हाउ विल यू प्लॉट द कॉम्लेक्स नंबर a + ब

ऑन द आर्गन प्लेन अजूम ए एंड बी आर बोथ

पॉजिटिव वी विल मूव ए यूनिट्स राइट ऑन द

रियल एक्सिस एंड बी यूनिट्स अप टू गेट टू

द पॉइंट a

बी वी कैन कॉल इट पॉइंट p एंड लेट्स से द

ओरिजिन इज पॉइंट ओ नाउ जस्ट फोकस ऑन पॉइंट

p इफ यू वांट टू आस्क योर फ्रेंड टू प्लॉट

पॉइंट p हाउ वड यू डू इट योर फ्र जस्ट हैज

एन एमटी ग्राफ इन फ्रंट ऑफ हर एंड यू वांट

टू आस्क हर टू प्लॉट पॉइंट p लेट्स मूव

दिस ग्राफ टू द लेफ्ट एंड हैव एन एमटी वन

ऑन द राइट सो वन थिंग यू कड टेल योर

फ्रेंड इज मूव a यूनिट्स टू द राइट एंड b

यूनिट्स अपवर्ड्स एंड यू विल गेट पॉइंट p

इज देयर एनी अदर वे इन व्हिच यू कैन आस्क

हर टू प्लॉट दिस पॉइंट लेट मी गिव यू अ

हिंट आई जॉइन पॉइंट्स o एंड p एंड कॉल दिस

एंगल थीटा एंड अज्यू दिस लेंथ इज r दिस

वाज अ बिग बिग हिंट व्हाट कैन यू टेल योर

फ्रेंड सच दैट शी प्लॉट्स पॉइंट p

करेक्टली वेल यू कैन टेल योर फ्रेंड टू

थिंग्स फर्स्ट द डिस्टेंस ऑफ़ पॉइंट्स p

फ्रॉम द ओरिजिन एंड सेकंड द एंगल एट व्हिच

लेंथ o इज प्लेस फ्रॉम द पॉजिटिव साइड ऑफ़

द हॉरिजॉन्टल

एक्सिस सो इफ यू टेल हर दैट पॉइंट p इज r

यूनिट्स लॉन्ग एट एन एंगल ऑफ़ थीटा

डिग्रीज फ्रॉम द पॉजिटिव x एक्सिस शी विल

बी एबल टू प्रोसाइज प्लॉट पॉइंट p सो दिस

वाज अनदर वे टू रिप्रेजेंट पॉइंट p यू कैन

इदर गिव हर a एंड b और r एंड थीटा इन इदर

केस शी विल बी एबल टू प्लॉट पॉइंट p दिस

थिंग यू सी ऑन द राइट इज द पोलर

रिप्रेजेंटेशन ऑफ अ

कॉम्प्लेक्शन वि जस्ट यूजिंग डिफरेंट

पैराट्स टू रिप्रेजेंट पॉइंट p वीव

वि दि concept.com

[संगीत]

Complex Numbers - Polar Form Part 1 | Don't Memorise

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consider the graphs of two different

sequences first I've graphed a random a

n that I made up and then I went and

graphed the absolute value of that same

sequence and then the top sequence a

there were some positive and negative

terms when I take the absolute value it

takes all of those negatives and makes

them all positives now this is a

sequence and in both of these sequences

at least visually it appears that the

limits going to be 0 but just because

the limit of the terms goes to 0 doesn't

say anything about what the series does

so let's investigate that let's compare

the series that adds up the a n to the

series that adds up the absolute value

of the a n now because that program is

saying I happen to know what this series

is this is the series minus 1 to the n

minus 1 all divided by square root of n

so the absolute value the effect it has

is just takes all that minus 1 stuff and

gets rid of it it just all turns it into

plus 1 now these two sequences are ones

that we can do by previous methods we've

seen the top one it's an alternating

thing with the minus 1 to the n minus 1

and then it multiplies it by 1 over

spurred n which is a positive decreasing

sequence with limit 0 it satisfies all

conditions of the alternating series

test so that top one indeed converges by

the alternating series test as for the

bottom one this is AP test it's AP with

N equal to 1/2 and so this is a p-series

that diverges so the lesson here is that

taking absolute values of the terms

makes a big deal in this example it

converts it from a convergent to a

divergent series no why might that be

the case if you think about the one with

positives and negatives when you take a

positive term and you add it to a

negative term it cancels maybe not

entirely all the way but the combination

of positives and negatives they don't

add up as much as only adding positives

or alternatively only adding negatives

together so when you take the absolute

value you get rid of any form of

internal cancellation that you have so

it would make sense that the absolute

value one would be the larger that it

might diverge even though the one with

the minus signs that converged and

indeed this is generally the case

so if here

that we have is that if the sum of the

absolute value of the ANS converges then

the original series with just the end

that converges as well it says that you

put this absolute value around it you

have a stronger condition so when you

know it converges with the absolutes

there then it converges without them

deep a little bit of terminology here

we'll say that absolute convergence that

expression is what we mean when we take

the sum of the absolute value of a ends

so another way to say this is that if it

converges absolutely then it converges

in the normal sense but the other

direction is not true if something

converges like in the example we saw

before that converged by the alternating

series test it does not necessarily

converge absolutely it might but it

might not and in that scenario where it

doesn't we call this conditional

convergence so conditional convergence

is where the original series converges

but does not converge absolutely it

doesn't converge when you wrap it in

absolute value signs indeed this example

that we started the video with this

minus 1 to the n minus 1 over root n

that was a conditionally convergent

series it converged but it didn't

converge absolutely so these are just a

couple of pieces of terminology that are

going to be very important for our next

test the ratio test and the root test

Absolute Convergence vs Conditional Convergence vs Convergence

We've seen regular convergence of a series before, but now we consider two special cases. Absolute convergence is when we take the series of the absolute value of the terms, which gets rid of any possible cancellation like what happened a lot in the Alternating Series Test. Absolute convergence implies regular convergence. Conditional Convergence is when it converges but does not converge absolutely.

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let's investigate this sequence we were

just looking at a little bit more

carefully because I want to give a

different view on what a sequence really

is now currently I have it written here

horizontally but I could also go and

switch it around and write it vertically

as well that shouldn't make any

difference it's still an infinite

ordered list and I also have another one

that we've seen before this is the one

two three four five six seven eight and

so on so we can think of these as two

different sequences but because I sort

of lined them up in this nice way we we

should have this association that like

the one over here is associated to the

three in the sense that they're both the

first terms and the two here and the

minus six are associated senses that are

both the second terms and I can think of

this as a sort of a dynamic way I can

think of it as the one goes to the 3 and

the 2 goes to the minus 6 and the 3 goes

to the not indeed the association

between here is precisely what that

explicit formula tells me if I choose a

different K it goes to the minus 1 to

the K times 3 K so what do I really have

here well if I think about my sequences

as being my arrow diagram when I've got

a domain here and I've got a co domain

and then the elements inside of my

domain are all Matt to things in my co

domain what I'm describing is a function

every single thing in my domain goes

somewhere and more importantly it obeys

the vertical line test there's there's

no element in my domain here that that

splits into multiple different things

because my association is to take one

element say the case 1 to the 8th one

over here not to two different ones so

really what a sequence is going to be

then is this function from this

particular set to some other set or in

other words I'm going to say this is my

formal mathematically precise definition

a sequence is a function where I'm

taking things from the positive integers

that was the 1 2 3 4

and I'm going to some codomain that we

haven't specified and the other example

look like it was integers as well but

it's going to be some codomain so as an

example let's take the one we just

looked at if I wanted to prescribe it in

this language the language of a function

with a domain and with a co domain what

I'm going to say is that I have a

function ass and it inputs things inside

of my Co domain so it's going to input

some K and then I had to tell you what

it does to the K well it does precisely

what we just saw I got that minus 1 to

the K times 3 K this could be an example

in other words if I think of the

function it inputs the K and it sits out

this a sub K now this formal definition

of a function I want us to note it and

to be there and for us to think that

sequences really aren't anything new

they're just a special type of function

however it's really intuitively useful

to us if we if we actually think of

sequences more in the informal sense but

there's this ordered list of numbers in

other words I typically don't use the f

of K notation at all when I'm just

noting a sequence I will say a sub K the

K term of my sequence and I don't give

the formula that depends on K I don't

make explicit that I'm talking about a

function I don't make explicit a

particular s however that is what is

going on behind the idea of a sequence

The formal definition of a sequence.

We have an intuitive picture of sequences (infinite ordered lists). But there is a formal definition of sequences based out of the idea of a specific function between sets, specifically from the positive integers to the real numbers.

