help-49

[wild glen]

Ohhh that makes sense

I can't really draw it very well but I think I can visualize it. I don't know why I wasn't thinking of it as starting at r(0,0)

Thanks for explaning

[pearl idol]

No problem! I think for multivariable calculus you should practice sketching things though

Otherwise it can get very confusing

Another way to think of this is via the right-hand rule for cross products, if you've learned about that

[wild glen]

Probably true

[wild glen]

Yes? Basically that the cross product is perpendicular?

And direction depends on the direction of input vectors ofc

How does it apply here?

[pearl idol]

Remember you calculate n by taking the cross product of dr/du and dr/dv (and then normalizing)

dr/du is the direction that you go on the surface when you increase u

In this example, that's upwards

Similarly, dr/dv goes counterclockwise around the cylinder

Using the right-hand rule, you can then figure out that the cross product points inwards

[wild glen]

Oh true, that makes sense

Thanks

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[paper ruin]

is the remainder -6 or -6/(x-4)?

[midnight plank]

@paper ruin Has your question been resolved?

[zenith cradle]

If you mean as in the division $\frac{2x^3-5x^2-16x+10}{x-4}$, then, by definition, $-6$ is the remainder

[grand pond]

Crystopher

[paper ruin]

I see

what did u do to go back to the expression?

[zenith cradle]

You mean find out the dividend?

[paper ruin]

yes

[zenith cradle]

Reverse-engineering. If that was the result of the division, then I can observe that -6 is a remainder and x-4 is the divisor. So I just multiply the result by x-4 and I should get the original, simplified rational expression.

[paper ruin]

tysm

is "reverse-engineering"

the name

[zenith cradle]

That's how I would call it, if it has some other formal name I don't know it. I just asked myself: what are some possible parts of the division I can see and what could the remaining parts be?

[paper ruin]

I see, thanks again this was very helpful

[paper ruin]

especially this

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[last slate]

I tried to reopen the last channel but I couldnt write stuff there anymore
Im trying to understand Vietas formula and I dont understand this part.
why is subtracting the roots from x and multiplying everything the same as ax^2+bx+c

I couldnt find anything usefull in my 15min of browsing the internet so maybe someone knows something I can look into.

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[sand bluff]

so this is kind of like "factored form"

all polynomials can be written in factored form as a(x-r1)(x-r2)...

[last slate]

thats the part I dont understand.
Ill try searching for factored form now
thanks

.close

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[final lagoon]

https://www.desmos.com/calculator/wvtx0ldbnw?lang=pl

Would it be possible to convert this equation to a function?

c variable is given in radians

I want this to be converted for something to be used in blender's nodes, and frankly I can't achieve 2 sided equation like this

Can it be simplified to a simple:

f(x) = (whatever would be written here)

If not I'll figure something else out, thank you in advance

[fresh sparrow]

do you need a semicircle?

sqrt(1-x^2)

[final lagoon]

Well basically it is a bit more complex than that

[grand latch]

It is already a function

[final lagoon]

well I know, basically I need a graph is a part of a circle, so I take a cutout of 45 degrees and turn it into a graph

lemme illustrate it

[fresh sparrow]

$\frac{c}{\pi}\cdot\sqrt{1-(x-1)^2}$

does the same thing

[final lagoon]

oh wow

well damn

lemme see if it works

[fresh sparrow]

sorry, not 0.5 but 1

[grand pond]

artemetra

[fresh sparrow]

fixed

[final lagoon]

no it doesnt work

it scales it

basically it is a cutout of a circle

[grand latch]

Doesn't seem like a circle to me

c=1

[final lagoon]

thus why this equation was so hard to achieve

and I had to ask for help here

[fresh sparrow]

it's stretched out

[final lagoon]

but the problem is that 2 functions as an equatation is not really possible in blender so I wanted to put it into a single function

so without the = in the middle

I am a math layman for apologies for ignorance

[fresh sparrow]

can't you take the 0.5*sqrt(csc^2... to the other side?

[final lagoon]

hmm I tried

[fresh sparrow]

what has failed?

[grand latch]

Now i've kept the ratio 1:1

[final lagoon]

I don't know

[fresh sparrow]

it's a semicircle at c=pi btw

[final lagoon]

oh shit wait

[visual tiger]

equation of an upper semicircle of radius r centered around (0,0) is y = sqrt(r^2 - x^2). Yours is centered around the x = 1 line so y = sqrt(r^2 - (x-1)^2) is the first step

[final lagoon]

ah no nevermind

[visual tiger]

then

[final lagoon]

yeah?

[visual tiger]

the coordinates of the upper point seem to be c = 1/r

so let's switch things up a bit

and write y = sqrt(1/c^2 - (x-1)^2) - c

[grand pond]

jesko

[final lagoon]

huh

[visual tiger]

I'm trying to tie things together

[final lagoon]

actually center changes

lemme show

oh okay I get what you meant

[grand pond]

jesko

[grand latch]

What I'm trying to say is, this is not a semi-circle

[visual tiger]

ok I got it backwards

so let's focus on how to get equation of a circle that goes through (0,0) and (1,0)

we know it should look like (x-0.5)^2 + (y-b)^2 = r^2

plug (0,0) in and you get 1/4 + b^2 = r^2
plug (1,0) in and you get 1/4 + b^2 = r^2

[grand latch]

A circle centred at (0.5,b) of radius r

[visual tiger]

yes exactly

[final lagoon]

hm

[visual tiger]

we know b^2 - (b-1)^2 = 0

[grand pond]

jesko

[visual tiger]

but this is strange

let me mull that over

[grand pond]

jesko

[visual tiger]

