help-43

[bleak dock]

yep!

it's funny how this GCSE knowledge becomes suddenly useful

[summer cedar]

oh so just t = 1?

[bleak dock]

no, that's not quite it

it's just 53 * 1

max value of sin and cos is 1, no matter what is inside the sin or cos function

[summer cedar]

ohhh

[bleak dock]

unless there's some domain restriction ofc but that's besides the point

[summer cedar]

where did 53 come from tho

oh right right

they got it back to one trig function with addition formula

ahh this part i also didnt get

I never done so bad on a trig modelling question before lol

[bleak dock]

we need to see the graph, Figure 4, for that

[bleak dock]

ahhhh, okay so the function 'ends' at the vertical asymptote

but the vertical asymptote is just when the denominator equals 0....

[summer cedar]

ah i dont get the right answer for some reason

u applu the 180 - theta interval right

[bleak dock]

yeah it should be that then

yeah it works, I checked

2x - arctan(28/45) = pi - arcsin(-35/53) = pi + arcsin(35/53)

[summer cedar]

oh

I j did what they did here in the ms

[bleak dock]

I mean you have your calculator so you could just graph and find root

[summer cedar]

i rly dunno what i did wrong here tho

last part i got from doing the 180- theta interval btw

[bleak dock]

ah you didn't do the 180 - theta bit

or like you didn't take arcsin on both sides properly

[bleak dock]

notice how it should be arcsin(-35/53), cause you apply arcsin on both sides

[summer cedar]

ah i got the right answer now

thanks, this was rly helpful

I got sooo many questions wrong on this paper tho, im gna uplaod a few more XD

[bleak dock]

no worries!

[bleak dock]

honestly just close this channel and open a new one, then ask

no one's going to check a channel this far down

[summer cedar]

got i t

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[summer cedar]

How do i do 4b?

[wheat pasture]

A general point on the line l1 is of the form (x, 2x - 1): what would it mean for the distance between that and the point A to be 2sqrt{13}?

[summer cedar]

uh u could use the distance formula

to work out x?

idk

could u explain the first bit tho

i dont get the (x,2x-1) thing

maybe it ssomething ill j remember acc

[wheat pasture]

The line's equation is y = 2x - 1; if you say that the point (x, y) is on that line, then y is 2x - 1, so you can write the point as (x, 2x - 1)

[summer cedar]

ohhhh

okkkkk

nice

lit

i got it

I dont rly get part c either

oh wait

wait

I think i might get it

ahhh

i j dont get

why do they sub the x roots into

y = 2x-1

oh

B lies on that line isnt it

that line is BC

ok thanks

my brains too fried rn to more

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[tawdry pivot]

not sure where to start 😓

[kind viper]

make a sketch of the cylinder and how you could fit a long pencil in there

[tawdry pivot]

something like this right?

with measurements ^

[next tide]

yessor

now use pythagorus uncle theorem

[tawdry pivot]

oh my god i overthought/overcomplicated this LMFAO THANK YOU 😭😭🙏

[bleak dock]

well 20cm is actually the length in your diagram

[tawdry pivot]

oh whoops

[bleak dock]

but yeah! you get it

[tawdry pivot]

thank you guys 🙏

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[strange pendant]

lets say I have a 3d plane
Π : x + y + z = 0
and a plane
π : x + y + z = k
where k ∈ R
I wanted to ask, how is plane Π parallel to π

recall the cartesian equation of the plane is
Ax + By + Cz + D = 0

[strange pendant]

they have same normal vector no? but what about D?

what I mean is how is a plane with cartesian equation

x+y+z=2
parallel to
x+y+z=3

[kind viper]

D controls the placement in space

try graphing them both in desmos

or otherwise observe algebraically that the two planes don't intersect

[strange pendant]

wdym?

[strange pendant]

only way for parallel planes to intersect is that they are same exact plane i pressume

[cerulean bramble]

Not quite

oh wait PARALLEL

[strange pendant]

[cerulean bramble]

yh you right

[strange pendant]

otherwise for example for
x + y + z = 2
and
2x + 2y + 2z = 2
and
2x + 2y + 2z = 3

[cerulean bramble]

Parallel planes mean they either intersect nowhere or they intersect everywhere

[strange pendant]

how would it be in this case?

[cerulean bramble]

Well all three are parallel

[kind viper]

these are x+y+z=k for k being 2, 1 and 1.5 respectively

[strange pendant]

ohh

all of them strictly parallel and not equal

I think i get the idea lowkey. I appreciate it

. lclose

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[deft tangle]

Hints

[deft tangle]

What operations should i use?

[pine osprey]

Diagonalize it and do the product of diagonal ig

[forest token]

i would do C3--->C3-C2-C1

[forest token]

oh C1-->C1-2C2 creates 2 1's

[next tide]

mechanical solving op

[dusky nymph]

do you mean columns instead of rows?

[forest token]

yes sorry

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@deft tangle Has your question been resolved?

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[normal orchid]

Can someone help me do this using the classic expansion because I prefer that method

[normal orchid]

As in 1+x to the power of n

And also I don't really understand the question in general

<@&286206848099549185>

[azure dust]

Hey what can i help

[normal orchid]

Hi can u help with this i have no idea what's going on

[azure dust]

Exponent is to large to expand by Hand it will take an hour

[normal orchid]

Ah ok I normally just use that

For binomials

And don't understand this

[azure dust]

Use the binomial expansion

to find the general pattern shown by it term

[normal orchid]

Yes but I use the other one not this one

[knotty basin]

you form a general term nCr a^(n-r) b^r, im not sure about the other method

[azure dust]

Yes

[knotty basin]

and you find the value of r for where you dont have an x term, i.e. x^0

[normal orchid]

Yh i use the bottom formula instead of ncr

But for this idk

[azure dust]

Use the 1 above

[normal orchid]

What does independent of x mean

[normal orchid]

Do I have to

[azure dust]

It means it has no x

[azure dust]

Yes

[normal orchid]

So it's the first term or what

Like why wouldn't it have an x

[azure dust]

Bro u are told it is not attached to x it is just a nun like 45 compared to 45x

[normal orchid]

Ok

Can u explain the method they're using

Like why 10-r and r

[knotty basin]

given the general formula

[normal orchid]

Ok how do u do that using a formula ?