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okay here we're going to find

25% of the number 76 whenever you're

finding a percentage of a number you can

read of basically just replace it with

multiplication and the only thing we

need to do is we need to turn 25% either

into a

decimal or a fraction before we

multiply so that's what I'm going to do

well 25% uh we can write that 25 over

100 of is going to be our

multiplication of 76 I'm even going to

write that as 76 over 1 and now we can

just start simplifying so 25 over 100

that would actually reduce to 1 over4 if

we divide top and bottom by

25 so multiplied by 76 over

1 well that's just going to give us 76

over

four and four will go into 76

evenly four will go into seven once 1 *

4 is

four drop down subtract we'll get our

three and drop down our six and then

four will go into 36 19 times so 76 / 4

is 19

so if you had

25% of say $76

you would have exactly $19

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding a Percent of a Number ( 25% of 76 in this video!).  Here we look at an example of finding the percent of a number.


so to give you a little bit of a flavor

for how these things can can get a

little bit more complicated look at this

statement here so we're going to have

not P or not q and I want to try to

figure out what does this mean it's

already getting a little bit complicated

of course I can keep on adding I could

add an r i could add more variables and

keep on combining these negations and

conjunctions and ores as much as I wish

so I want to have this systematic way of

doing it and we're going to use the

truth table here so let's pull up our

template

and you're notice that it's going to be

a little bit bigger than it was last

time so if I go way off to the left

again we're going to have the same sort

of beginning Parts here so I want to

have an initial input variable

p and another initial input Q so those

are my same two inputs that we've had

before and I can go down and fill out

the different possibilities that we have

exactly we've done before I could have

true and true I could have true and

false I could have false and true and I

could have false and false so the same

sort of beginning moves here now I don't

want to jump right to the final column

over here I don't want to jump right to

this final answer I want to give some

intermediate ones that are going to help

me along so what I'm going to do in

these ones is I'm going to put not P

here and I'm going to put not Q over

there and I'm going to try to figure out

what those are and then find finally

after I'm done with everything I can

come down here and I can come and put

this inside of

there all right so not p is just exactly

the opposite of whatever p is if if p is

going to be true then not p is going to

be false p is true not p is false p is

false not p is true and P is false not p

is true all right so I'm able to fill

that in pretty easily same with not Q

not Q I just look at Q q and then I take

the negation of every one so if Q is

true then not Q is going to be false

false is going to turn into a true true

is going to turn into a false and false

is going to turn into a

true so now comes the tricky part I want

to put these two things together because

I have that it's going to be not P or

not q and so that tells me if one of the

two of these columns the not p and the

not Q columns is true then my output is

going to be true as well so what do I

have here I have a false and a false so

neither of them is true it's not the

case that one or the other one is going

to be so neither of them is true and so

therefore it is going to be

false down here one of the two is true

this one's true so one or the other is

true yeah that works that's going to be

a

true true and false one or the other

that's going to be another true and true

and true one of them is going to be true

in fact both of them true and so that's

going to be a final true so what do I

get false true true and true so these

truth tables are very useful at taking

these complicated compounded logical

statements and just trying to interpret

when are they false and when are they

true depending on the initial parameters

the P's and the q's

Truth Table Example: ~p V ~q

Learning Objective: We investigate the truth table for the more complicated logical form ~p V ~q

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because sets in the set operations like

Union and intersection and complement

are all defined in terms of some logic

condition like or or and or not in it

means that all of those rules of logic

that we got all those logical

equivalences can be applied to the

language of set and therefore we get a

lot of equalities of sets so for example

consider this claim I'm claiming that a

union B equals B Union a so how could we

prove a claim like this again I want you

to notice inequality so I want to show

if it's in the Union a union B then it's

in the B Union a and vice versa

if it's in B Union a then it's an a

union B I want to show both sides of

that so I want to say show that what my

proof is going forward everything is in

biconditional

and if and only if so what I'm going to

use the element method mainly I'm going

to start with an element X which is

inside of the a and the B inside of a

union B but the definition of a union B

and by the way all definitions are

inherently if and only if you don't have

to worry about that if you're applying a

definition it is always if anomia so X

is an a union B is the same thing as X

is in a or I'm going to put this symbol

for my disjunctive or X is inside of B

but now we can use a property that we

had for disjunctive we know that two

properties that are connected with an or

that I can reverse that order that I can

say that this is the same thing as X

inside of B or X is inside of a indeed

back in the day we would have verified

this logical property by looking at it

at level of truth tables this is P or Q

and this is Q or P and you can verify

that they have the same truth table so

this is a logical equivalence and then I

can go backwards X and B or X and a this

is precisely the same thing

saying that X is in B Union a and so I

have a chain of if and only if and I get

if I start in a union B I end up in B

Union a and the key thing to note here

is that the important property is

something about logic the the reversing

the order came down here is a logical

property and then because the Union was

defined in terms of this or statement

you could sort of bring it out of the

logical property into this statement

about unions but the same is true in a

much larger domain for instance we have

a de Morgan's law for logic we have a de

Morgan's law for sets and that idea is

going to continue for all of the

different logical equivalences

Proving equalities of sets using the element method

In this video we prove that we can commute unions, that is change the order of a union of two sets. This is just a stand in for a larger class of equalities between sets. To prove this we use the element method. We need to show an element in the left hand side is in the right hand side, AND vice versa.

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all right let's do one more example what

we have here is now the integral of e to

the 2x on the top all divided by 1 plus

e to the 2x on the bottom so they figure

out what's my f here and what's my you

now one of the things that I can sort of

know right off the top is that when I

take the derivative of 1 plus e to the

2x I'm going to get twice eetu 2x and

there's this factor of 2 there that's a

bit confusing but but it sort of looks

like the top here is related off by a

factor of 2 but it's related to the

derivative of the bottom so let me

experiment with setting u to be that

denominator ok so how does that work

U is the 1 plus e to the 2x it's

derivative us and we said was that twice

either 2x DX

well if that's my you what's my app look

this looks like the function 1 divided

by u times D U but we're off by this

factor of 2 all right so that's the U

and the D U but what is going to be in

that case the F now if I think about

this expression the place where you

occurs is on the denominator so we think

my F looks like 1 divide it out by U and

if they set my F to be 1 over u that

means that they capital F the

antiderivative of the lowercase F is

just gonna be the natural logarithm of

the absolute value of U now why is this

the case well if we take the derivative

of the natural logarithm of the absolute

value of U we're going to get 1 divided

by u so indeed capital F is an

antiderivative of a lower case of F ok

so now if I take what my expression is

notice how I'm going to relate it so we

have that syndra we want to compute but

what I've done here is I've divided out

by a 2 and I've multiplied by a 2 and I

can multiply and divide by 2 that's just

x 1 that doesn't doesn't do anything but

the reason why I do it is because now

when I've got this 2 on the top even

though is cancelled down here to e to

the 2x DX 2 e to the 2x DX that is just

D u that is what I have done is managed

to write this as

1/2 the integral of 1 over u du u so we

get the 1/2 that just comes along for

the ride the integral of 1 over u

becomes the natural logarithm and I

evaluate it at U where u is this 1 plus

e to the 2x and finally I add in the

plus C so that is going to be my general

antiderivative to this funky expression

and it's worth noting that integration

in general is actually quite a bit

harder than differentiation most of

these functions that you can combine

with quotients and products and

compositions and exponents of the

elementary functions we've been talking

about e to the ax and sine and

polynomials we can just apply our rules

and take the derivatives of all of them

but the other way around is not quite so

simple some integrands when you take the

integral of some functions it's able to

be done because it can be manipulated to

look into one of these nice forms like

we have here with the substitution rule

but that is in general not the case and

there's all sorts of integrals where we

don't have a nice rule like this to be

able to apply

Adjusting the Constant in Integration by Substitution

In this example, the derivative of the inside doesn't exactly line up in the integrand, we have to adjust the constant.

Now it's your turn:
1) Summarize the big idea of this video in your own words
2) Write down anything you are unsure about to think about later
3) What questions for the future do you have? Where are we going with this content?
4) Can you come up with your own sample test problem on this material? Solve it!

Learning mathematics is best done by actually DOING mathematics. A video like this can only ever be a starting point. I might show you the basic ideas, definitions, formulas, and examples, but to truly master calculus means that you have to spend time - a lot of time! - sitting down and trying problems yourself, asking questions, and thinking about mathematics. So before you go on to the next video, pause and go THINK.

This video is part of a Calculus course taught by Dr. Trefor Bazett at the University of Cincinnati.

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welcome back to logic 101 I'm William

Spaniel and this lecture is on the rule

of inference known as the constructive

dilemma this one is a little bit more

complicated than previous rules of

inference we've seen because it involves

three three premises we have two

implications to start off with P implies

q and R implies s notice that we have

four simple sentences in total there and

they're split evenly across those two

premises so we have just p and Q up top

and just R and S down bottom there is no

overlap between those two the third

premise is the disjunction between the

antecedence of those two implications so

we have P or R and if we have all of

that then we can conclude that the

disjunction between the second half of

those two implications QRS is true as

well essentially what's going on here is

that we're applying modus ponin across

the various different cases that part

three gives us so part three premise 3

tells us that P is true or R is true or

both P and R are true well in the first

case suppose p is true then we have q

being true through line one and modus

ponents if we have the second case where

just R is true then through modus

ponents in line two we have S being true

and if both of them are true then

through modus ponents and lines one and

two we have both q and s being true and

that's exactly what we see with the

conclusion there we see Q or S being

true which means Q could be true or S

could be true or both of those things

could be true to see this in action with

an example suppose that premise one said

if I am running I am happy and premise

two said if I am sleeping I am dreaming

and finally premise three said I'm

running or I'm sleeping well if I'm

running then I'm happy if I'm sleeping

then I'm dreaming and if I'm running and

sleeping perhaps I'm running in my sleep

then I'm both happy and I'm Dreaming

which is exactly again what the

conclusion says I'm happy or I'm

Dreaming or both now normally what I've

been doing for these rules of inference

is then showing you that this is true on

a truth table but as I mentioned at the

top we have four different simple

sentences here and in the past we've

explored what it looks like when you do

a truth table with four simple sentences

and of course we have 1 2 3 four complex

or compound sentences here which are

only going to make things more

complicated so I'm actually going to

hold off on doing that truth table

because again it's pretty complicated

something that you might want to do on

your own and instead of showing you how

to prove it using a truth table instead

later on in this course when we've

learned how to do proof by contradiction

I will use proof by contradiction to

prove the constructive dilemma through

what actually is going to turn out to be

a much simpler process than a truth

table so of course you don't know how to

do this quite yet but by the end of this

course when we start getting into proofs

you will learn how to do this and I will

take you through this step by step so

you can see that constructive dilemma is

true not using of course constructive

Dilemma to arrive at that conclusion all

right I hope you enjoyed this and I hope

to see you here next time when we talk

about the modus tolan's version of the

constructive dilemma which is known as

the destructive dilemma hope you enjoy

this again and I hope to see you next

time take

care

Logic 101 (#31): Constructive Dilemma

http://gametheory101.com/courses/logic-101/

Constructive dilemma is a logical rule of inference that says if P implies Q, R implies S, and P or R is true, then Q or S is true as well. Essentially, the constructive dilemma passes the disjunction through two conditional statements. For example, if the statements

