all right so here we're going to look at

some examples of the ratio test so part

A we're just going to figure out whether

the series converges or diverges n = 1

to Infinity 4 n * n factorial 2 over n +

factorial so I think the thing that'll

probably make this one most confusing is

just the factorials so we'll just have

to be a little careful when we simplify

those so the limit as n goes to Infinity

we would have 4 the n + one um we would

have n +1

factorial squared we would have we would

replace the N with n + one so we would

have n + 1 + 2 or n + 3

factorial we can multiply by the

reciprocal n + 2

factorial over 4 n * n factorial

2 so all right so now just kind of the

simplification part um I'm just going to

sort of reline everything up just to you

know you certainly don't have to do this

but to maybe make things a little

clearer we have 4 the n + 1 I'm going to

put the four to the n in the denominator

we have an N +1 factorial squared so

that means we have n +1 factorial times

another n +1

factorial we've got the N factorial

squared so I'm going to put an N

factorial here and another n factorial

there and then our n + 2 Factor

factorial over n +

3 factorial how often do you get to say

factorial um all right so I think the

first part's easy um we just have 4 the

n + 1 over 4 the N that'll just be a 4

to the 1st in the numerator n + 1

factorial Over N

factorial again if you imagine expanding

this out you would have n + 1 * n * n -1

* N - 2 dot dot dot in the denominator n

factorial would be n * nus1 * nus 2 time

dot dot dot so everything would cancel

out except for an n + one so we would

have an n+ one in the numerator well

same thing is going on with our third

Factor so we would have n + one over 1

and then n + 3 over n + 2 or excuse me n

plus 2 factorial over n+ 3

factorial well again if you imagine

expanding this out we would have n + 2

one smaller would be n + 1 then time n *

n minus one dot dot dot in the

denominator we would start with n + 3

then we would subtract one from that and

have n + 2 n + one dot dot dot notice

everything's going to cancel out except

for the n + 3 in the

denominator okay so now we're taking the

limit of this stuff but we've got the

limit as n goes to

Infinity notice in if we multiply n+ one

and n + one we would get n 2 + 2 n + 1

over n + 3 well this is a case where

we're taking the limit as n goes to

Infinity the degree of the numerator is

larger than the degree of the

denominator so this is just going to go

off to positive Infinity well that's

certainly greater than one and when it's

greater than one the ratio test tells us

that the um original series will

diverge

The Ratio Test , Another Example #3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Ratio Test , Another Example #3.  Just another example showing that series converges or diverges using the ratio test.


all right in this video we're going to

look at finding some arc lengths of

parametric curves

so in part a we've got the absolute

value of 6 minus t

equals x and then y equals t we're going

to evaluate this from

0 to three so we'll have to integrate

from zero to three

we'll take the derivative of x with

respect to t squared

plus the derivative of y with respect to

t squared

dt so notice

for x equals the absolute value of 6

minus t

we can actually just say this is the

same thing as 6 minus t

for t on the interval 0 to 3 because uh

if you take any value between 0 and 3

the 6 minus t will stay positive so we

can just remove the absolute value

well that says the derivative of x with

respect to t then is just going to be

negative 1.

and that tells us the derivative of x

with respect to t

squared is just going to equal positive

the derivative of y with respect to t

that's also just going to be

positive one so the derivative of y with

respect to t

squared will also just be positive one

easy enough so we've got the integral

from zero to three

we just have one plus one dt

well that's the integral from 0 to 3 of

just the square root of 2

times or square root of 2 dt

so when we integrate we'll get the

square root of 2 times t we'll have to

evaluate that from zero to three

and that's just going to leave us with

three root two minus zero

so our arc length will simply be three

times the square root of two

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Arc Length of a Parametric Curve, Another Example #1.  In this video, I find the arc length of a parametric curve.


all right in this video we're going to

look at finding some limits of vector

functions so in part A we have the

vector function R of tal 4 minus t * I

plus < TK of 12 + T * J minus cosine of

pi T / 8 * K we want to find the limit

as T approaches

4 well all we have to do in this case

kind of like you know with

a you know limits that you see in

calculus one um so we're taking limit as

T approaches 4 all we're going to do is

just plug in um tal

4 and get our solution you know if we

end up with something undefined um you

know maybe we'll have to do something a

little bit

different but in this case you know this

is kind of a very simple one I don't

think we'll run into that issue at all

so notice if we plug in tal

4 well we'll just get 0 * I plus we

would have the < TK of 12 + 4 so we

would get the < TK of 16 * J and then

we'll get minus cosine of I guess we'll

have 4 pi over

8 time

K but if we simplify this this is just

going to be 0 I + 4 * J um let's see

this is going to be cosine of Pi / 2

when we simplify cosine of Pi / 2 is uh

going to give us Z

so 0

K so really uh we could rewrite this as

0 comma 4 comma 0 that will be the limit

of uh the vector valued function as T

approaches pos4

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Limit of a Vector Function - Another Ex 1. In this video, I find the limit of a vector valued function.


okay so now we're gonna do another

example using implicit differentiation

so in Part B here we're gonna find DS DT

for the function S to the third minus T

to the third equals eight and sometimes

people get it liking a little confused

when you have different variables

besides x and y so the main idea to

remember you know like in the first

example when we take the derivative with

respect to X that's what's in you know

the bottom part here that's the variable

that we kind of treat like normal so in

this case we're gonna treat the T sort

of like the X in the sense that we'll

treat it like normal so what I mean is

okay so when we take the derivative here

of s to the third minus T to the third

equals eight so s to the third minus T

to the third equals eight again we're

taking the derivative with respect to T

of both sides is what we're doing and

normally people don't write this step

but again the idea is so in this case

when we take the derivative of s to the

third we'll just get well three s

squared but then we have to tack on a DS

over DT and that's what we're trying to

solve for in this problem and then we

would have - ok so the derivative of T

cubed would be 3t squared and in this

case you could think about you're really

tacking on like a DT over DT which is

just 1 so we don't write it so again the

idea is whatever is on the bottom that

variable we just kind of do it like

normal we don't have to tack on any you

know DS DT or dy DT or any other extra

little bit but for the S variable in

this case we do and so the derivative

with respective T of 8 is just 0

and now just the same thing as before we

have to solve for D s DT well we can add

3t squared to both sides that will leave

us with three s squared times DS over DT

and now all we have to do is just divide

both sides by 3s squared so 3s squared

and then we'll be left with our solution

DS DT the 3s will cancel so we can just

simply write that as T squared over s

squared

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Implicit Differentiation - Basic Example 2 / 3 .  Just a fairly straight forward example of finding a derivative using implicit differentiation.


all right so one more example here

regarding inconsistent systems and again

what we're trying to do is just

determine if a system of linear

equations and three variables has no

Solutions or if it does have a solution

and again we're just using elimination

by addition so we've got this system of

equations x - 4 y + 2 Z = -2 x + 2 y - 2

Z = -3 and then our last equation just x

- y = 4 so you could write in plus z z

if you if you want uh you know if you

actually want to see that

term so I think what I'm going to do

really again though if you think about

this last equation it it's already uh

down to two variables in the sense that

it's just X and Y so what I'm going to

do is I'm going to take my uh my first

two equations I think and I'm going to

again just eliminate the Z from those

two so let's see so we've got x - 4 y +

2 Z = -2

we've got x + 2 y - 2 Z =

-3 and again if we add these up

immediately we'll get 1 x + 1 x which is

2x we've got -4 y + 2 y which is 2 Y 2 Z

uh plus -2 Z will be 0 Z and then we've

got -2 + -3 which

is5 so now what I'm going to do do is

I'm going to use this equation that

we've just found so

2x - 2 y = -5 and then I'm going to use

my third equation that we had originally

which said uh x - y = 4 and I'm going to

do the same thing I'm going to try to

eliminate one of my variables uh it

looks like to me we could multiply this

uh this bottom equation we could

multiply both sides by -2 so -2 and

-2 so let's see I'm going to copy down

my my 2x - 2

Yus or excuse me equals 5 just like it

was and then our second one if we

distribute we'll get -2X we'll get -2 *

Y which is pos2 y we get 4 * -2 which

is8 and well if we add these together

now we get 0x

we'll get 0 y so we're left with zero on

the left side we have -5 +8 which is -13

again we're at this place where clearly

uh you know 0 doesn't equal

-3 so this tells us that there is no

solution to this system of

equations and we say that it is an

inconsistent we would say it's an

inconsistent system of equation a ations

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Systems of Linear Equations - Inconsistent Systems Using Elimination by Addition - Example 3


okay so just another little example of

writing a uh an infinite sum using

series or Sigma notation so here we've

got the series 2 over 13 uh - 4 over 13

2 + 8 over 13 cubed - 16 uh over 13 to

4th Etc so usually to me the first thing

is to kind of you know spot the pattern

and then uh hopefully we can figure out

a nice little formula from that so one

thing I notice immediately is its

alternating sign it's positive negative

positive negative so at some point in

there I'm going to Le like a um you know

a negative one to some power um so I'm

thinking like a negative one to the N

let's don't put a equal sign or -1 to

the n + one we'll need something like

that at some point and again that will

just simply make it alternate signs um

the N or the N plus one just to make it

start positive or negative depending on

our index

okay so I noticed that and then I think

about kind of the numbers on top you

know we've got two to the 1st well four

that's really just 2 to the

2 um and then we've got eight which is 2

to the

3 and then we've got 16 which is 2 to

the 4th so hey I'm just seeing really

powers of two on top again we talked

about the alternating signs and then in

the bottom it's pretty clear we've got

you know powers of of of 13 13 to the 1

13 SAR 13 to the 13th 13 to the 14th Etc

so to me now um it's starting to look

like something where I can put in my my

uh series

notation so let's see so we've got a two

to a power um of n if you want to the

alternating part the negative one uh

instead of having negative 1 to the end

we can just make it

negative -2 to the N because again as n

Al you know increases uh the sign will

still Al alternate between positive and

negative so perfectly valid to instead

of having this negative one term just

include it with the uh you know the

Power of Two And in the bottom we also

have

13 well you know the the uh the

exponents are always the same so again

this the exponent to me is kind of

what's changing so that's where I put an

n generically and notice whatever the

power is on the two The Power of the 13

is exactly the same so if you've got an

n on top we should have an n on the

bottom as

well let's see we want this to start at

an exponent of one so we'll just have

well Nal 1 up to infinity and I'm going

to rewrite this one more time so Nal 1

to Infinity I'm going to write this as

just one big fraction -2 over

13 all raised to the N power and now

we've kind of got a a nice uh clean

little formula to uh to go with our

infinite series here

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Geometric Series using Sigma / Summation Notation, Ex 2. This video shows how to write an Infinite geometric series .using sigma / summation notation. I do not find the actual sum for this particular convergent geometric series.