I find it hard to believe the center of the circle is always (1/2,1/2)

Oh I know what I did wrong

[grand latch]

b is changing isn't it?

the centre moves on the line x=1/2

[visual tiger]

edited this

we only have one equation, 1/4 + b^2 = r^2

so

upper half :

(y-b)^2 = r^2 - (x-1/2)^2

y = b + sqrt(r^2 - (x-1/2)^2)

since b is negative

y = sqrt(r^2 - (x-1/2)^2) - sqrt(r^2-1/4)

there you go

(r >= 1/2)

[final lagoon]

so wait you are trying to make a function based on the fact that you just change location and radius of that circle because it seems to just go from 0 to infinite?

or wait no

apologies I tried getting it from many angles for like few hours and I may be getting dumber with every passing minute

ahaha thank you regardless

[grand pond]

jesko

[final lagoon]

oh wow

o it is radius based?

honestly amazing, now just to figure a way to somehow put in radians as input instead of radius hmm

but this is a step in the right direction

[grand latch]

by restricting the domain we can make it a function

[visual tiger]

wait r = 1/(2sin(c/2))

[grand pond]

jesko

[grand latch]

Don't you get the same thing

[grand latch]

refer

[final lagoon]

degrees or radians all the same

[visual tiger]

say you have angle alpha, radius r

then

[final lagoon]

yup range 0 to 180

[visual tiger]

we know the center lies at (0.5,b) where b = -sqrt(r^2-1/4)

we found that previously

so

notice

that if we let O that center

A = (0.5,0)

B = (1,0)

OAB

is just a rectangle triangle

with sides OA unneeded, OB = r and AB = 1/2

but then

alpha/2

is exactly angle AOB

how do we get it?

[final lagoon]

ohh okay I finally got what you all were talking about

okay honestly the remaining solution is simple enough that even I can solve it

thank you this was a coherent introduction

Okay yep successfuly implemented! Thank you @visual tiger for explaining this by steps because I actually understood all of this! bless you

[midnight plank]

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[grave locust]

hello

Im not sure how to start

idk how to do this

[last slate]

,start

or smthn

[sage olive]

elementary matrices correspond to row operations

remember doing row reduction / gaussian elimination?

what elementary row operation was applied to the matrix A to get the matrix B? your three row operations are scaling a row, swapping two rows, and adding a scalar multiple of one row to another row

[grave locust]

oh ok

[sage olive]

to find the elementary matrix corresponding to a row operation, start with the appropriate sized identity matrix (in your case, the 2x2 identity matrix), and apply that same row operation to the identity matrix

[grave locust]

they swapped row 1 and row 2

[sage olive]

the resulting matrix is your elementary matrix E

[grave locust]

so i just swap row 1 and row 2 of an identity matrix order 2?

[sage olive]

yes

[grave locust]

ah ok

thank you very much

[sage olive]

happy to help

you can double check by multiplying it out to see that it works

[grave locust]

ok that makes sense

.close

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[untold lagoon]

so, i have an exam in roughly 8h and i have lots of studying and understanding to do. first things first, i need to more or less understand about half of what's here. i know some are related to the order of numbers but thats really it the rest im really lost

[rose trout]

So I suppose you at least know about complex numbers to some extent?

[untold lagoon]

no

like

0

im a complete newbie

[rose trout]

I think your best bet is going through like Khan academy or something

It's a lot of explaining to go from i = sqrt(-1) to complex roots and all

[untold lagoon]

man

ok ill try

ty

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[last slate]

About proving there is no rational solution for x^2 +10x -3
Without rational root theorem, or quadratic formula or completing the square

About properties of numbers

About number theory?

About geometry?

About graphing relations

[sage olive]

you've listed like five things here, is the actual problem showing that there are no rational solutions to x^2 + 10x - 3 = 0?

[last slate]

It's deeper than that. I have so many questions

[sage olive]

hm well i probably don't have the time to answer all of them but to answer your question about rational solutions, it has to do with rational root theorem

[last slate]

I want to prove without rational root theorem or completing the square or using sqrt(b^2-4ac)

I want to prove using another method

[sage olive]

do you have a method in mind?

some problems are just really difficult to show without using particular theorems

[last slate]

Im not really sure. Allow me to type.

I've tried using properties of numbers

Like k^2+k is always positive and even

And 2(anything) is even

But it doesn't help because the other side of the equation is also usually always even

Many times I've gotten to a point like this where I need to show that an expression can't be odd

If I could prove that that expression can't be an odd integer, then I could prove there is no solution for k,p and hence there is no rational solution

Btw k,p are integers

And k=/=p

Can't equal 1 or -1

.

Why does graphing (3a/b)-(b/a)=10 and (10a/b)[(3a/b)-10]=10 show the same thing but graphing (3a/b)-(b/a)=(10a/b)[(3a/b)-10] shows something different

How can we show that 10(a/b)[(3a/b)-10]=(3a/b)-(b/a) has no integer solutions?

<@&286206848099549185> :-(

[midnight plank]

@last slate Has your question been resolved?

[last slate]

no =(

ive been working on this problem for over 30 hours

[last slate]

someone....? =(

[quartz otter]

hello

i have an idea

let's take the inverse of y = x²+10x-3

=> y = x²+10x+25-25-3 = (x+5)²-28

=> y = (x+5)²-28
=> y+28 = (x+5)²
=> √(y+28)-5 = x
and
=> -5-√(y+28) = x

well, let's take the derivative when y = 0,

derivative of -5+√(y+28) = 1/2√(y+28)

derivative of -5+√(y+28) when y is 0 = 1/2√28

and the derivative of -√(y+28)-5 is -1/2√(y+28), when y = 0, it is -1/2√28

[midnight plank]

@last slate Has your question been resolved?