"If I am running, I am happy."

and

"If I am sleeping, I am dreaming."

and

"I am running or sleeping."

are true, then it must also be true that I am happy or I am dreaming.


in this video we're going to take a

closer look at the idea of relations

which are something that we've

previously seen for now you'll recall

that a relation is to let it involve two

different set you had a set a and you

had a set B and I'm put a bunch of

elements inside of a and I put a bunch

of elements inside of B then you could

visualize sort of an arrow diagram for a

relation which is connections from the

set a to the set B so for example I can

have these connections where the the a

one is going to go down to the B 2 and

the a five is goes to the B 2 and the a

3 is going to be 1 and the B 4 and so on

and then you would be able to sometimes

say that a relation was a function if

additionally it had the property of

passing the vertical line test and that

everything in the domain went somewhere

we've previously discussed that and then

our formal definition for what a

relation was is going to be the

following

it is a set of the Cartesian product

that is I look at all pairs something in

a something in B that's what the

Cartesian project pairs we choose a

first thing in a and a second thing in B

and then we say that some pair like say

how about a 1 B 2 that is an element of

your relation if when you look at this

arrow diagram a 1 really is related to B

2 and then the nomenclature we had is

we're going to say that an element in R

is written as a relates to be precisely

lame little a comma little B was an

element of your Cartesian product or

there was an arrow up here in your arrow

diagram now if I have a relation I can

also define something referred to as the

inverse relation and the idea of an

inverse relation is it's exactly doing

what the relation from A to B did but it

flips it around so for example let me

put up my relation R that I saw before

for the inverse relation we're going to

try to take the a and the B and we're

going to try to swap them so I

put them up I'm not going to fill in the

arrow just for the moment but for the in

Reverse relation first of all the the

starting set your domain if you will is

the be your codomain is the age so we

flip that order and then if I think

about how I should be connecting these

well for instance if I go over here the

B 1 is connected down to the a3 the the

B 2 is connected up to the a1 and the a5

and so on so I can fill in my line like

this where these connections are exactly

this picture but it's the mirror image

of that picture

Relations and their Inverses

We review the idea of a relation, and show how to express the inverse of a relation. A relation from A to B i a subset of the cartesian product A x B and we write an element in the relation as aRb to mean that a relates to b.

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now let's investigate what it means to

negate a conditional so in this case I

have this conditional P implies Q and

then the question is what happens if I'm

going to negate that particular

conditional now what I'm going to do is

I'm going to use just a bunch of logical

equivalences here I'm going to get some

result and then we're going to see

whether that result makes sense to us so

the first thing that I can say is that

we know that that I can take this P

implies Q and we've just shown that this

is logically equivalent to I'm not

changing the knot here at all but we

know that the inside the conditional is

logically equivalent to not P or Q so

that was one step and now I have a

disjunct of an or statement and Center

than a knot in front of it and we know

that one of de Morgan's laws applies for

a scenario like this that I can take

them not and I can distribute it through

but it does two things it takes the

order to an and and it takes a knot in

front of both of the two different sides

so in other words I'm going to claim

that this is logically equivalent to the

first thing is I'm going to bring the

knot to the inside so goes naught naught

P and then it flips the order to an and

so I go to an and and then I take the

knowledge and I put it in front of the

queue as well and so I go naught Q and

this was de Morgan and then I also note

that here I have this this double

negative and we've previously seen that

I can manipulate the double negatives

not not P is the same thing logically as

just P so I can say that this is the

first thing and not Q now this should

make some sense to us if we think about

what an implication was there was only

going to be one way that the implication

was not going to be true when the

Assumption the hypothesis was true was

that the conclusion was going to be

false

that was the only one in her

truth table that gave a value of false

for the application and that really has

here the first one has to be true the

assumption and it has to be the case

that the conclusion needs to be false

that was the one scenario the other

scenarios like for example when P was

false

those were bumped into the vacuously

true scenario and of course when P was

true and Q is true that was also going

to be true for the implication as well

so what we managed to prove here by

showing a bunch of logical equivalences

is the result is the exact same result

we would have gotten if we just gone to

the truth table and negated the column

for P implies Q so you've got the same

result which is good

Negating a Conditional Statement

Learning Objectives: Take the negation of a conditional not by using truth tables, but by using known logical equivalences.
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*****************************************************
YOUR TURN! Learning math requires more than just watching videos, so make sure you reflect, ask questions, and do lots of practice problems!

****************************************************
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*****************************************************

This video was created by Dr. Trefor Bazett


let's recall the fundamental theorem of

calculus that we learned in kindergarten

so the formulation i'm going to take

here is that the integral

of a to b of f prime of x dx

is equal to f of b minus f of a

so if we look at the graph of f prime of

x the fundamental theorem of calculus

tells us that if we take two points a

and b then the area underneath

this graph is determined by the value of

another function namely

f at these boundary points alone

now a priori we need to know every point

between a and b

to determine the area but the

fundamental theorem of calculus tells us

that it's actually

only determined by the values of a

separate function namely its

antiderivative

at the endpoints now to get further

insight on the fundamental theorem of

calculus

what we'll do is we'll rewrite the

integral from a to b of f prime of x

dx as the integral over the closed

interval

a b f prime of x dx for us this is going

to be the same

and we'll also write f prime of x as df

dx

also write a b as the boundary

of the closed interval a b to denote the

boundary of a set

we use this partial symbol so the set

containing only the elements a b

is denoted by partial of the closed

interval a b

so now what the fundamental theorem of

calculus says is that the

integral over the closed interval a b of

df dx

is equal to the integral over the

boundary of the closed interval

of a b of f dx now just to make this

subsequent remark more transparent what

we can do is it can actually replace

df with partial f so from what we've

done here the way we've written the

fundamental theorem of calculus what we

see is that it's not that derivatives

and

integrals are the opposites of each

other it's that derivatives

are dual to the boundaries

of regions of integration and the

integral is the dual pairing here

you

Why you don't understand the Fundamental Theorem of Calculus | Geometric Algebra

In this video, we discuss how the derivative is not the "opposite" of the integral. We offer a perspective on the fundamental theorem of calculus from the point of de Rham cohomology. This is part of an earlier video that I made concerning Green's theorem: https://www.youtube.com/watch?v=XEDtm09ol5o The audio quality was not the best, so I fixed the audio (I hope), and focused only on the part of the video which treats the fundamental theorem of calculus. This perspective has been growing in interest, and commonly goes under the umbrella of Geometric Algebra in the YouTube community. For a swift introduction to geometric algebra, see the following video: https://youtu.be/60z_hpEAtD8

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Hi, my name is Kyle and I'm currently doing my doctoral mathematics degree in complex differential geometry under the supervision of Professor Gang Tian and Professor Ben Andrews.


the function e to the X is a very

important very famous function within

mathematics and indeed it has this

interesting property which is that the

derivative of e to the X is just the

same function that is if I take a point

like how about 1 here all right

so 1 e to the 1 is going to have the

value just e to the 1 is e and so if I

figure out what the slope is that the

height of this which looks like a little

bit over 2 that's basically the value of

e here is the same thing as the slope of

this particular function they have the

same function value and slope and indeed

if I take in my slope down here or my

slope up here it doesn't really matter

it's always gonna have the same slope as

a function value indeed you could ask a

broader category like there's all these

different exponents like 2 to the X and

3 to the X and PI to the X and 100 to

the X e to the X is defined to be or one

way you can define it to be is defined

to be the number such that the

derivative of e to the X is itself we

define e to be the number where this

property is true so we really are gonna

like Exponential's as we go forward in

our course and they have this really

nice property indeed what you can do is

take the derivative of the derivative of

e to the X first wrote it takes UVs

second river takes you to the X you take

a hundred rooms in a row you get back to

e to the X and so forth so where are we

at we now know how to take the

derivatives of two different types of

functions the derivatives of an

exponential like e to the X and the

derivative of a power like X to the n so

we've got those sort of two like

baseline kind of functions and then the

additive 'ti in the scalar

multiplication those are ways that I can

combine them so for example I could have

a polynomial which is a bunch of power

rules or I can have like e to the X plus

x squared and I would be able to combine

them by these different rules so I

haven't got all functions by any means

but we're developing so far a decent

list

The derivative of e^x.

Description:
The derivative of e^x is....itself! e^x.

Learning Objectives
1) Compute the derivative of e^x

Now it's your turn:
1) Summarize the big idea of this video in your own words
2) Write down anything you are unsure about to think about later
3) What questions for the future do you have? Where are we going with this content?
4) Can you come up with your own sample test problem on this material? Solve it!

Learning mathematics is best done by actually DOING mathematics. A video like this can only ever be a starting point. I might show you the basic ideas, definitions, formulas, and examples, but to truly master calculus means that you have to spend time - a lot of time! - sitting down and trying problems yourself, asking questions, and thinking about mathematics. So before you go on to the next video, pause and go THINK.

This video is part of a Calculus course taught by Dr. Trefor Bazett at the University of Cincinnati.