all right here we're going to do uh look

at some examples related to the Limit

comparison test so in part A here we

have the series from 1 to Infinity 5

over 2 n^2 +

3 when I decide what what series to

compare this to again I just kind of

keep

uh I keep again you know what I call

dominant terms so in the numerator you

know kind of the only term that is there

is the five um and when it's a constant

it doesn't even matter whether you keep

the constant or not um in the

denominator as n gets large well again

you know three is a constant the the 2 N

squar is going to be the the dominant

term that's what's going to get the

largest the

fastest um so I'm going to compare um

whoops I'm going to compare um our our

Series 5 over 2 n^ 2 + 3 I'm going to

compare that to the series 5 over 2 n^ 2

and this series again we can Factor just

the 5 over2 out

front and then we would be left with 1

over n^ 2 and again uh 1 over n^2 is a

convergent P

Series so again that simply means it has

some finite value well if you multiply

it by 5 over two it's still going to be

finite so this is a convergent Series so

if we do our limit comparison uh test

we'll do the limit as n goes to Infinity

so I'm going to take our original series

5 2 n^ 2 + 3 and I'm going to divide

that by this new series uh I'm going to

divide that by this series that I'm

comparing it to 5 over 2 n

s well that's going to give us the limit

as n goes to Infinity we've got five

over 2 n^ 2 +

3 if we uh you know flip and multiply we

have 2 n^2 over 5 but what we're doing

here is we're forcing you know the

degree of the numerator to equal the

degree of the denominator so um let's

write this one all the way out we've got

10n SAR usually at this point I just

kind of stop um we would have again a 10

n^ 2ar +

15 well as n goes to Infinity since the

degree of the numerator equals the

degree of the denominator that limit is

simply going to equal the ratio of the

leading

coefficients so the uh limit's going to

equal 10/ 10 or in this case one and

recall the uh the uh limit comparison

test simply says that if you have a

positive finite number both series do

the same thing so in this case it would

mean that our original series must also

converge so this series also converges

again just because we're comparing it to

a series

that converges

Limit Comparison Test for Series - Another Example 1

(https://youtu.be/Hg5RwzLAqRs)
(https://youtu.be/xesQnFWw8f8)
(https://youtu.be/Hg5RwzLAqRs)
(https://youtu.be/gNSWa3VDx9k)
(https://youtu.be/dX_bwYp-lVc)
(https://youtu.be/9kZrKC9ueEE)
(https://youtu.be/QAZZgGDnnCg)

In this video, we dive into the Limit Comparison Test for evaluating series, with a specific example to demonstrate the process.

What You Will Learn:
How to apply the Limit Comparison Test to a given series.
Choosing a suitable comparison series.
Evaluating the limit and interpreting the result to conclude about convergence.
If you find this video helpful, please consider liking, sharing, and subscribing. Don’t forget to leave a comment if you have any questions or suggestions, and share the video with friends or classmates who might benefit from it!

Support me on Patreon for more math content:
https://www.patreon.com/patrickjmt?ty=c

Hashtags:
#LimitComparisonTest #Convergence #Divergence #Series #Calculus #InfiniteSeries #PatrickJMT #MathHelp #LearnMath #Mathematics


all right here we're going to do an

example of multiplying mixed numbers so

we've got 2 and 4/7 times 1 and 1/2 what

I like to do is just make my mixed

numbers back into fractions and we've

seen how to make a mixed number back

into a fraction so so for example 2 + 4

/ 7 again what we do is we take the

denominator which is 7 we multiply it by

whatever numbers out front which is 2

and then we add whatever numbers in the

numerator of our fraction which is 4 and

we divide that by our original

denominator of 7 so if we do the

arithmetic 7 times 2 is 14 plus 4 is 18

it says 2 and 4/7 we can actually write

that as 18 over 7 well 1 and 1/2 will do

the same thing we'll do 2 times 1 and

then we'll add to it

another one that's all over - 2 times 1

is 2 plus 1 looks like we'll have 3 over

so all I'm doing is I'm taking my mixed

numbers and I'm just turning those back

into fractions and then I'm going to

multiply the same way I did a second ago

so it says we're going to do 18 over 7

times 3 over 2 okay well all I'm going

to do now is just cancel just like

before

so let's see 18 and 2 have a common

factor - 18 divided by 2 is going to

leave me with 9 2 divided by 2 is going

to leave me with 1 and now I think we

can just multiply 9 times 3 is 27 7

times 1 is just 7 we could leave it like

this of course if you typically if I

start with mixed numbers you know I'll

finish with mixed numbers so let's see

if we can do this so 7 goes into 27 I

think 3 times let's see 3 times 7 is 21

so we would have 6

as a remainder and our original

denominator which is seven so we can

write 27 over seven as the mixed number

three and six seven

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplying Fractions and Mixed Numbers, Example 3.


okay in this video we're gonna look at

multiplying and dividing complex numbers

in trigonometric or polar form so I've

got two complex numbers here in polar

form it says all you do to multiply them

is we just multiply the r1 times the r2

and then it says we add the angles we

take cosine of the angles added together

plus I times the sine of the angles

added together to divide we just divide

our R sub 1 by R sub 2 and then we

simply subtract the angles instead of

adding them so nothing nothing too bad

at all here so let's do an example here

so we've got 6 cosine of 60 degrees plus

I sine of 60 degrees over 3 times cosine

of 90 degrees plus I sine of 90 degrees

so in this case we're doing division so

it says what we'll simply do is we just

take our r1 over r2 and then it says we

simply subtract we'll take our theta 1

minus our theta 2 so in this case well

if we do the division we'll get 6

divided by 3 whoops I'm already getting

the answer there too

so 6 divided by 3 and then it says we

simply take cosine of 60 degrees minus

90 degrees

plus I times sine of 60 degrees minus 90

degrees and well 6 over 3 is 2 this will

be cosine of negative 30 degrees plus I

sine of negative 30 degrees and boom

that's it we're done

so pretty straightforward I think you

know kind of put another random formula

to remember you know it's always good to

understand the derivations of these two

because then you're not just memorizing

random formulas you're kind of you know

hopefully understanding where they come

from let's go ahead and do Part B here

as well here we're multiplying two two

complex numbers here in polar form and

again it says when you are multiplying

it says all we do is we multiply the r1

and the r2 and then we add the angles

together instead of subtracting them so

in this case we'll simply take our two

times our five which will give us 10 and

then in our parentheses we would have

cosine of 45 degrees and then I would

have to add the angle 30 degrees and

then we just do our plus I sine

we do the same thing we just take 45

degrees and add to that the value of 30

degrees here and in this case we'll just

be left with 10 times cosine of 75

degrees plus I times sine of 75 degrees

and again in this case we'll just leave

them in our polar form again we've now

got our solution here so not a lot to do

Part C which we'll do in another example

that'll be a little a little more

complicated just because the numerators

not in polar form so we'll first have to

put that in polar form and then it'll

just turn back into a problem like we

did here in part eight

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Complex Numbers: Multiplying and Dividing in Polar Form, Ex 1. In this video, I give the formula for multiplication and division of two complex numbers that are in polar form.


