foreign

start with a length one line segment and

use a straight edge Encompass to

construct a perpendicular length one

line segment connect the endpoints like

this and label the length of the

hypotenuse of this right triangle by a

because the triangle satisfies the

Pythagorean theorem we see that a equals

the square root of two

now use the Straight Edge Encompass to

draw a line segment of length one that

is perpendicular to the diagonal of

length root 2. connect the endpoints

again and now we have a new length a and

this time by the Pythagorean theorem we

can determine that a squared is 3 so

that a is the square root of 3. let's

repeat this process one more time use

the Straight Edge Encompass to create a

circle of radius 1 and then a line

segment of length one that is

perpendicular to the line segment of

length root 3. connect the endpoints

like this and use the Pythagorean

theorem to determine that the length of

the hypotenuse of that new right

triangle is the square root of four

using a straight edge Encompass we have

constructed line segments of length

square root of 2 square root of 3 and

square root of 4 but there's no reason

to stop there we can keep applying the

exact same process over and over again

to the new links that we create and we

see that we can create line segments of

length square root of 5 square root of 6

and square root of 7.

how far can we take this process can we

just keep going forever the object shown

here is the first six stages of the

Spiral of theodorus

the spiral can be extended indefinitely

but just for our purposes let's take it

out to I don't know how about n equals

one thousand we'll see the first 1000

levels of the Spiral of theodorus and

along the way we're constructing the

square root of n for every single n

value between two and a thousand

once we get to the end of our spiral as

we've done right here we see that the

final line that we've drawn has a length

of size square root of one thousand

for appearances we can construct the

circle with a radius of square root 1000

and draw a radius in the same direction

as our original length one line segment

while this may not be the most efficient

way to get the length of square root of

1000 using this process we can construct

the square root of n for any positive

integer n this means that the spiral of

theodorus constructs square roots and

shows us how to prove that square roots

are constructable using Straight Edge

Encompass using proof by induction

Spiral of Theodorus Constructs Square Roots (visual proof; straightedge and compass)

This is a short, animated visual proof demonstrating how to construct square roots of any positive integer using the Spiral of Theodorus #manim #math #mathshorts #mathvideo  #trisect #construction #geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath

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

___________________________________________________________
Music in this video:
Glimpsing Infinity by Asher Fulero


[Music]

foreign

[Music]

start with a triangle with sides of

length a b and c

the angle bisectors of this triangle

meet at a point

this point is known as the in center of

the triangle and it's also the center of

a circle known as The Inn circle of the

triangle which is a circle that is

tangent to all three sides and the

largest Circle that fits inside the

triangle if we draw the radii that are

perpendicular to each of the sides we

get a shape that looks like this and we

can fill the six regions with different

colored triangles

now let's unfold these triangular

regions so that they all lay out in a

line of triangles like this with bases

that are all parallel

if we take a second copy of this shape

flip it upside down and then shift the

triangles as we see here we end up with

a nice rectangular shape

in fact we can find the dimensions of

this rectangle using the original labels

of the triangle

as shown in the picture

we see that the length of the rectangle

is a plus b plus C while the width of

the rectangle is the in radius length R

but now we see a nice fact 2 times the

area of the original triangle must be

equal to R the length of the in radius

times a plus b plus C the length of the

perimeter but this means that the area

of the triangle is equal to R times a

plus b plus C over 2 which is R times s

where s is known as the semiperimeter so

we've proved that the area of a triangle

is the product of the radius in the

semiperimeter

Triangle Area from Inradius and Semiperimeter (visual proof)

This is a short, animated visual proof showing that the area of a triangle is given by the product of the inradius and the semiperimeter of the triangle. #manim #math #mathvideo #mathshorts #geometry #circle #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof #incircle #inradius #incircle #semiperimeter

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:
Luminous Rain 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=USUAN1100169
Artist: http://incompetech.com/


here's an intriguing way to use a

rectangle to find the sum of various

geometric series with ratio

1/2 first let's start with a 1x1 Square

the diagonal length in this square is

the square otk of two by the Pythagorean

theorem so if we draw a line segment to

the midpoint of the square on that

diagonal it has a length ofun of 2 over2

now we can draw a horizontal line back

to the square that has a length of 1/2

but then we can repeat this exact

picture in the upper right corner we

take half of the diagonal line to be 2 4

half of the horizontal line to be 1/4

and we can repeat this process over and

over each time we take half of the

previous diagonal and half of the

previous horizontal along the way each

of these line segments has a length of

size 1/ < tk2 raised to the K where K is

some integer starting at one and going

on forever now the infinite sum

represents a limit so we need to think

of the limiting process in this diagram

in the limiting process we see that the

diagonal and horizontal lines fill in

all the way up to the top right corner

of this square and so the sum of the

lengths of this zigzag diagram represent

our infinite sum but now we perform our

trick we rotate the horizontal line

segments 90° about their end point

filling in the right side of the square

and then we shift the diagonal line

segments over to the main diagonal of

the square filling that entirely as well

and this means that the sum of the

lengths of the zigzag diagram must be

equal to the sum of the side length of

the square plus the diagonal of the

square that means our infinite sum is

equal to 1 + the < TK 2 so from this

diagram we see that the sum where K

ranges from 1 to Infinity of 1K 2 raised

to the K power is equal to 1 + < TK of 2

as a bonus fact we note that the number

1 + the of two is known as the Silver

ratio let's see this process in action

again with a 1/2x one rectangle instead

the diagonal length in this rectangle is

the < TK of 5/2 by the Pythagorean

theorem if we take half of the diagonal

and then Traverse back down to the side

of the rectangle we end up with a length

of < TK 5 4 + 1/4 and now we repeat this

process on the lower one quarter by 1/2

rectangle we scale the diagonal length

by half and the vertical line by half

giving us two line segments that sum to

> of 5 8 + 1/8 and then we repeat this

process again and again each time taking

1/2 of the previous sum of two lengths

once again we are interested in the

infinite sum of the lengths of the

zigzag diagram shown here and that means

we're interested in a limiting process

we can think about this limiting process

geometrically as the zigzag diagram

converging in on the lower left corner

filling in entirely again we can rotate

the vertical pink segments 90° and they

fill in half of the side length of one

we can also shift the blue diagonal line

segments up to the main diagonal of the

rectangle filling in the entire diagonal

and that means that the sum that we're

interested in the infinite sum is the

sum of these two lengths so it's 12 + <

TK 5/ 2 this is a geometric series with

first term < TK 5 4 + 1/4 and ratio 12

and we see from this diagram that the

infinite sum where K ranges from 0 to

Infinity of the < TK of 5 + 1 over 4 * 2

K is 12 + > 5/2 this latter number is

known as the golden

ratio now it's your turn why don't you

try this process on a rectangle with

generic side lengths A and B what

infinite sum do you find the sum of

geometrically using this particular

argument this visual proof is due to

anhel Plaza if you liked this video

check out my playlist on geometric sums

that show other ways to find the sum of

geometric series

Sum Geometric Series with Ratio 1/2 in Rectangles (visual proofs)

In this video, we show an animated version of a recent proof without words that finds the sum of the geometric series with first term 1/root(2)+1/2 and ratio 1/2 using the unit square. Then we show how to find other geometric series sums with ratio 1/2 using different sizes of rectangles. Careful, the final bonus question might require a different technique than rotating :)

If you like this video, consider subscribing to the channel or consider buying me a coffee: 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 Ángel Plaza in the June 2024 issue of Mathematics Magazine (https://doi.org/10.1080/0025570X.2024.2336443 page 328).

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

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


suppose we have a right triangle with

legs of length A and B and a hypotenuse

of C we can draw an A by a square off

the side of length a we can draw a b by

b square off the side of length B and we

can draw a c by C Square Off the

hypotenuse though we leave the C by C

Square unshaded the total shaded area in

this diagram is a^2 + b^2 let's perform

a few area preserving Transformations

first we Shear each of the two squares

along the opposite side of the triangle

we rotate these sheared regions and then

we Shear them down again filling the C

by C Square since shearing and rotating

doesn't affect the area we see that a^2

+ b^2 must equal to c^2 so here we have

a visual proof of the Pythagorean

theorem which says that the sum of the

squares of the legs of a right triangle

must equal the square of the hypotenuse

Pythagorean theorem: Euclid’s proof reimagined

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) following essentially Euclid's proof. 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.

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

For a longer version of this video with a bonus dissection proof based on this proof, see
https://youtu.be/g3C3hs8_nzU

You can also see other Pythagorean theorem proofs here:
https://youtu.be/o2mbN452ywA (compilation)
https://youtu.be/XFHm0Ot3PP0
https://youtu.be/q28H5NY5UVU
https://youtu.be/yfGtbNgcrQ8
https://youtu.be/JRjPsuJ7S-g
https://youtu.be/TsKTteGJpZU
https://youtu.be/ZlGaQdNRdqA
https://youtu.be/uqGjs_n-f7U
https://youtu.be/wWEZ-rUlBDQ
https://youtu.be/I75HYC8TRNY
https://youtu.be/LWVog8hg74I?si=4GSzTLZP7XIWeaxJ

or check out my Pythagorean theorem playlist:

https://www.youtube.com/playlist?list=PLZh9gzIvXQUvlRlnYryejWGHYQquM92rB

This animation is based on a proof due to Euclid. For a static version of this proof, see Roger Nelsen's first "Proof Without Words: Exercises in Visual Thinking" compendium (page 5) : https://bookstore.ams.org/clrm-1. You can also check out Howard Eves' "Great Moments in Mathematics (Before 1650)" page 31-33.