[azure dust]

The general pattern is as follows nCr a^(n-r) b^r

[normal orchid]

Ah ok

[azure dust]

this is more like Tn

[knotty basin]

and knowing n=10, a=sqrt(x), b=-k/x^2

[azure dust]

like similarly to finding explicit formula of a sequence

[knotty basin]

you get an expression in terms of x^r

upon expansion

[normal orchid]

And then the bit I highlighted they got that by simplifying ??

[knotty basin]

"the term independent of x" means x^0

[azure dust]

and then since it is independent of x the exponent of x has to be zero

[knotty basin]

yep indices

x^a/x^b = x^(a-b)

etc etc

[normal orchid]

Ah ok

Yh that makes more sense

Thanks

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[acoustic solstice]

How to solve this

[acoustic solstice]

Can we say lim x -> inf tan^-1(x+n) - tan^-1 (x) = 0?

So sec^2 (0) = 1?

[median prism]

Process is right but It depends how rigorous you need to be

[acoustic solstice]

If I have to prove it rigorously what can I try?

[pine osprey]

Don't you have to do it for + inf ?

[acoustic solstice]

Wdym

[pine osprey]

You say lim as x goes to 0 but its written +inf on paper

[acoustic solstice]

Yes sorry

[pine osprey]

Anyway, it isn't changing anything to the result

arctan in +inf is pi/2

Since they are finite limit you can substract

And so sec²(0) indeed

[acoustic solstice]

Thanks

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[sand heron]

I need help on these I don’t understand it

[sand heron]

The teacher told us to use the formula above hold on it cut off

[compact pewter]

@sand heron Has your question been resolved?

[eternal pulsar]

What is L?

[sand heron]

Length

[compact pewter]

@sand heron Has your question been resolved?

[compact pewter]

@sand heron Has your question been resolved?

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[idle haven]

for d

i dont understand how to find f(t)

like the t chart on the bottom

how to find those values

im cooked my test id tomorrow

huys pls

[minor pebble]

do you know how to find abs minima and maxima?

or in general minima and maxima?

[idle haven]

yea get the canidates so its critical numbers (f’=0) and also fhe endpoints

but im confused like after that

[minor pebble]

its when f"=0 i thought

am i coooked

[craggy sluice]

minima OR maxima happens when f'=0

[minor pebble]

oh

[minor pebble]

so then its a minima if it goes from negative to positice

[craggy sluice]

actually no i am wrong

[minor pebble]

and maxima if it goes from positive to negative

[craggy sluice]

this is only true if f' is a polynomial

[craggy sluice]

yes

[idle haven]

yeah except

im wondering how

[craggy sluice]

and minima when negative to positive

[idle haven]

you find f(x) given grah of f’(x)

or is it possible?

[craggy sluice]

you probably need to do integral

actually no

you dont need to

because

it told you f(2)=1 XDDDD

[idle haven]

how fo i find the others

[craggy sluice]

why do you need to find the others?

the other 2 zeros ain't minimas

[minor pebble]

oh

[minor pebble]

the others would be a POI and a relative max

[craggy sluice]

exactly

[idle haven]

i dont but i want to know how to do it

because in case its on test

i guess

[craggy sluice]

um, you will have to know integral

do you know integral?

[minor pebble]

lmao 😭

im cooked

i dont know differences bettween max and min

[craggy sluice]

oof

[idle haven]

i think i got it maybe

[craggy sluice]

alright so for this, you will need to break it into piecewise functions

for [-2,0)

it looks to me like f(x)=-2x-2

therefore the integral of that is -x^2-2x+C

you cannot determine C however, thus you cannot find a value of f(x) in that range

[idle haven]

wait

so is this right

that

to find f(6)

[craggy sluice]

that is impossible

[idle haven]

no its not i just

[craggy sluice]

you just?

[minor pebble]

because you only have the f' graph, you cannot determine f(6)

[idle haven]

i did it can checj work???

or is it right

[minor pebble]

whay

@craggy sluice HELP

[craggy sluice]

?

[minor pebble]

what did she do

[idle haven]

integral is area under the curve

[minor pebble]

you found it from 2-6

yeah

[idle haven]

yes

i was given one of them

[minor pebble]

you found the area under the curve from 2-6

[idle haven]

yed

[craggy sluice]

ok

yours is not correct

what you did

you assumed that f(x) is continous

however it might be a not continous function

thus that integral will be improper

[idle haven]

its continuous on [2,8]

i wm closing this now byeee

[craggy sluice]

bye

[idle haven]

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[fallen frost]

In this definition:

[fallen frost]

Is the symbol "there is/exist" the correct symbol to replace that use of "for some"?

[hidden relic]

yes, though if you're using that symbol you'd usually write it the other way round

like there exists v in V, w = T(v)

[fallen frost]

gotcha. Thank you.

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[gilded kayak]

i forgot to change the limits here but then when i tried to fix it idk how to change limtis like infinity

[floral raven]

You can leave the limits in terms of x (just write x =), and in this last step rewrite the integral back in terms of x and then try to evaluate.

[gilded kayak]

do i revert to x before or after integrating

[floral raven]

After integrating.

