help-38

[last falcon]

hold on

if undefined -> not included
if zero -> included

this is the rule right

so 4 and 3 ,s undefined

so we dont include these

[near vessel]

correct

,,(-\infty,-4)\cup[-2,-1]\cup(3,+\infty)

[solid kiln]

Renz

[near vessel]

yep

[last falcon]

okay thanks

one more question if ur available

[near vessel]

this is the rule if the inequality is $\le$ or $\ge$

[solid kiln]

Renz

[near vessel]

this

[last falcon]

how does it change when it is < or >

[near vessel]

if its only $<$ or $>$ we dont include anything at all

[solid kiln]

Renz

[last falcon]

oh okay

[last falcon]

.

[near vessel]

sure

[last falcon]

so sometimes some numbers have both positive and negative numbers

between the zeros

should i include them as - or +

[near vessel]

between the intervals?

no thats not possible

[last falcon]

dont mind my bad handwriting

for example 2x+5

this is another question alr

when -2 its +

when -4 its -

[near vessel]

oh

-5/2 is your zero right

[last falcon]

yep

[near vessel]

thats

-2.5

-4 is between -5 and -2.5

[last falcon]

oh okay

-2 isnt in there

oh dont mind me lol

[near vessel]

while -2 is between -2.5 and 3

yea

[last falcon]

okay lol sorry

[near vessel]

its on another interval

[last falcon]

that was a dumb one

[near vessel]

np

[last falcon]

but it cant happen

right

[near vessel]

yea

[last falcon]

alr

thank u so much

[near vessel]

no prob

[last falcon]

have a great day im closing the chat

.close

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[graceful zenith]

how do u do this

[graceful zenith]

i used this

ignore the thing in the way btw

[pseudo grotto]

draw out a triangle

with legs 1 and sqrt 3

and then another with legs 1 and 1

[graceful zenith]

is there a way to do it using the formula

[graceful zenith]

this one

[subtle lava]

Both of those values are common values of arctan

Whereas when you combine them you get a not so nice value

It would make sense splitting it up using the formula, but recombining makes no sense

[pseudo grotto]

i mean yes but who would just know the arctan of that argument off the top of their heads

[graceful zenith]

u mean the formula?

[subtle lava]

Yeah but the resulting value is more complicated than the two started with

[graceful zenith]

i just remember the tan (x+-y) thing

[pseudo grotto]

i was simplyn answering the question "is there a way to do it using the formula", not suggesting for him to do it this way

[subtle lava]

I'm guessing this question is part of a question where that formula is suggested but to go the other way.

[strange ice]

can't u just use exact value table

[subtle lava]

Yeah sqrt(3) and -1 are nice values of arctan

[strange ice]

ok thats pretty simple with exact values

remember to always use general solutions

[solid kiln]

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

[graceful zenith]

would arctan (surd 3 + 1)/ 1 - surd 3 be this?

[pseudo grotto]

they are asking you to evaluate the arctans and give your answers not containing arctan

[pseudo grotto]

so you don't want to use this to evaluate

[graceful zenith]

oh i do

[graceful zenith]

this is my ans but im getting -5/12 pi

when the actual answer is 7/12 pi

@pseudo grotto

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[odd galleon]

Let f be the function given by f(x) = 2e^(4x^2). for what value of x is the slope of the line tangent to the graph of f at (x, f(x)) euqal to 3

[wraith hinge]

what have you tried?

[odd galleon]

IVE tried derriving it

[wraith hinge]

thats a good start

[odd galleon]

the derrivative is 8xe^

right

^everythignesle

ok 8xe^(4x^2)

right

[wraith hinge]

that's incorrect

[odd galleon]

yeah wait

[wraith hinge]

you are off by a factor of 2

[odd galleon]

16

[wraith hinge]

yeah that's right now

[odd galleon]

now what

i set it equal to 3

[wraith hinge]

yeah

[odd galleon]

now what

3/16 stuck

[wraith hinge]

solve for x

[odd galleon]

dunno how to

xe^4x^2 = 3/16 stuck

[wraith hinge]

yeah that doesn't sound like it can be solved without some numerical approximation. Can you show a screenshot of the original problem maybe?