all right in this video I'm going to do

another example of using the remainder

theorem so in this case we've got 4x to

the 4th plus 2x squared plus 1 divided

by X minus 3 so the function that we're

that's being divided we'll call that f

of X so that's 4x to the fourth plus 2x

squared plus 1 and again since we're

dividing by X minus 3 again according to

the remainder theorem whatever number

you're subtracting away it says if we

plug that into our function that will

give us the remainder so ok so since

we're subtracting X minus 3 since we're

subtracting the number 3 and says all we

have to do is plug 3 into the function

to get the remainder so that'll be 4

times 3 to the fourth plus 2 times 3

squared plus 1 3 to the fourth is going

to give us 81 3 squared is 9 a little

more arithmetic so 4 times 81 that would

be 324 2 times 9 is 18 plus 1 would be

19 and if we add those together we'll

get our remainder of 343

The Remainder Theorem - Example 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Remainder Theorem - Example 2


okay in this video I'm going to talk

about what are called arithmetic

sequences and recall a sequence is just

a listing of numbers um a regular just a

planal sequence doesn't have to have any

pattern at all to the numbers that

appear but an arithmetic sequence is

going to be a sequence where succeeding

terms in the sequence differ by a

constant amount so um in this video I

just want to give a real quick

introduction to what arithmetic

sequences are and another video I'll

talk about finding the general term a

and then also about um finding sums of

of partial arithmetic uh uh

series um so the only thing we're going

to do here determine which of the

following sequences are arithmetic if

they are arithmetic we're going to give

the value of D um so we'll talk about

this D here so the idea of an arithmetic

series is is simple um notice okay so to

go from three to eight we would have to

add five to go from 8 to 13 we would

have to add five to go from the number

13 to 18 hey we add five and notice this

pattern

continues um we'll just keep adding five

adding five adding five and you know

assumedly if this pattern continues well

then it would certainly be what we call

an arithmetic sequence because every

term is uh is differing by constant

amount and this value that you have to

add to get to the next term is going to

be positive five so we say our D value

is positive

five notice in my second example here I

start at .7 then we go to - 1.7 -2.7 3.7

-4.7 Etc in this case every term is

differing um you have to subtract one to

get to the next term so in this case we

would say yes this is arithmetic and our

d value equals

1 and our third example okay well let's

just test um one of my numbers here so

um so 1.6 to two uh 2.2 it looks like

we've added on 6 2.2 to 2.8 we've added

on6 oh 2.8 to 3.3 though we've added on

positive

.5 and once you find um two numbers that

are different it doesn't matter what

happens from then on out

um since they're not differing by the

same amount we say it's not

arithmetic okay so you know the only

real thing here to check if a if a

sequence is arithmetic is just to be

careful with your arithmetic and in my

last one here 4/3 5/3 2 7/3 8/3 3 well

notice we could change two if if we make

that into a fraction we could make it 2

over one multiply top and bottom by

three hey we would get 6/3 like wise we

can make three into 9/3 and we go from

1/3 5/3 6/3 7/3 8/3 9/3 so it looks like

we're adding the value of 1/3 um each

time to get to the term after that one

so again we would say this one is

arithmetic and our d value would equal

positive3 so the first one yes it's

arithmetic the second one is AR Matic

the third one we said wasn't um and the

fourth one we also said was arithmatic

so that's the basic idea of an

arithmatic um sequence that's all there

is to it so finding again a formula for

the general term and using that can

sometimes be a little tricky also

sometimes too they'll have you add up um

finitely many terms and we'll in uh some

other videos I'll talk about doing both

of those things

Quick Intro to Arithmetic Sequences

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Quick Intro to Arithmetic Sequences - Just a quick idea of what an arithmetic sequence is!  I give a few examples!


all right here we're going to look at

finding vertical asmp tootes of rational

functions and kind of the thing that um

you know that you're looking for if you

figure out you know values of X or

whatever your independent variable are

if uh it makes the denominator zero but

the numerator not zero um in that case

you're going to have a vertical ASM

toote at that x coordinate it turns out

if you get 0 over zero what's going to

happen is you would actually just have a

little hole in the graph so

all right so let's find the vertical

asmp tootes of these functions we'll do

uh part A here first um usually what I

try to do is uh I try to factor

everything so um you could just set the

denominator immediately equal to zero

and solve but if you factor let's see so

I think the denominator does Factor

nicely so two numbers that multiply to

-6 but add up to positive one I think we

would need some sort of combination of

three and two and I think

pos3 and -2 would do that for

us so it's kind of easy to see what

makes the denominator equal to zero um

so if we set the denominator equal to

zero we would just get x = -3 and X =

pos2 but notice if we plug uh you know

so if we plug -3 in in the numer

numerator we would have -3 -2 which

would be -5 and if we plug -3 into the

first uh you know the first Factor here

we would get

zero and notice if we plug two in well

clearly we're going to get 0 over 0 in

that case so it turns out at xal 2 we

actually have again a little hole in the

graph but at xal -3 that's where we have

our vertical ASM

toote so sometimes what people will do

is they'll just you know sort of factor

and cancel everything out and here

you're left with 1 /x + three and once

everything's simplified in this case

whatever makes the denominator

denominator equal to zero um that'll be

your vertical ASM toote because again

you you've already kind of gotten rid of

these common factors so you're getting

rid of those uh values of X that would

give you 0 over 0

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Vertical Asymptotes of Rational Functions: Quick Way to Find Them, Another Example 1.  Just another example of finding vertical asymptotes of rational functions.


all right in this video we're going to

look at integrating some rational

functions using

substitution so um to start off with

here we have X over x^2 + 4 so again you

know if I was kind of looking at this

problem just sort of you know U out of

the blue and I didn't know it was U

substitution again the thing that I

would think about is I see something to

the second degree in the denominator and

I see something to the first degree in

the numerator

and that to me makes me think you know

maybe a little U substitution will work

here so if we let U =

X2 + 4 the derivative of X2 is just

going to be 2x uh and then we'll tack on

our DX well notice we have an X DX in

our problem so I'm just going to

multiply both sides by2 so we have2 du

equals x

DX so the X DX that's what I'm going to

replace with

the2 um and then in the denominator

that's what we're calling

U okay so now we've got it down to a

formula that we can uh use so the

anti-derivative of 1/ U is the natural

logarithm of the absolute value of U and

now we just have to plug back in our U

substitution which was x^2 +

4 and now we have our um our

anti-derivative

technically for any value of x this

quantity on the inside would be positive

so you could even write it without

absolute value but in general you know I

just leave it an absolute value and that

way I don't have to really even worry

about

it um let's look at our other example

here so we have 24x 3r - 4 over 3x 4 -

2x +

one so I'm going to do the same thing

here I notice we have an X X to 4th in

the denominator so if I take the

derivative of that I would get something

cubed and then we have an x to the 1st

if I take the derivative of that um I'm

just going to be left with a constant so

hopefully by picking U to be the

denominator we can um manipulate it so

that our U substitution will work so if

we take our DU we would get four * 3

which would be 12 and then X the 3 and

then minus 2 and then DX

well that's almost what we have in the

numerator we have 24x Cub minus 4 well

if I just doubled both sides so 2 du and

If I multiply the right side by

two well now it says that 2 du and if we

distribute hey that would give us our 24

x 3r minus

4 DX just like we want so now we can do

our U substitution okay so it says the

quantity in the denom denominator that's

what we're calling our that's what we're

calling U so we'll just have U

downstairs and then um in the numerator

24x 3r minus 4 DX it says that's going

to be the same thing as 2 du I'm going

to pull the two out front and just leave

the one

inside and again now if we integrate we

just get the natural logarithm of the

absolute value of U but I'm just going

to plug our U substitution back in

that's 3x 4th - 2x +

1 and again we've now found our

anti-derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Substitution Rule / U-sub Basic Example 2


okay in this example we're going to look

at a problem related to percents so

suppose Patrick's going to put down 20%

of the cost of a bike and that bike

sells for

$179 14 we want to know roughly how much

money he's going to put down so since

we're estimating here

roughly well

1794 that's pretty close to

$180 and we we want to find

20% we want to find 20% of

180 well 20% we can turn that into a

decimal that's

0.20 of turns into

multiplication and then we just multiply

that by

180 so let's see um notice that 10% of1

180 would simply be $18 So 20% of 180

would be

$36 so we can say that Patrick will

roughly put down $36 for the bike

Estimating the Percent of Another Number

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! In this video, I plan on putting down 20% of the cost of a bike that costs $179.14.  How much will I put down approximately?


okay in this example we're going to

simplify some algebraic expressions by

combining like terms so again when I see

an expression like this and it says

simplify okay first you know I basically

see that there's addition and

subtraction at least in Part A I start

looking for like terms that's that's the

first thing I think about so I see 4

times X to the third power X cubed I

start looking for other terms that have

an X to the third power

again terms are separated by pluses and

minuses so we have an x squared well

that's not X to the third power then I

say okay we do have negative 2x to the

third and I don't see anything else

involving X to the third so in my head I

kind of block out the other everything

else and to me I think about well if it

said four X to the third minus two X to

the third what would I get

well four X to the third minus two x to

the third that's just gonna leave us

with two x to the third power so we've

combined those like terms and now there

are no other like terms we have an x

squared and then we have a constant so

typically and it already is in that

order

we write things in descending powers so

I'm just gonna drop those terms right

down and that would now be simplified so

the only thing we could really do we had

a couple like terms that we were able to

condense okay let's go ahead and do Part

B as well so the difference here in Part

B at least kind of the the first extra

step is we're gonna get rid of the

parentheses once we get rid of the

parentheses will basically be back to an

expression like we had in Part A and

then I'm just gonna start thinking what

are the like terms okay so what we're

gonna do here I'm gonna drop my 10 X to

the fifth right

I see a positive 3 positive 3 is going

to get distributed to the 2x to the 5th

and also to the negative B squared so

let's see positive 3 times positive 2 is

gonna get positive 6 then we just stick

on or X to the fifth positive 3 times

negative 4 will give us negative 12

again we have to still multiply by B

squared and now I think okay what are

the like terms well I see something

involving an X to the fifth and again a

number with exactly X to the fifth so we

can combine those two terms well if

you've got 10 X to the fifth and

somebody gives you another 6 X to the

fifth right you're adding them you would

have 16x to the fifth

and again our minus 12 B squared will

just drop down and at this point there's

nothing else we can do

so sometimes when I remember when I

first saw these types of problems I

would want to keep simplifying I would

want to somehow just get a single thing

and definitely that doesn't happen so

there may only be really a minut amount

of simplification that you can do

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplify an Algebraic Expression by Combining Like Terms.  Here we simplify a couple of algebraic expressions by combining like terms by adding, subtracting, and using distribution.