[gilded kayak]

itd be just swapping dummy variables what would that do

[floral raven]

x = tan(theta) was the substitution you used.

and then theta = arctan(x)

[gilded kayak]

better presentation

still arctan(inf) gives multiple values and i cant write it nicely

forget it new thread

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[gilded kayak]

what does correlation exactly mean?

[gilded kayak]

im gonna take a shower

Can I let U = number of heads in last 2 tosses, then T = U + S/3

Wait no S/3 is a probability
Do I need an indicator function in this

<@&286206848099549185>

<@&286206848099549185>

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[vast pelican]

can anyone help me with this? i don't really understand it too well or know how to do it

[lime juniper]

do you know the law of sines for triangles

[vast pelican]

is that the a/sinA = b/sinB ect?

[lime juniper]

yeah

you can switch the numerator and denominator of all.terms same thing

[vast pelican]

yeah

[vast pelican]

how do i do this then?

[lime juniper]

so, look at the first one

you have A and a and B

so are missing side b

[vast pelican]

so i could uses the law of sines?

[lime juniper]

yeah, except, one will be unknown

sinA/a =sinB/b

you know 2 out of three

[woeful schooner]

You can use A and B to get C

And then get c

[vast pelican]

so far i have sin37/8.5 = sin81/b

[lime juniper]

first inthink dogra should solve for A,a,B, and b Hiten

[woeful schooner]

Yes

[lime juniper]

great, solve for b

[woeful schooner]

Oh that's what you meant, mb

[vast pelican]

i dont know how to get the b on its own

[lime juniper]

if you had something like

1/2=7/x

would you know how to solve for x?

[vast pelican]

times the 1/2 by 7?

[lime juniper]

well i would cross multiply

turns the equation into 1•x=7•2

[vast pelican]

what do the dots mean?

[lime juniper]

times

×

[vast pelican]

so it would be sin37 x b = sin81 x 8.5

[lime juniper]

[vast pelican]

yay

whats the next step? (sorry for being so dumb)

and tsym for helping

[lime juniper]

its okay!

if you have 2 angles in triangle, do you know the third?

[vast pelican]

no

[lime juniper]

sum of interior angles of a triangle is always 180⁰

[vast pelican]

ohhh yeah

[lime juniper]

that should be enough

[vast pelican]

it would be 62

[lime juniper]

then you can use law of sines on the missing side

[vast pelican]

but we still dont know "b"

[lime juniper]

divide both sides by sin 37⁰

[vast pelican]

okkk

[lime juniper]

gn

[vast pelican]

oh good night

thanks for helping me

[compact pewter]

@vast pelican Has your question been resolved?

[compact pewter]

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[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

(x1,x2,x3,x4) = a(1,2,-1,0) + b(0,1,2,-1)

(x1,x2,x3,x4) = (a,2a,-a,0) + (0,b,2b,-b)

(x1,x2,x3,x4) = (a, 2a+b, -a+2b,-b)

i) x1 = a

ii) x2 = 2a+b

iii) x3 = -a+2b

iv) x4 = -b

1. x2 = 2x1 - x4

2. x3 = -x1 -2x4

,, \mathcal{S} \cap \mathcal{T} : \begin{cases} 2x_1 - x_2 - x_4 &= 0 \ -x_1 - x_3 - 2x_4 &= 0 \ x_2 + x_3 - x_4 &= 0 \end{cases}

[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

,w nullspace {{2,-1,0,-1},{-1,0,-1,-2},{0,1,1,-1}}

[boreal girder]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

,, \mathcal{S} \cap \mathcal{T} = \langle (4,7,-6,1) \rangle

[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

v = (2,1,-2,3)

,, (\mathcal{S} \cap \mathcal{T}) + \mathcal{W} = {\mathbf{st} + \mathbf{w} \mid \mathbf{st} \in \mathcal{S} \cap \mathcal{T} \text{ and } \mathbf{w} \in \mathcal{W} }

[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

(4,7,-6,1) + (w1,w2,w3,w4) = (2,1,-2,3)

(w1,w2,w3,w4) = (2,1,-2,3) - (4,7,-6,1)

(w1,w2,w3,w4) = (-2,-6,4,2)

,, \mathcal{W} = \langle(-2,-6,4,2) \rangle = \langle (1,3,-2,-1) \rangle

[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

(2,1,-2,3) = a(4,7,-6,1) + b(1,3,-2,-1)

i) 2 = 4a + b

ii) 1 = 7a + 3b

iii) -2 = -6a -2b

iv) 3 = a - b

a = 3 + b

2 = 4(3+b) + b

2 = 12 + 4b + b

-10 = 5b

b = -2

a = 3 + b

a = 3 - 2

a = 1

1 = 7 -6

-2 = -6 + 4

(a,b) = (1,-2)

(2,1,-2,3) = a(4,7,-6,1) + b(1,3,-2,-1)

(2,1,-2,3) = (4,7,-6,1) -2(1,3,-2,-1)

(2,1,-2,3) = (4-2,7-6,-6+4,1+2)

(2,1,-2,3) = (2,1,-2,3)

W = <(1,3,-2,-1)>

SnT = <(4,7,-6,1)>

.close

[compact pewter]

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[delicate musk]

How would I create an extention field of $\mathbb{Q}$ containing $\sqrt3 + \sqrt5$

[boreal girder]

BOSS

[compact pewter]

@delicate musk Has your question been resolved?

[delicate musk]

for more persepctive im trying to understand this example:

[compact pewter]

@delicate musk Has your question been resolved?

[coral lichen]

@delicate musk Did you get x^4 - 16x^2 +4 as the minimal polynomial?