(I may receive a small commission at no cost to you for these affiliate links):

https://amzn.to/48mjeX1
https://amzn.to/3Tr0OjW

#math #pythagoreantheorem #pythagorean #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof  #mathshorts #mathvideo

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


we are to prove Pythagoras Theorem c² =

a sare +

b² that is hypot square is equal to sum

of squares of other two

sides we will prove it by the Yan

property this property says that square

a

square will be changed to rectangle

whose area will be a square and square

with area b square will be changed to

rectangle whose area will be b s

now see the animation

process first

slider ch ches scares to

parallelogram and these have area same

because we have applied rule on the

right hand side given area and

parallelogram area of the both sides

figures are same because they have the

same base and same

height now see animation

these have three steps so off slider one

off slider two offsider

three you can repeat it first second

third third second 1 1 2

3 C animation again

step third second

one step first second

third third second 1 1 step two step

three therefore a square + b square

are merged in the big Square

c² therefore c² = to a sare + b square

Please Subscribe and share my channel

Pythagoroas Theorem ,Pythagoroas  Theorem Proof, Mathematical Visual proof ,Animation In Geogebra

Thanks for whatching
kataria math animation,
Contact Me :-  rattanlalkataria@gmail.com
On YouTube :- @KatariaMathAnimationOfficial
 Your Queries :-
Pythagoroas Theorem
 Pythagoroas Theorem Proof
 Kataria Math Animation
Animation In Geogebra
Geogebra In Mathematics Teaching
Pythagorean Theorem Animation
Exterior Angles Theorem
Exterior Angles Properties
Prove Exterior Angles Of Triangle = Sum of  Opposite Interior Angles
Maths Animation
Teaching of  Maths
Learning of Maths
Sum Of angles of triangles

Geometry
 Algebra
GeoGebra
Animation
Easy Maths
On Line Teaching
Math Teacher
Triangles
Square
Quadrilateral
Property of Triangles
Property of Circle
Area
Volume
Find Nets of Solid Figure
Student's favorite method
Before classroom
In The Classroom
#Geogebra
#Angles
#Sumofangles
#favorite
#Method
#Maths
#GeoGebra
#viralmath
#onlineteaching
#Quiz For Kids Visual Addition
#Quiz On Geogebra#
# Visual Math
# Animation
Pythagoroas Theoram
Easy Math
 Mathematical Visual proof
 Only 2 Minutes
#mathvisualproofs


imagine we have a triangle with sides of

length a b and c and the angles across

from the sides are labeled capital a

capital B and capital c respectively

drop in altitude from the vertex where

angle a is to the side of length a and

label it h this altitude cuts the

triangle into two right triangles from

the first one we see that the sine of

angle b equals H divided by c as H is

the opposite side and C is the

hypotenuse from the other right triangle

we see that the sine of C is equal to H

divided by B as H is the opposite side

and B is the hypotenuse of this triangle

as H is part of both of these equations

we can see that H equals B times sine of

C and H equals c times sine of B

from this we see that b times the sine

of angle C equals c times the sine of

angle B dividing both quantities by B

times C we see that sine of angle C

divided by C is equal to the sine of

angle B divided by B

now we can do a similar thing for the

angles A and B but this time we have to

be a little more careful this time when

we try to drop an altitude from the

vertex at angle C we see that we get a

new triangle created outside the

original like this we can label this

altitude by H and note that the newly

created outside triangle has an angle of

measure 180 degrees minus a the

supplementary angle to a

from this newly created right triangle

here we see that the sine of angle B

must be the opposite side length H

divided by the hypotenuse a on the other

hand from the small right triangle

created here we see that the sine of 180

degrees minus a must equal the opposite

side H divided by the hypotenuse B

once again both of these equalities

involve the quantity H so we see that H

is equal to B times the sine of 180

minus a and H equals a times the sine of

angle B

we can divide both of these quantities

by a times B so that sine of 180 degrees

minus a over a is equal to sine of angle

B divided by little B but now we use the

fact that sine of 180 degrees minus a is

equal to the sine of angle a this means

that the sine of angle a divided by side

length a is equal to the sine of angle B

divided by the side length little B when

we couple this result with our first

computation we see that the sine of

angle a divided by a equals the sine of

angle B divided by B and that equals the

sine of angle C divided by C where a B

and C are the side lengths of a triangle

with opposite angles capital a capital b

capital c respectively this is known as

The Law of Sines for triangles

Law of Sines Visual Proof

This is a short, animated visual proof of the Law of Sines using the definition of sine and by dropping appropriate altitudes of the triangle for given angles. The altitude may intersect the opposite side or an extended segment of the opposite side.

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 Cosines, check out:
https://youtu.be/NHxJ3Z_58Lw or
https://youtu.be/vNCe7ciFwVc

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

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


here's an amazing geometric structure

called the fanel plane start with seven

points laid out like this by a line we

meet a collection of three points so

draw these seven lines note that the

final one is actually going to look like

a circle even though it's just a

collection of three points we see that

there are three points on each line and

there are three lines through each point

we can realize this structure using the

three-dimensional Vector space over Z2

the points correspond to the non-zero

vectors in the vector space so they

correspond to the onedimensional linear

subspaces the lines correspond to the

two-dimensional linear subspaces and

that means each line consists of three

vectors that coordinate wise add up to

zero mod 2 you can check that the three

vectors on each line as highlighted here

do in fact add up to zero mod 2 when you

perform coordinate wise addition this is

the world's smallest projective geometry

and it shows up in a lot of places in

mathematics

Fano Plane: The World’s Smallest Projective Geometry

This is short animation of the Fano Plane, which is the smallest projective geometry. We also show how to realize the geometry as the set of linear sub spaces of a three dimensional vector space over the field with two elements.

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

#mathshorts #mathvideo #math #geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #vectorspaces #projective #projectivegeometry #shorts #fanoplane #fano

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


here are two standard infinite series

that are related sure you can find the

sum of these series using classical

techniques but let's see a geometric way

that computes the sum of both of these

infinite series using regular polygons

start with an equilateral triangle the

interior angle of this triangle is pi

over 3 which can be Rewritten as 2 piun

3 * 2 now draw a square around that

equilateral triangle with side length

equal to the base the interior angle of

the square is 2 Pi / 4 and so the angle

between the square in the triangle is 2

piun 4 - < 3 which simplifies to 2 piun

/ 4 * 3 Let's extend this again by

drawing a pentagon with side length

equal to the square length again the

interior angle of the Pentagon is 3 pi

over 5 so the difference in angle

between the Pentagon and the square is 3

piun / 5 - 2 piun 4 and this simplifies

to 2 piun / 5 * 4 we can continue this

extension process by drawing larger and

larger polygons around so we start with

the hexagon and then we move to the

heptagon and so on using the interior

angle theorem we see that the angle

difference between the N gon and the

nus1 gone is n - 2 * piun / n minus the

quantity n- 3 * piun / nus1 and this

simplifies to 2 piun / n * nus1 so the

angle between the regular n gon and the

regular n -1 gone is always 2 piun / n *

nus1 as we extend this picture out but

now we can just keep extending this

picture drawing more and more regular

polygons where the number of sides

increases indefinitely in fact we can

consider a limiting shape of this

process the limiting outer shell of the

resulting shape is a circle by zooming

back into the original equilateral

triangle we get an amazing fact the

angles that we've determined can be

added together so we see that on one

hand the full angle measurement is 2

piun 3 * 2 + 2un 4 * 3 + 2 piun over 5 *

4 and so on each time adding up angles

of the form 2 pi/ n * nus1 on the other

hand all of these angles together form

the full angle measurement of a

horizontal line since this line is

tangent to the circle and therefore the

sum of the angles must equal Pi but now

we can cancel the pi and the numerators

of the fractions on the left with the pi

on the right and we see that 2 3 * 2 + 2

4 * 3 + 2 5 * 4 and so on must equal 1

if we then add 2 over 2 * 1 on the left

and 2 over 2 * 1 on the right which is

one then we get our first famous

infinite sum notice that each fraction

on the left is 1/ n choose 2 where n

choose 2 is a binomial coefficient or a

triangular number this means that the

infinite sum where n ranges from 2 to

Infinity of 1/ n choose 2 is equal to 2

we can also divide both sides of the

equation by two producing another famous

infinite sum the one where n ranges from

1 to Infinity of 1 n * n + 1 and this

sum equals 1 these infinite sums are

usually computed by recognizing the

infinite series as a telescoping series

but here we've shown how to find this

infinite sum using a limiting process on

an amazing diagram constructed by larger

and larger regular polygons if you like

this video check out my playlist on

infinite series there I've animated

classic visual proofs that Compu the sum

of infinite Series in interesting ways

Two Infinite Series Sums from Regular Polygons (visual proof)

This is a short, animated visual proof computing the sums of two series - one of reciprocals of triangular numbers (i.e., certain binomial coefficients) and the other a classic series that is used to demonstrate telescoping series.