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continuing in this fashion always

cutting up the rightmost piece into

those same proportions you'll find that

1 - p + P * 1 - P plus p ^ 2 * 1 - p on

and on always adding P to the next power

* 1 - P equal 1 dividing both sides by 1

- P we get this nice

formula in this formula the universe has

offered a weird form of nonsense even

though the way that you discovered it

only makes sense for values of P between

0 and 1 the right hand side still makes

sense when you replace ke with any other

number except maybe for one for instance

plugging in -1 the equation reads 1 - 1

+ 1us one on and on forever alternating

between the two equals 12 which feels

both pretty silly and kind of like the

only thing that it could be plugging in

two the equation reads 1 + 2 + 4 + 8 on

and on to Infinity = -1

this is MF DOOM (ft. 3blue1brown)

original video: https://youtu.be/XFDM1ip5HdU?si=dUjEakmKfR3wJI7x
(just cut it up a teeny bit to make it fit more :3)


这是贝叶斯定理主要视频的脚注。

如果您的目标只是从数学角度理解为什么它是正确 的，

那么实际上有一种非常快速的方法可以基于分

解单词 AND 在概率中的工作原理来了解它。

假设有两个事件 A 和 B。

这两种情况同时发生的概率是多少？

一方面，您可以首先考虑 A 的概率 ，

即 A 为真的所有可能性的比例 ，

然后将其乘以 B 也为真的事件的 比例，

这称为给定 B 的概率A。

但奇怪的是，A 和 B 中的公式看起来不对称。

据推测，我们还应该能够将其视为在所有可能性

中 B 为真的情况的比例乘以 A 也为真 的情况的比例，

即 A 给定 B 的概率。

它们都是相同的，而且它们相同的事实为我们提供了 一种方法，

可以用给定 A 的 B 的 P 来 表达给定

B 的 A 的 P，或者反之亦然。

因此，当其中一个条件比另一个条件更容易用 数字表示时，

例如，当更容易考虑在给定假设 的情况下看到某些证据的概

率而不是相反时， 这个简单的恒等式就成为一个有用的工具。

尽管如此，即使这在某种程度上是理解公式的

更纯粹或更快速的方法，我选择在主视频中用

证据更新信念来构建一切的原因是为了帮助第 三个层次的理解，

能够识别当这个公式在众多 可用的数学工具中恰好是正

确使用的时候。

否则，很容易只是看一眼，点头，然后很快就忘记了。

你知道，当我们在这里时，有必要强调一个常见的误解，

即 A 和 B 的概率是 A 的 P 乘以 B 的 P。

例如，如果您听说四分之一的人死于心脏病，

您 很容易认为这意味着您和您的兄弟死于心脏病的

概率是四分之一乘以四分之一，或十六分之一。

毕竟，连续两次抛硬币出现反面的概率是 1/2 乘以 1/2，

一 对骰子上掷出两个 1 的概率是 1 6 乘以 1 6，

对吧？

问题在于相关性。

如果你的兄弟死于心脏病，考虑到 某些遗传和生活方式的联系，

你死 于类似疾病的可能性就会更高。

像这样的公式，虽然看起来很诱人、很干净，但其实是完全错误的。

抛硬币或掷两个骰子等情况的情况是 ，

每个事件都独立于上一个事件。

因此给定 A 时 B 的概率与 B 的概率相同。

A发生的事情不会影响B，这就是独立性的定义。

请记住，许多介绍性概率示例都是在非常游戏化 的环境中给出的，

例如骰子和硬币，真正的独立 性，

但所有这些示例都可能会扭曲您的直觉。

具有讽刺意味的是，概率的一些最有趣的应用 ，

大概是使用这些游戏化示例的课程的全部动 机，

只有当事件不是独立的时才具有实质性。

贝叶斯定理就是一个完美的例子，它准确地

测量了一个变量对另一个变量的依赖程度。

The quick proof of Bayes' theorem

Including some added words on independence.
Main video: https://youtu.be/HZGCoVF3YvM
Help fund future projects: https://www.patreon.com/3blue1brown
An equally valuable form of support is to simply share some of the videos.
Special thanks to these supporters: http://3b1b.co/bayes-thanks

------------------

These animations are largely made using manim, a scrappy open-source python library:  https://github.com/3b1b/manim

If you want to check it out, I feel compelled to warn you that it's not the most well-documented tool, and it has many other quirks you might expect in a library someone wrote with only their own use in mind.

Music by Vincent Rubinetti.
Download the music on Bandcamp:
https://vincerubinetti.bandcamp.com/album/the-music-of-3blue1brown

Stream the music on Spotify:
https://open.spotify.com/album/1dVyjwS8FBqXhRunaG5W5u

If you want to contribute translated subtitles or to help review those that have already been made by others and need approval, you can click the gear icon in the video and go to subtitles/cc, then "add subtitles/cc".  I really appreciate those who do this, as it helps make the lessons accessible to more people.

------------------

3blue1brown is a channel about animating math, in all senses of the word animate.  And you know the drill with YouTube, if you want to stay posted on new videos, subscribe: http://3b1b.co/subscribe

Various social media stuffs:
Website: https://www.3blue1brown.com
Twitter: https://twitter.com/3blue1brown
Reddit: https://www.reddit.com/r/3blue1brown
Instagram: https://www.instagram.com/3blue1brown_animations/
Patreon: https://patreon.com/3blue1brown
Facebook: https://www.facebook.com/3blue1brown


در مربعی به ضل ا از هر رسش دایره به شع ا

رسم شده اس رو پیدا بکنید در یه مربع جدید

د تا از کمانها رو رسم نموده و از نقطه

برخوردشون به دو رس پایین وصل می‌کنیم

مثلث متساوی الاضلاع تی به وجود میاد زیرا

اضلاعش شعاع‌های دایران مطابق شکل ناحیه

ان همراه با مثلث تی یک ششم دایره‌ای به

شعا a رو تشکیل میده پس ان مساوی ی ششم

مساحت دایره منهای تی خواهد بود از طرفی

ناحیه‌های ان تی و ان همراه با ناحیه ام

یک چهارم دایره پس ام مساوی یک چهارم

دایره منهای د ان منهای تی خواهد بود در

شکل بالا سطح مربع توسط ۱s و ۴ ام پوشانده

شده در نتیجه ناحیه s برابر میشه با سطح

مربع منهای ۴ ام با جایگذاری ام در عبارت

فوق رابطه جدید حاصل میشه که مساحت مربع

a۲ تی رادیکال س۴ a۲ و مساحت دایره پی۲

است با ساده کردن و فاکتورگیری نتیجه

نهایی حاصل میشه که به طور تقریبی ۳۱ و

نیمد از سطح مربع خواهد بود

find area of the shaded region #maths #explore #3blue1brown #animation


draw two points on a circle and connect

them obviously this divides the circle

into two different regions if you put a

third point on the circle and then draw

lines connecting it to the previous

points now the circle's been divided

into four separate regions at a fourth

Point add lines connecting it to all the

previous ones and again count up how

many regions the circle's been divided

into this time we get eight if you start

to see the pattern let's do one more at

a fifth Point connect it to all the

previous ones count up the regions and

this time we get 16 again maybe you can

guess what's coming next but would You

Bet Your Life on it well let's try I'll

add a sixth Point connect it to all of

the previous ones and this time if I go

through and meticulously count all the

regions you can pause and do it yourself

I'm not pulling any trickery here you

actually get 31 be careful of

extrapolating patterns without proof in

this case there's a very good reason why

it looks like powers of two until

suddenly breaking at the sixth iteration

it's a little too long for a short form

explanation but before watching the full

video on YouTube I highly encourage you

to try thinking it through yourself

Don't let it fool you! (Link above explains what's happening)

A link to the full video is at the bottom of the screen.
Or, for reference: https://youtu.be/YtkIWDE36qU

Thanks to Dawid Kołodziej for editing together this short


this is a gton

board maybe you've seen one before it's

a popular demonstration of how even when

a single event is chaotic and random

with an effectively unknowable outcome

it's still possible to make precise

statements about a large number of

events namely how the relative

proportions for many different outcomes

are

distributed more specifically the gon

board illustrates one of the most

prominent distributions in all of

probability known as the normal

distribution more colloquially known as

a bell curve and also called a gaussian

[Music]

[Music]

distribution

Order from chaos

A link to the full video on the Central Limit Theorem is at the bottom of the screen.
Or, for reference: https://youtu.be/zeJD6dqJ5lo

Thanks to Dawid Kołodziej from long-to-short editing


hello think about a forward gain

transfer function represented as 2 l/ s

+ 2 * s + 3 integrated to a system

featuring a Unity gain feedback

path the transer function for the closed

loop system can be written as GS over 1

plus GS and can be further simplified as

follows when we use a step input as our

reference signal the output YT will have

the shape of this form to achieve better

control of the output we can introduce a

controller to the system let's consider

adding a controller for our system

represented as s+ 2/ s+

10 now the close loop transfer function

for our system is gscs over 1 plus

gscs and can be further simplified as

follows

Animated Guide to Discrete Control Systems #3blue1brown#animation #shorts #manim#controlsystems

@BuddhiWijenayake2000
#youtubeshorts #discretemathematics #manim #maths #electricalandelectronics #electronic #engineering


[Música]

SA

[Música]

at

[Música]

Taylor & Maclaurin Series Explained with an Animated Example ! #3blue1brown #mathshorts #animation