[quartz otter]

but these are not the values of dy/dx's

just dx/dy's

so we need to upside down

to find the dy/dx's

which are

2√28 and -2√28

so we will take the derivative of x²+10x-3 which is 2x+10

and we will just solve the equations 2x+10 = 2√28 and 2x+10 = -2√28

2x+10 = 2√28

2x = 2√28-10

x = √28-5

2x+10 = -2√28

2x = -2√28-10

x = -√28-5

let's plug in and see if these are correct

(√28-5)²+10(√28-5)-3 = 28-10√28+25+10√28-50-3 = 28+25-53+10√28-10√28 = 0+0 = 0, it's correct

(-√28-5)²+10(-√28-5)-3 = 28+25+10√28-10√28-50-3 = 53-53+10√28-10√28 = 0+0 = 0

it's also correct

[last slate]

wellll

i dont understand calc, also i didn't want you completing the square

y = x²+10x+25-25-3 = (x+5)²-28

=(

[quartz otter]

it's just for taking the inverse

though

you can reach the answer

directly

[last slate]

atp you could just sqrt both sides

yeah

[quartz otter]

but i didn't do

i used this for different purpose

[last slate]

rules are rules >=(

[quartz otter]

completing the square is the most natural thing in universe dude.. what kind of rules are that..

[last slate]

i know is it easy when you can complete the square im trying to understand if you are able to look at

(a/b)[(3a/b)-10]=1 and notice due to properies of numbers that there is no integer solutions

[quartz otter]

how did you get this

[last slate]

if there x^2+10x-3 is factorable by integers, then there is some f,g such that fg=-3 and f+g =10

there are no solutions

[quartz otter]

and isn't it +10 inside

3a/b+10 or am i wrong

[last slate]

if x^2+10x-3 is factorable by rational numbers then there is some (fa/b)+(gb/a)=10 and abfg/ab=-3

i set f and g to 3, -1

[quartz otter]

abfg/ab??

what was that

[last slate]

we know fg=-3

[quartz otter]

yes

[last slate]

so if i do abfg/ab i get the same answer

[quartz otter]

why abfg/ab

[last slate]

but (fa/b)+(gb/a)=10 is different from f+g=10

so that we get this equation (fa/b)+(gb/a)

[quartz otter]

i don't get it

what are a and b

where they came from

[last slate]

so

we know there are know integer solutions

for the factoring

so to consider rational set i mutiplied f and g by rational numbers

[quartz otter]

(x-f)(x-g) = x²+10x-3

[last slate]

a/b

right

exactly

[quartz otter]

no need to a/b or smth

[last slate]

its useful

[quartz otter]

why is needed

[last slate]

because we know 3*-1=-3

so if we set f and g to 3 and -1 we simplify the problem imo

(fa/b)+(gb/a)=10

[midnight plank]

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[last slate]

where f and g are 3, -1

.reopen

[midnight plank]

✅

[quartz otter]

i know 3/2*-2 is also -3

why 3*(-1)

specifically?

or not

[last slate]

this can be expressed as 3(1/2) -(2/1)

i belive it makes the problem easier to deal with

[quartz otter]

can you show me steps

[quartz otter]

.

f+g = -10

[last slate]

so, a.b.f.g are all elements of Z

[quartz otter]

or if you say (x+f)(x+g) = x²+10x-3, then f+g = 10

[quartz otter]

impossible?

[last slate]

(x+f)(x+g) = x²+10x-3, then f+g = 10 do you understand this part or need explaination

[last slate]

what is?

[quartz otter]

you just said there are no integer factors

[last slate]

yes exactly

[quartz otter]

how can f and g be integer

[last slate]

a rational number is defined as an integer/integer

[quartz otter]

i understand, no need

[last slate]

(fa)=integer

b= integer

so (fa)/(b) = rational number

[quartz otter]

is fa integer?

[last slate]

so a,b,f,g are all integers

integers are closed under mutiplication

integer*integer=integer

[quartz otter]

why f is integer

i didn't understand

why f and g are integer

[last slate]

because definition of rational number is Integer/integer

so f,g need to be integer

my rational number is composed of fa/b

[quartz otter]

they're roots

they can't be integer

[last slate]

and gb/a

what do you mean?

[quartz otter]

is there any integer like f+g = 10 and fg = -3

?

[last slate]

theres no solution

[quartz otter]

yes

[last slate]

but there are different questions

[quartz otter]

so how did you say they're integer

[last slate]

there are no integer solutions for that

but there may be integer solutions for (fa/b)+(gb/a)=10 and fg=-3

[quartz otter]

okay but aren't -f and -g are roots?

[last slate]

?

[quartz otter]

didn't you define f and g like

(x+f)(x+g) = x²+10x-3

so -f and -g are the roots here

and they can't be integer

since they can't be integer

f and g can't be integer

[last slate]

this was when i was only considering the integer set

what we are looking at now is rational numbers

[quartz otter]

olay how do you define the f, g,b and a

[last slate]

(x+(fa/b))(x+(gb/a))=x^2+10x-3

f and g are integers

but i just set them to 3 and -1

because i know 3*-1=-3

[quartz otter]

okay

fa/b+gb/a = 10

[last slate]

yes

but let's just set f and g to 3 -1

[quartz otter]

wait

can't we just say

(x-(fa/b))(x-(gb/a))

because it seems more good

[last slate]

what does that do?