all right so just a couple more examples

here of simplifying down some trig

expressions using identities so here

we're gonna simplify cosecant squared X

minus 1 over cosecant squared X and I'm

gonna get make use here I'm gonna write

so we've got a tangent squared X plus 1

equals secant squared X we don't need

that one just yet

the other identity is cotangent squared

X plus 1 equals cosecant squared X and I

think that's the one we're going to need

in this case so notice if you subtract 1

from this second identity will get

cosecant squared X minus 1 equals

cotangent squared so that's what I'm

gonna write in the numerator

it says cosecant squared X minus 1 would

be cotangent squared X and that's all

being divided by the original cosecant

squared X and at this point you know

probably what I would do is I would

rewrite these so let's see cotangent is

gonna be cosine X over sine X but again

they're being squared so I'm gonna put

the squares on both of them cosecant

squared that's 1 over sine squared X so

really I've got cosine squared over sine

squared all over 1 over sine squared so

that's going to give me cosine squared X

over sine squared X and now I can take

the bottom fraction and just flip it

we'll have sine squared X over 1 well we

can cancel the sine Squared's out and

we'll simply be left with cosine squared

X

all right so Part D I would say that

reduce is simply two cosine squared X

alright so here we've got cosecant

squared theta minus cotangent squared

theta and tangent squared theta minus

secant squared theta

notice that looks an awful lot like

these first two identities I put down

that's why I put them down there let's

see if we can't sort of manipulate these

identities to get these expressions

let's maybe deal with the numerator

first we've got tangent squared theta

minus secant squared well I could

subtract the secant squared over and

that would give me tangent squared X

minus secant squared X plus 1 equals 0

well then I can simply subtract the 1

from both sides and I'll get tangent

squared x minus secant squared x equals

well negative 1 so in the denominator it

says we'll be left with a negative 1 and

I'm going to do the same thing to figure

out an expression for cosecant squared

minus cotangent squared well that's our

second identity here so again we want

the cosecant squared to stay positive so

what I'm going to do is just simply

subtract cotangent squared from both

sides and that'll give us cosecant

squared X minus cotangent squared X well

that's the same thing as positive 1 so

that's what I'm going to stick in the

numerator so I really have 1 over

negative 1 which is simply going to give

us negative 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplifying Trigonometric Expressions Using Identities, Example 3. In this video, I simplify a trigonometry expression using some trig identities.


all right in this video I just want to

do a couple examples about determining

Symmetry and using these definitions so

if f ofx = f ofx we say it's symmetric

about the Y AIS or it's an even function

if f ofx equals the negative of f ofx we

say it's an odd function and it's

symmetric about the origin so we've got

a couple formulas here we're just trying

to figure out what type of symmetry um

if any these functions possess so again

the idea is we just replace all the X's

with negative X X Sox

4x^2 + 5 um well if you take a negative

number and raise it to an even power

it's going to be positive x to the 4th

would still be X to 4th so this would

just simplify to 3x 4th likewise x * X

is postive

x^2 POS 5 still just hanging out there

so notice we plugged in Nega X we got

got back the very original function we

started with so it says F ofx does equal

F ofx so in this case we would say that

this is what's called an even

function okay so it's symmetric about

the Y AIS maybe one other one real quick

suppose we have f ofx = x Cub - x well

if we plug in Negative X we'll

getx cubed minus X

well we can get rid of the parentheses

if you Cube 1 three times will get

Negative we still have X cubed two

negatives make a positive X and notice

we could Factor the negative out we

would have X Cub - x but again that's

the negative of the very original

function that we started with that's the

negative of f ofx so in this case we

would say hey this is an odd function

okay it has the Symmetry about the

origin and usually I kind of stop um

honestly I stop at this step right here

if you look at the signs on the original

notice the original X Cub was positive

and now it's negative the original uh

next sign was a negative now it's a

positive if all the signs flip flop

that's again I mean you can basically do

the step of factoring the negative out

but if you recognize that all the signs

have uh just switched that's going to

imply that it's an odd function

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Symmetry - A Quick Discussion for Testing if a Polynomial is Even / Odd.  In this video, I give the definition of an even function and an odd function and use it to discuss if two given polynomials possess any symmetry.


okay in this video we're going to do a

rough sketch of the function f ofx equal

the quantity x + 1 4us 1 so the thing

I'm going to do is I'm just going to

think about some graph shifting so x to

the

4th it looks a lot like x to the

2 it has that same U shape to it so

that's going to be the graph of x to 4th

well the plus one inside of the

parentheses that's going to shift our

graph of X to 4th it's going to shift

the graph one

unit left so again when it's in the

parentheses it's I always remembered

it's sort of the opposite right it feels

like it should you know you see a

positive one which makes you think to

the right but it actually goes one unit

to the left the negative one on the

outside is going to have the effect of

Shifting the graph it's going to shift

one

unit down okay so if it's on the outside

you see a negative hey do go down if

it's a positive do go up so it's

basically going to take our graph of x

to the 4th one unit to the left one unit

down and again if you're not sure you

can always plot points that's one thing

that I think people sometimes are forget

about doing so instead of passing

through the origin 0 0 since it's one to

the left and one down it'll pass through

the point -1 comma -1 notice if we plug

in X = 0 we'll get 1 to 4 which is 1

minus 1 will get 0er likewise at -2 by

symmetry it it will equal

Zer notice if we plug in x = 1 we would

have 1 + 1 which is 2 2 4th is going to

be 16 minus 1 it says at at xal 1 we

should actually already be way up here

at yals 15 so my

graph I'm going to make it get pretty

deep uh pretty

quickly and again that's going to be

just a rough sketch of our of our new

graph the quantity x + 1 4 minus 1

❖ Horizontal and Vertical Shifts Using f(x) = x^4 - Fast Example! ❖

Horizontal and Vertical Shifts Using f(x) = x^4

In this video, we'll explore how to perform both horizontal and vertical shifts on the parent function f(x) = x^4. This quick example demonstrates how shifting transformations work, moving through the process a bit faster for those who are looking for a quick refresher on graph transformations. Ideal for anyone familiar with basic graphing techniques who wants to brush up on shifting transformations.

What You Will Learn:

How to shift a function vertically and horizontally.
Understanding transformations of the parent function f(x) = x^4

 .Quick tips on graphing and interpreting shifted functions

Support my work on Patreon!
https://www.patreon.com/patrickjmt?ty=c

#Math #Algebra #GraphTransformations #ShiftingFunctions #Graphing #VerticalShift #HorizontalShift #ParentFunctions #Precalculus #Mathematics #FunctionGraphing #GraphShifts #QuadraticGraph #AlgebraReview


all right here we're going to look at

some very basic differential equations

and again a differential equation is

simply an equation that involves

derivativ so here we've got the

differential equation dydx = x 4th we

want to know which one of these uh three

functions a b or c are solutions to this

differential equation and maybe there's

uh more than one solution but again all

this really says is is which derivative

is gives us just x to the 4th so we can

check part A Dy DX here will'll get 1/5

* 5 x 4th + 1 well that leaves us with X

4th + 1 but that's certainly not just x

to the 4th so part A is not our solute

the not our solution Part B if we take

the derivative of 4 x + 5 the derivative

of 4 x is going to be 4 x * the natural

logarithm of 4 the plus five just turns

into a constant again that's definitely

not just x to the 4th

and likewise we can check part C here

well the same thing the derivative of y

with respect to x uh we'll get 1/5 * 5x

4th which will give us X 4th the

derivative of Pi is just zero so hey we

found our solution uh uh of these so

part C would be our solution to this

differential

equation let's maybe look at uh one

other very similar idea and now in inad

of just a first derivative we've got a

second derivative so the second

derivative with respect to Y is -4 y we

have these two functions we want to know

which one of these uh again potentially

both uh could be Solutions here so all

right we've got y = cosine of

4X well the first derivative Dy DX the

derivative of cosine is negative s you

leave the inside alone have to chain

rule it so the derivative of the

inside's going to be four

so we'll get --4 sin of

4X well when we take a second

derivative we'll get -4 * cosine of 4X *

4 that's going to give us -16 * cosine

of

4X and you know definitely that doesn't

look like -4 y we can certainly replace

the starting function right we started

with yal cosine of 4X so we could always

plug that back in to the cosine 4X so we

could rewrite this as -16 y but again

that's not the solution that we want so

part A doesn't

work let's try Part B here s of

2x so the first derivative dy over DX I

was going to write y Prime which is

certainly legal but uh you know I'm just

going to stick with the notation that

we've got here the liance notation so s

of 2x the derivative will be cosine of

2x * 2 or hey just 2 cosine of

2x the second derivative with respect to

Y well let's see the derivative of

cosine is going to be negative s we'll

leave the 2x alone but then again the

chain rule so we'll have to multiply by

another two and let's see that's going

to give us -4 * s of 2x but again the

same thing as before we can replace S of

2x uh with just y so we're left with ne4

Y that's exactly what we wanted to be

left with so now we found our solution

here uh Part

B

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Differential Equation : What Does It Mean to Be a Solution.
Just a quick example of what it means when a function is a solution of a differential equation!