[odd galleon]

typed it exactly but illget pic

[wraith hinge]

i mean none of those are right since the answer is seemingly irrational (one of them approximates the actual answer though)

do you know some numerical methods to solving this type of problems?

[odd galleon]

i dont know

this is on a past ap ab exam

1997

[odd galleon]

if i knew i would do

[celest cloud]

,rotate

[rough goblet]

@odd galleon what did you get as the derivative?

[odd galleon]

16xe^(4x^2)

[rough goblet]

hm

[solid kiln]

ren

[rough goblet]

...

[odd galleon]

is that right?

[rough goblet]

yes

[wraith hinge]

you could Taylor expand e^(4x^2)

that does get u the answer

[rough goblet]

huh

didn't think of that

so hang on

[odd galleon]

what is taylor expand

[rough goblet]

ah

nvm then

there's something called a taylor series, basically

[wraith hinge]

do you know linearisation?

[odd galleon]

maybe

[wraith hinge]

if not this is basically impossible for you to solve

[rough goblet]

its an infinite series (and a polynomial) that represents a function

[odd galleon]

i dont think so

[wraith hinge]

,w taylor e^(4x^2)

[solid kiln]

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[wraith hinge]

you can approximate e^(4x^2) = 1

[rough goblet]

...??

ah

[odd galleon]

1 ?

[wraith hinge]

then your problem reduces to 16x = 3 which is close enough to one of your answers

[odd galleon]

alright

A?

[wraith hinge]

,w 16xe^(4x^2) = 3

[solid kiln]

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[wraith hinge]

yeah seems like it

,calc 3/16

[solid kiln]

Result:

0.1875

[wraith hinge]

its off a bit but what can you do

that equation is unsolvable without some sort of approximation I'd think

[odd galleon]

okay

well i dont know how to do that

thank you though

goodbye...

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[wraith hinge]

uh

[solid kiln]

Akti

[wraith hinge]

divide by a^2 + b^2

and multiply by the same

you get a new formula

[solid kiln]

Akti

[wraith hinge]

yes

[solid kiln]

Akti

[wraith hinge]

right

now since a > b > 0

if you see carefully your f(x) is an even function

for ab > 0

even function means f(x) = f(-x)

[solid kiln]

Akti

[wraith hinge]

this

if ab we have same sign

equality holds

[solid kiln]

Akti

[wraith hinge]

there will be solutions at multiple points if a,b belongs to R

you can draw plot both sides of equation

on graph

you will get periodic solutions

we can't commend with signs of a and b

but if a and b have same sign equality will always hold is what i meant

solve it then

[solid kiln]

Akti

[wraith hinge]

solve the trigonometric equation

equate both

write it in terms of cosine

square

but thats not efficient

i suggest you solve it graphically

a b belongs to R ?

or theyre greater than zero

ah

there will be cases

i guess

yup

easier case

[solid kiln]

Akti

[wraith hinge]

yes i guess

but still i am not good with proofs , maybe theres some direct way

to do this

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[wraith hinge]

Can anyone help me out with this one

[regal bough]

.rotate

,rotate

[wraith hinge]

Like i solved it but ans not matching

[regal bough]

the range of values for x that are impossible to attain in this function

u find that

can i see ur working?

[wraith hinge]

Yeah yeha just a min

[regal bough]

,rotate

[wraith hinge]

,rotate

[wraith hinge]

.rotate

[regal bough]

chill

i already did it

https://cdn.discordapp.com/attachments/908078008609431674/1226874158240239636/855026437198577695.png?ex=66265a99&is=6613e599&hm=ed63a53eb24a55eef745ace303e4d60ae14b144d38d3e184869060f6d747c6f2&

when solving for the inequality

u must consider the denominator as well

because the critical points (zeroes) of the denominator influences the sign too

[wraith hinge]

This is the answer like if we condier the denominator as well

We can just not have -4 included and if -4 is there denominator becoms 0 and function becomes invalid

<@&286206848099549185>

[regal bough]

yes but the behaviour around the function can change the sign

what u did with the numerator for the inequality, with the sign diagram

do the same with the zeroes of the denominator

[wraith hinge]