I posed a certain puzzle in a recent

video and based on a couple comments I

realized that I could have been clearer

about what makes this puzzle tricky the

question starts by supposing that you

have a disc with radius one and you

cover it with a whole bunch of strips

when I say strip here I mean a region

that's Bound by two parallel lines and

the quantity that you'll care about is

the width of that strip the distance

between those lines the question is

supposing that your strips completely

cover the disc what is the smallest

possible value for the sum of all of

those widths for example if you just

used a bunch of parallel strips then the

sum of all of the widths would be the

diameter of the circle which is two Now

spoiler alert it turns out that two is

the smallest that you can get and a

couple commenters claimed that they

don't see the point this is obvious

parallel strips should be optimal CU

otherwise you would have some kind of

overlap and overlap is clearly wasteful

that would be true if we were trying to

minimize the total area that these

strips have in that case overlap is

necessarily a waste of area but that's

not the challenge the tricky part here

is that the width of a strip is not

proportional to the area it's perfectly

possible to have a very fat strip near

the edge of the circle with a bigger

width but a smaller area than a thinner

strip near the center of the circle so

for all you know there might be a clever

covering that has a little overlap here

and there which trades off an

inefficient use of area for a more

efficient use of the total width the

challenge is to find find a rigorous way

to prove that this is not possible that

the total width of the strips really

cannot be lower than the diameter of the

circle it's really not an obvious fact

and if you appreciate this it makes the

solution that we cover in the main video

all the more beautiful

This puzzle is trickier than it seems

From this full video: https://youtu.be/piJkuavhV50


when a large language model generates

text there's a parameter that can be set

called temperature which in a manner of

speaking controls how creative it is to

understand temperature you first have to

understand that large language models

are originally trained to take in a

passage of text and predict what word or

other common character combination comes

next this prediction looks like some

kind of probability distribution when

these models generate new text like an

answer to a user question with a chatbot

what's going on is that a full context

describing the interaction is input into

the model the model predicts what word

comes next it takes a random sample from

the distribution that it generates it

appends that random choice to the full

context and then it runs it all again on

this extended text to make another

prediction and take another sample and

so on and so on over and over the

temperature is a way to modify the

probability distributions that it

generates if you set a high temperature

it gives more weight to the less likely

Words which essentially gives the model

a better chance of selecting unusual

phrases and ideas whereas a low

temperature makes it more likely to

choose the most predictable words

there's more to say but as always the

full video goes into the full details

Check out @3blue1brown for more


option e to the t is to ask what

properties does it have the most

important one and from some binding

property is that it is its own

derivative together with the added

condition that inputting zero returns

one it's actually the only function with

this property and you can illustrate

what this means with a physical model if

e to the T describes your position on a

number line as a function of time then

you start at the number one and what

this equation is saying is that your

velocity the derivative of position is

always equal to that position the

farther away from zero you are the

faster you move so even before knowing

how to compute e to the T exactly going

from a specific time to a specific

position this ability to associate each

position with a velocity paints a very

strong intuitive picture of how the

function must grow you know that you'll

be accelerating and at an accelerating

rate with an all-around feeling of

things getting out of hand quickly

and if you add a constant to that

exponent like e to the 2 times T the

chain rule tells us that the derivative

is now two times itself so at every

point on the number line rather than

attaching a vector corresponding to the

number itself first double the magnitude

@3blue1brown e^t #maths #viral


[Music]

[Music]

keep

to iimp

[Music]

Pythagore Demonstration Using Geometry ! #3blue1brown #mathshorts #animation #demonstration #maths


[Music]

[Music]

keep

[Music]

gang SL get that Tempo just up keep

tight make f

What is a derivative ? #3blue1brown #animation #mathshorts #derivative #explained #definition


这是一个非常好的谜题。

假设你在三维空间中以随机朝向放置一个立方体，

然后观察其阴影区域。

该区域的预期面积是多少？

或者换句话说，如果你一遍又一遍地重复这个过程，

随机扔那个立方体并统计你得到的面积，

那么当你这样做越来越多次时，所有这些测得面积的经验

测量平均值会接近什么？

严格来说，在提到阴影时，这将取决于光源在哪里，

但在这里我们以典型的数学家的思维方式，将光源视为无限远，

所以阴影只是一个平面投影，比如说，到 xy 平面上。

这绝对是一个难题，我实际上很好奇在评论中你会如何解决它。

我迄今为止制作的最长的视频探讨了这个问题。

事实上，这是一个关于两种不同的解决问题的风格的传奇故事，

以及为什么数学的普及往往会对它们所代表的风格产生偏见。

事实证明，影子谜题是这个故事的一个非常完美的场景。

The cube shadow puzzle

A link to the full video is at the bottom of the screen. Or, for reference: https://youtu.be/ltLUadnCyi0

Thanks to these viewers for their contributions to translations
Chinese: ZstringX
French: GiveMeChocolate, Yoyodotpy
German: Josh, dlatikay
Hindi: VaMErYT, rajeshwar-pandey
Korean: tebaioioo


در مثلث یکی از ظلها به چهار قسمت مساوی و

ضل دیگر به ۵ قسمت مساوی تقسیم شدند با

رسم دو پاره خط مطابق شک سطح آبی رنگ

برابر ه۸ است ا رو بیابید تا توجه کنید

اگر از رأس مثلثی پار خطی زل مقابلش را به

نسبت ام به ان قطع کند نسبت مساحتها هم ام

به ان است با توجه به نقطه فوق چون این ضل

به نسبت یک به سه تقسیم شده پس نسبت

مساحتها اس به سه اس و چون این ضل به نسبت

دو به سه تقسیم شده پس نسبت مساحتها دو

اسپریم به سه اسپریم است بناب ب این سطح

خواسته شده a برابره با دو اسپریم به

اضافه s خواهد بود حال در مثلث بزرگ چون

نسبت قاعده‌ها یک به سه است از رابطه زیر

اسپریم برابر ۸ پونزدهم به دست میآید و در

این حالت که قاعده‌ها به نسبت دو به سه

تقسیم شده مساحتها هم به همین نسبت تقسیم

می‌شود و از رابطه زیر s برابر چ سوم

خواهد بود با جایگذاری s و اس پریم a

برابر ۳۶ پونزدهم خواهد

شد

Area Of A?#maths #3blue1brown #animation #ریاضی #مساحت


the surface area of a sphere is 4 pi

times its radius squared exactly four

times the area of a circle with the same

radius but why Archimedes had a very

beautiful proof that this surface area

is the same as the area of a cylinder

that encloses that sphere if you

disregard the circular caps of that

cylinder the idea is that if you shine

some light from the z-axis perpendicular

to that axis then if you compare the

area of a small rectangle drawn on that

sphere to the area of the shadow cast on

that cylinder those two areas turn out

to be the same this isn't obvious but

when you work it out that shadow is a

copy of the original Little rectangle

but squished down in One Direction and

stretched out in another when you

analyze the relevant geometry the two

effects actually cancel out perfectly

treating the sphere as a sum of all of

these little patches and the cylinder as

a sum of all of those Shadows we can

infer that the surface area of the

sphere is the same as the area of that

cylinder now the cylinder is something

you can unwrap into a rectangle where

one side length corresponds to the

circum

@3blue1brown #mathematics projections

Content @3blue1brown refer for full video.
Area of sphere
#mathematics
#concept
#geometry


Has it ever bothered you how with exponents, logs, and roots,

we have three completely different types of notation to write the same fact?

If you write 2 cubed equals 8, it's the relative

position of the 3 over the 2 that indicates the operation.

But when you write the cube root of 8 equals 2,

which is the same fact, you introduce this new squiggly radical symbol.

And then writing log base 2 of 8 equals 3, again the same fact,

you write out a word for the operation.

This weird discrepancy in notation isn't just counterintuitive,

I think it's counter-mathematical.

Rather than making seemingly different facts look the same, which is what math should do,

it makes three facts that should obviously be the same look artificially different.

I came across a really nice suggestion on math

stack exchange for a more symmetric notation here.

In our example, the way this would work is you write a triangle with

a 2 in the lower left, a 3 on the top, and an 8 on the lower right.

To express the operation 2 cubed, you remove that bottom right corner,

and the symbol as a whole represents the value that should go in that missing corner.

To express log base 2 of 8, which is asking the question 2 to the what equals 8,

what you do is remove the top number.

Again, the symbol as a whole represents the value that should go in that missing corner.

You can guess the third case.

If you want to express the cube root of 8, you remove the lower left corner.

Once again, the symbol as a whole represents the

value that should go in the missing corner.

This much more clearly expresses the relationship of all three operations.

The definition alone is mildly pleasing, but where it becomes useful is in

seeing how the rules for exponentiation logs and radicals are all really the same.

The most extreme example might be how with our current notation, there are six,

six distinct ways to express the various inverse operations,

and it looks like a complete mess.

If you use the triangle of power, all of these

operations follow the same basic aesthetic pattern.

Admittedly, it looks a little bit weird when your brain has already been trained with

the traditional notation, but our brains are really good at picking up on patterns.

And in this case, once you get used to it a little,

you only need to remember one pattern which unlocks all six of these cases.

Essentially, any rule that's associated with exponents logs and radicals

becomes essentially three times faster to learn and to recognize.

If you have a pencil and paper, I highly encourage you to just take a

moment and mess around with what some of these rules for exponents look like.

It's weird at first when you're trained with the traditional notation,

but it's super satisfying once it all clicks.