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

For some related videos see these videos:
https://youtu.be/moxeqJx2YsY
https://youtu.be/I0juicS3SIc

This animation is based on a visual proof by Paul Stephenson from the December 2022 issue of Mathematics Magazine (https://doi.org/10.1080/0025570X.2022.2126173 page 572 ).

#mathshorts #mathvideo #math #infiniteseries #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #triangularnumbers  #animation #theorem #proof #iteachmath #series #infinitesums #infiniteseries #pascaltriangle #binomialcoefficients #binomials #binomialtheorem #telescoping

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


aah

[Music]

Volume of cone by ingration visual proof #shorts

Discover how to calculate the volume of a cone with integration! Watch this short video to see the mathematical process and graphical proof that brings the formula to life. Perfect for math enthusiasts and students alike!
#trending #shorts#viralshort #MathWeb


let's use integral calculus to find the

area under the curve Y equals X plus a

half

according to the fundamental theorem of

calculus the integral from 0 to n of X

plus a half DX is equal to the quantity

x squared over 2 plus X over 2 evaluated

at n and zero when we plug in n and 0

and subtract we get N squared over 2

plus n over 2 for the integral

on the other hand we can also compute

this integral by breaking up the region

into a bunch of different pieces like

this each of the pieces will then be the

definite integral from I minus 1 to I

under the function X plus a half so we

can add up all of these integrals

according to properties of integrals

once we do this we again get that each

integral is x squared over 2 plus X over

2 evaluated at the endpoints and thus we

get that the integral in question is

given by the sum from I equals 1 to n of

I

but now that we've computed the definite

integral in two separate ways we see

that the sum of the first n positive

integers is equal to N squared over two

plus n over 2.

Summing integers using calculus!

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

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

This proof is adapted from a static diagram due to Jaime Gaspar: Proof without words: using trapezoids to compute triangular numbers, Math. Mag. 91 (2018), no. 3, 206–207. (https://www.jstor.org/stable/48665541)

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

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


[Music]

[Music]

in a previous video

we saw a visual proof of this formula

for the sum of the first

n positive squares

what happens if we instead consider the

alternating sum

of squares is there a nice formula now

let's take a look

a square array with side length n

represents the value n squared

removing the upper n minus 1 by n minus

1 square

corresponds to subtracting n minus 1

squared

we can add back an n minus 2 by n minus

2 square and then

remove the upper right n minus 3 by n

minus 3 square

we can continue this process alternating

between adding a square back

and removing the next square array until

eventually

we add or subtract the two by two square

and then subtract or add back the one by

one square

the resulting l shaped gnomons can then

be squished together

to produce a triangular array with base

n

which has t n squares where t

n is the nth triangular number therefore

the alternating sum of the first

n squares is the nth triangular number

you

Alternating Sum Of Squares (visual proof)

This is a short, animated visual proof demonstrating how to visualize the alternating sum of squares. #mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #triangularnumbers

This animation is based on a visual proof by Steven L. Snover from the April 1992 issue of Mathematics Magazine (https://doi.org/10.2307/2690486 - page 90).

Compare this formula to the regular sum of squares formula visualized here: https://youtu.be/-tJhH_k2LaM

Here is an alternate visual proof of this same fact: https://youtu.be/stSftarO82s

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


[Music]

here's an isosceles right triangle with

legs of length 1 so that the hypotenuse

has a length square root of 2. if we

walk from the upper corner to the lower

corner going one unit East and then one

unit South we travel a total of two

units but we can keep poking in the

corners of these paths so that we walk

east south east south east south all the

way down the diagonal to the corner if

we do this process indefinitely have we

produced a proof that the square root of

2 which is the length of the hypotenuse

is exactly the same as 2 which is the

length of the path that we walk how

could this be let's take a zoom in and

see what's going on

foreign

so no this isn't a visual proof that

square root of 2 equals 2. but can you

explain what's going on

if you want to see some honest visual

proofs of some real facts then go check

out my channel

Is square root of 2 equal to 2? (Spoiler : no)

This is a short animation of an oft-claimed visual proof that the square root of 2 is equal to 2. Can you see why the two lengths aren't equal? Can you give a good, brief explanation? #math #manim #mathshorts #mathvideo #visualproof #proofwithoutwords #false #mathfail

To learn more about animating with manim, check out:
https://manim.community
_________________________
Music in this episode:
Prelude No. 5 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/


have you ever wondered if e to the pi or

Pi the E is the larger number here's a

way to figure it out without a

calculator first plot the function y =

1X then plot the x coordinate e and the

x coordinate Pi notice e is less than Pi

we can find the area under this curve

between e and pi as the definite

integral where X ranges from E to Pi of

DX we can also draw this rectangle

living over e to the pi and we see that

the area under the curve is less than

the area of the rectangle which is 1 E *

piun - e according to the fundamental

theorem of calculus the definite

integral is the natural log of Pi - 1

and we can simplify the area of the

rectangle to Pi eus 1 from this we see

that the natural log of Pi is less than

Pi e we can multiply both sides by E and

then apply two natural log rules to see

that Pi the E is less than e to the pi

so using the curve y = 1X we've proved

analytically that pi to the E is less

than e to the pi

e^Pi vs Pi^e : which is larger? (pi day battle)

This is a short, animated visual proof demonstrating which of the values e^pi or pi^e is larger (where pi is the circle constant and e is Euler's constant) .

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 Bikash Chakraborty from volume 41 issue no. 56 of the Mathematical Intelligencer in 2019 (https://doi.org/10.1007/s00283-018-9816-4)

Here is wide format, high def, slightly longer version: https://youtu.be/I7wiS9rH2h0

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

#math #inequality #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #mathshorts #mathvideo #mtbos #euler #pi #piday

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


here's something interesting the area of

a regular octagon equals the product of

its longest and shortest diagonal here's

a visual proof

we can split the Octagon like this

and get these triangles

let's rearrange these parts

[Music]

all the parts combine to form this

rectangle

so the area of the Octagon is same as

the area of this rectangle

and we know that the area of the

rectangle equals the product of its

length and breadth

now let's go back to the octagon

[Music]

the length of the rectangle will be same

as the longest diagonal and the breadth

of the rectangle will be same as the

shortest diagonal

so the area of the Octagon is same as

the product of its longest and shortest

diagonals

[Music]

A cool way to find the area of an octagon

The area of a regular octagon equals the product of its shortest and longest diagonals.

"Learn more about the Cuemath Way of learning Math:
VISIT our Website: https://s.cuemath.com/QwbIxn

Fall in love with math with more fun content on all social platforms
- LIKE us on Facebook: https://www.facebook.com/Cuemath
- FOLLOW us on Twitter: https://twitter.com/cuemath
- FOLLOW us on Instagram: https://www.instagram.com/cuemath"

#math #geometry #animation #education #learning #proof #shorts #visual


start with two angles alpha and beta

both in the interval zero to pi over two

draw a triangle with horizontal base and

an altitude that cuts the triangle into

pieces with angle alpha and angle beta

let's suppose that the legs on this

triangle are called a and b

the area of this triangle can be

determined using the side angle side

formula for the area of the triangle and

is thus one half a times b times the

sine of alpha plus beta

suppose the altitude from the angle

alpha plus beta has length y

this altitude cuts the triangle into two

sub-triangles that are both right

triangles and both with a side length of

y

according to the left triangle y is

adjacent to alpha so y is equal to a

times the cosine of alpha and in the

right triangle y is adjacent to beta so

y is equal to b times the cosine of beta

but the value y is part of both

triangles so we can use opposite values

on opposite triangles like this

now the original triangle must have area

given by the sum of the areas of the two

sub triangles

but these two areas can be computed

using the side angle side formula for a

triangle again so we have that one

triangle is one half times a times b

times the cosine of beta times sine of

alpha and the other one is one half b

times a cosine alpha times the sine of

beta

we can cancel the one half a times b

from both sides of this equation and

that results in the formula sine of

alpha plus beta equals the sine of alpha

times the cosine of beta plus the cosine

of alpha times the sine of beta

this is the sine of a sum formula

Sine of a Sum I (visual proof; trigonometry)

This is a short, animated visual proof of the sum formula for the sine function using the Side-Angle-Side triangle area formula. This theorem relates the sine of a sum to the sum of products of sines and cosines. #mathshorts #mathvideo #math #trigonometry #sine #sinofsum  #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #sideangleside #trianglearea

This animation is based on a proof due to Christopher Brueningsen from the April 1993 issue of Mathematics Magazine (https://doi.org/10.2307/2691126) page 135.

To learn more about animating with manim, check out:
https://manim.community
____________________________________________________
Background music:
Cinematic Ambient Beat | STEPS by Alex-Productions | https://www.youtube.com/channel/UCx0_M61F81Nfb-BRXE-SeVA
Music promoted by https://www.free-stock-music.com
Creative Commons Attribution 3.0 Unported License
https://creativecommons.org/licenses/by/3.0/deed.en_US


[موسيقى]

[موسيقى]

quadratic formula#proof#visual#amazinn#reel#for you


can be represented by a triangular array

of dots as pictured the first row has

one dot and the nthrow has n dots we can

also add the N plus first row like this

where there are n plus one dots now we

can build a one-to-one correspondence

between the dots in the Triangular array

representing the sum of the first n

positive integers and pairs of dots in

the N plus first row the dot in Row 1

maps to the pair of dots in row N plus

one that are n units apart the two dots

in row two map to the two pairs in row

one plus one that are n minus one units

apart this process continues all the way

down until we see the dots in row N

correspond to the pairs of dots in row N

plus one that are one unit apart sure

enough we've mapped to every possible

pair of two dots in row N plus one

therefore the sum of the first and

positive integers must be n plus one

choose two which is n plus one times n

divided by two

Bijective sum!

