 2 = 1

2 + 1^ 2 - 2 * 1 * 1 * the cosine of 2

thet dividing this equality by two and

isolating the cosine of 2 Theta we see

that cosine of 2 thet = 1 - 2 * sin^2 th

we have used one triangle diagram along

with with the law of ss and the law of

cosin to prove that sin of 2 thet = 2 *

the sin of thet * cosine thet and cosine

of 2 thet = 1 - 2 * the sin^2 of thet

these are known as the double angle

identities for S and cosine implicit in

this diagram is that the angle Theta is

acute do you think the identities hold

when Theta is obtuse can you prove it is

there a visual proof showing the double

angle identities in that

case

A proof to remember: Double Angle Formulas I (visual proof)

This is a short, animated visual proof of the Double angle identities for sine and cosine. To get the formulas we employ the Law of Sines and the Law of Cosines to an isosceles triangle created by from a reflected copy of a given triangle.

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 the classic visual proof the fact but most likely not the one Euclid used.

For visual proofs of the Law of Sines and Law of Cosines, check out:
https://youtu.be/TU0043SuGsM
https://youtu.be/NHxJ3Z_58Lw or
https://youtu.be/vNCe7ciFwVc

This animation is based on a visual proof by Sidney H. Kung  from the March 1996 issue of The College Mathematics Journal (https://www.jstor.org/stable/2687407 - page 155).

#mathshorts #mathvideo #math #trigonometry #lawofsines #lawofcosines  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #obtuseangle #acuteangle #angle #trigidentities #identity #righttriangle #isoscelestriangle #doubleangle

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


[موسيقى]

h

[موسيقى]

#trending #mathematics Visual Solve Algebraic Equation 2^x=8-x^2

#trending #mathematics Visual Proof: Factors of Quadratic Equation 2^x=8-x^2

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


consider the function f of x equals x

squared where the output is obtained by

multiplying the input by itself the

graph of this function looks like this

if we plot a point on the curve we can

draw a line called the tangent line

which looks like the curve locally then

we can compute the slope of this line at

that point and now let's compute the

slopes of the tangent lines at every

point along the curve like this we see

that as we move along the curve the

slope of the associated tangent line

changes we can go one step further and

create a new set of axes on this set of

axes we're going to plot the y

coordinate as the slope for the

corresponding x coordinates in the curve

above when we do that we see a linear

function map out in particular the slope

function for f of x equals x squared is

the line y equals 2x we use the notation

F Prime of x equals 2x for this slope

function and say the derivative of x

squared is 2x because the derivative

function is the function that outputs

the slopes of the tangent lines to

verify this fact requires more work but

at least we can see that the formula

holds visually

Visual derivative of x squared

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

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


now it is a first

position where there are two squares

with area a square and B

Square now it is a second position

where two triangles are

selected whose sides are ab and

hypotenuse is

C now it is a third

position where triangles are

moved to the new position

check by sliders triangles are moving as

Arrow

shows it is a third

position where figure is

moved to one

square with area C

sare first there were two

squares now it is a single square with

area C

sare therefore theorem is

proved a square + B sare is = to

c² if you like Please Subscribe it share

it like it

subscribe

it thanks for watching

Thanks for whatching
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let X be a positive real number and

assume that X is greater than or equal

to 1x consider a right triangle with one

leg of length 2 and the other leg of

length x -

1X as this is a right triangle we can

use the Pythagorean theorem to find the

length of the hypotenuse in terms of the

legs in this case we see that the

hypotenuse length is the < TK of 2^2

plus the quantity x - 1x^2 and this

simplifies to x + 1 overx but in any

right triangle the hypotenuse is longer

than the leg lengths and so in this

situation we see that the quantity x +

1X is greater than or equal to 2 this

inequality is true for any positive real

number X because we can interchange X

and 1 /x can you see from the diagram

when equality holds

Sum of positive number and its reciprocal from a right triangle

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/zBUK8C6wSqs?si=dQIql-H09Q-G1Xt7

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 me show you a geometric

visualization of the classic difference

of squares identity which says x^2 - y^2

always equals x + y * x - y here's an X

by x² and if I cut a y by y² out of the

corner the area left is X2 - y^ 2 these

little lengths are y with these longer

ones being x - y Y and if I cut this

rectangular strip off of the side making

this side now a whole X and that an x -

y this x - y fits perfectly with that

one to add together to an x - y * x + y

rectangle

A Visual Proof of the "Difference of Squares" Rule... #shorts

This short is a few months old. (It was previously on my other channel but most of you haven't seen it. Sorry if you already have)... Make sure you've also seen my latest main @ComboClass  episode about numerical opposites here: https://youtu.be/VNH_PaUsnJE?si=4s8eySoXyrLljU0J

Special thanks to Evan Clark and to all of my Patreon supporters:
Max, George Carozzi, Peter Offut, Tybie Fitzhugh, Henry Spencer, Mitch Harding, YbabFlow, Joseph Rissler, Plenty W, Quinn Moyer, Julius 420, Philip Rogers, Ilmori Fajt, Brandon, August Taub, Ira Sanborn, Matthew Chudleigh, Cornelis Van Der Bent, Craig Butz, Mark S, Thorbjorn M H, Mathias Ermatinger, Edward Clarke, Christopher Masto, Joshua S, Joost Doesberg, Adam, Chris Reisenbichler, Stan Seibert, Izeck, Beugul, OmegaRogue, Florian, William Hawkes, Michael Friemann, Claudio Fanelli, The Green Way, Julian Zassenhaus, Bailey Douglass, Jan Bosenberg, Brooks Boutwell, David Irvine, qe, George Sharabidze, Jack Dwyer, Fredrik, and Dave Brondsema!.. And if I'm missing any more recent names, thanks to you too, and I'll update this list soon :)
To join that list of people supporting this channel, and get some cool bonus content, check out the Combo Class Patreon at https://www.patreon.com/comboclass

If you want to send me anything (including old extra clocks/dice/etc.), here's a private mailbox address which I check every couple weeks for mail from viewers. If you're going to send anything, please watch this short video first: https://youtu.be/_-gSvbvO8W4
Domotro
1442 A Walnut Street, Box # 401
Berkeley, CA 94709

Links:
Main channel: www.youtube.com/comboclass
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Subreddit: https://www.reddit.com/r/comboClass/

If you want to reach out to possibly collaborate or help with Combo Class somehow, my email is: combouniversity (at) gmail (dot) com

Combo Class, taught by me (Domotro), is a crazy educational show where you can have fun learning rare things about math, science, language, and more! This is actually kind of my "bonus" channel for my livestreams, shorts, and extra content - make sure you're also subscribed to the main @ComboClass channel where my main episodes go!

DISCLAIMER: any use of fire, tools, eating unexpected things, or other dangerous-seeming actions in this series are experiments that are conducted safely and professionally, and/or are accidents that I include in the videos for educational purposes. Do NOT try to copy any experiments (or accidents) that you see in these videos yourself.