[quartz otter]

i will say fa/b = -5-something and gb/a = -5+something

when added together they will be -10

because the axis of symmetry of this parabola is x = -5 actually

because now fa/b will be root itself

and gb/a will be root itself

so i can say that

fa/b = -5-smth

gb/a = -5+smth

and if f = 3

a/b = (-5-smth)/3

and if g = -1

b/a = 5-smth

idk what to do now

but i have another idea

but probably you won't use so

[last slate]

what's that

[quartz otter]

using focus point of a parabola

do you know about it?

[last slate]

no

[quartz otter]

well, there is a focus point and directrix for every parabola

directrix is something like y = c

it's a line which has 0 slope

the distance of the line segment drawn from the focus point to the parabola is equal to the distance between that point of the parabola and the directrix

that's all basically

the coordinates of focus point is (-b/2a,(4ac-b²+1)/4a) for every parabola

and the equation of directrix is y = (4ac-b²-1)/4a for every parabola

and the point on a parabola can be represented as (x,ax²+bx+c)

the distance between the point (x,ax²+bx+c) and (-b/2a, (4ac-b²+1)/4a) is equal to |ax²+bx+c-(4ac-b²-1)/4a|

√((x+b/2a)²+(ax²+bx+c-(4ac-b²+1)/4a)²) = |ax²+bx+c-(4ac-b²-1)/4a|

[last slate]

can you draw it

or show me an example

[quartz otter]

when we square both sides it will be

(x+b/2a)²+(ax²+bx+c-(4ac-b²+1)/4a)² = (ax²+bx+c-(4ac-b²-1)/4a)²

[last slate]

uh

[quartz otter]

x+b/2a = 4ax+2b/4a

[last slate]

what happened to the right side

it doent look squared

[quartz otter]

sorry

[last slate]

also

[quartz otter]

ax²+bx+c-(4ac-b²+1)/4a = (4a²x²+4abx+4ac-4ac+b²-1)/4a

[last slate]

(3a/b)-(b/a)=10 doesnt graph to a parabala

[quartz otter]

it's a graph for a parabola

[last slate]

?

[quartz otter]

yes, try it

you will see

ax²+bx+c-(4ac-b²-1)/4a = (4a²x²+4abx+4ac-4ac+b²+1)/4a

so it becomes

((4ax+2b)/4a)² + ((4a²x²+4abx+b²-1)/4a)² = ((4a²x²+4abx+b²+1)/4a)²

let's cancel the 16a²'s on the denominator

(4ax+2b)² + (4a²x²+4abx+b²-1)² = (4a²x²+4abx+b²+1)²

well i will just say u = 4a²x²+4abx+b² here

(4ax+2b)²+(u-1)² = (u+1)²

16a²x²+16abx+4b² + u²-2u+1 = u²+2u+1

=> 16a²x²+16abx+4b² = 4u

[last slate]

can you explain the graph thing

[quartz otter]

u = 4a²x²+4abx+b²

as we said

[last slate]

how does 3a/b........ show a parabla

[quartz otter]

so the equation is true

[quartz otter]

a/b = x, b/a = 1/x

3x-1/x = 10

(3x²-1)/x = 10

[last slate]

oh

[quartz otter]

3x²-10x-1 = 0

so we can just find the distance between the point where y = 0 and the focus point

and then the distance between directrix and the point where y = 0

then we can equalize them

for the x²+10x-3

[last slate]

sounds super interesting!

[quartz otter]

so

let's say the intersect point of x is

(a,0)

and the other one is (b,0)

because there are two

since it is a parabola

the focus point is

(-b/2a,(4ac-b²+1)/4a)

which is

(-5,-111/4) here

and directrix is

y = (4ac-b²-1)/4a which is

y = -113/4

[last slate]

ah

not an integer

[quartz otter]

doesnt matter lol

[last slate]

shhh

xd

[quartz otter]

okay the distance between (a,0) and y = -113/4

[last slate]

im trying to follow along

[quartz otter]

is ofc 113/4

because you will just subtract the y's

to find the perpendicular distance

that's how we find the distance between a line and a point

we find the perpendicular distance between them

and which is directly 113/4 here

because y = -113/4 is something parallel to x axis

did you get it

[quartz otter]

this represents the distance between the point on parabola and directrix

[last slate]

ok

[quartz otter]

okay now we will find the distance between focus point and the point on parabola

(-5,-111/4) and (a,0)

√(a-(-5))²+(0-(-111/4))²

= √(a+5)²+(111/4)²

and this is equal to 113/4

by the thing that we solved

[last slate]

YOURE LOSING ME

but keep going

[quartz otter]

okay..

√(a+5)²+(111/4)² = 113/4

[last slate]

hard for me to understand without a visual here

[quartz otter]

(a+5)²+(111/4)² = (113/4)²

(a+5)² = (113²-111²)/16

a+5 = √(113²-111²)/4

well it's enough to find a+5

because if a+5 is something, a is also something

[last slate]

right

[quartz otter]

in terms of set like integer rational irrational

[quartz otter]

113²-111² = (113-111)(113+111)

which is

[last slate]

√(113²-111²) is irattional im gussing

[quartz otter]

2(224)

[last slate]

which shoes a is irattion

[quartz otter]

√2(224)/4

is √2(224)/√16

and this is equal to

√28

or 2√7

which is irrational

[last slate]

is this the same a in 3a/b.....