This is a short, we explore the famous formula for the sum of the first n positive integers via a bijective technique.

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

This proof comes from a compilation video showing 12 ways to prove this formula: https://youtu.be/eHbtc50-qXo

This animation is based a proof from Loren Larson's article in the last issue of the 1985 College Mathematics Journal : Loren C. Larson, A discrete look at 1 + 2 ++ n, The College Mathematics Journal 16 (1985), no. 5, 369–382 (https://doi.org/10.2307/2686996)

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

#sumformula #sumintegers #integers #mathvideo #math #mtbos  #animation #theorem #pww #proofwithoutwords #proof #iteachmath #mathematics #combinatorialproof #combinatorics #bijection


[Music]

[Music]

the

going

to I

[Music]

[Music]

Geometric Series (Not my video, Mathematical Visual Proofs on Youtube) #facts #maths #geometry


[Music]

in the cold

alone

there are times when

I'm almost letting it go

the cards were stacked against us both

[Music]

I know

[Music]

I know it's a member

[Music]

North/East Lattice Paths

This enumeration shows all of the northeast lattice paths on the 4x4 integer grid. The number of such lattice paths to a point (a,b) is counted by the binomial coefficient {a+b choose a}, as is discussed in this video: https://youtu.be/KYpOoA2D63w. So as you watch this video, Pascal's triangle (also known as Halayudha's triangle or just the arithmetic triangle) appears as the counts.

Here is a link to a longer "synthwave enumeration" of the same idea (but up to the 5x5 grid and a bit slower): https://youtu.be/9YUl0k2FYzc

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

#pascaltriangle #pascalstriangle #mathvideo #math #mtbos #manim #animation #theorem #visualproof #proof #iteachmath #mathematics #binomialcoefficients #latticepaths #discretemath #combinatorics #enumeration #synthwave

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


suppose that the > of 2 is rational so

there exist positive integers A and B

with a^2 equal to 2 b^2 that means that

we have an a by a square that has the

same area as 2 B by B squares according

to the carpet's theorem if we overlay

the 2 B by B squares into the a by a

square since the areas are equal we must

have that the area of overlap which is a

2 b - A by 2 b - a square is equal to

the uncovered area in the a by a square

which is 2 Aus B by A- B squares but

this means that the quantity 2 b - A 2

is equal to 2 * the quantity a - b^ 2 so

we've produced a pair of smaller

positive integers satisfying the same

equation x^2 = 2 y^2 we can repeat this

process over and over again producing a

list of smaller and smaller positive

integers but such an infinite list of

decreasing positive integers can't exist

so we've produced a contradiction this

means that a squ can't equal 2 b^ s with

A and B integers so the square root of

two is irrational

Irrational Root 2!

In this video, we introduce and prove the "Carpets Theorem" and then utilize the theorem to prove that both the square root of two and the square root of three are irrational by a visual infinite descent argument.

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 from Stanley Tennebaum. If you are interested, I recommend this article from John Conway and Joseph Shipman about the irrationality of square root of two (and others):

http://dev.mccme.ru/~merzon/mirror/mathtabletalks/files/irrational-conway.pdf

#irrationalnumbers #realnumbers #manim #math #mtbos  #animation #theorem #visualproof #proof #iteachmath #mathematics #irrational #carpetstheorem #proofbycontradiction #root2 #algebra #infinitedescent

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


[Music]

back

Difference Identity| Visual Proof | Mathletes |Sunny Sir  #maths #algebra #foryou #satisfying

Difference Identity| Visual Proof| Mathletes | Sunny Sir  #maths #algebra #foryou #satisfying  #cbse #funnyshorts

Do Like Share and Subscribe For more interesting videos

Join our telegram channel: https://t.me/Mathletes_29

Follow us on Instagram: https://www.instagram.com/mathletes19?igsh=MWo2ZzdlY3cxdjg1MA==


the sum of the powers of one-fifth is

one-fourth but why is that well consider

a square whose set length is one and

connect each vertex to the midpoint of

the following Edge like this the Central

Square will have an area of one-fifth

because look I can build five identical

squares and therefore also this red

triangle we have an area of one-fifth

and I can repeat the process in the

Central Square like this and I'll get a

new red triangle whose area is one-fifth

times one-fifth that is one over five

squared and I can keep repeating the

process like this till I get this shape

that fits four times in the Square

and therefore

Genius Visual Proof. Sum of Powers.

Today's math video is about a genius visual proof that the sum of the powers
of 1 fifth is 1 forth.

This animation is based on a proof by Rick Mabry from his website
https://lsusmath.rickmabry.org/rmabry/fivesquares/fsq2.gif
and featured in Roger B. Nelsen's third Proofs Without Words compendium
https://bookstore.ams.org/clrm-52/

New Puzzles every Sunday and Thursday.
I post math puzzles and their solutions with animations.
Subscribe: https://www.youtube.com/@guzmat?sub_confirmation=1

If you prefer you can watch the videos in
🇪🇸 Spanish: https://www.youtube.com/@guzmat-espanol

🇮🇹 Italian: https://www.youtube.com/@guzmat-quesiti

English:  https://www.youtube.com/@guzmat

             🌟My Links🌟
Twitter (english): https://twitter.com/GuzMatPuzzles
FaceBook (english): https://www.facebook.com/GuzMat
FaceBook (italian): https://www.facebook.com/GuzMatQuesiti
FaceBook (profile): https://www.facebook.com/GuzMatBase
Instagram (italian): https://www.instagram.com/guzmat_puzzles/

Personal Website: http://www.guzman.it
Personal Blog: https://guzman-tierno.blogspot.com/


foreign

[Music]

suppose we have a right triangle with

sides of length A and B and a hypotenuse

of length C

we can take a second copy of this

triangle and place it here which forms a

square shape like this

we can also fill out the two triangles

to rectangles in this manner here

all three of these new shapes the yellow

Square the blue rectangle and the green

rectangle all have a side of length C

we can use the altitude of the original

triangles to then create the final

missing square up in the upper right

corner here

the full diagram that we have just

created is a square with a side length

of C plus h

therefore the area contained in this

square is the quantity C plus h squared

but now move the four triangles used to

create the two rectangles in this

fashion here

the resulting shape is now two squares

the upper right square has a side length

of H

and the new square that is diagonally

tilted has a side length of a plus B

therefore the area contained is the

quantity a plus b squared added to the

quantity H squared

this formula implies the Pythagorean

theorem let's see how by revisiting the

original triangle

this triangle had sides of length A and

B and a hypotenuse of length C and an

altitude of length h

we can expand the formula algebraically

like this and then we compute the area

of the triangle in two ways one-half

times the base times the height to

realize that a times b equals c times h

therefore in the equation the 2 times c

times H cancels with the 2 times a times

B and the H squares cancel from both

sides resulting in the equation c

squared equals a squared plus b squared

this ladder formula is known as the

Pythagorean theorem it says that the

square of a hypotenuse of a right

triangle is equal to the sum of the

squares of the leg lengths

foreign

[Music]

Pythagorean Theorem VII (visual proof)

This is a short, animated visual proof of an extended version of the Pythagorean theorem (the right triangle theorem) that implies the Pythagorean 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, and we prove the theorem using rigid translations of objects and the triangle area formula. #mathshorts #mathvideo #math #pythagoreantheorem #pythagorean #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #leonardodavinci #davinci

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

This animation is based on a recent proof by Roger B. Nelsen and Claudi Alsina from the January 2022 issue of The College Math Journal (https://doi.org/10.1080/07468342.2022.1997306).

For other visual proofs of this same fact check out:
https://youtu.be/q28H5NY5UVU
https://youtu.be/yfGtbNgcrQ8
https://youtu.be/JRjPsuJ7S-g
https://youtu.be/TsKTteGJpZU
https://youtu.be/ZlGaQdNRdqA
https://youtu.be/uqGjs_n-f7U

To learn more about animating with manim, check out:
https://manim.community
___________________________________
Music in this video:
Tranquil Fields (Tense) 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/


[Music]

thank you

[Music]

Why is π = 3.14…??? #pivalue #circle #circles #geometry  #maths  #mathematics


let A and D be fixed real numbers in N

be a non-negative integer then consider

the sum a plus a plus d plus a plus 2

times D and so on up to a plus n times D

we can represent this sum of an

arithmetic sequence using a staircase

diagram consisting of a bunch of

rectangles the first level has a

rectangle with area a the next one has

two rectangles in closing area a plus d

and so on up to the last level which has

an area in closing a plus n times D but

then imagine a second staircase diagram

rotated and fit together with the

original to form a rectangle as shown

here this rectangle has a length given

by 2 times a plus n times d and a width

given by n plus one so the area enclosed

in this rectangle is n plus 1 times 2A

plus n times D and that must be equal to

2 times the original arithmetic sum this

means that the sum where I ranges from 0

to n of a plus I D must be n plus 1

times 2A plus ND all divided by 2.