[delicate musk]

Ill agree with it there, it is not irreduceable so im guesing its minimal

[coral lichen]

I suppose you know to construct it

[delicate musk]

just set it equal to x right

then find out when x would be zero

so like

$\sqrt3 + \sqrt5 = x$

[boreal girder]

BOSS

[coral lichen]

right, and essentially square it and do some algebra

[delicate musk]

yup

[coral lichen]

where are you stuck here?

[delicate musk]

im completely honest im new to this and having trouble grasping everything

so to start slowly

why does this hold?

[coral lichen]

This is the notation for a degree of a field extension

You may have proved earlier the relation between the degree of a field extension and minimal polynomial

[delicate musk]

my book didnt have this proof i did write it down

The notation just means that Q(sqrt3+sqrt5) is a vector space over Q of dimention 4 right

[coral lichen]

There you go.

[delicate musk]

which importantly tells us it can be the linear combination of 4 elements in Q times some vectors in the vector space

deg of \alpha in that case is just the degree of its minimal polynomial in F right

I learned a lot of definitions lol just double checking

[coral lichen]

The degree of the irreducible polynomial for over F is called the degree of alpha over F

[delicate musk]

which would be the degree of its minimal polynomial in F[x] right

in this case x^4-16x^2+4

[coral lichen]

Let me give you a simple example

Let C denote the complex numbers

[delicate musk]

ok

[coral lichen]

[C : R] = 2

[delicate musk]

because c = a+bi | a,b in R

the basis is {1,i}

[coral lichen]

right

or take f(x) = x^2 - 1

[delicate musk]

in what field

[coral lichen]

this is irreducible over R

[delicate musk]

hmm yeah

[coral lichen]

We may view C as R(i)

[delicate musk]

Ah

x^2 - 1 minimal polynomail for i

[coral lichen]

so essentially pick alpha = i

[delicate musk]

deg 2

[coral lichen]

As a recap,

Given a Field F, if we construct a field extension F(alpha)

Suppose alpha has a degree 'n' over F, ie the minimal polynomial as degree 'n'

then [F(alpha) : F] = n

[delicate musk]

yes

i agree

sure

so far so good

[coral lichen]

same reasoning here

[delicate musk]

Yeah, i didnt realize why its always the same degree as the minimal polinomial

ok either way

continuing

[coral lichen]

You may the prove the theorem for it

[delicate musk]

already in my notes lol

okok so

I get that {1,sqrt3} is a basis for Q(\sqrt3) and similarly {1,sqrt5} for Q(\sqrt5)

what does this notation mean

smallest field containing Q, sqrt3, and sqrt5?

[coral lichen]

right

[delicate musk]

I dont get this equality:

[coral lichen]

You may need to try using some field properties to understand different types of elements here

[delicate musk]

wdym

[coral lichen]

ie what is the general form of elements in here ?

[delicate musk]

smallest field containing sqrt3 and sqrt5?

im not sure tbh

[coral lichen]

Does it contain k*\sqrt(3) , where k in Q?

[delicate musk]

yes

inverces right

it contains sqrt 3

so then it contains 1/sqrt3

so then it contains k times 1\sqrt3

oh

wait

i misread ur question holy

[delicate musk]

id say yes but im not sure

because it has to have 1*sqrt3

[coral lichen]

I can restate my question.

What is the general form of elements in Q(\sqrt(3), \sqrt(5))

this will help you to understand which are there and which aren't

[delicate musk]

thats what im trying to figuer out

I know what it is seperately but not together

[coral lichen]

let's keep it simple

let's say alpha is some element in F

what are the elements of the field F(alpha) ?

adjoining only one this time

[delicate musk]

a+b\alpha

for a and b in F

[coral lichen]

does (alpha)^2 belong to F(alpha) ?

remember it's a field

[delicate musk]

oh sorry yeah

a+b \alpha + c\alpha^2 ........

[delicate musk]

no i know this i was just tired and working with sq roots where the second power goes away as its just part of F

[coral lichen]

You may see

for F(alpha)

The elements are of the form

a0 + a1alpha + a2(alpha)^2 + ... + a_n(alpha)^n , where ai in F

essentially these are the only types of elements in F

sorry for not typsetting math

[delicate musk]

nw

ues

yes

ok wait

is this how ur suposed to read it

if so is it just like

${a+b\sqrt5 \mid a,b \in \mathbb{Q}(\sqrt3)}$

[coral lichen]

right side you may read it as

Q(sqrt(3)) adjoint (sqrt(5))

[boreal girder]

BOSS

[delicate musk]

ok yeah

[delicate musk]

so something like this

[delicate musk]

and thats true notation for like

F(a,b,c,d) = (((F(a))(b))(c))(d)

for example

[coral lichen]

right

[delicate musk]

cool

[delicate musk]

so this is correct?

[coral lichen]

yes

[delicate musk]

ok so then we have

[coral lichen]

I actually have to go now.

[delicate musk]

(a+b\sqrt3)+(c+d\sqrt3)\sqrt5

[coral lichen]

I'm sorry

[delicate musk]

ok os this the right path tho

nw

thank you so much :D

[coral lichen]

I can share a pdf with you. All these things are done it a sequentially order here

[delicate musk]

that would be appreciated

[delicate musk]

tyty

[coral lichen]

There might be errors in the pdf, be careful

the orginal reference is Micheal Artin Algebra

[delicate musk]

ok

tysm

[compact pewter]

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[tame fox]

Could you please help me find the areas of the red and green colored regions SEPARATELY?

[winged dock]

What have you attempted ?

[tame fox]

It's easy to find the red and green areas together, but I have no idea how to find their separate areas.

[winged dock]

We first need to find BC

[tame fox]

I don't know how to do it too

[winged dock]

How will you find AC ?