all right in this video we're going to

do an example related to the law of

signs and all the law of signs does it

relates uh angles and the sides opposite

them by this equation so little a little

B little c those will be the side

lengths uh a b and c again capital A B

and C are just going to be the angles so

it says we get little a over sin a

equals little B over sin B equals little

C over sin C so let's do an example here

suppose we're given that one angle of a

triangle is 120° and another angle is

40° we know the side opposite the 40°

angle measures 12 M we want to find the

side opposite the unknown angle so just

going to make a little picture

here try to make it somewhat to scale

hopefully so um so here's let's make

this our 120° angle this will be our 40°

angle the side opposite the 40° angle

that has length 12 well this would our

missing angle uh in our triangle and

we're simply trying to find the length

opposite that so maybe I'll call that

c all right well we know that the angles

inside of a triangle add up to 180 so

120 + 40

is60 um which means the other angle must

be simply 20° those will now all add up

to 180 and now we can just set up our

equation we can do 12 over s of 4

40° that's going to equal C over s of

20° and now to solve for C we'll simply

just multiply both sides by S of

20° so on the left we'll be left with 12

* s of

20° over s of

40° that's going to be our value for C

and now just imag kind of plugging it

all into a calculator and simplifying it

down here a little bit

so all right so let's do s of 20

degre I'm going to round a three decimal

places just sort of arbitrarily so

that's

0.342 um if we do s of 40° I'm getting

that to be

0.643 after rounding so if we take

0.342 and if we multiply that by 12 I'm

getting that to be

4104 uh over our

0.643 so now if we take uh

4104 and simply divide that by

0 whoops uh so

4104 divided by

by sorry I keep punching it in wrong

here so

4104 we'll divide that by

0.643 I'm getting this to be roughly

equal to

6.38 and again after rounding I'm

getting 6. 383 so um just a touch over

uh well a little bit over six meters is

what we're getting

and you know certainly that seems

reasonable um you know if the side

opposite 40° is 12 certainly the side

opposite 20° would have to be you know

significantly smaller um so certainly

that agrees with my intuition I think

you know that's in the ballpark at least

um so that's all there is to do we're

just simply setting up a little ratio uh

solving for our unknown value by using

these law of sign this law of signs

The Law of Sines, Example 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Law of Sines, Example 1.
  This video talks a bit about the Law of Sines.  I give the formula and proceed to do an example using it. Very exciting!! (it actually is quite useful though)


all right in this video I'm going to do

an example related to oblique asymptotes

and recall that oblique asymptotes occur

when the degree of the numerator is one

larger than the degree of the

denominator so that's certainly what we

have in this case first off I'm going to

find any vertical asymptotes so vertical

asymptotes basically I just set the

denominator equal to zero that would

give me in this case x equals one and I

know that if you plug one in you get

zero on the bottom I also have to make

sure that if you plug one in we don't

get zero on the top I notice if you plug

one end when we get one plus four which

is 5 plus seven which is 12 over zero we

have something non zero over zero that

tells me that x equals one in this case

is a vertical asymptote again since the

numerator is larger degree than the

denominator that tells me that there are

no horizontal asymptotes

so no horizontal asymptotes since the

degree that of the numerator is larger

than the degree of the denominator

okay so last but not least we said there

is an oblique asymptote since the

numerator is one larger than the

denominator and to find it I basically

just have to do the division well to do

the division since there's just a linear

term on the bottom I think I'm going to

use synthetic division so there's our 1x

squared our positive 4x and our positive

7 again we see X minus 1 but we use +1

so if I drop the 1 down 1 times 1 is 1

we add we get 5 1 times 5 is 5 this will

give us a 12 again we start we started

with something squared so this would be

our X term and this would be our

constant and this would be our remainder

so it says really we can write x squared

plus 4x plus 7 divided by X minus 1 it

says we can rewrite that as 1x plus 5

with a remainder of 12 or we could have

written it as X plus 5 plus 12 over X

minus 1 but the idea is once you've done

the long division forget about the

remainder whatever is out front whatever

you get when you do the long division if

you omit the remainder that's going to

be your oblique asymptote so remember

our oblique asymptote is just a line so

it says in this case our oblique

asymptote

would be the line y equals x plus five

so there is one vertical asymptote at x

equals positive one again we said there

were no no horizontal asymptotes at all

and then we've got our oblique asymptote

of y equals x plus five

Find Asymptotes of a Rational Function (Vertical, Horizontal and Oblique/Slant) - Ex 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Find Asymptotes of a Rational Function (Vertical, Horizontal and Oblique/Slant).  In this video, I find the vertical asymptotes and a slant / oblique asymptotes of a rational function.


[Music]

the square Ro of

[Music]

two the golden racial

fire the square to

three we lose

comen the silver

R

scorpion

sub son elen

[Music]

2 the plastic raal

r l Master on constant Dem

Ten Mortal Math Constants

In this video, we show visualizations of ten classic mathematical constants:

1) The square root of 2 as the length of the diagonal of a 1 by 1 square.
2) Pi as the area of the circle with radius 1.
3) Tau as the circumference of the circle with radius 1.
4) The golden ratio as the number satisfying (x+1)/x=x/1.
5) The square root of 3 as the altitude or area of an equilateral triangle with side length 2.
6) Euler's number e as the ending x-coordinate yielding the area under the curve of 1/x starting at x=1.
7) The silver ratio as the horizontal length across the regular octagon with side length 1 or the area of the largest rectangle inside the same regular octagon.
8) The natural logarithm of 2, ln(2),  as the sum of the convergent alternating harmonic series.
9) The plastic ratio as the limit of the ratio of side lengths in a spiral of equilateral triangles with side lengths dictated by the Padovan sequence.
10) The Euler-Mascheroni constant as the difference between the harmonic series and the area under the curve y=1/x from 1 to infinity.

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

 #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #constants #goldenratio #silverratio #plasticratio #squareroot #squareroot2 #squareroot3 #euler #eulerconstant #eulermascheroni #ln2 #logarithm #octagon #square #pi #tau

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


connecting these two line segments of

length one produces a line segment of

length root2 draw a circle of radius one

at the end of that line segment draw

perpendicular to that line segment and

this produces a line segment of length

root three you can continue repeating

this process over and over again drawing

a circle of radius one drawing a

perpendicular to the previous line

segment and then the new line segment

will have a length given by the square

root of the next integer so here we've

constructed the length < TK of two up to

< TK of 7 repeating this process over

and over produces a spiral shape known

as the spiral of theodoris we can use

this spiral to construct lengths of sare

root of n for any n that we want where N

is a positive integer just using a

straight edge and a compass for example

as we keep spiraling out here we'll end

with a line segment whose length is < TK

of 200 do you think you could construct

this by hand

Spiral of Theodorus

This is a short, animated visual proof demonstrating how to construct square roots of any positive integer using the Spiral of Theodorus

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

#manim #math #mathshorts #mathvideo
 #construction #geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #spiral #theodorus #squareroot

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


if you have a collection of objects that

have five different possible

characteristics you need a proper five-

set vend diagram like this one notice in

this diagram there are 32 possible

regions created by the five sets there

are five regions shown here that lie in

exactly one set after that we see 10

distinct regions pictured here that lie

in exactly two of the sets by symmetry

there are also 10 regions as shown here

that lie in exactly three of the sets

the five regions shown here lie in

exactly four sets and finally we have

one region that lies in all five sets

and one region that lies outside of all

five sets so this is a beautiful proper

five set vend diagram can you draw one

with six sets

Five set Venn diagram

In this video, we show an example of a five-set Venn Diagram by highlighting all 32 possible regions.

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

Here is a longer, wordless video, showing the same thing along with four set Venn diagrams as well : https://youtu.be/WBot4kqc3ro

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

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

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


this is one of the coolest things I've

ever seen for two non-negative numbers

their arithmetic mean or simple average

is always greater than or equal to their

geometric mean or square root of their

product equality holds exactly when the

two numbers are equal there's an amazing

way to see this visually construct a

circle with diameter X Plus y the radius

of the circle is half the diameter so it

is the arithmetic mean now construct a

perpendicular chord where X meets y if

each semicard has length G then by the

chord chord power theorem G times G is

equal to x times Y which means G is

equal to the square root of x times y or

the geometric mean we can now

geometrically see the semi chord is

always less than or equal to the radius

with the quality if and only if x equals

y and that is precisely why the GM is

always less than or equal to the am and

thus we have proven the amgm inequality

Visual proof every student should see

This is one of my favorite visual proofs that uses geometry to demonstrate an algebraic inequality. The arithmetic mean is always greater than or equal to the geometric mean, with equality if and only if the numbers are equal. (Requires the numbers to be non-negative). #shorts

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

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

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

Book ratings are from January 2022.