Visual Arithmetic Sums

This is a short, animated visual proof demonstrating how to visualize the sum of an arithmetic sequence, producing a closed formula for such a sum.

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 wordless version (with epic music), check this link : https://youtu.be/uYaFE-1Lm5Y

This animation is based on a well known visual proof technique (and it can be found in Roger Nelsen’s book Icons of Mathematics). Specializing to a = 0 and d = 1 gives the sum of the first n integers (triangular numbers) formula. Specializing to a = 1 and d = 2 gives the sum of the odd integers (which yields the squares). In general, we can get any polygonal number formula from this fact (in addition to many others).

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

#mathshorts  #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #triangularnumbers #arithmeticsum #arithmeticsequence #finitesums


let's learn a fascinating fact about an

amazing curve

imagine a circle lying on a flat line

let's label one point on the circle

that's tangent to the line and let's

start rolling this circle over and over

and over again

if we Trace out the path traversed by

the Red Dot we get this curve here

this curve is known as the cycloid and

it will continue forever as long as we

let the circle roll but we can see that

the cycloid is really one curve repeated

over and over and over again so let's

study this one section of the cycloid

and see if we can find something

interesting about that particular curve

so assume that we have a circle with a

radius of R and we label one point right

at the bottom of the circle tangent to

the surface on which we roll

we roll This Circle exactly one full

turn to create one portion of the

cycloid as shown here one question of

interest about this curve is can we find

the area contained underneath the

cycloid and above the line upon which we

rolled that is can we find this area in

terms of the radius of the circle that

we rolled we know that the distance

between the two endpoints of the cycloid

is 2 pi r because it corresponds to the

circumference of the circle we rolled

we can take this circle we rolled and

place it in the middle like this and the

circle covers some of the area it also

splits the rest of the cycloid into two

symmetric pieces on either side

but now we do something interesting we

apply the two-dimensional cavalieri's

principle

to do this we will go along each of the

two symmetric shapes on either side of

this circle and we will take the width

of the shape at a given point and shift

it over to the left as shown here

because we keep the widths the same at

each point and don't change the height

the resulting shape that arises out of

this process will have the same area as

the shape we started with

it might not seem any easier to find the

area of these resulting shapes but now

we can apply another property the

symmetry of the circle if we draw the

lines coming from the center of the

cycloid to the top of these new shapes

we see that because of the symmetry of

the circle this area above the line

matches the missing area below the line

and likewise this happens on the other

side as well

after applying cavalieri's principle and

the Symmetry argument we see that we've

decomposed the area under the cycloid

into a circle of radius R and two

triangles of height 2R and base pi times

r

the area of the circle is pi times r

squared the area of both of the

triangles is also pi times r squared so

the area under the cycloid is 3 times pi

times r squared where R is the radius of

the circle rolled to create the cycloid

this is a truly amazing fact the area

under the cycloid is three times the

area of the circle rolled to create the

cycloid

The Amazing Cycloid Area (Visual Proof via Cavalieri)

In this video, we show how to find the area under one arch of the cycloid, which is a curve generated by tracing the path of a single point on a circle as the circle rolls on a flat surface. The cycloid area has a surprisingly nice connection to the area of the rolled circle.

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

This animation is based on a visual proof by Richard M. Beekman from the February 1993 issue of Mathematics Magazine (https://www.jstor.org/stable/2690472 - page 39).

This animation is based can also be found in Roger Nelsen’s book Proofs without Words (I may receive a commission from Amazon, at no cost to you, if you click the link below):

https://amzn.to/456nvfg

#cycloid #cavalieri  #cavalieriprinciple #manim #math #mtbos #animation #theorem #visualproof #proof #iteachmath #mathematics #area #circle #dissection #algebra #triangle

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


[موسيقى]

Visual proof of (A + B)^2 #shorts #maths #cbse #algebra  #academicdiary

think of a square when you see x^2. What about x^3?? Cubes :)


thank you

[Music]

suppose we have a right triangle and off

each of the legs we have a semicircle

with diameter equal to the leg of the

triangle

furthermore imagine a semi-circle that

has a diameter along the hypotenuse

according to the Pythagorean theorem the

area of the large semicircle on the

hypotenuse equals the sum of the areas

of the other two semicircles

so in this picture the Blue Area

counteracts the yellow area now when we

reflect the semicircle on the hypotenuse

about the hypotenuse and cancel the blue

and yellow areas of overlap we see that

the right triangle is filled and we've

created two loons off of the legs of the

right triangle like this

but from the Pythagorean theorem we know

that the blue shaded area and the yellow

shaded area are still equal

this means that the combined area of the

two loons constructed on the legs of a

given right triangle must be equal to

the area of that right triangle

this is a theorem due to Hippocrates of

chios

[Music]

Lunes off a Right Triangle (visual proof)

This is a short, animated visual proof showing that the area of a square inscribed in a circle is equal to the area of the four lunes created by drawing semicircles off of the sides of the square. #manim #math #mathvideo #mathshorts #geometry #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #root3 #pww #proofwithoutwords  #proof #algebra #areas #mathematics #mathvideo #mtbos

Check out these related videos:
https://youtu.be/-IKyUjRcFDw
https://youtu.be/5VaQVXECaVM
https://youtu.be/6DnnJlzbCC4
https://youtu.be/PNQxS6antBY
https://youtu.be/vq853dCS1Fs

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

This animation is based on a proof by Eugene A. Margerum and Michael M. McDonnell from the December 1997  issues of Mathematics Magazine (https://doi.org/10.2307/2691177 ) page 380. The letter to the editor in the  June 1998 issue of the same journal pointed out that Hippocrates of Chios proved this in the 5th century BC E(https://www.jstor.org/stable/2691213 ) page 224.

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Cambodian Odyssey 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=USUAN1100585

Artist: http://incompetech.com/


[موسيقى]

y

[موسيقى]

#trending #mathematics Visual Proof: Factors of Quadratic Equation Part 1 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


why is the infinite sum of the powers of

one-third called a geometric series how

can we make sense of the value of this

sum and what does it have to do with

geometry imagine a rectangle with area

one that can be subdivided into three

congruent rectangles each similar to the

original if we shade the top rectangle

we've shaded one third of the area now

we can repeat this process each time we

shade one of the rectangles and we take

the central rectangle and decompose it

into three congruent rectangles shading

one and continuing indefinitely if we do

this process forever or infinitely it

turns out that we have shaded exactly

the infinite sum one-third plus

one-third squared and so on of the

rectangle

but amazingly we see that this shaded

sum fills in exactly half of the

original rectangle since the original

rectangle had an area of one the

infinite sum has an area of one-half

Geometry behind the geometric sum of powers of 1/3!

This is a short, animated visual proof demonstrating the sum of the infinite series of the powers of 1/3. #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/RmTZmNrkqss

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

This animation is based on a proof by Rick Mabry from the January 2001 issue of The College Mathematics Journal page 19 ( https://www.jstor.org/stable/2687214).

For an alternate series decomposition for the same series, check out
https://youtu.be/vfEDDI3vfHU

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


let's see an approximation of a

fascinating curve called the virr

function this function is continuous

everywhere but differentiable nowhere in

particular that means that at every

point on the curve there is no uniquely

defined tangent line as the curve is

essentially an infinite collection of

spikes however there are also no breaks

in the curve it's one continuous

function if you zoom in you see a type

of self-similarity exhibited by this

function that allows it to have these

amazing properties

to get the full vros function replace

the finite sum shown here with an

infinite sum instead

Continuous everywhere but differentiable nowhere : Weierstrass Function Visualization!

This is a visualization of an approximation of the Weierstrass function, which is a function that is continuous everywhere but differentiable nowhere. The curve is defined as an infinite sum and there are several curves like this given different parameters.

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

#manim #math #mathshorts #mathvideo
 #construction #calculus #mtbos #manim #animation #theorem  #iteachmath #spiral #weierstrass #continuouseverywhere #continuity #differentiation #differentiablenowhere #function

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


here's an interesting fact about the 3 4

five right triangle find the in Center

as the intersection of the three angle

bis sectors draw the in radius to each

of the sides and draw the in circle do

you know the area of the in circle of

the 345 right triangle one way to figure

this out is to compute the area of the

triangle two ways first of all the area

is 1 12 * 3 * 4 but the area of the

triangle can also be found as the sum of

the areas of these three subt triangles

each with a height of R and with bases

of length three four and five so the

area is also 1 12 3 * r + 1 12 4 * r +

12 5 * R from these two computations

then we see that 6 = 6 * R but that

means the in radius R is of length 1 so

that the in circle has a radius of one

giving it an area of Pi it's amazing

that we get PI from the 345 right

triangle

Incircle area in 3-4-5 triangle?

This short demonstrates an intriguing fact about the area of the incircle of the 3-4-5 right triangle. Using two different methods of computing the area of the 3-4-5 rectangle, we find that the inradius has a length of one so that the area of the incircle is Pi!