[quartz otter]

and the other root is the other solution of a actually

[quartz otter]

nope

[quartz otter]

like here a can take 2 values

we just solved for a+5 one

it could be -a-5

and then it would be -a-5 = √28

a = -5-√28

[quartz otter]

and here the a is √28-5

you can watch a video about focus point of a parabola if you want

[last slate]

yeah yeah yeah

[quartz otter]

thus you can understand easily

[last slate]

=D

[quartz otter]

we just found the distances and equalized them, that was all

[last slate]

so, can the focus point be arbitray?

[quartz otter]

no it has a rule

btw, unfortunately i didn't see any proofs about focus point

so i wrote the proof in my notebook

not finished yet but

it helped me to understand it better

[last slate]

well

is there anything to proceed with or are we finsihed?

[quartz otter]

nothing really

we finished

[last slate]

ah

the coordinates of focus point is (-b/2a,(4ac-b²+1)/4a) for every parabola

thats the rule

ok thank you so much

.close

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[bleak pier]

Riemaan sum so how to solve it

Integration limits will be?

r/n=x

1 to 2??

Log(1+x)dx

[nova yoke]

yes

[bleak pier]

Why 1?

We divide

2n/n so 2

[nova yoke]

(n+1)/n = 1 + 1/n, so it's the right endpoint of the first interval

the first interval of the partition, namely [1, 1+1/n]

[bleak pier]

But why are we including 1?

first point is 1+1/n no?

[nova yoke]

you can use any point in the interval [1, 1+1/n]

and any point in the other intervals as well

this sum is using the right endpoints of every interval

but in any case they are a partition of [1,2]

so those are the integral bounds

[bleak pier]

I see

Tq very much

How to integrate it?

[nova yoke]

integrate log(1+x)?

just do a change of variable u = 1+x

[bleak pier]

logu du

Which is ulogu

[nova yoke]

yea

now just eval at the endpts and subtract

[bleak pier]

Opps

ulogu-u ?

[nova yoke]

oh yea, don't forget the -u

your x went from 1 to 2

so what are the bounds for u?

[bleak pier]

xlogx-x ?

U=1+x
U=2,3

[nova yoke]

yep

[bleak pier]

3log3-3-2log2+2

???

Am I on right track?

3log3-2log2-1

log9-log4-log10
log9/40?

[nova yoke]

wait where did log 10 come from?

also log 9 is wrong

[bleak pier]

opps no it should be loge

[nova yoke]

yea

[bleak pier]

3 log3-2log2-loge

[nova yoke]

and 3^3 is not 9

[bleak pier]

Log27/4e

[nova yoke]

yes

that is correct

[bleak pier]

Ahhhh yeee

Tq bungooooo

Bungoooo

[nova yoke]

sure cheers

[bleak pier]

Yeah i am happy to learn like this

I meant some extraordinary person helps on the way

[nova yoke]

ABCD

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[rotund junco]

need help with basic probability

[tulip mason]

Both parts?

[rotund junco]

yeah

for a) im pretty sure it's nPn right?

[tulip mason]

Yep

[rotund junco]

alright how do i approach b) then?

[tulip mason]

For b) we want to exclude the overcounted arrangements

[rotund junco]

idk what that means

does not distinguishable mean that order doesnt matter?

[tulip mason]

No the order still matter

But the chairs forms a circle

So you can rotate them

Not distinguishable here means if you can rotate one arrangement to get another, they are considered the same

[rotund junco]

still lost

[tulip mason]

Let's consider the case when n=3

..O
O O
We have three chairs like this

[rotund junco]

ok

[tulip mason]

The arrangments
...1
2 3
And
..2
3 1
Are considered same since they are off by a rotation

Ignore the dots btw

[rotund junco]

ok but how do i use that to solve the rest of the problem

[tulip mason]

How many times can you rotate until you get the same arrangement?

[rotund junco]

n?

[tulip mason]

Correct

So every n arrangments are treated as 1

So we divide nPn by n

[rotund junco]

ok we do that because we're trying to find distinct seating arrangements?

[tulip mason]

Yes

[rotund junco]

alright i get it now thank you

[strange jewel]

Respect to a UCLA student 🔥

[midnight plank]

@rotund junco Has your question been resolved?

[midnight plank]

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[blissful drum]

help me solv

e

[blissful drum]

integration question

[icy flower]

i gotchu

ok

[blissful drum]

using substitution

[icy flower]

first step

ok wait i dont got this

[blissful drum]

bruh

[icy flower]

no im just bad at math nvm

[hollow oyster]

Let tan2t be u

[blissful drum]

tan2t?

k

wait can you just make t x

t is confusing

so dt = du/sec^2(2x)

right

i mean

2sec^2(2x)

[hollow oyster]

Then $$\frac{du}{dt}=\sec(2t)^2\cdot 2 $$

[blissful drum]

isnt it 2sec^2(2x)?

[hollow oyster]

Yes

[blissful drum]

wait what is this

[blissful drum]

alr alr

[icy flower]

$du=\sec(2t)^2\cdot 2 $

i think latex yeah!

[blissful drum]

kk

[icy flower]

here, this is what it should look like

[hollow oyster]

Yes it is that

[icy flower]

oh you did it 😔

[blissful drum]

so do i divide

sec^2(2x)

[icy flower]

wait loge are you almost done with your course?

[blissful drum]

so it becomes u^5 * sec^3(2x) * (du/2)?

[icy flower]

like is this the second last chapter?