My Books (worldwide links)
https://mindyourdecisions.com/blog/my-books/#worldwide

My Books (US links)
Mind Your Decisions: Five Book Compilation
https://amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 224 reviews
http://amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4/5 stars on 24 reviews
http://amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.1/5 stars on 38 reviews
http://amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.2/5 stars on 76 reviews
http://amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.3/5 stars on 30 reviews
http://amzn.to/XRm7M4

Mind Your Puzzles: Collection Of Volumes 1 To 3
https://amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 87 reviews
http://amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.1/5 stars on 24 reviews
http://amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.2/5 stars on 22 reviews
http://amzn.to/1NKbGlp

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

My Blog
https://mindyourdecisions.com/blog/

Twitter
https://twitter.com/preshtalwalkar

Instagram
https://www.instagram.com/preshtalwalkar/

Merch
https://teespring.com/stores/mind-your-decisions

Patreon
https://www.patreon.com/mindyourdecisions

Press
https://mindyourdecisions.com/blog/press #math #maths #mathematics


here's one of my favorite Pythagorean

theorem proofs start with a right

triangle with legs of length A and B and

a hypotenuse of C use four rotated

copies to create a c by C Square the

region in the middle is an A minus B by

Aus b square as the side length is equal

to the difference of the leg lengths of

the triangle the area enclosed in this

diagram is c^ s but now we can compute

the area of this diagram by moving

around the shapes inside if we place the

triangles next to each other like this

then we get an interesting shape with

Dimensions listed here but we can then

view this shape as a b by b square next

to an a by a square and therefore the

enclosed area is a 2 + b^ 2 because

these areas are the same we conclude

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

the leg lengths of a right triangle with

hypotenuse C this visual proof can be

attributed to bascara an Indian

mathematician from the 12th century

Pythagorean Theorem - Behold!

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

Here is a bit longer version: https://youtu.be/I75HYC8TRNY

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 due to Bhāskara. For a static version of this proof, see Roger Nelsen's first "Proof Without Words: Exercises in Visual Thinking" compendium (page 4). You can also check out Howard Eves' "Great Moments in Mathematics (Before 1650)" page 29-32.

Check out Roger's book on Amazon (affiliate link to follow; I may receive a small commission): https://amzn.to/3SjsqEh

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

#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


the N plus first triangular number t

subn + 1 can be represented by an n + 1

by n +1 triangular array of unit cubes

if we place n such triangular Rays side

by side we end up with an N by T sub n

+1 triangular staircase diagram when we

removed the back n by n +1 rectangle

we're left with an N by T subn staircase

diagram and we can decompose the N by

n+1 rectangle into two more n byn

triangular arrays we place these two

arrays next to the remaining St St case

diagram to produce an n + 2 by T subn

staircase diagram this means that n * t

subn + 1 is equal to n + 2 * T subn from

this we can conclude the recursive

formula that t subn + 1 = n + 2 / n * t

subn

A 3D Triangular Number Recursion

This is a short, animated visual proof demonstrating how to multiplicatively obtain the (n+1)st triangle number from the nth triangular number.

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

To see a wide screen version of this proof check out A Multiplicative Triangular Number Recurrence (visual proof)
https://youtu.be/69TvfNA0Lxc

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

#manim #math #mathvideo #mathshorts #triangularnumbers  #numbertheory  #triangle  #animation #theorem #pww #proofwithoutwords #visualproof #proof

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


[موسيقى]

y

[موسيقى]

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

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

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


start with a pentagon of area one take a

scaled copy that's size 1 six and place

it in the center leaving the outer ring

of size 56 we can cut the outer ring

into five congruent trapezoids each with

Area 1 six when we shade all five we've

shaded 5 six of the Pentagon do this

again on the inner Pentagon of Area 1 16

shading 56 of the 1 16 area therefore

shading a total area of 5 over 6^ SAR we

can repeat this process over and over

again each time shading 5 over 6 to the

N of the outer Pentagon if we can

continue this process indefinitely that

means if we consider a limiting process

we can look at the limiting shape and

the limiting shape is the filled

Pentagon therefore the Shaded area given

by the infinite sum matches the area of

the Pentagon which is one one

interpretation of this is that point

five repeating in base 6 is equal to 1

another one is that 5 * the infinite sum

of the positive powers of 1 six equals 1

or that the infinite sum of the positive

powers of 1 16 is 1/5

0.5555… = 1 (in base 6)

This is a short, animated visual proof showing the sum of the infinite geometric series with first term 5/6 and ratio 1/6, which in turn allows us to compute the sum of the series of powers of 1/6 and determine an interesting base 6 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/o0AyHmRV8RA?si=2fExvp1WqrmZp5SJ

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

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

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

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


let's see a visual proof of an

interesting inequality using the unit

circle in the cartisian plane label the

origin o start with a radius of length

one and draw a unit circle centered at o

like this let's label the point on the

xais by a imagine a point p on the unit

circle in the first quadrant

corresponding to rotating an angle of T

if we drop a perpendicular from P to the

x-axis and label the intersection Point

M then the length of the line p m is the

S of T by the definition of s now use

the line pm to generate a circle

centered at M with radius s of T call

the intersection point of that Circle in

the x-axis B and label the other

intersection point between the two

circles Q notice that the point Q lies

directly below the point P so that

they're

antipaladin segment O M plus the length

of the line segment MP is equal to the

length of the line segment o and this is

greater than or equal to the length of

the line segment O A from these facts we

can see that the arc length of the Ark

pbq is greater than or equal to the arc

length of the Ark P AQ and that in turn

is greater than or equal to the length

of the chord PQ by construction the ark

pbq is the semicircle of a circle with

radius sin T and therefore has a length

of Pi * sin T The Arc P AQ is 2 two

copies of the Ark PA a subtending the

angle T and therefore the ark PA a q has

a length of 2 T and finally the length

of the chord PQ is equal to 2 * the

length of the line p m which is s of T

and therefore the chord length PQ is 2 *

the S of T from this argument we see

that Pi of s of T is greater than or

equal to 2T which is greater than or

equal to 2 * the S of T from the latter

inequality we see that 2 * the S of T is

less or equal to 2 T so that s of T is

less than or equal to T from the first

inequality we see that 2 * T is less

than or equal Tok * the S of T so that

2T / piun is less than or equal to S of

T when we glue these two together we see

that 2T over Pi is less than or equal to

S of T Which is less than or equal to T

notice this is only true if the angle T

is in the first quadrant so if T lies

between 0 and < / 2 this inequality is

known as Jordan's inequality thanks for

watching if you liked this video check

out my playlist on other mathematical

inequalities and consider subscribing to

the channel

Jordan's Inequality Visual Proof

This is a short, animated visual proof demonstrating Jordan's inequality, which is an equality providing upper and lower bounds for the sine function over the interval [0, pi/2].

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

For other inequality visual proofs, see my inequality playlist:

https://youtube.com/playlist?list=PLZh9gzIvXQUuKcFHdtjFkwAVELDxu9nH4&si=R5wj4VJ1DCfI4Wbh

This animation is based on a visual proof by Feng Yuefeng from the April 1996 issue of Mathematics Magazine (https://www.jstor.org/stable/2690669 page 126 ).

#math #inequality  #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #areas #mathematics #jordaninequality #jordan #algebraicidentity #mathshorts #mathvideo #mtbos #circle #unitcircle

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


what is the sum one plus a half plus a

third plus a quarter plus so on to

figure this out let's plot the function

y equals one over X over The Domain C to

c squared then the area one over C plus

one plus one over C plus two plus one

over C plus three and so on up to one

over c squared can be represented by the

area of the rectangles depicted here but

this geometric region contains a

rectangle with area c squared minus C

divided by c squared pictured here so

the sum of one over C plus one up to one

over c squared is greater than or equal

to c squared minus C divided by c

squared which is 1 minus one over C we

can use this fact to see that the sum of

one over C plus one over C plus one up

to one over c squared is greater than or

equal to one applying this result

repeatedly to the harmonic series shows

us one over one is greater than or equal

to one the sum from one half to

one-fourth is greater than or equal to

one the sum from one-fifth to one

twenty-fifth is greater than or equal to

one and so on continuing this process

indefinitely shows us that the harmonic

series diverges as it's greater than an

infinite sum of ones

Visual Harmonic Series Divergence from Bernoulli!

This is a short, animated visual proof showing Bernoulli's method of determining that the harmonic series diverges. This technique is similar to, but different from, the technique used by Oresme. #manim #math #mathvideo #series #infiniteseries #mathshorts #geometry  #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #harmonicseries #pww  #proof #areas #mathematics #mtbos #harmonic #divergence #bernoulli #oresme

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

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

Here is an alternate video showing how to prove the harmonic sum diverges: https://youtu.be/hvIusavrRf8

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


let's see a formula's visual proof a

square plus b square

here

a square and b square

then

we form a plus b and a plus b square

now i'm forming two rectangle

a b and a b

two a b

then replace

square

and a square

and subtract

first rectangle a b and

again subtract second rectangle

a b

that is a plus b square minus two b

let's see

the shape is same

a plus whole square minus two a b n

a square plus b square is same

Visual proof of a formula

some formula of algebra must be memorise for solving general algebra
but students sometime dont know the geometrical proof of these formula
here you go a visual proof of a formula
this called geometrical representation of a algebraic formula
please subscribe
thank you


[Music]

foreign

[Music]

[Music]

Restricted Lattice Path Enumeration #catalan

This synthwave enumeration short shows all of the northeast lattice paths to the points (n,n) that don't pass below the line y=x where n ranges from 0 to 6. The number of such lattice paths is counted by the Catalan numbers.

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

To see a longer version of this that comes with a visualization of a proof for the recurrence relation, see https://youtu.be/__w_MDQeUYc .

To find out how to find a closed form for the Catalan numbers, check out https://youtu.be/kaInaIUABzY

To become more familiar with lattice path enumeration, check out https://youtu.be/9YUl0k2FYzc and https://youtu.be/KYpOoA2D63w.