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

To see a related video, showing the general connection between the area of a triangle and the inradius, see https://youtu.be/lRDtdIyFhpE?si=QcluFxSoJzEZMcmR

#manim #math #mathvideo #mathshorts #geometry #circle #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof #incircle #inradius #incircle #semiperimeter #incenter #pi #righttriangle #area

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


[Music]

start with a right triangle

with legs of length a and b

and hypotenuse of length c

now draw an a plus b by a plus b square

we can fit four rotated copies of the

triangle

in the square like this

notice that the space in the square not

covered by the triangles

is a square of side length c therefore

this unshaded area is c squared

we can also move the same colored

triangles together

to form two different a by b rectangles

now the unshaded area is given by an a

by a square

and a b by b square so that the unshaded

area

is a squared plus b squared consequently

we have seen that c squared must be

equal to a squared plus b

squared this is the pythagorean theorem

Pythagorean Theorem II (visual proof)

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using a dissection of a square in two different ways. 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 one of the oldest known visual proofs of the fact. According to page 1 of Roger Nelsen's first compendium (Proofs Without Words : https://www.amazon.com/Proofs-without-Words-Exercises-Classroom/dp/0883857006), this proof is adapted from the Chou pei suan ching.

For another proof of this same fact check out: https://youtu.be/q28H5NY5UVU

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


here's an interesting infinite sum one

plus two times a half plus three times a

quarter plus four times an eighth and so

on do you know how to find the sum of

this infinite series what if I showed

you this geometric picture would this

help you figure it out

[Music]

can you figure out the sum now let me

know in the comments

What's this infinite sum?

This is a short, animated visual proof computing the sum of a differentiated geometric series with ratio given by r=1/2. The proof uses a dissection proof of a 2 by 2 square.

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 bit longer/slower version with the final sum, check out this wordless video: https://youtu.be/2d4CTG1uJs8

This animation is based on a visual proof by Roger B. Nelsen from the December 1989 issue of Mathematics Magazine (https://www.jstor.org/stable/2689487 p. 332-333).

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

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


let's see how to use geometry to find

the exact trig values of multiples of

18° draw an isoceles triangle with base

angles given by 72° vertex angle given

by 36° base length of 1 and a side

length of x to construct this triangle

note that it appears naturally inside a

regular pentagon with side length one

use the angle bis sector of the lower

left 72° angle to create two subt

triangles one of them the one pictured

in yellow has angles of 36° in 72° so

the other angle must also be 72° that

means this yellow triangle is isoceles

so its other leg length is one but then

the other subt triangle the one shown in

purple here has two 36° angles which

means that it's also isoceles and that

means that its other leg length is one

as well and this fact allows us to see

that the base length of the yellow

triangle must be x -1 notice that this

yellow triangle is similar to the

original triangle because they both have

the three same angle measurements but

that means that the ratio x / 1 is equal

to 1 / X - 1 this equation implies that

x^2 - x - 1 must be 0 and the positive

root of this equation is x = 1 + > 5/2

this number is the golden ratio and is

often denoted by the Greek letter Fe

that means the side length of the

original triangle is the golden ratio

and the base length of the yellow

triangle is one less than the golden

ratio here's how we can use use this

diagram to find interesting trig values

first notice that the supplementary

angle to the 72° angle is 108° as shown

here when we BCT the 108° angle like

this cutting the other side in half we

create a right triangle with angles 54°

and 36° and a side length that is half

the golden

ratio we can also bisect the lower left

36° angle like this creating a right

triangle with angles 18° and 72° and a

leg length of the golden ratio - 1 / 2

notice that both these triangles

pictured in blue and purple have a

hypotenuse of length one using the blue

right triangle we see that the S of 54°

is equal to the cosine of 36° and they

are both the golden ratio divided 2

which ends up being the < TK 5 + 1 / 4

from the purple right triangle we see

that the sign of 18° which is equal to

the cosine of 72° is equal to 1 less

than the golden ratio / 2 and this

simplifies to the < TK 5 - 1 4 we've

been able to use this interesting Golden

Triangle to find trig values for various

multiples of 18° can you use the diagram

or perhaps other trig identities to find

the missing trig functions of the

multiples of 18° in the first quadrant

for instance can you find the cosine of

54° or cosine of 18° or the S of 36° and

the S of

72° let me know in the comments and

check out my playlist on trigonometry if

you like this kind of video

Exact Trig Values for Multiples of 18 Degrees Geometrically (visual proof)

This is a short, animated visual proof demonstrating how to compute the various exact trigonometric values for multiples of 18 degrees using the geometry of the Golden Triangle.

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

Check out these related videos about trigonometry:
https://youtu.be/88-uO2RFm78 (tangents of 22.5 degrees and 67.5 degrees)
https://youtu.be/A0ZvO_F-LZo?si=9HmF52L34P3LzhEx (tangents of 15 and 75 degrees)
https://youtu.be/XIC7SIdl7pI?si=vM3inxCJHEFqo70f (half angle tangent formula that you can use to get this result)
https://youtu.be/XJ-GaUrY-LE?si=vIpTUyAjHs5wzxvK (arctangents)

This animation is based on a visual proof from Brian Bradie in the September 2002 issue of The College Mathematics Journal ( https://www.jstor.org/stable/1559057 page 318--319)

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

 #mathshorts #mathvideo #math #trigonometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #pi #triangle #root5 #squareroot #18 #18degrees #36 #54 #72 #degrees #isoscelestriangle #sine  #cosine #sinefunction #cosinefunction #goldenratio #goldentriangle


let's see an elegant semicircle proof of

the double angle formulas for S and

cosine in the XY plane imagine the

origin as a point labeled

O then imagine a point labeled a that's

exactly distance one from o the line

segment O A has a length of one so we

can use it to create a semicircle of

radius one let's label the opposite

point of the semicircle as B we can take

the radius and rotate to some point C on

the semicircle let's imagine that this

Arc AC subtains an angle of 2 * T we can

create a right triangle by dropping a

perpendicular from point C down to the

x-axis and labeling the intersection

Point D in the right triangle OD d c the

leg opposite to the angle 2T has a

length of sine of 2T and the leg

adjacent to the angle of 2T has a length

of cosine

2T now we draw another triangle

connecting the points a c and B like

this by ph's triangle theorem this is a

right angled triangle and the right

angle occurs at Point C now consider the

triangle B pictured here the angle B is

supplementary to the angle 2T and

therefore has a measurement of 180 minus

2T because two of the side lengths of

this triangle are radi of the circle

they both have a length one and that

means that this triangle is isoceles so

the other two angles in this triangle

must be equal equal and therefore they

must both be T so that the full angle

measurement of the triangle is

180° notice that the right triangle ACB

has a hypotenuse of length two since it

consists of two radi of the semicircle

the leg opposite to angle T in this

triangle therefore has a length 2 * the

S of T and the leg adjacent to the angle

T pictured has a length 2 * a cosine of

T now we have enough information to

prove our double angle identities this

yellow shaded triangle right here is a

right angled triangle with one angle T

and this red shaded triangle here is

also a right triangle with one angle T

so these two triangles have the same

angle measurements and that means that

they're similar right triangles using

similarity we know that the ratios of

the short leg to the hypotenuse in the

two triangles are equal from this we can

conclude that the S of 2T / 2 * the

cosine of T is equal to 2 * the S of T /

2 but now the twos cancel on the right

side and we can rearrange this equation

so that it reads s of 2T = 2 * the S of

T * the cosine of T we can also use

similarity to see that the ratios of the

long leg length to the hypotenuse are

equal in both the

triangles from this we see that 1 + the

cosine of 2T / 2 * the cosine of T must

equal 2 * the cosine of T divided Again

by the hypotenuse of the big triangle

which is 2 2 we can cancel the twos on

the right side again and clear

denominators to see that cosine of 2T =

2 * the cosine s of tus 1 the identity

on the left sin of 2T = 2 * the S of T *

the cosine of T is the double angle

identity for the sign function and the

identity on the right cosine of 2 T = 2

* cosine 2 of tus 1 is the double angle

identity for the cosine function

together these form the double angle

identities for S and cosine and we prove

them using a semicircle diagram this

proof is due to Roger B Nelson check the

description for more information about

this proof or for a link to another

proof of the same facts that uses the

law of signs and the law of cosines in a

triangle

Trig Double Angle Formulas from Semicircle (visual proof)

This is a short, animated visual proof of the Double angle identities for sine and cosine. To get the formulas we use a semicircle diagram and rely on similarity of two right triangles formed inside.

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

For another visual proof of this fact using the laws of sines and cosines, check out this animation:
https://youtu.be/0ggGQ96eaiE

This animation is based on a visual proof by Roger B. Nelsen  from the January 1989 issue of The College Mathematics Journal (https://www.jstor.org/stable/2686819 - page 51).

#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


start with a line segment of length a

and cut it into two segments of length a

plus b/2 and a minus b/2 use the Aus b

over2 line segment as the radius of a

circle connecting these two points

creates a right triangle with hypotenuse

of length OT of a s + b^ S ID 2 by the

Pythagorean theorem the hypotenuse is

greater than the leg length so a plus b/

2 is less than or equal to the < TK of a

+ b s / 2 now draw this radius so that

connecting these two points creates a

tangent to the circle Again by the

Pythagorean theorem the side length here

isun of a * B so theun of a * B is less

than or equal to a plus b/ 2 this yellow

shaded triangle is similar to this

larger green shaded triangle and by

similarity that means this side length

is 2 a b / a plus b and that side length

2ab ID a plus b is less than or equal to

the hypotenuse square root of a any

choice of non- negative real numbers A

and B leads to a similar diagram except

in the case when a equals B then all the

line segments collapse to the same so we

have equality throughout if and only if

a equals B we used one diagram to

produce an inequality relating the

harmonic geometric arithmetic and

quadratic means

Mean Inequalities

This short animated proof demonstrates the two variable mean inequalities for the arithmetic mean, geometric mean, harmonic mean and root mean square (or quadratic mean).