[blissful drum]

wdym

[icy flower]

like this calc class you're in

is it almost over?

this is like where calc one ends irrc

[blissful drum]

well im in like a accelerated summer program for IB (international baccalerate) we just started integrals today

[icy flower]

Ew IB

i mean good job

[grand pond]

convergence

[blissful drum]

ay no disrespect IB

lol

[blissful drum]

ye

then

can you help me on the next steps

[icy flower]

IB is great, IB schools just make me erm

[hollow oyster]

Sure

Wait a minute i have to do something

[blissful drum]

are u a levels?

[icy flower]

Nah CC(common core)

[blissful drum]

what is that

[icy flower]

However I did 10th boards with IGSCE

[blissful drum]

which grade u in?

[icy flower]

12th

[blissful drum]

ic

gl

[icy flower]

thanks

[blissful drum]

are you able to help w/ this question or nah

[icy flower]

I can try i think

[blissful drum]

next step?

[icy flower]

one sec

[hollow oyster]

mate you didnt do the question properly

[blissful drum]

how come

i havent finished

btw

[hollow oyster]

its $\int \tan(2t)^5\sec(2t)^2dt$

[grand pond]

convergence

[hollow oyster]

.

[blissful drum]

oh crap

mbmb

[hollow oyster]

then we use it to get

[blissful drum]

oh that makes it so much

easier

[hollow oyster]

$\frac{1}{2}\int u^5du$

[grand pond]

convergence

[hollow oyster]

now you can integrate this

[blissful drum]

yeye

wait i first make into 1/2(integral) u^6/6 right

the integral symbol stays?

[hollow oyster]

this is equal to$\frac{u^6}{12}$

[grand pond]

convergence

[hollow oyster]

which is equal to: $\frac{\tan(2t)^6}{12}+c$

[grand pond]

convergence

[hollow oyster]

do you understand now?if yes them close

[blissful drum]

so when you apply

the intergal rules

like u^6/6,

do you remove the integral symbol

[hollow oyster]

yes

[blissful drum]

alr alr

thanks man

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[twilit field]

could someone verify this ?

[twilit field]

\textbf{theorem 1.3}:- If a set $S$ satisfies the following conditions, it consists of all the positive integers
\begin{enumerate}
\item It consists of the integer 1
\item If k is a positive integer k, belonging to the set, as must $k+1$
\end{enumerate}

\begin{proof}
Let $T$ consist of the set such that it consists of any element not in $S$.

let $a$ be an arbitrary element of $T$ that doesn't belong to $S$ , and let $a-1$ belong belongs to $S$ .
this implies that $a$ belongs to $S$ ,.

QED

\end{proof}

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

\textbf{theorem 1.3}:- If a set $S$ satisfies the following conditions, it consists of all the positive integers
\begin{enumerate}
\item It consists of the integer 1
\item If k is a positive integer k, belonging to the set, as must $k+1$
\end{enumerate}

\begin{proof}
Let $T$ consist of the set such that it consists of any element not in $S$.

let $a$ be an arbitrary element of $T$ that doesn't belong to $S$ , and let $a-1$ belong belongs to $S$ .
this implies that $a$ belongs to $S$ ,.,we have thus arrived at a contradiction and must conclude any arbitrary integer $a$ belongs to S

QED

\end{proof}

sory this

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

Let T consist of the set such that it consist of any element not in S

what is this supposed to mean?

[runic hamlet]

why should such an a exist

[twilit field]

I mean any integer not in S

[runic hamlet]

write T = {....}

not everything can or should be done in words

[twilit field]

$T={a}?$

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

what even is a?

[twilit field]

an arbitrary element of T

[dreamy lichen]

kinda circular definition

[twilit field]

T is just assumed to be non-emptyy

[dreamy lichen]

T = {a}
a is arbitrary element of T

[twilit field]

ok, T is just a set consisting of those integers not in $S$

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

That's much better

[runic hamlet]

write that sentence in symbols

T={...}

[dreamy lichen]

this is also a good thing to do

[runic hamlet]

also you are imprecise. do you mean integers or positive integers

[twilit field]

positive integers

[twilit field]

you mean in set builder notation?

$T={x|x \notin S, x\in \mathbb{Z^+}}$

[dreamy lichen]

the theorem talks about positive integers

are you sure you want to have x in Z?

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

okay that's better

this set is guaranteed to exist

Now regarding your choice of a,
You said "let a be an arbitrary element of T, that doesn't belong to S".
First of all, how do we know that such element exists? You should explicitly state the assumption that T is non-empty, and later contradict it.

Next,
"let a - 1 belong to S". At this point, a is already some fixed element that was established in the previous sentence, so a-1 either belongs to S or it doesn't. You can't just say "let a - 1 belong to S"

[twilit field]

hmm, ok. So I have to use the well ordering principal I guess?

[dreamy lichen]

Correct

[twilit field]

ok, let me re-write a proof that uses that then

thanks !

both of you!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

\textbf{theorem 1.3}:- If a set $S$ satisfies the following conditions, it consists of all the positive integers
\begin{enumerate}
\item It consists of the integer 1
\item If k is a positive integer k, belonging to the set, as must $k+1$
\end{enumerate}

\begin{proof}
Let $\mathbb{T}$ consist of the non-empty set such that it consists of any integers not in $ \mathbb{S}$

let $a\in \mathbb{T}$

let $a$ $\in \mathbb{T}$ and let it be the smallest element of the same
$\implies$ (a-1)\ in $\mathbb{S}$ which by definition implies a$\in$ $\mathbb{S}$

but this contradicts the definition of $\mathbb{T}$
thus $\mathbb{T} $ must be empty and all positive integers must be in $\mathbb{S}$

QED

\end{proof}

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is this proof fine ?