[tame fox]

Oh it's 3√3 AC, then BC is 6 too

Rest of the question really out of my range tho

[winged dock]

Now that you have BC = 6

The triangle OBC is equilateral

[tame fox]

Thanks man I got what you re trying to say 👍🏼

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[plush swallow]

<@&286206848099549185>

[potent berry]

!15min

[compact pewter]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[compact fractal]

have you tried drawing a diagram

[plush swallow]

It failed

Especially point F I failed to find where to put it

[compact pewter]

@plush swallow Has your question been resolved?

[compact pewter]

@plush swallow Has your question been resolved?

[compact pewter]

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[neat fossil]

I really do not understand

[subtle helm]

Note: ${x = r\cos\theta, y = r\sin\theta}$

[boreal girder]

k

[neat fossil]

So if I apply that

Here

[subtle helm]

yes

So, the boundary of the graph is ${8\cos\theta}$. The question asks for the region enclosed, so we include all radii from ${r = 0}$ to ${r = 8\cos\theta}$. Hence, ${0 \leq r \leq 8\cos\theta}$.

[boreal girder]

k

[neat fossil]

Got that part

But what about the other part

[subtle helm]

A circle is a simple shape that includes all angles from ${\theta = 0}$ to ${\theta = 2\pi}$ intuitively.

[boreal girder]

k

[neat fossil]

Yess

[subtle helm]

so

to draw a complete circle

u have to go from theta = 0 to theta = 2pi

[neat fossil]

So why does it say this is wrong?

[subtle helm]

my bad gang

theta = 0 to theta = pi

[neat fossil]

No problemo

And what is that reason can you explain?

Why only pi

And not 2pi

Oh wait

Look

It says -pi/2 to pi/2

I am confused

[subtle helm]

this should still work...

[neat fossil]

It marked it as wrong for some reason

[subtle helm]

but the thing about ${r = 8\cos\theta}$ is

[boreal girder]

k

[subtle helm]

${r < 0}$ for some ${\theta}$

[boreal girder]

k

[neat fossil]

Hmmmmm

So it has to be positive?

[subtle helm]

say

after the pi/2

cos gets negative

[neat fossil]

Yes

[subtle helm]

so

instead of pointing to the 2nd quadrant

[neat fossil]

R becomes less than 0

[subtle helm]

it flips and points to the 4th quadrant

[neat fossil]

I see

[subtle helm]

in here

they use -pi/2 to pi/2

because they want radius to be positive

to remove all the stupidity

[neat fossil]

Oh

So do I just need to know it has to be positive

If there was sin it would be 0 to pi?

[subtle helm]

sin?

id say

0 to pi

[neat fossil]

Yess that makes sense

Thank you so much

[subtle helm]

i should brush off my polar coor knowledge

😭

[neat fossil]

I am learning this rn hahaha

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[strong nymph]

Hey

[strong nymph]

i have a test on calculus soon its Trig & Exponential Functions i have an iep so im allowed a cheatsheet can i have some help or insite on what might be good to write/remeber for the test

i alr have the formulas for derivatives

[frigid holly]

idk if this is the best place for this question

[strong nymph]

😩

[frigid holly]

like this is specific math question help stuff

[strong nymph]

okay thanks for lmk 👑

[molten coral]

usual wisdom is that relying on a cheatsheat isnt as helpful as spending time making it yourself

because then you get to spend time not only reviewing all of the material, but boiling it down to the important stuff to remember

for what its worth

[strong nymph]

true true

but it helps me better understand the way to solve the question so eventually i dont need it

[molten coral]

would recommend magic hexagon i guess

[molten coral]

fair

[strong nymph]

thanks thanks ❤️

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[carmine garden]

Let $a$ and $b$ be positive integers,not both zero, and let $d$ be their $GCD$, the positive integer that generates the subgroup $S= \Za + \Zb$. So $\Z d = \Z a+ \Z b$. Then

[compact pewter]

Please don't occupy multiple help channels.

[boreal girder]

What a wonderful world !