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

This animation of the inequalities is based on a visual proof by Roger Nelsen from the June 1987 issue of Mathematics Magazine (https://www.jstor.org/stable/2689561) p. 158.

If you like visual proofs of inequalities, see these links:

AM-GM inequality via rectangles/squares: https://youtu.be/62G9fak1vyk

AM-GM inequality via circles: https://youtu.be/Pt8IX_Q5U1A

AM-GM-HM-QM via Moments: https://youtu.be/XJ3643WhKjY

#math #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #inequality #iteachmath #mathematics #pythagoreantheorem #similartriangles #means #meaninequality #mathshorts #mathvideo #mtbos

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


the Pythagorean theorem says that the

area of the square should be equal to

the sum of these two squares but we can

really understand why it's true by

looking at these three triangles the

large green triangle is a reflection of

the original triangle the pink one is a

scaled down copy a similar triangle and

the blue one is similar again because

the triangles are similar these angles

must be equal so we can fit this pink

triangle nicely inside the big one at a

right angle to the hypotenuse

blue triangle works the same way it must

form a right angle with hypotenuse which

means that blue and pink areas must add

to give the original triangle cool right

but how does that prove that the sum of

the squares is equal well remember again

that the smaller triangles are just

scaled down copies of the big one so if

one square is K times the area of one

triangle that factor K must apply to all

squares and triangles and that is enough

to prove that a squared plus b squared

is equal to c squared

Understand WHY the Pythagorean Theorem is True - Nice Visual Proof

The Pythagorean Theorem is one of the most famous theorems in Geometry and all of Mathematics. This is a really nice proof using similar triangles.
https://medium.com/nice-math-problems

Music:
Island by Luke Bergs | https://soundcloud.com/bergscloud/
Music promoted by https://www.chosic.com/free-music/all/
Creative Commons CC BY-SA 3.0
https://creativecommons.org/licenses/by-sa/3.0/


[संगीत]

a

[संगीत]

Factorial Product Formula (visual proof)

This is a short, animated visual proof showing an identity involving various factorials by taking the products of all the integers in a certain n x (n+1) grid in two different ways. #manim #math #mathvideo #mathshorts #sequences #triangles #animation #theorem #pww #proofwithoutwords #visualproof #proof #factorials  #product #pww #proofwithoutwords  #proof #algebra #factorial #mathematics #mathvideo #mtbos

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

This animation is based on a visual proof by Edward T.H. Wang from the March 1989 issue of The College Mathematics Journal (https://www.jstor.org/stable/2686271- page 152).

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

_____________________________
Music in this video:
Reaching The Sky (2020) 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/


I've got a right angled triangle having

happiness 10 centimeters side B is 8

centimeters and side C is six

centimeters we are going to prove the

Pythagoras Theorem so in five seconds

theorem we are to prove that hypotenuse

10 centimeter square is equal to the sum

of 6 centimeters square and eight

centimeter Square so let's prove the

bicycle theorem I have got four copies

of this trendal ADC we are going to plot

these four triangles in this Square p q

r s Having side 14 centimeters so first

we are going to create the square on the

hypotenuse so you can see that by using

the four copies of the triangle ABC we

have created a white square inside 10

centimeters that is the hypotenuse of

the triangle ABC area of this y Square

hundred centimeter Square now let's

prove the another part of the particular

theorem so this is the second part of

the particular theorem here we have b

square is 6 into 6 is 36 centimeter

centimeter Square whereas the other

white part is 8 into 1864 64 square

centimeters we see that 64 plus 36 is

100 square centimeters so uh before we

have 100 square centimeters as C square

and now we have 64 plus 36 that is 100

square centimeters as b square plus C

Square so a square equals to b square

plus C square and hence Pythagoras

Theorem is proved

Visual Proof of Pythagoras Theorem #pythagorustheorem #pythagoreantheorem #mathshorts

The Visual Proof of the Pythagoras Theorem
a²=b²+c²
How to prove pythagoras theorem practically?
Enjoy this video and learn something!

Comment your doubts and suggestions.

*You may enjoy my other videos as follows:-*
My Maths Shorts playlist:-
https://youtube.com/playlist?list=PLOPjB5QEzAe2z6q6Jmvl9g1HDp9XBXykS

My Maths Problem and Solution playlist:-
https://youtube.com/playlist?list=PLOPjB5QEzAe3Nk93GXyHhBhI_uarZZL-v

Hashtags:-
#pythagorustheorem
#pythagorastheorem
#youtubemath
#pythagoras_theorem
#proofofpythagorastheorem
#pythagorastheoremproof
#youtubemaths
#geometry
#rekhaganit
#umeshkhetan
#viralmathshorts
#ytmathshorts

Proof of the pythagoras theorem,Pythagoras theorem,Pythagorean theorem,derive pythagoras theorem,how to prove pythagoras theorem,how to prove pythagorean theorem,pythagoras theorem class 10,pythagoras theorem proof,pythagorean theorem explained,pythagorean theorem proof,pythagorus theorem,what is pythagoras theorem,How to prove pythagoras theorem practically?,How to prove pythagorean theorem practically?,Pythagoras pramey ko siddh karen,पाइथागोरस प्रमेय का सूत्र


let's see the geometry behind the

infinite sum of the powers of one-fifth

start with a one by one square if we

connect the vertices to the midpoints

like this it turns out that we divide

the square into five equal pieces the

four Triangles around the outside and

the inner Square all consist of

one-fifth of the area of the square so

this triangle is one-fifth but then we

can repeat the process on the inner

Square to get one-fifth of one-fifth and

so on we can continue this process

indefinitely each time taking one-fifth

of the inner Square which is one-fifth

of the previous Square we spiral this

down forever and it turns out that we've

shaded one-fifth plus one-fifth squared

and so on of the entire Square

but this shape can be rotated three more

times to fill in the entire Square

therefore the infinite sum of powers of

one-fifth must be one-fourth

foreign

Geometry of sums of powers of 1/5

This is a short, animated visual proof demonstrating the infinite sum of the powers of 1/5.

Here is the longer form (wordless) version of this video: https://youtu.be/IguRXWNwrn8

Here is an alternate dissection proof for this series: https://youtu.be/yp7afEXYeC4

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

If you like these videos, please consider subscribing.

This animation is based on a proof by Rick Mabry from his website https://lsusmath.rickmabry.org/rmabry/fivesquares/fsq2.gif and featured in Roger B. Nelsen's third Proofs Without Words compendium (https://bookstore.ams.org/clrm-52/).

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

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


you've probably heard of the Pythagorean

theorem a squared plus b squared equals

c squared but what is it actually and

why is it true let's draw this out A and

B are these two sides and C is the

hypotenuse if you squared say the

hypotenuse then you would get the area

of a square with a side length C the

Pythagorean theorem states that for any

right angle triangle the square of the

hypotenuse C is equal to the sum of the

squared of the other two sides A and B

also called the feet but don't just take

my word for it let's prove it using a

visual proof this is a right triangle

these are four identical right triangles

you can arrange them in a way that they

form a square of side length a plus b as

you can see from the side length the

square in the top left has an area of a

squared and the one in the bottom left

has one of B squared however these

triangles can be rearranged to form an

identical square of side length a plus b

now however the middle square has formed

an area of c squared hence from this

proof a squared plus b squared equals c

squared now there are many other

numerical ways to prove this however I

find this visual proof so elegant hope

you've learned something stay curious

and I'll see you in the next one

An elegant visual proof of the Pythagorean Theorem #shorts

Welcome to DeSTEMber, a series where I answer your questions about math and science throughount the month of December. Do stay tuned, for more knowledge is on the way!

In this video, I'll be describing the Pythagorean Theorem and introducing a famous visual proof for it.


e o jogo por favor parte agora inteira

que é que tu Square of here a esse

Guanambi Square Orange is the new Black

2 and White is a Long skirt and Lines of

Range of the Idiot widgets Will of Seeds

Where does cancros da pele da s4devs

Where e depois do Circular 1

Visual Proof of Pythagorean Theorem | Mathematics | JEE Mains - Advanced | NEET | KVPY | Olympiads

Pythagoras Theorem - Visual Proof

A² + B² = C²

Comment your views

Tag a friend to show this
Turn on post notifications

Follow : @championsacademy.in

Tags :

#Pythagoras #Theorem #Math #Triangle #Proof #PythagorasTheorem #Trigonometry #RightAngle #Hypotenuse #ProofOfPythagorasTheorem #VisualProof #StudyFromHome #MathAnimation #MathTutorial #MathematicalCreativity

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


foreign

visual proof of (a+b)² | algebra formula solution trick| #maths #algebra #jee #engineering

visual proof of (a+b)² || #mathematics #proofing #formula1 #video #mathstricks #shorts

@akaash.sain04 @PhysicsWallah
#mathstricks  #viral #physics
@ABHINAYMATHS

math formula class
a plus b ka whole 2wise


foreign

[Music]

[Music]

Visual Proof of Maths Formula

#Hukabi #HukabiBD
Puzzles and math tricks can have a number of benefits, including:
1. Boosting brain power: Puzzles and math tricks can help to stimulate the brain and improve cognitive function, memory, and problem-solving skills.
2. Improving concentration: When working on puzzles or math problems, individuals need to concentrate and focus, which can help to improve overall concentration and attention span.
3. Enhancing creativity: Puzzles and math tricks can help to foster creativity by encouraging individuals to think outside the box and approach problems in new and innovative ways.
4. Developing critical thinking skills: Puzzles and math tricks require individuals to think critically and logically, which can help to develop these important skills.
5. Reducing stress: Engaging in puzzles and math problems can provide a welcome distraction from everyday stressors, helping individuals to relax and unwind.
6. Building confidence: Successfully completing puzzles and math problems can help individuals feel a sense of accomplishment and build self-confidence.
7. Improving math skills: Math tricks can help individuals to develop a deeper understanding of mathematical concepts and improve their overall math skills.
8. Providing entertainment: Puzzles and math tricks can be a fun and engaging way to pass the time, providing hours of entertainment for individuals of all ages.