The Triangle Of Power

The referenced stack exchange post by usename 2'5 9'2,  http://math.stackexchange.com/questions/30046/alternative-notation-for-exponents-logs-and-roots


مجموع ریشه‌های معادله زیر کدام است توجه

کنید که اگر ای ۱ و ا ۲ و ا۳ و ا ۴

ریشه‌های معادله درجه چهارم زیر باشند که

لزوماً همه اون‌ها حقیقی نیستند در این

صورت این معادله که به کمک ریشه‌ها نوشته

می‌شود با معادله درجه ۴ بالا هم عرض است

و اگر جملات معادله بالا رو بر آ تقسیم

کنیم این دو معادله یکسان خواهد شد از

ساده کردن معادله اخیر و متناظر قرار دادن

اون با معادله بالا ار رابطه به صورت زیر

به ریشه‌های معادله و ضرائب معادله به دست

می‌آید که اولین رابطه مدنظر برای حل این

مسئله

است حالا اگر معادله اصلی رو ساده کنیم به

یک معادله در ۴ می‌رسیم که ا و بی به

ترتیب ۱۱۶۰ و منفی ۳۲۰ هستند پس مجموع

ریشه‌های معادله برابر د خواهد بود

تست کنکور 1402 ریشه های معادله #maths #mathematics#3blue1brown #animation #ریاضی #کنکور

برای دیدن ویدیوهای جذاب آموزشی کانال ما رو سابسکرایب کنید


take a look at this little grid of

values which involve some positive

numbers on the left and some negative

numbers on the right which I'll color

with blue and red

respectively all in all this little

process basically detects wherever

there's variation in the pixel value As

you move from left to right and so it

gives you a kind of way to pick up on

all the vertical edges from your

image and similarly if we rotated that

grid around so that it varies As you

move from the top to the bottom this

will be picking up on all the horizontal

edges

which in the case of our little pie

creature image does result in some

pretty demonic

eyes this smaller grid by the way is

often called a kernel and the beauty

here is how just by choosing a different

kernel you can get different image

processing effects not just blurring

your Edge detection but also things like

sharpening for those of you who have

heard of a convolutional neural network

the idea there is to use data to figure

out what the kernels should be in the

first place as determined by whatever

the neural network wants to detect

Deep learning math: Convolution #deeplearning #machinelerning #ai Credit:@3Blue1Brown


the international math Olympiad consists

of six problems broken up over 2 days

the three problems given on each day are

supposed to get progressively harder but

on the test from 2011 problem number two

was much more challenging than the

organizers intended for it to be here's

what it states let s be a finite set of

at least two points in the plane assume

that no three points of s are collinear

a windmill is a process that starts with

a line L going through a single point P

and S the line rotates clockwise about

the pivot P until the first time that

that line meets some other point

belonging to S this point Q takes over

as the new pivot and the line now

rotates clockwise around Q until the

next time it meets a point of s this

process continues indefinitely show that

you can choose a point p in s and a line

L going through P such that the

resulting windmill uses each point of s

as a pivot infinitely many times it's a

really beautiful problem and in the full

video we cover a very clever way that

you can solve it

A beautiful international math olympiad problem

The link to the full video is at the bottom of the screen. For reference, here it is: https://youtu.be/M64HUIJFTZM


inventing math is no joke and there is a

difference between being told why

something's true and actually generating

it from scratch but at all points I want

you to think to yourself if you were an

early mathematician pondering these

ideas and drawing out the right diagrams

does it feel reasonable that you could

have stumbled across these truths

yourself in this initial video I want to

show how you might stumble into the core

ideas of calculus by thinking very

deeply about one specific bit of

geometry the area of a circle maybe you

know that this is pi * radius squ but

why is there a nice way to think about

where this formula comes from well

contemplating this problem and leaving

yourself open to exploring the

interesting thoughts that come about can

actually lead you to a glimpse of three

big ideas in calculus integrals

derivatives and the fact that they're

opposites

Essence of Calculus by 3Blue1Brown #math #calculus #nerd #computerscience


imagine you're at a restaurant you order

your favorite dish and in return you're

given a buzzer that will ring when your

order is ready this buzzer is like a

JavaScript promise it represents a

future value in this case your delicious

dish just like the buzzer can either

alert you that your meal is ready or

something went wrong maybe they're out

of your favorite dish a promise in

JavaScript can either be fulfilled or

rejected what's a promise A promise is

an object representing the eventual

completion or failure of an asynchronous

oper operation it's a way to handle

asynchronous operations in JavaScript

allowing you to write cleaner more

readable code a promise has three states

pending the initial State the operation

is still ongoing fulfilled the operation

completed successfully and the promise

now has a resolved value rejected the

operation failed and the promise has a

reason for the failure if you found this

content helpful don't forget to

subscribe to my channel

Promises in JavaScript #3blue1brown #javascript #learningcoding #animation #asynchronousjavascript

Dive into the world of JavaScript Promises through an engaging and relatable comparison to a restaurant buzzer system. This short video takes you on a journey from the suspense of waiting for your favorite dish, to understanding how Promises work in programming. Whether you're a coding enthusiast eager to tidy up your asynchronous operations, or just curious about the magic behind code execution, this analogy will illuminate the concept of JavaScript Promises in a fun and accessible way. Discover the states of a Promise, why they're a game-changer compared to traditional callback methods, and how to create your own Promise to manage asynchronous tasks more efficiently.

#javascript  #programmingtips  #techtutorial  #coders  #async  #promises  #codingforbeginners  #webdevelopment  #viral #technology


I'd like to show you a simple argument

for the fact that Pi is equal to 4 we

start off with a circle say with radius

1 and we ask how can we figure out its

circumference after all Pi is by

definition the ratio of this

circumference to the diameter of the

circle we start off by drawing the

square whose side lengths are all

tangent to that Circle it's not too hard

to see that the perimeter of this square

is eight then the argument proceeds by

producing a sequence of Curves all of

whom also have this perimeter of eight

but which more and more closely

approximate the circle and there are all

sorts of examples throughout calculus

when we talk about approximating one

thing we want to know as a limit of a

bunch of other things that are easier to

understand so the question at the heart

here is why exactly is it not okay to do

that in this

example

The limit of limiting arguments

A link to the full video is at the bottom of the screen.
Or, for reference: https://youtu.be/VYQVlVoWoPY

That video gives multiple examples of lying with visual proofs

Editing from the original video into this short by Dawid Kołodziej


The surface area of a sphere is 4π times its radius squared,

exactly 4 times the area of a circle with the same radius.

But why?

Archimedes had a very beautiful proof that this surface area is the same as the area of

a cylinder that encloses that sphere, if you disregard the circular caps of that cylinder.

The idea is that if you shine some light from the z-axis perpendicular to that axis,

then if you compare the area of a small rectangle drawn on that sphere to the area

of the shadow cast on that cylinder, those two areas turn out to be the same.

This isn't obvious, but when you work it out, that shadow is a copy of the original

little rectangle, but squished down in one direction and stretched out in another.

When you analyze the relevant geometry, the two effects actually cancel out perfectly.

Treating the sphere as a sum of all of these little patches,

and the cylinder as a sum of all of those shadows,

we can infer that the surface area of the sphere is the same as the area of that cylinder.

Now the cylinder is something you can unwrap into a rectangle,

where one side length corresponds to the circumference, 2 pi r,

and the other corresponds to the height of the sphere, 2 times its radius.

This gives us the area we want, 4 pi r squared.

And if you're also familiar with the trick of showing a circle's area,

by unwrapping it into a triangle, one whose height is r and whose base is 2 pi times r,

you can see how four of those unwrapped circles fit perfectly into this unwrapped shape.

Sphere surface area proof sketch

Full video: https://youtu.be/GNcFjFmqEc8


है आई एम रुद्रांश इन टुडेस वीडियो वे बिल

लुक एट एन ग्रेट अपरोक्ष और मठ ऑफ

थर्मोडायनेमिक्स

लेट स्टार्ट की ए फैक्टोरियल

[संगीत]

थर्मोडायनेमिक्स और स्टैटिसटिक्स

मैकेनिक्स

बट थे फैक्टोरियल का 70 इस इन डी ऑर्डर ऑफ

गूगल आईटी इसे वेरी डिफिकल्ट तू कंप्यूटर

[संगीत]

[संगीत]

[संगीत]

आफ्टर सम सिंपलीफिकेशन

और सब्सीट्यूशन

[संगीत]

[संगीत]

वे कैन कंपेयर डिसइंटीग्रेशन विदर्भ रोशन

इंटीग्रेशन

[संगीत]

और आफ्टर सम मोर सिंपलीफिकेशन वे विल गेट

एन फैक्टोरियल =

[संगीत]

नॉन स्टर्लिंग अपरोक्ष

यू कैन सी डी ग्राफ ऑन डी स्क्रीन दत पर

लार्जर वैल्यू ऑफ एक्स आर अपरोक्ष इस

गेटिंग मोर एक्यूरेट

आईएफ यू आर इंटरेस्ट रेट इन दिस टॉपिक यू

कैन अलसो वॉच सम वीडियो डटेस्टेड और

डिस्क्रिप्शन और रीड एन आर्टिकल फ्रॉम

वुल्फ फ्रॉम मठ वर्ल्ड

Factorials and Physics : Revving up Thermodynamics with Stirling's Approximation @3blue1brown

In this video , you will learn about Stirling's relation and math of thermodynamics. suggest me some more topics and improvement in future.
Music by Vince Rubinetti:
Dr. Trefor Bazett
 Flammable math
Michel van Biezen
blackpenredpen

Wolfram Mathworld
@DrTrefor
 @papaflammy
@MichelvanBiezen
 @blackpenredpen
 @3blue1brown

math
physics
math and physics
thermodynamics
statistical mechanics
video
permutation
combination
csir net
prmo
maths
factorials

Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "fair use" for purposes such as criticism, commenting, news reporting, teaching, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, educational or personal use tips the balance in favor of fair use.