Let $a$ and $b$ be positive integers,not both zero, and let $d$ be their $GCD$, the positive integer that generates the subgroup $S= \Za + \Zb$. So $\Z d = \Z a+ \Z b$. Then
```Compilation error:```! Undefined control sequence.
l.1422 ...eger that generates the subgroup $S= \Za
                                                   + \Zb$. So $\Z d = \Z a+ ...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[carmine garden]

F

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[quartz yoke]

The question is to determine a condition on |x-1| that will assure that |x²-1|<1/2

[quartz yoke]

This is what I've done

Can someone tell me if it's right or not?

[wheat pasture]

I'm happy with it
One comment is that you may find it easier if e.g. you notice that |x + 1| = |x - 1 + 2| <= |x - 1| + 2, you get the same bound but a bit less to "worry about", but that's more a minor comment than anything

[quartz yoke]

Ohhhh yes

Thank you @wheat pasture !

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[orchid sequoia]

I'm learning multivariable calculus and I'm a little confused...
In multivariable calculus, when theres a function of f(x,y,z), would that function be in the fourth dimension?

[brazen quiver]

yup!

[tiny hamlet]

like f(x,y,z)=t or smth

yes

[brazen quiver]

you have 3 input dimensions and 1 output dimension

so in order to fully visualize it, you need 4 dimensions

[orchid sequoia]

hmmm ok, and like even though we can never visualize it, we could still apply the same theorems and rules of calculus?

[brazen quiver]

Exactly, though, I think with a little bit of effort it's possible to visualize 4 dimensions. 4d golf is fun, for instance.

[rigid perch]

most of the same rules apply as to functions of 3 variables, yes

[silver canopy]

you can go up to 1000 dimensions if you want

[brazen quiver]

Honestly, neural networks can be considered to be optimization functions of billions of variables.

and there's a LOT of space when you're considering billions of variables. For instance, if you have a billion degree unit cube, it's possible to travel 31,000 units and still be in the cube.

[eternal pulsar]

Fun fact, the entire point of linear algebra is to also be able to perform things like this on matricies, etc.

Most of upper level mathematics is abstract in the sense that we cannot actually visualize or feel it (e.g. dimensions greater than 3) but we are still able to use mathematics on it based on the fundamentals built up in lower level mathematics

[orchid sequoia]

and also I have 1 more question, idk if it makes sense but ok in regular calculus, when you have a basic function its often representing the location of an object with respect to time and then if you take it's derivative, you get the velocity of an object with respect to time. I'm learning partial differentiation and I'm wondering what my original equation represents and what my derivative equation represents

[brazen quiver]

So let's reframe time

let's think of it in terms of space and other space.

So we have a standard 2d, one input one output function

the derivative in the input direction measures how quickly the output changes as a proportion to the input, in other words the slope, right?

[orchid sequoia]

right

[brazen quiver]

So if we have 2 input variables, the partial derivative with respect to one of them at a point is if you draw the same tangent line, but now because you're in 3d, there are multiple tangent lines, and entire tangent plane, to the function at any point.

the partial derivative of a function with respect to a variable is the tangent line such that the line only varies with the input variable we're taking the derivative of, and the output

so it's like we sliced the function

and we're doing the standard derivative.

[orchid sequoia]

so is there any concept of taking a derivative of the entire 3d function? or is that not possible?

[brazen quiver]

That's possible. We generally talk about the gradient in the case.

which is a vector that has all of the partial derivatives packed into it.

it makes sense that this is required, because our derivative in 2 input directions is a plane, which has 2 degrees of freedom, so we need 2 pieces of information to specify its slope.

[orchid sequoia]

right ok, so then I should just think of the gradient as the rate of change for the original function?

[brazen quiver]

I mean, there is no "the" rate of change anymore.

if you pick any direction to walk along the function, that direction will have a (generally) different rate of change.

[orchid sequoia]

so are you saying there are multiple "rates of change" to the differential function? or would you say that I should stop trying to apply "rate of change" to multivariable functions because its entirely different?

[brazen quiver]

hmm... let me do a little math and make a demo

[lime juniper]

did you take lin alg

[orchid sequoia]

no

[brazen quiver]

https://www.desmos.com/3d/pqm70ih1cp

[lime juniper]

okay, no problem

[lime juniper]

the standard derivative tells you how to represent the function at each point as a tangent line, right?

[brazen quiver]

The above demo shows you an example of different lines tangent to the function.

We formalize this as a "directional derivative" which you will learn about a little bit later.

[orchid sequoia]

ok so right now my gradient can be in any direction?

[brazen quiver]

exactly

[lime juniper]

so the higher dimensional versions are related to the idea that if you're in a small enough area of a point on the curve/surface/whatever

[brazen quiver]

if you have 3 inputs, you'll have a tangent volume (to your hyper surface), for instance

[lime juniper]

the object youre analyzing is approximately "flat" in some sense

and a line with a slope of a number is no longer sufficient to describe

[brazen quiver]

this is because a surface in 4d is 3 dimensional

[lime juniper]

this is what omni is calling a tangent volume for example

tangent line, tangent plane, then it gets harder to visualize

[orchid sequoia]

ohhhhhh ok i think i'm getting it now, thank you so much for the explanation and demo!

[compact pewter]

@orchid sequoia Has your question been resolved?

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[signal pollen]

hello

[signal pollen]

i am learning about complex numbers and there i came up with the term iota, our teachers say that i(root of -1) is iota, but when i asked chatgpt it says that "i" (root of -1) is actually imaginary unit and not actually iota, pls clear what actually "i" is?

thank you

This is what chatgpt sent:
That's a very thoughtful question — and you're onto something with the idea of smallness, but let’s clarify:

---

🧮 In math, “iota” and "i" are not the same — though they sound similar.

---

❌ Misconception:

The "i" in math (as in √−1) is not "iota."

✅ Reality:

In mathematics:

• “i” stands for the imaginary unit, defined as:

  $$
  i = \sqrt{-1}
  $$

• It's called “imaginary”, not because it's tiny, but because it doesn’t exist on the real number line. It was invented to extend math beyond real numbers, creating complex numbers.

---

🧠 So then… where is "iota" used in math?

"Iota" is rarely used in math, but in logic, computer science, or set theory, it can refer to:

• The iota operator (𝜄), meaning “the unique x such that...” — a very advanced and specific use.

But in everyday math, "iota" isn't really used — it’s the imaginary unit “i” that we usually see.

---

🧾 In short:

Term	Meaning	Field
iota	Very tiny amount (language use)	English, Logic
i (√−1)	Imaginary unit	Mathematics

---

Would you like to explore why i = √−1 is useful in physics or engineering next?

[boreal girder]

Hagrid
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rigid perch]

iota is the name of a greek letter ι which is sometimes used instead of i to represent the same number

it's less common than i, though

[signal pollen]

so iota isn't actually the root, but rather just the name of the letter?

[rigid perch]

well it's the name of the letter which is sometimes used to refer to the root, the same way pi is the letter π but also refers to the number which is approximately 3.14

[signal pollen]

so more commonly we just say "imaginary unit" rather than saying iota or we just say "i", am i right?

[rigid perch]

yes

[signal pollen]

ohh, do u know why we do this?

like why make it more confusing?

just use what the letter says

[lime juniper]

(j is the usual symbol in contexts dealing with electricity and in some programming languages)

[rigid perch]

i don't know why people would write it as i but call it iota because that's something i have only seen it used that way on this server, not in any of my own classes

[trail cave]

humans moment

[rigid perch]

just a weird mix of two conventions which became a habit for some people

[signal pollen]

ohh

[lime juniper]

never heard the imaginary unit called iota

[signal pollen]

u mean j or that was a mistype?

[rigid perch]

i've only ever seen it on this server

[lime juniper]

I mean j

[signal pollen]

maybe it's a regional thing

[signal pollen]

alright

Well, i think i got the answer, if u have anymore insights, pls tell

[lime juniper]

in contexts with electricity i is already used for current

so j is used instead

[signal pollen]

but that's the capital i

[lime juniper]

not always

[signal pollen]

not always? rlly?

[rigid perch]

often capitals are used for constant quantities and lowercase are used for variable quantities

so I is current if the current is constant, and i is the current if it varies

[signal pollen]

agreed, but for greek, it could be different ig

[lime juniper]

or, capital is the dimension and lowercase is the variable with that dimension

[signal pollen]

is that a guess or a known insight, cuz' that seems legit

[rigid perch]

that's the convention i've seen specifically in my circuit analysis classes

[signal pollen]

could u explain that more a bit?

[signal pollen]

so they say for current things: iI ?

[lime juniper]

like

if t is a variable in seconds or minutes or hours

its dimension is time, T

I've seen

wikipedia

[signal pollen]

ohh

well i did learn something here, thanks guys 😄

i think my work is done

[compact pewter]

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[compact pewter]

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[strange adder]

Hello! Can someone help me to demonstrate the following:

the limit of $(u_n){n\in\N}$ is a solution of $l = l(1 - l)$
(with $u_n$ defined as follow $\begin{cases}u_0 \in ]0;1[ \ u{n+1} = u_n(1 - u_n)\end{cases}$

[boreal girder]

Badr
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[compact pewter]

@strange adder Has your question been resolved?

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@strange adder Has your question been resolved?

[copper sierra]

u_n is monotone (why?)

[strange adder]

this was already proved before (monotony)

[copper sierra]

what theorem involves monotone sequence?

[compact pewter]

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[strange adder]

sorry i figured it out in the meantime, I should have closed the channel, thanks though!

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[snow mica]

Hi hi

[snow mica]

Can anyone explain me how to get the first step

I m confuse

D

[odd birch]

So the motivation behind the first step is that negative exponents are difficult

What can you say about negative exponents?

[snow mica]

Idk

[odd birch]

Well, have you learned laws of exponents?

[snow mica]

You can add they if they r multiplying each other

And so on

[odd birch]

Dig further into that “and so on” hole

[snow mica]

Minus when they r dividing

[odd birch]

Ok

So you know that $a^{m-n}=\frac{a^m}{a^n}$, yes?

[boreal girder]

CST (reply ping for help)

[odd birch]

Consider what happens if m=0

[snow mica]

Yup

1/ a^n

?

[odd birch]

Yes, so $a^{-n}=\frac{1}{a^n}$

[boreal girder]

CST (reply ping for help)

[odd birch]

Now, that’s step 0.5, converting all those negative exponents using this

[snow mica]

Ahh

Got it

Thx

[odd birch]

np

[compact pewter]

@snow mica Has your question been resolved?

[snow mica]

Oh wait

Another question

[odd birch]

Yes?

[snow mica]

@odd birch

But

X and y is connected

How do I like

What I mean is x-uy

X^-1 - Y^-1

I can’t split them right

???

Oh wait nvm

I got it hehe

I got confused

Thx

@compact pewter

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[compact pewter]

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[upbeat badge]

using maxima, i need to plot a graph of s =t^4 +2t^3 −6t^2 +1, where t ≥ 0 however the input im using is not giving me a graph

[silver canopy]

Your question is better suited in #computing-software perhaps

[upbeat badge]

okay

.close

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[young flame]

HI GUYS!!!!!!!!!!!!!!!!!!!!!!

[young flame]

given three points in space, how I get the area of triangle they form

with determinant!

with determinant

and how it works!

;lets goooooooooooo

[subtle helm]

Get two vectors

Find determinant

[subtle helm]

Half the determinant

[young flame]

how it work

[young flame]

it is a parelellpiped

[subtle helm]

Or that

[subtle helm]

That’s why I halved it

[young flame]

half of parelellpiped? still not the area of triangle