#Math #Maths #Trick #Tricks #IQ #IQTest #Quiz #Riddle #MathChallenge #MathRiddle #Technic #MathTricks #Magic #MathMagic #MagicMath #MathMemes #MathSkills #MathStudent #ScienceMath #Mathematician #MathProblems #MathTeacher #Teacher #Motivation #ScienceMemes #Puzzle #MathPuzzle #Learning #Building #Education #Study #Studygram #Chemistry #Physics #Biology #Engineering #Art #BabyToy #LearningToy #EducationalToy #Reels #Shorts #ViralVideo #Hukabi

Like, Follow & Subscribe!!!
Instagram ▶ https://www.instagram.com/hukabi.bd
Facebook ▶ https://www.facebook.com/hukabi.bd
Tweeter ▶ https://twitter.com/HukabiBD
TikTok  ▶ https://www.tiktok.com/@hukabi.bd
Linkedin ▶ https://www.linkedin.com/company/hukabi

💜 Helpful? Do Subscribe (#Hukabi)

------------------------------------------------------------------------------
👉For business inquiries: hukabi.bd@gmail.com
------------------------------------------------------------------------------


I know

[Music]

yeah

[Music]

I can do this every morning every

evening if you screaming straight back

to Sunrise

foreign

[Music]

[Music]

visual proof of sum of first n natural numbers #viral #youtubeshorts #maths #shorts #formula1


start with a square of side length one

cut the square into nine smaller

congruent squares and shade eight of

them like this shading a total of 8 9ths

of the square now repeat the process on

the unshaded Square shading 8 9th of

that 1 nth thus shading an additional 8

over 9 squar of the outer square if we

repeat this process indefinitely we

shade 8 over 9 cubed and then 8 over 9

4th and so on at each stage we shade 8

over 9 to the N if we then think of this

as an infinite process or a limiting

process we can then think of the

limiting diagram which is the entire

shaded square and therefore the infinite

sum that we're interested in which is

the Shaded area is equal to the Shaded

area of the full square or one one

interpretation of this is that 0.8

repeating in base 9 is 1 another

interpretation is that 8 * the infinite

sum of the positive powers of 1 9th is 1

or that the infinite sum of the positive

powers of 1 nth is 1/8

0.8888… = 1 (in base 9)

This is a short, animated visual proof showing the sum of the infinite geometric series with first term 8/9 and ratio 1/9, which in turn allows us to compute the sum of the series of powers of 1/9 and determine an interesting base 9 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 (more dramatic) of this animation see
https://youtu.be/C4t_ps3VKvI

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

This animation is inspired by the a proof by Elizabeth M. Markham from the October 1993 issue of Mathematics Magazine page 242 (https://doi.org/10.2307/2690738).

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

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


suppose A and B are two positive real

numbers draw a right isoceles triangle

with side length sare otk of a so that

its area is/ 12 * a off the top vertex

draw a right isoceles triangle with s

side length square root B so that it has

an area of 1 12 B these two triangles

form the boundary of a rectangle that

has Dimensions squ < TK of a and squ <

TK of B so it has an area of the squ <

TK of a * B in this diagram imagine

folds along the top and left of the

rectangle so that when we fold in the

two isoc right triangles they meet along

their diagonal like this because the

areas of the two triangles cover the

area of the rectangle we see that the

square root of a * B which is the area

of the rectangle is less than or equal

to 12 a + 1 12 B this inequality is

known as the arithmetic mean geometric

mean inequality note that we get

equality if and only if the Shaded area

is a square so that A and B are equal

Fold the AM-GM inequality

This is a short, animated visual proof demonstrating the arithmetic mean geometric mean inequality using paper folding.

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

For other visual proofs of this fact see:

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

This animation is based on a visual proof by Yukio Kobayashi from the 2002 issue of the Mathematical Gazette (page 293).

#manim #shorts #mathshorts #mathvideo #math #mathematics #amgminequality #mtbos #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #calculus #inequality

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


start with a right triangle with sides

of length A and B and a hypotenuse of

length C now draw squares off the side

lengths enclosing an area of a squared

and B squared respectively then draw a

square off the hypotenuse in closing an

area of c squared here's a proof that

the area contained in the C by C square

is equal to the sum of the areas in the

other two squares imagine these two

lines of length C pass through the

center of the B by b square then they

cut the side length of the B by b square

into lengths of B plus a over 2 and B

minus a over 2. we can then take the

four resulting pieces and shift them to

the C by C square like this the leftover

middle piece is also a square and you

can check that the side length is a so

we can shift to a by a square inside

therefore we've seen that c squared

equals B squared plus a squared where A

and B are the leg lengths of a right

triangle with a hypotenuse of length C

this is the Pythagorean theorem As

proved by Henry dudney

Pythagorean Theorem XI (Dudeney’s Dissection)

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using a Henry Dudeney's dissection argument (congruent-by-addition proof).

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

This animation is based on a proof from Henry Dudeney and is discussed in Howard Eve's book "Great Moments in Mathematics (before 1650)." You can also find this in Roger Nelsen's first PWW compendium on page 6.

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
https://youtu.be/ZlGaQdNRdqA
https://youtu.be/uqGjs_n-f7U
https://youtu.be/wWEZ-rUlBDQ
https://youtu.be/I75HYC8TRNY

#math #manim #pythagoreantheorem #pythagorean #triangle #animation #theorem #pww #proofwithoutwords #visualproof #proof  #mathshorts #mathvideo

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


here's a beautiful visualization for the

alternating sum of squares starting with

n squared and ending at 1 squared N

squared can be represented by an N by n

array of boxes then we can remove the

lower n minus 1 by n minus 1 array

re-add the lower n minus 2 by n minus 2

array and so on all the way down until

we either add or subtract the bottom one

by one square array this process has

resulted in a collection of L-shaped

gnomons which we can shift together to

form a triangular array with base length

n this means that there are t sub n

boxes in the resulting array where T sub

n is the nth triangular number therefore

the alternating sum of squares starting

with n squared and ending at 1 is given

by n times n plus 1 over 2 by the

formula for the Triangular numbers

Alternating sum of squares is triangular!

This is a short, animated visual proof demonstrating how to visualize the alternating sum of squares.

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 Steven L. Snover from the April 1992 issue of Mathematics Magazine (https://doi.org/10.2307/2690486 - page 90).

Compare this formula to the regular sum of squares formula visualized here: https://youtu.be/-tJhH_k2LaM

Here is an alternate visual proof of this same fact: https://youtu.be/5u34ifloAB4

#mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #triangularnumbers

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


here's an alternate interesting way to

find the sum of the first n positive odd

integers visually start with a square

array of cells as pictured here if we

imagine both dimensions to be n then we

have a total of n squ cells in this

array now let's go through and place two

tokens on every cell in the

array because there are n s cells and

two tokens per cell there are a total of

2 * n^ 2 tokens in this diagram

but now let's count the number of tokens

in a different way we can instead divide

the square array into triangular and

trapezoidal shapes each containing a

single diagonal line of tokens the first

one contains one token the next one

contains three tokens followed by five

tokens and so on each trapezoidal strip

contains exactly two more tokens than

the previous one so we get an ascending

list of consecutive odd integers ending

at the odd number 2 N minus one

but notice that this part of the diagram

can be repeated on the bottom half as

well and that means that the total

number of tokens is 2 * the sum of the

first n odd positive integers 1 + 3 + 5

and so on up to 2 nus 1 but now if we

cancel the two from both sides we get a

formula for the sum of the first n odd

positive integers in particular we see

that n^2 is equal to the sum where K

ranges from 1 to n of 2K minus 1 by

Counting tokens on a square array of

cells we've seen that the sum of the

first n odd positive integers is equal

to n^ S if you liked this visual proof

check out the description for other

visual proofs of the same fact and let

me know in the comments which one is

your favorite

Summing Odds using Tokens (visual proof III)

This is a short, animated visual proof showing one method to find the sum of the first n odd positive integers by stacking tokens in a square array of cells.

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/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 Timothée Duval that appeared in the Mars-Avril 2007 issue of Tangente no 115 (link.  page 10).

You can also find this diagram in Roger Nelsen's book on page 114: (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