#catalannumbers #catalan #mathvideo #math #mtbos #manim #animation #theorem #visualproof #proof #iteachmath #mathematics #binomialcoefficients #latticepaths #discretemath #combinatorics #enumeration #synthwave #recurrence #quadraticrecurrence

This animation is based on the well-known interpretation of Catalan numbers as counting restricted 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).

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


hello everyone in this video we will

prove Pythagoras Theorem

visually in this Square we draw a

rectangle with one side a and the other

side is B with its diagonal as C now we

copy paste this over here so this these

all lengths are a and these all lengths

are B now from here this unshaded area

is equal to a square while this unshaded

area is equal to b square

we have to prove that this unshaded area

is equal to c² now let move the figure

like this one we move this triangle to

this region and this this unshaded

region is a square with its side length

C and thus this area is equal to C

Square by this figure we have a square +

B sare is = c² and that is the answer

Visual Proof of Pythagoras Theorem

In this video, we prove through visual the existence of Pythagoras theorem
#Shorts #shortsfeed #pythagoras_theorem #mathematics #geometry #area_problem


James Garfield the 20th president of the

United States gave this proof of the

Pythagorean theorem start with a right

triangle with legs of length A and B and

hypotenuse of length C place a rotated

copy so that complimentary angles fit

together creating a right isoceles

triangle with leg length C the area of

the newly created triangle is 12 c^ 2

and the area of the original right

triangle is 12 a * B so the area

enclosed in this trapezoid diagram is

c^2 + a * B on the other hand using the

formula for the area of a trapezoid we

see that the is 12 a + b * A + B and

this simplifies to 1 12 * the quantity a

2 + 2 a + b^ 2 because these two

quantities measure the same area we

conclude that c^2 + a = A2 + 2 A + B ^2

applying some Elementary algebra shows

us that c^2 equal A2 + b^2 When A and B

are the legs of a right triangle with

hypotenuse C this is Garfield's

trapezoid proof of the right triangle

theorem

Garfield’s Pythagorean Proof

This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using the trapezoid that is now attributed to President James Garfield. This theorem states the square of the hypotenuse of a right triangle is equal to the sum of squares of the two other side lengths.

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

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

You can also see the original on MAA Convergence: https://maa.org/press/periodicals/convergence/mathematical-treasure-james-a-garfields-proof-of-the-pythagorean-theorem

For other proofs of this same fact check out the following playlist:

https://youtube.com/playlist?list=PLZh9gzIvXQUvlRlnYryejWGHYQquM92rB&si=o_im7-_NHp1zhkcD

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

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


thank you

[Music]

[Music]

thank you

[Music]

[Music]

[Music]

[Music]

[Music]

[Music]

thank you

foreign

Viviani's Theorem II (visual proof via reflection and translation)

This is a short, animated visual proof of Viviani's theorem, which states that the sum of the distances from any interior point to the sides of an equilateral triangle is equal to the length of the triangle's altitude. #math #geometry #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #polygons #triangle #mathshorts #mathvideo

This animation is based on a visual proof by James Tanton (@JamesTantonMath ) from the October 2001 issue of Mathematics Magazine (https://www.jstor.org/stable/2691103 page 313).

Here's an alternative proof using rotations:
https://youtu.be/5r2jLt0_PLM

Here is a cool interactive GeoGebra applet to play with this theorem yourself:
https://www.geogebra.org/m/Jgv9N6gR#:~:text=Observe%20the%20sum%20of%20the,or%20on%20an%20equilateral%20triangle.

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Cologne 1983 by Josh Kirsch/Media Right Productions


let's find a closed formula for the sum

of triangular

numbers the M triangular number T subm

is the sum of the first M positive

integers this number can be represented

by a triangular stack of squares like

this one here where there's one square

in the first row all the way down to M

squares in the bottom

row let's start with a triangular stack

of squares corresponding to the

Triangular number T subn that means

there's n blocks in the bottom row in

the first row place one token so we'

placed T1 tokens in the second row Place

one token in the first square and two

tokens in the second Square so that

we've placed T2 tokens in the second row

then Place T3 tokens in the third row as

1 + 2 + 3 continue this pattern in each

row placing I tokens in the I Square in

the row then the number of tokens in

this triangular diagram is given by the

sum of the first n triangular numbers T1

plus T2 and so on up to TN to find this

sum we perform a trick we create two two

more triangular arrays like this in the

first one we place the same number of

tokens in each row but this time we

place the tokens in reverse order

starting with the largest number of

tokens in a row and going down to one in

the third triangular array we place the

same number of tokens but this time we

go column by column starting from the

end so the last column has one token the

second from the last column has T2

tokens and so on so that the First

Column has TN tokens so the total number

of tokens spread across these three

three triangular arrays is 3 * the sum

of the first n triangular numbers now we

perform the trick stack all three of

these triangular arrays together

aggregating the number of tokens at any

spot it turns out that in any single

spot we always have exactly n plus 2

tokens on that spot and therefore 3 *

the sum of the first n triangular

numbers must be equal to n + 2times the

number of squares in the Triangular

array which is T subn and that

simplifies to n * n + 1 over2 so we used

three stacks of triangular arrays with a

triangular number of tokens in each row

to see that the sum of the first n

triangular numbers is n * n + 1 * n + 2

all divided 6 in his first proof without

words compendium Roger Nelson attributes

this proof to Richard K guy can you see

other identities you can prove using

stacks of tokens on certain arrays check

the linked video where we show how to

use this technique to find the sum of

the first n squares

Tri-Nums Sums! (visual proof IV)

This is a short, animated visual proof showing how to find the sum of the first n triangular numbers (which themselves are sums of the first n integers) using stacks of tokens on three triangular stacks of squares.

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

Here are some related videos with sums of triangular numbers:
https://youtu.be/HrmOQJ3HBhI
https://youtu.be/N0ETyJ5K6j0
https://youtu.be/pucFDbdEyuE
https://youtu.be/LBls3jRfxeY

Here is a sum of squares visual proof using a similar technique:
https://youtu.be/UqVmocdLFGc

This animation is based on a visual proof attributed to Richard K. Guy by Roger B. Nelsen in his first Proofs without Words compendium (I may receive a small commission at no cost to you for these affiliate links):
https://amzn.to/48mjeX1

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

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


start with four positive real numbers a

b c and d and suppose that a times D is

greater than b times C then draw a right

triangle with legs of length a d and AC

rotate a copy of this triangle and scale

by the value B over a creating a right

triangle with legs of length b c and b d

place the two triangles so they meet at

complementary angles connect the apexes

of these two triangles with a straight

line to create a third triangle as shown

here we can use the Pythagorean theorem

on the original two triangles to find

two of the side lengths of this new

triangle the first is B times the square

root of c squared plus D Squared while

the second is a times the square root of

c squared plus D Squared because the

first two triangles fit together along

complementary angles we know that the

blue triangle is also a right angle

triangle so we can then use the

Pythagorean theorem to find the final

side length or the hypotenuse of the

blue triangle to be this square root of

a squared plus b squared times the

square root of c squared plus D Squared

the resulting diagram is a trapezoid we

can then draw another right triangle in

the upper right corner to complete this

trapezoid to a rectangle one side length

of this new right triangle is a D minus

BC because it's equal to the difference

of the two parallel side lengths of the

original two triangles the other side

length of the new right triangle is AC

plus BD because that length is equal to

the sum of the two adjacent side lengths

of the original two right triangles

but now we can apply the Pythagorean

theorem one last time to this new right

triangle from this theorem we see that

the square root of a squared plus b

squared times the square root of c

squared plus D Squared all squared must

equal the quantity AC plus b d squared

plus the quantity a D minus BC squared

if we Square the left side we get the

resulting formula that the quantity a

squared plus b squared times the

quantity c squared plus d squared equals

the quantity AC plus b d squared plus

the quantity a D minus BC squared this

theorem says that the product of two

sums of squares must be a sum of squares

itself the result is of particular

interest when all the numbers a b c and

d involved are actually integers and not

just real numbers in this case we call

this diaphanes's sum of squares identity

Diophantus's Sum of Squares Identity I (visual proof)

This is a short, animated visual proof of Diophantus's sum of squares identity (also sometimes called the Brahmagupta–Fibonacci identity) using "Garfield's trapezoid" and scaling of right triangles.

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

To see a related proof of the Cauchy-Schwarz inequality, check out:
https://youtu.be/OWR6QCPYjqI

This animation is based on a visual proof by Roger Nelsen from the April 2017 issue of Mathematics Magazine (https://doi.org/10.4169/math.mag.90.2.134 - page 134).

#math #inequality  #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #areas #mathematics #diophantus #algebraicidentity #mathshorts #mathvideo #mtbos #triangle #scaling #garfieldtrapezoid #trapezoid #identity

To learn more about animating with manim, check out:
https://manim.community
_____________________________
Music in this video:
Kiss the Sky by Aakash Gandhi


the square pyramidal numbers represent

the number of unit cubes in a pyramid of

stacked Square arrays with sides of

length one up to n for instance this

diagram shows the 24th square pyramidal

number because the bottom square has

side length 24 and the top square has

side length 1 it turns out that this

particular square pyramidal number 4900

is special other than the trivial

Solutions of 0 and 1 it is the only such

number that can also be arranged in one

large Square array we can lay down the

square array levels like this only

splitting up the 4x4 and the 6x6 squares

to fill gaps once we've done this we

have an entire 70x 70 square filled in

thus 4900 is the only non-trivial

solution to the so-called Cannonball

problem which asks for numbers that are

simultaneously square pyramidal and

square

Cannonball Problem!