[dreamy lichen]

Use the set builder notation

also, "T consist of the non-empty set such that it consist of any integers not in S" is wrong anyway

consist of the non-empty set

I'd probably interpret it as T contains some non-empty set

meaning there is non-empty A, such that A ∈ T

[twilit field]

will make that change

[dreamy lichen]

there are also some cases that weren't considered

what if a-1 isn't positive integer?

then it's not in S

ik it's trivial to deal with those, but it should be mentioned anyway

[twilit field]

so I have to mention such that $a-1$ is positive?

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

or non -negative

rather

[dreamy lichen]

positive.

0 doesnt belong to Z+, or at least you dont want it to be there

[twilit field]

👍

[dreamy lichen]

but a-1 isnt necessarily positive

thats the problem

what if a = 1

then a - 1 = 0

you need to show that a =/= 1

which is trivial

=> (a-1) in S

what I mean is that this is unjustified

We know that (a-1) isn't in T.

that means that (a-1) isn't a positive integer or (a-1) isn't in S

by definition of T

which should be defind like this

[twilit field]

I see

[dreamy lichen]

But once you manage to exclude the possibility that (a-1) isn't a positive integer, you'll be able to conclude that it's not in S

[twilit field]

a-1 can be 0 though, when a=1, no?

[dreamy lichen]

Well you need to show that a can't be 1.

Keep in mind that a is the least element of T

[twilit field]

oh yeah

[dreamy lichen]

meaning that a isn't in S...

[twilit field]

right

so the least element will be 0?

[dreamy lichen]

Wait what

[twilit field]

I'm confused

[dreamy lichen]

okay you wrote this

why should that be true?

how does it imply (a-1) in S?

dw, the idea is correct, it indeed implies it, im just asking why

[twilit field]

a is the smallest element in T

[dreamy lichen]

okay, and?

that definitely shows that a-1 is not in T.

Does it show that (a-1) is in S?

If so, why?

this is your definition of T

[twilit field]

yes

[twilit field]

as it's not in S it has to be in T

[dreamy lichen]

Do you mean that it's not in T so it has to be in S?

(a-1) is not in T so it has to be in S?

[twilit field]

yes

[dreamy lichen]

that's almost correct, but it misses one detail

if (a-1) is not in T, then we can conclude that it's either not true that (a-1) ∉ S or it's not true that a-1 ∈ Z+

by definition of T

do you agree with that?

[twilit field]

yes

[dreamy lichen]

just rephrasing it, either (a-1) ∈ S or (a-1) ∉ Z+

[twilit field]

yea

[dreamy lichen]

now the problem is, you immidiately jumped to the conclusion that (a-1) ∈ S, without considering the case when (a-1) ∉ Z+

[twilit field]

I see

[dreamy lichen]

since a is in Z+, the only way for (a-1) to not be in Z+ is that a = 1

so we need to show that a = 1 can't be the least element of T

or simply that 1 can't be the least element of T

Simplest way to do that is to show that 1 isn't even in T

Can you do that?

[twilit field]

let me try

[dreamy lichen]

Try using as much information about S as you can, you only utilized one part of it in your proof

[twilit field]

$\mathbb{S}$ contains $1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I use that?

[dreamy lichen]

indeed

this immidiately implies that 1 can't be in T

because T consists of those positive integers that are not in S

[twilit field]

hmm, ok, got it

so is the rest of the proof fine?

[dreamy lichen]

I think that the rest is quite good, it has few stylistics mistake, such as this unnecessary repetition

also you could mention that you are applying WOT when you let a be the smallest element in T

because without WOT, you wouldn't really know whether such element even exists

[twilit field]

ah, right

Thanks a lot!

I'll make those changes to my proof and post it here?

[dreamy lichen]

sure

[hard umbra]

\mathbb users

[dreamy lichen]

lol what's wrong about mathbb

[hard umbra]

its so overused for sets

[subtle blaze]

mathbb should only be used to emphasize things

[twilit field]

I just used it for fun

sorry

[dreamy lichen]

what

can you give example? lol

[subtle blaze]

$\mathbb R$

[grand pond]

Frosst

[hard umbra]

its okay you can have fun

[subtle blaze]

for real numbers

and i know snow doesn't agree but $\mathbb E$ for expectation

[grand pond]

Frosst

[twilit field]

begin{proof}
Let $\mathbb{T}$ consist of the non-empty set such that it consists of any integers not in $ \mathbb{S}$

let $a\in \mathbb{T}$

$a\neq 1 $ as $1$\in $\mathbb{S}$

let $a$ $\in \mathbb{T}$ and let it be the smallest element of the same
$\implies$ (a-1)\ in $\mathbb{S}$ which by definition implies a$\in$ $\mathbb{S}$

but this contradicts the definition of $\mathbb{T}$
thus $\mathbb{T} $ must be empty and all positive integers must be in $\mathbb{S}$

QED

\end{proof}

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[subtle blaze]

\begin

$1$\in $\mathbb{S}$man what is this

[dreamy lichen]

\begin{proof}
Let $\mathbb{T}$ consist of the non-empty set such that it consists of any integers not in $ \mathbb{S}$

let $a\in \mathbb{T}$

$a\neq 1 $ as $1\in \mathbb{S}$

let $a$ $\in \mathbb{T}$ and let it be the smallest element of the same
$\implies$ (a-1)\ in $\mathbb{S}$ which by definition implies a$\in$ $\mathbb{S}$

but this contradicts the definition of $\mathbb{T}$
thus $\mathbb{T} $ must be empty and all positive integers must be in $\mathbb{S}$

QED

\end{proof}

[grand pond]

MæthIsAlwaysRight

[dreamy lichen]

Definition of T was unchanged

[subtle blaze]

what do you mean by consist of the non-empty set

shouldn't it say let T be a non-empty set such that...