در مربع ب زل ۴ دو کمان یکی به مرکز رسی

از مربع و دیگری به مرکز وسط یک زل مربع

رس شده‌اند سطح محصول اس کدام است از

مرکزهای نقطه برخورد دو کمان پار خط‌هایی

رسم می‌کنیم این پار خط‌ها شعاع‌های مربوط

به کمانهای رسم شده‌اند پس طول یکی از

آنها چار و طول دیگری برابر دو است حالا

اگر دو مرکز را به هم وصل کنیم دو مثلث

قائم و زاوی هم نهشت حاصل می‌شود در شکل

روبهرو ناحیه s فصل مشترک دو قطاع رس شده

می‌باشد پس مساحت مثلث‌ها با مجموع مساحت

های دو قطاع منهای مساحت ناحیه محصول s

برابر است اگر زاویه کوچیک‌تر مثلث را

آلفا در نظر بگیریم این زاویه ۲ آلفا و

این زاویه پی منهای دو آلفا است مساحت

مثلث‌ها و قطعها با توجه معلوم بودن اجزای

آن‌ها قابل محاسبه

اند با توجه به اینکه آلفا همان تانژانت

اینورس ۱ دوم است اس به دست آمده و به طور

تقریبی برابر این عدد خواهد شد

AreaBetweenTwoArcs #maths #rectangle #3blue1brown#olympiad


sometimes it feels like the universe is

just messing with you I have up on

screen here a sequence of computations

and what I want you to notice is how the

sequence follows a very predictable if

random seeming pattern and how each

computation happens to equal Pi you

might think that this was a pattern that

would go on forever but it doesn't at

some point it stops and instead of

equaling Pi you get a value which is

just barely barely less than

pi and before you go thinking this is

the result of some numerical error maybe

because we're doing something with

floating Point arithmetic if you work

this out more precisely here is the

exact value of that last integral which

is a certain fraction of Pi where the

numerator and the denominator are absurd

they're both around 400 billion billion

billion

These integrals all equal π, until...

A link to the full video is at the bottom of the screen.
Or, for reference: https://youtu.be/851U557j6HE
These are known as Borwein integrals

Editing from long-form to short by Dawid Kołodziej


first we need to figure out where ihat

goes after applying M1 the new

coordinates of ihat by definition are

given by that First Column of M1 namely

1 one to see what happens after applying

M2 multiply The Matrix for M2 by that

Vector 1

one working it out the way that I

described last video you'll get the

vector

21 this will be the first column of the

composition

Matrix likewise to follow J hat the

second column of M1 tells us that it

first lands on

-20 then when we apply M2 to that

Vector you can work out the Matrix

Vector product to get 0 -2 which becomes

the second column of our composition

Matrix

Linear Algebra for #Machinelearning #mathematicsformachinelearning Credit:@3Blue1Brown


given a convex polygon how do you find

the area of it

let be zero be some fixed point we can

now divide the polygon into triangles

the area of each triangle can be found

by the length of the cross product of

these two vectors

you just sum up the area for all of the

triangles this algorithm Works in all

fan where n is the number of points of

the poly

Area of a Convex Polygon #3blue1brown #animations


good day everyone you're welcome to my

YouTube channel today we are giving that

X Plus Y is equal to 4 and X Y is equal

to one all over two then we're going to

evaluate X raised power 3 plus y raised

to power 3. so without finding the value

of X and Y so how do we do this we're

going to use the identity that a raised

to square 3

plus b square 3 is equal to

a plus b raised to square 3 minus 3 a b

Open brackets a plus b so now using this

identity here

we have

X respiratory plus y raise power 3 equal

to

X Plus y raised by 3 minus 3 times x y

then times

X Plus y so this is equal to X Plus y

from the given equation is four so for

raise power 3 minus 3 times x y is one

all over 2 times X Plus Y is what four

so here two one you're here two so we

have this to be

4X battery is 64 minus 3 times 2 and

this is equal to 64 minus 6 and this is

giving us what

58 and this is the right answer to the

question thank you for viewing and make

sure you subscribe to my YouTube channel

for more interesting videos

Algebra.


hello good viers welcome back once again

we're going to solve for the value of x

in the following nice looking

exponential equation 9^ x + 15^ x this

is equal to 25 to ^

x now let's go ahead and divide both

sides each St from both sides by 9 to^

of

X and again we record this property a a^

m/ B of M this is same as a / B all the^

of M but for here we have just one plus

here we have 15 / 9^ X this is will be

equal to 25 ra the power sorry 25 9 then

to the power

x so from here we have 1 plus if you

divide if you reduce this to its slope

term we just have 5 over 3 to the^ X

this is equal to 25 / 9 is same as 5 /

3 2 to the^ of X let us recall that M

power of n z we can interchange the

powers as M of Z of n so here we have 1

+ 5 / 3 to the^ X this is equal to

5/ 3 to^ X2 so make a substition we let

5/ 3 to the power of X = Y from the

right hand side you have y^ 2 minus here

y - 1 this is equal to zero now for real

value of x here y needs to be greater

than zero right so which actually mean

that we're not interested in the

negative value of y here but the only

Val positive value of y is 1 +un 5 all/

2 using the general quadratic formula

right if Y is negative here x will have

a complex solution if Y is equal to 0

then there will be no solution but when

Y is greater than Z we will have a r

solution so now from there 5 3^ x is = y

y is 1 + root

5/ 2 let us recall that the

logm of m to the base of M isal to 1

right and again the logarithm of M of B

to the power sorry to the base m is

equal to B so from here we can log both

sides but the logi must be to the base

5/ 3 so here we have the

logarithm of 5/ 3 to the power of x to

the base of 5 / 3 this is equal to the

logarithm of 1 +un 5 / 2 to the base of

5 / 3 so using this property we have

your X here this is equal to the

logarithm of 1 +un 5 / 2 to the base of

5 / 3 and this is the solution which is

approximately equal to

0.94 2

785 and this is the solution thank you

for watching please kindly subscribe to

my YouTube channel also like comment and

share bye-bye

Olympiad Math | Interesting exponential equation | #maths #olympiad


let's solve the

following radical equation we have theun

of X+ the otk of X this is equal to X so

here we're going to be solving for the

values of X so let's get started so from

the left hand side we have theare root

of x plus the < TK of X this is equal to

X so we know the otk of X+ the root of x

from

here this is equal to 2 * the squ of X

then this is equal to X now the next

thing to do let's Square both sides of

the equation right so we have 2un

x^ 2 this will be equal to x^ 2 now

let's use this identity we know that a *

m the^ b is = a the^ of B * m to power

B so from here this will be 2^ 2 * < TK

of x x to^ 2 then this is equal to x^

2 now we can change the sides of this

equation so let's make the right hand

side the left hand side so we get x2 is

equal to here 2 2 is 4 multip by square

we cancel square root here and we have

just X now there's a mistake students

always make when they arrive at this

point most students always divide both

sides by X thereby getting X is just

equal to 4 right but if you do so you

will miss a solution so the best way to

solve this problem from here is to

subtract 4X from both

sides so subtract 4X from both sides so

from the left hand side we have x^2 - 4x

this is equal to 4 x - 4x is 0 so we can

see that from the left hand side now we

have something common which is X so

let's Factor it out so we have x - 4

left then this is equal to Z so this

implies from here that X is equal to 0

or x - 4 is = to0 so from here we have X

is = 4 so you can see that we have this

solution of x is equal to0 which also

satisfy the equation thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting radical equation | Can you solve ?


hello welcome back once again today

we're going to solve this interesting

exponential equation with cube root of x

as power we have 2 to the^ cube root of

x this equals 2 ^ of x / 2 - 12 and here

we're going to solve for x values right

so let us start with a substitution

first so we're going to

let the cube root of x

M and x / 2 = N

right so from the equation we're going

to have 2 to the power of M will be

equal to 2 ^ of N -

12 so from here we're going to write

this as 2 ^ nus 2 the ^ M this is equal

to

12 so from here this 12 right over here

we're going to write it as this frence

of two numbers right so here we have 2^

of n minus 2^ M this will be equal to 12

is the same as 16 - 4 right okay I know

that this method may not be

mathematically always true right but I

think this is something really nice and

something we can can easily memorize

right so if we can have the difference

of two numbers in this form and here we

have basis of two then we can either try

relating this with some powers of two as

follow here we have 2^ of N - 2^ n and

this is equal to 2 to the^ of 4 right

then - 2 to the^ 2 right so if you have

something like this it is awesome right

because we can easily State now that n

is = 4 and the m is equal

2 remember we're solving for natural

numbers right for x m and n right so

here we found that n is equal 4 and M is

equal to 2 now let us recall again that

the cube root of

x is equal to M and the x / 2 is = n

now when m is equal to 2 so we get the

cube root of

x is = 2 when n is equal 4 we get x / 2

is = 4 now here CU both sides we get x =

2 cubed which is 8 and from here cross

multiply we also get that xal 8 and

therefore in the above solution we can

see that X is = to 8 is the only integer

solution thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation | Can you solve ?