In this short, we show the only nontrivial solution to the cannonball problem, which asks for numbers that are simultaneously square pyramidal and square. The problem gets its name because cannonballs stack nicely in square pyramidal arrays.

Can you prove that 0,1, and 4900 are the only such solutions to the problem?

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

#math #manim #animation #theorem #iteachmath #mathematics ##mathshorts #mathvideo #mtbos #mathematician #squarepyramidal #numbertheory #cannonballproblem #cannonballrun

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


let's see two formulas for the sum of

the first n cubes the number I cubed can

be represented by an I by I by I cubical

stack of unit cubes so this stack

represents 1 cubed this one represents 2

cubed this one represents 3 cubed and so

on where we can end at n cubed in this

case we've pictured Nal 4 when we put

these Stacks together we see we have a

representation for the sum of the first

n positive cubes now if we lay out each

of the cubes flat each Cube forms an I

by I squ rectangle so the first Cube

forms a 1x1 rectangle the next one a 2x4

rectangle then a 3x9 rectangle all the

way down to the final one which forms an

N byn squ rectangle the first rectangle

represents the number one the second

rectangle represents the sum of the two

odd numbers 3 + 5 as shown here the

third rectangle represents the sum of

the odd numbers 7 + 9 + 11 so each of

these rectangles can be decomposed into

a sum of consecutive odd numbers the

last rectangle can be done so by

decomposing it into n * n -1 + 1 + n * n

-1 + 3 and so on up to n * n -1 + 2 N -

1 but this last number simplifies to n *

n + 1 -

1 from these rectangular diagrams then

we see that the sum of the first n

positive cubes is equal to the sum of

the odd integers starting from one and

ending at n * n + 1us 1 this formula is

amazing but we can use these rectangles

to say even more we can break each

rectangle into two separate rectangles

where one has dimensions and by the nth

triangular number we can take this

smaller rectangle rotate it and glue it

onto the other side of the rectangle

forming an L-shaped nomon in each Case

by construction these gons all fit

together perfectly to form a

square because because of the rectangle

Dimensions we see that this square has a

side length given by 1 + 2 up to n or

the sum of the first n positive integers

this means the total number of unit

cubes which was given by the sum of the

odd numbers from 1 to n * n + 1 - one

must also equal the square of the sum of

the first n positive integers but then

we can finish this off by connecting the

first and last formulas so that we see

that the sum of the first n positive

cubes must equ the square of the sum of

the first n positive integers this

formula is attributed to the ancient

Greek philosopher nomus the visual proof

we just presented is due to aleno Flores

but there are plenty of other visual

proofs of the same fact you can see

others on my Channel or you can find a

variety of them in Roger Nelson's books

proof without words volumes 1 2 and 3

have a look and let me know which visual

proof of nomiss theorem is your

favorite

Sum of Cubes as Sum of Odds and more (visual proof VI)

This is a short, animated visual proof demonstrating two formulas for the sum of the first n positive cubes, one in terms of consecutive odd numbers and the other in terms of the square of a sum of consecutive positive integers. We end with a bonus visualization of a direct version of Nicomachus's theorem.

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

Here are five other visual proofs of sum of cubes formulas:

https://youtu.be/jWpyrXYZNiI
https://youtu.be/YQLicI8R4Gs
https://youtu.be/Ye9OPNqV9FA
https://youtu.be/NxOcT_VKQR0
https://youtu.be/d1yM6Rq7Tfw

This animation is based on a visual proof by Alfinio Flores from the January 1998 issue of The College Mathematics Journal  (https://www.jstor.org/stable/2687639 page 61).

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

#mathshorts #mathvideo #math #calculus #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #finitesums #discretemath #calculus #sum #induction


[Music]

to make sense of this visual proof

we first need to see that the number n

squared which is typically represented

by a square array

can be decomposed into two triangular

arrays with bases differing by one

and these arrays can be morphed into

arrays of triangles

that fit together like this to form a

triangular tiling of an equilateral

triangle by equilateral triangles

so each square can be visualized this

way

starting with the square of a multiple

of 3 let's say 3k

we can consider the trapezoid tile in

the lower left

with bases 2k and k

three copies of this trapezoid tile can

be used to cover the entire triangle

in this way so that 3k squared

is three times the size of the trapezoid

but the trapezoid is a difference of two

triangular tilings so it contains two k

squared

minus k squared triangles next

start with a square of a number that is

one less than a multiple of three

say three k minus one again we consider

the trapezoid tile in the lower left

this time with bases two k minus one and

k minus one

using three copies of this tile this

time covers all but

one of the small triangles in the tiling

therefore

the quantity 3k minus 1 squared is equal

to 1

plus 3 times the quantity 2k minus 1

squared minus k minus 1 squared

as the trapezoid is again the difference

of two squares

finally if we start with a square of a

number one more than a multiple of three

say three k plus one then we can again

use three copies of a trapezoid tile

but this time the bases are two k plus

one

and k plus one and again these cover the

entire triangle

except for one middle triangle

therefore we conclude that three k plus

one squared

is one plus three times the quantity two

k plus one squared

minus k plus one squared as again the

trapezoid covers the difference of

squares

therefore we've seen that n squared is

congruent to 0 mod 3

if n is congruent to 0 mod 3 and n

squared is congruent to 1 mod 3

if n is not congruent to 0 mod 3. since

every number is either a multiple of 3

one more than a multiple of 3 or 1 less

than a multiple of 3

we've seen that every square is

congruent to 0 or 1 mod 3.

you

Squares Modulo 3 (visual proof)

This is a short, animated visual proof demonstrating how to visualize the congruence classes of squares modulo 3. #mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #squares #modulararithmetic #numbertheory

This animation is based on a visual proof by Roger B. Nelsen from the September 2013 issue of The College Mathematics Journal (https://doi.org/10.4169/college.math.j.44.4.283 - page 283).

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]

oh e

[Music]

[Music]

[Music]

Proof of the Completing the Square Method

Completing the square might sound like an intimidating term, but it’s actually a fascinating way to connect algebra and geometry. This method transforms a quadratic equation into a perfect square form, giving you a visual and intuitive understanding of the solutions. It’s not just about finding the answer—it’s about appreciating the elegance of mathematics. While it’s a classical method that dates back centuries, it still feels fresh and exciting when you see it in action. So, let’s explore this timeless technique and uncover the joy of solving quadratic equations with a geometric twist!

Here are some links for those who want to learn more about educational social media and support:
https://sociabuzz.com/philosophizeus

A website with a collection of free learning materials (all you need is dedication and an internet connection):
https://philosophizeus.org/

Stay motivated,
Suntar Jono, S.T., M.Si.

Note:
Content Creator: Suntar Jono
Animations and opening music created using CS 3 user

For feedback, please contact via email:
Suntar@philosophizeus.org


I Got That good that you

want let me be your push push

push I Got That good ST that

you let me be push push Push

[Música]

poio

[Música]

Can you explain the pattern?

This enumeration shows all the tilings of a circular grid divided into n pieces from n = 1 to n = 8 using tiles that cover one or two cells. Can you tell the pattern? Can you prove why or how the Lucas numbers appear? If you are looking for more information about this, I recommend the wonderful textbook Proofs that Really Count: The Art of Combinatorial Proof from Arthur Benjamin and Jennifer Quinn: https://bookstore.ams.org/dol-27

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

If you enjoy this enumeration, check out the longer form: https://youtu.be/4S85bHkmM_Y
or check out a Fibonacci tiling video : https://youtu.be/AFAcKDTmYXI

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

#dominotilings #fibonacci #lucas #circulargrid ##countingtiling  #count #mathvideo #math #mtbos #manim #animation #iteachmath #mathematics #discretemath #combinatorics #enumeration #synthwave


[Music]

start with a semi-circle of radius one

then take the radius and rotate it a

fixed angle

the radius has length one and let's call

the angle created t drop a perpendicular

chord from the endpoint of the radius on

the circle down to the horizontal

diameter the length of this chord is the

sine of t

the third leg of the triangle formed by

the radius and this chord has length

cosine of t so that the remainder of the

horizontal radius has length one minus

the cosine of t

[Music]

the line connecting the other end of the

horizontal diameter to the endpoint of

the radius creates this triangle which

is isosceles because it has two

congruent sides

the supplementary angle to t is 180

minus t and because the other two angles

must be equal they must be

both t over two

but then

t over two is an angle inside

of this highlighted right triangle

so that the tangent of t over 2

is equal

to the sine of t

over 1 plus the cosine of t

furthermore if we create this line

segment we have built a large right

angle triangle by thaley's triangle

theorem and consequently the two

highlighted subtriangles must be similar

equating the leg ratios then tells us

that the sine of t over one plus the

cosine of t

is equal to one minus the cosine of t

over the sine of t

the semicircle has given us two formulas

for the tangent of half an angle

[Music]

you

Half Angle Tangent Formulas (visual proof; trigonometry)

This is a short, animated visual proof of the half angle formula for the tangent using Thales triangle theorem and similar triangles. This theorem gives two ways to compute the tangent of a half angle, like tan(t/2). #mathshorts #mathvideo #math #trigonometry #triangle #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #circle #thalestheorem #similartriangles #semicircle #identity #halfangle #trigidentity #trig

Please consider subscribing to the channel if you enjoy the content.

This animation is based on a proof due to R.J. Walker from the May 1942 issue of The American Mathematical Monthly (https://doi.org/10.2307/2303100) page 325.

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