[young flame]

Could you explain it

why the formula?

[covert yew]

its a parallelogram not a parelellpiped

[kind viper]

ain't that for R^2 tho

[young flame]

what is r^2

[subtle helm]

2d

[kind viper]

big R not small r

[kind viper]

yes basically

[young flame]

It is actually included in 3b1b's video

[kind viper]

you wanted to ask about the area of a triangle in $\bR^3$

[winged lion]

watch a video with nice animations

[boreal girder]

Ann

[winged lion]

it wasnt specified but i am sure it can be extended

[young flame]

I want a short explanation rather than tracing back to the video everytime when I forget how it works

[kind viper]

he said space

[young flame]

yes

[subtle helm]

Ahh right

Wouldn’t cross product work

[kind viper]

and it extends to the volume of a simplex, so in 3d space you can get the volume formula for a tetrahedron at best

[winged lion]

hmm okay looks like i am wrong and you proved your point

[young flame]

I used to understand the formula

I m so sad

[winged lion]

yes i would go with that

[young flame]

how

cross product requires three vectors to operate

[subtle helm]

Then u find magnitude of this

[young flame]

that's nostalgia.

3b1b explains it with the concept of duality

and i just love the series so much

[young flame]

This explains how cross product works

and cross product is actually the orthogonal with its magnitude equivalent to the pellelogram spanned by the two vectors.

That's so sweet

I used to have the note that I spent hours to take down until I lost them all in a misconducting of my phone.

Thank you guys, I have no question now

It is so sweet to recall the memory

[winged lion]

||u x v|| = ||u|| • ||v|| • sinθ
now the area of a parallelogram is base × height, and the height is ||v|| • sinθ

and then for a triangle you half it

[winged lion]

the cross product can also be interpreted as computing a determinant

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@young flame Has your question been resolved?

[compact pewter]

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[carmine garden]

Prove the nth roots of unity form a cyclic subgroup of $\C^{\cross}$ of order $n$.
\
Proof: The nth roots of unity are given by $e^{ \frac{i j \pi}{n}}; 0≤j≤n-1$.
We now verify that the axioms for this to be a subgroup are met
\begin{enumerate}
\item setting $j$ to $0$, we get $1$, which is the identity of $\C^{\cross}$.Thus the identity is in the subgroup
\item for any element $e^{ \frac{ i j \pi}{n}}$, the inverse is $e^{ \frac{ i (n-j-1) \pi}{n}}$, the product of which is $1$, the set is thus closed under inverses.
\item As exponents multiply associatively, it follows that this set is associative.
\end{enumerate}
The set thus forms a group,
\

[boreal girder]

What a wonderful world !

[azure vault]

and... where is the proof that the product of two is still in the set of nth roots

[timid lodge]

unrelated but I feel like I've seen you ask questions about 50 different math subjects lol

[carmine garden]

Oh yea, have to prove closure

[carmine garden]

I used to be a JEE student

So makes sense

[kind viper]

this set is associative

and other such minor wording hiccups

also the inverse of exp(iπj/n) is not exp(iπ(n-j-1)/n)

also like idk i dont think even mentioning associativity is worth

can you take a photo of your book's defn of subgroup?

[carmine garden]

yes

one minute

[kind viper]

if your footing's as uncertain as you said a few hours ago then you should be sticking to the book's defns for everything instead of trying to write vaguely related shit that does not actually prove what you want to prove

[carmine garden]

I think I accidentally used the subgroup axioms instead

*group axioms

[kind viper]

ok and can you write down the definition of an "n'th root of unity" that does not rely on complex exponentials

nor on trig

(yes there is one)

[carmine garden]

A number is said to be a root of unity if $z^n=1$

[boreal girder]

What a wonderful world !

[carmine garden]

Hmm, ok, I didn't know that

[kind viper]

you just wrote it out

if this were a phoenix wright game i would yell OBJECTION! at this moment

anyway yes ok so

to give some names to it

let $U_n$ be the set of all $n$'th roots of unity in $\bC$, in other words $$U_n := {z \in \bC \mid z^n = 1}.$$

[boreal girder]

Ann

[kind viper]

prove that U_n satisfies closure, identity & inverses based on this defn

(or don't, who am i to stop you from making your own life 1000 times more difficult)

[carmine garden]

1. Closure : Let $z^a$ and $z^b$ be two roots of unity, then $z^a \cdot z^b = 1\cdot 1$. So $z^{a+b} = 1$. This proves closure.
   \
2. We now prove that it has the identity. Setting $n$ to $0$, we have $z^0=1$. The set thus has the identity , 1.
   \
3. Let $z^a$ be a root of unity. Then $z^{-a}$ is also a root of unity as $z^{a} = 1 \implies z^{-a}=1^{-1}=1$. $z^{a} \cdot z^{-a} = z^{a+-a}=z^{0}=1$. Therefore $z^a$ has an inverse in $U_n$.

[boreal girder]

What a wonderful world !

[azure vault]

I understood none of that

"let z^a and z^b be two roots of unity"?

also we're investigating n-th roots of unity

also z^a = 1???

[carmine garden]

yes, two nth root so unity would have been more accurate

uugh, yea

messed up big time

1. Closure : Let $z^a$ and $z^b$ be two nth roots of unity, then $z^a \cdot z^b = 1\cdot 1$. So $z^{a+b} = 1$. This proves closure.
   \
2. We now prove that it has the identity. Setting $z$ to $1$, we have $z^n=1$. The set thus has the identity , 1.
   \
3. Let $z^n$ be a nth root of unity. Then $z^{-a}$ is also a nth root of unity as $z^{a} = 1 \implies z^{-a}=1^{-1}=1$. $z^{a} \cdot z^{-a} = z^{a+-a}=z^{0}=1$. Therefore $z^n$ has an inverse in $U_n$.

[azure vault]

again

[boreal girder]

What a wonderful world !

[carmine garden]

I should re-write the whole thing probably

[azure vault]

why are those seemingly arbitrary nth roots of unity both powers of some z

and what is a and b

[carmine garden]

yea, I got confused

[azure vault]

use only this definition

[carmine garden]

1. closure, Let $z_1 z_2$ be two nth root of unity. We then have $z_1^n \cdot z_2^n = 1 \cdot 1$. So $(z_1 \cdot z_2)^n=1$. This proves closure under multiplication.
   \
   2)Let $z=1$, $1^n=1$. The identity element is thus in $U_n$
   \
   3)If $z$ is a root of unity, then $1/z$ is one too , $(1/z)^{n}=z^{-n}$ ,$( z^{n}=1 \implies z^{-n}=1^{-1}=1$.We consider the product $z^n \cdot z^{-n} = z^{0}=1$. Thus $U_n$ is closed under inverses too

[azure vault]

again

it's z_1 and z_2 that are roots of unity

z_1^n and z_2^n are both 1

[carmine garden]

yea

oh

[boreal girder]

What a wonderful world !