[twilit field]

I mean it's not empty

my bad

[hard umbra]

i dont like that phrasing

[dreamy lichen]

the fact that its non-empty should be assumed and later contradicted

[hard umbra]

you should define T first and then assume it is non-empty

[subtle blaze]

well i dont know what he's writing tbh i haven't read that far

[hard umbra]

the fact that T is not empty shouldnt be part of its definition

[dreamy lichen]

and define it like this

[hard umbra]

same ngl

[subtle blaze]

it sounded like he wanted it to be non-empty

so that's what my chatgpt brain told me he should say instead

[dreamy lichen]

well this is the first thing to be changed

second, you defined a at 2 different places

also you can just say "let a be the smallest element of T"

its a bit shorter and imo also more readable

[subtle blaze]

also

S is not defined

i dont know what's in T besides it's got element(s)

[hard umbra]

dw it is

[subtle blaze]

or is it above

oh

is this S the same S as mathbb S

[twilit field]

yes

[subtle blaze]

please for the love of god use whatever notataion they give you

dont invent your own for the sake of "fun"

[hard umbra]

god doesnt love you

[subtle blaze]

[twilit field]

of that much I'm sure

[twilit field]

ok, will do. sorry

[subtle blaze]

it sounds like you want to start with

let T be a subset of S

[dreamy lichen]

start by changing the definition of T please

[subtle blaze]

(im not sure i haven't read that far yet)

[hard umbra]

no its the complement

[subtle blaze]

oh

[dreamy lichen]

lol

this here was fine

from this it's clear that T is subset of Z+, hence we will be able to apply WOT

if we assume its non-empty

[subtle blaze]

ok so $S\subset \mathbb Z^+$, and $T = \mathbb Z^+ \setminus S$

[grand pond]

Frosst

[dreamy lichen]

im not sure whether the first one holds

its not specifically stated anywhere

there could be some extra random elements probably

[hard umbra]

its kinda silly in that regard

[hard umbra]

wlog let S = S intersect Z^+

[dreamy lichen]

it might be, but its not wrong

[hard umbra]

yeah

[twilit field]

is this better?

[hard umbra]

i agree

[dreamy lichen]

Ohh I love the proper class T

Does T contain the element {{1}, {2}}?

[twilit field]

I think I messed up the ordering ?

[dreamy lichen]

Because it might not be in S

[twilit field]

no

[dreamy lichen]

Well by your definition it could

just use the set builder notation

it makes stuff easier

if you want to avoid it anyway, mention that T contains exactly those positive integers that are not in S.

[subtle blaze]

why is (a-1) not in math mode

why is the a in a \in \mathbb S not in math mode

[hard umbra]

everything is in math mode if you believe hard enough

[subtle blaze]

i think it makes no sense to talk about "elements not in S", not in S in terms of what

[fleet moss]

let it be a non empty set
not sure if you can do that

[subtle blaze]

is "apples" not in S?

i feel like you need let $T = \mathbb Z^+ \setminus S$

[grand pond]

Frosst

[fleet moss]

in general the set of elements not in S is a contradiction, because you end up including "the set of all sets"

[dreamy lichen]

What if the set of all sets is in S

[subtle blaze]

^^something something doesn't work sounds like what he wrote

[fleet moss]

actually whats the context here? proving the inductive construction of the positive integers?

[dreamy lichen]

proving induction assuming WOT basically

[twilit field]

is this better or am I forgetting something?

The TeX is still messed up

[fleet moss]

what is wot

[dreamy lichen]

You need to make the assumption that T is non-empty

[dreamy lichen]

well ordering theorem / principle

[fleet moss]

ah

[subtle blaze]

you'd write $T={x\in \mathbb Z^+|x\notin S}$

[grand pond]

Frosst

[dreamy lichen]

both can be used ig

[subtle blaze]

the set you're taking it from goes in the first slot

well at least that makes it easier to read

[fleet moss]

firstly, whats the definition of $\mathbb{Z}^+$?

[grand pond]

esca (@ with reply)

[subtle blaze]

otherwise you just have a bunch of conditions after the |

[dreamy lichen]

{1, 2 ...}

[subtle blaze]

im using $\mathbb Z^+$ to represent the "all positive integers" set

[grand pond]

Frosst

[fleet moss]

well i mean what is there to prove? the ellipses are eliding away the induction, which is how i know Z+ to be defined in the first place

[subtle blaze]

idk if physicsrocks wants to get into how you construct Z so i've just used Z

[dreamy lichen]

What

[dreamy lichen]

ellipses?

[subtle blaze]

the ...

[fleet moss]

...

idk how to spell it

[subtle blaze]

ellipsis

[fleet moss]

ah ok thanks

[twilit field]

\begin{proof}
Let $\mathbb{T}$ = ${x\in \mathbb Z^+|x\notin S}$
where T is non-empty
let $a$ be the smallest element of $\mathbb{T}$
$\implies$ (a-1)\ in $\mathbb{S}$ which by definition implies a$\in$ $\mathbb{S}$

$a\neq 1 $ as $1$$\in$ $\mathbb{S}$

but this contradicts the definition of $\mathbb{T}$
thus $\mathbb{T} $ must be empty and all positive integers must be in $\mathbb{S}$

QED

\end{proof}

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I think I've still forgotten something

[dreamy lichen]

where T is non-empty