hello good day everyone welcome back

once again to my YouTube Channel please

kindly support this channel by

subscribing to this

Channel please subscribe to this Channel

and share my videos today we have this

continued fraction here Z is = to 1 + 2

all over 1 + 2 all over 1 + 2 and so on

we are asked to find the exact value of

this continued fraction right okay let's

get started so here we have Z = 1 + 2 1

+ 2 over 1 + 2 and so on so if we

carefully look at this continued

fraction here you can see here we have Z

= 1 + 2 over and right over here we can

see here 1 + 2 over 1 + two so we can

actually

call

this Z right because it is still

repeating and be going we can see Z = 1

+ 2 all over right over here we have Z =

1 + 2 then we have another Z here which

goes on forever right so now we derive

this

equation Z is = to 1 + 2 all over Z

right so now the next idea here is to

multiply both sides of this equation by

Z right so here we have to multiply by z

z * and here we have to multiply by Z

Open Bracket 1 + 2 all over Z right so

here Z * Z is Z 2 is equal to Z * 1 is Z

Plus Z * 2 all over Z you know Z will

cancel Z we're left to it two so now let

us subtract Z + 2 from both sides so

here we get z^

2 - Z - 2 is equal to 0 right so this

quadratic expression at the left hand

side of this equation is factorizable so

if you factorize this this is going to

give us Z + 1 and what Z - 2 this is

equal Z

so then by solving here we have Z + 1 =

0 or Z - 2 = 0 you know here Z is = -1

or Z is equal to 2 but we need to know

that this continued fraction yelds a

positive value and not a negative value

so this is disregarded right so the

right answer is z equal to 2 so we can

actually say that this continued

fraction is exactly equal to two and

this is the right answer thank you for

watching do wait to subscribe to my

YouTube channel and share my videos

Continued Fraction Problem. From Chidex Hosea.  Very interesting problem


hello good viewers welcome back once

again let's find solution to the

following given exponential equation 3^

of root of x this is equal to 9 ^

x we have here 3^ of root X this is

equal to 9 is the same as 3 S right then

ra of X now let us record this

property a to the power of B then in

bracket of C this is same as a^ of B * C

so from here we have 3^ ofun X this is

equal to 3 ^ of 2 x right now here we

have equal bases from both sides so we

can actually equate the powers right so

here we have root of x which is equal to

2 x now let's go ahead and square both

sides so we have root of

X2 this is equal to 2x^ 2 so from here

this root and the square we cancel out

we have here x is = to 2x2 will give us

x^2 now so many students makes mistake

dividing both side by X if you divide

both side by X from here you're going to

lose a solution which is X = 0 so

whenever you encounter this kind of

equation all you have to do is to

rewrite it as 4x^2 - x then = 0 from

here from here we can factor out X from

the left hand side we have X into

bracket 4x - 1 right then this is equal

to Z so from here we have X is equal to

0 or then in the bracket there which is

4x - 1 is = 0 so from here add 1 to both

side we get 4X is = 1 divide both side

by 4 we get X is = 1 / 4 so from here we

have the following solution X is equal

to0 and also we have another solution X

is = 1 / 4 thank you for watching please

kindly subscribe to my channel turn on

the notification Bell so that you be

notified anytime I upload new videos

bye-bye

Interesting exponential equation | step by step solution | #maths #mathematics


hello welcome back once again today

we're going to solve this exponential

equation from the left here in the

bracket we have root3 which is same as 3

to the^ 1 / 2

then of X this is equal to here 27 is

the same as 3 to the^ 3 then here we

have power ofare root of

x now

recall a to the power of B in bra

C this is equal to a to the power B * C

so from the left here we get 3 power of

1 / 2 * X the same as x / 2 then this is

equal to here 3 power of 3 * root X

which is same as 3un

X from here we have the same basis right

so we're going to equate the powers so

this implies that x / 2 = + 3un

x so from here this is over one so let's

go ahead and cross multiply so here x *

1 here will give us X this is equal to

here 2 * 3un X will give us 6 < TK

X okay so from here let's Square both

sides to get rid of this root sign so

here this is squared and from the right

here the whole of this is squ so X2 is

X2 right then this is equal to here we

have 6 * x 2 first of all 6 2 will give

us 36 and here root X oh root X2 I mean

Square we cancel the roots so we have

just

X now let's subtract 36 from both sides

that is x² - 36 x this is equal to Z

factor out X so X into bracket here x² /

X will give us xus 36 x / X will give us

36 and this is equal to Z so from here

this implies that X is equal to Z or x -

36 = 0 of which we solve to get x = 36

and here we have the solution and also

here we have a solution so we only have

two R Solutions x = 0 and x =

36 thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Exponential equation | Can you solve ?


hello welcome back once again today let

us solve this equation and this is

indeed a very nice equation we have the

root of 2^ x / 2 = the cube root of 4^

of X and here we're going to solve for x

values okay now let's get

started from the left right over here

let us take note of this that the end

root of M is same as M to the^ of 1 / n

so from the left right over here that

will become 2 to the^ of x / 2 then with

that square root that will be to the^ 1

/ 2 and this is equal to from here that

will be 4^ X then the cube root will

become to the^ of 1 / 3 let us take note

of this when we have a to the power of B

to the power of C this is same as a to

the power of B * C right

so from here we're going to have 2 ^ x /

2 * 1 / 2 will give us x 4 then this

will be equal to here 4^ of x * 1 / 3

will give us x /

3 okay now remember that this four right

over here is the same as 2 s right so

here we have 2 to the power of x / 4

this is equal to 2^ 2 then to the power

of x ID 3 so from here we get 2^ x / 4

this is equal to 2^ of 2 * x / 3 will

give us 2x / 3 so from here we can see

that we have the same bases right so

we're going to equate the

powers and that will imply

that x / 4 that is by equating the

powers will be equal to 2X

divided

by 3

right

okay so from here let us cross multiply

so 3 * X will give us

3x and this is equal to 4

* 2x will give us 8 x right okay so from

here many may think that there's no

solution but obviously there solution do

not make the mistake of canceling X from

both sides because that will result to

no solution to the question but the best

thing we can do is either we say 8 x -

3x is = 0 right so that we get here

5x is equal to Z then dividing both side

by 5 we arrive at X is equal to Z and

obviously Z is the only solution to this

exponential equation thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye


hello welcome back once again today we

have another interesting equation you

know why this equation is interesting

we're going to solve for both real and

complex values right so let's get

started so from the left hand side we

have x * root X and this is equal to 8

now let's Square both sides so we get x

* root X squ this will be equal to 8 squ

now recall when we have 8 is B all to

the^ n this is same as a^ NTI b^ n so

here we have X to the^ 2 * otk x^ 2

which is X then this is equal to 82 is

64 so from here this is X cubed - 64

this is equal to

0 from here x cubed - 64 is 4 Cub this

is = to 0 Let's make use of the

difference of two cubes formula remember

a cubus b Cub this is equal to a - b

into braet a 2 + a b + b

2 so from here this is going to be x -

4 into braet

X2 + 2 I mean that is 4 * X yes so 4 x

this is a here here then plus 4 2 which

is 16 then this is equal to 0 right okay

so from here we have these two products

which is equal to Z so we have either

this is equal to Z or this is equal to Z

so this implies from here x - 4 is = 0

of which X is = 4 so this is the only

real solution and here we have X2 + 4x +

16 this is equal to 0 we're going to

solve this using the general Quadra

quadratic formula so we get xal to will

be 4 plus or minus otk of b 2 is 4 2

which is 16 - 4 * a * C 4 * 16 is 64

then the whole of this / 2 * a which is

2 so from here we get X is = -4 +

orus < TK of -48

then / 2 from here we get X is =

to -4+ orus of-1 is

Iota then we know that 48 here is 16 * 3

so of 16 is 4 then we have

root3 the all of this divid two which

when you split this denominator 4/ 2 is

-2 plus or minus 4 I3 / 2 is 2 I <

tk3 so here we come up with the solution

so we have the only real solution at X

is equal to 4 and we have two complex

Solutions -2 plus or minus 2 I root of 3

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye


hello good viewers welcome back once

again let's find a solution to this

given equation we have here 4 xot of 27

this is equal to

4X now the left hand side of this

equation we will write it make use of

this property we know the end root of M

this is equal to m^ of 1 / n right now

from the left we have 27 to the^ of 1 /

4X then from the right hand side this is

equal to

4X now to get rid of this power here 1

over 4X we shall raise both sides of

this equation to the power of 4X right

so from here we have 27 ^ of 1 / 4X in

bracket rais power of 4X

this is equal to from the right hand

side 4X to the power of

4X now from the left hand side let us

recall another property here when we

have a to the power of B in bracket of C

this is same as a^ B * C now from here

when we multiply these Powers together 1

/ 4X * 4X that will give us 1 and here

we left with 27 right to the^ of 1 and

from the right hand side we have 4X ra

power 4 x now let us recall 27 is the

same as 3 to the^ of 3 then this is

equal to

4X power of 4X now from here this

implies that 4X is equal to 3 right from

the base we have 4X and from the base

the left hand side we have three three

right and the power here corres

responding to what we have as the basis

4X which is equal to 3 right so from

here soling for X we divide both sides

by

four and from here we come up with the

solution that X is = to 3/ 4 and this is

the right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Japanese | Interesting exponential equation | #maths #mathematics


hello welcome back once again today

we're going to evaluate 18 /un x given

that 9^ ofun X is equal to root3 now

let's get started from the left hand

side this base right over here which is

9 we can write it as 3 2 so here we have

3 S then ra to the power of root X then

this is equal to now theare root of 3 is

the same as 3 to the^ of 1 all over

2 okay now from here let us take note of

this when we have a to the^ B then to

the power C this is equal to A to the^ B

* C so from here we have 3 to the^ 2 *

root X which will give us 2un X then

this is equal to 3 to the^ of 1 /

2 okay now from here we have equal bases

right so we can actually equate the

powers so here we get 2un X is = 1 / 2

now let's go ahead and divide both side

by

two and by two the reason for this is to

solve for root

X after getting the value for root X so

we can easily divide 18 by the value for

root X and there we have our answer

right so here this cancels this so here

we have root X is equal to 1 / 2 2 / 2

will give us 1 over 4 here we have the

value for root X and therefore we can go

ahead and evaluate this expression 18 /

otk X which should be equal to 18 / 1/ 4

which we can write as 18 / 1 / 1 4 which

is equal to 18 / 1 if we change this

multiplication then we take the

reciprocal of 1/ 4 which is 4/ 1 and

from here we get 18 * 4 which is 72 and

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Interesting algebra problem