I did but idk its not going can ya solve and show

<@&286206848099549185>

[past umbra]

Are you familiar with the Wavy curve method?
You'll find it easier to solve inequalities that way

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[wraith hinge]

Yeh actually im trying that only can u just illustrate it once

<@&286206848099549185>

<@&286206848099549185>

[wraith hinge]

.rotate

,rotate

[raven pagoda]

Find the value of x where the function doesn't exist

@wraith hinge

You know that

$\frac{k}{0} = \infty$, where k is a constant

[solid kiln]

Tittom_123

[raven pagoda]

And

$\sqrt{-k} \text{ does not exist}$

[solid kiln]

Tittom_123

[heavy quarry]

?

[wraith hinge]

I did that I placed the denomiantor in the ineq >0 and numbertor to be >=0

But dude my doubt is

This is the answer

I am getting this but just not the [-4,0] interval but again I think it should not be there as if -4, is included denominator becomes 0 and function becomes invalid

<@&286206848099549185>

@raven pagoda

[raven pagoda]

You can always try the function for values in the interval you found to confirm your result

[wraith hinge]

Wait one of the terms in denominator is in even power i guess i applied the wavy cury method wrong

[wraith hinge]

@raven pagoda i got the answer but still one things that not fitting it according like all the correct ways -4 is counted as one of the inputs for x but if we put -4 denominator is beconing 0 either the interval [-4,0] should be(-4,0]

[shrewd vessel]

yh should be (-4,0], answer made a mistake

[raven pagoda]

It's not rare that mistakes are present in the correction of exercises in exercise books

[wraith hinge]

Yes man

@raven pagoda and @shrewd vessel
Thanks guys

.close

alr

.close

[shrewd vessel]

dont worry bout closing it, the bot is down atm

[wraith hinge]

Alright cool

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[light karma]

do you know the

newton-leibnitz

formula

[shrewd vessel]

?

[regal bough]

use fundamental theorem of calc

[shrewd vessel]

this is Fundamental Theorem of Calculus

[dapper swift]

It's the chain rule

[light karma]

they asked derivative

[regal bough]

evaluate the function over the bounds provided

[light karma]

so

leibnitz

formula

[dapper swift]

So you have F'(x^2 + 5x + 3) - F'(5x + 5)
And also F'(x) = sec^2 x

[regal bough]

yes

[calm quarry]

Yea right

[regal bough]

that

[light karma]

its leibniz

[fervent prairie]

ah ive never seen leibniz

[regal bough]

neither have we

he is dead

[dapper swift]

lmao

[light karma]

https://www.cuemath.com/jee/newton-leibnitz-formula-indefinite-integration/

[dapper swift]

If you understand how to do the chain rule you should get how to do this

[light karma]

:)

[fervent prairie]

ohhh its kinda what ive been doing but in diff format

[dapper swift]

True

[fervent prairie]

never knew there was eqn for it

ok

[light karma]

here's the formula

[fervent prairie]

ty

[light karma]

well, something new in that case!

:)

[fervent prairie]

ok i got it thanks math genuises

geniuses

[dapper swift]

Npnp

[fervent prairie]

same ting

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[hasty bobcat]

how would I solve this?

[hasty bobcat]

I knew but I forgot

with factorial

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[scenic jacinth]

Random thought, is this true?$\ \lim_{k \to \infty} \sum_{n=a}^{b \cdot 10^k} f(n \cdot 10^{-k}) = \sum_{n \in \mathbb{R}}^{a \geq n \geq b} f(n)$

[solid kiln]

7aman

[subtle lava]

uh

the sum on the right should be an integral right?

because otherwise it is undefined

[scenic jacinth]

Abuse of notation

But yeah that should be an integral

Wait

You're right

Lmao I just remade the reinmann sum

[austere cedar]

Kind of. You also need the lengths of the rectangles

For some f this won't be defined

[scenic jacinth]

Damn

Nice

Thanks guys

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[wraith hinge]

he’ll

[wraith hinge]

how do i solve the top

i thought maybe use a triangle but im no closer to solving

[ionic pendant]

then you have it that $\theta = \cos^{-1} \ab(\frac 1x)$

[solid kiln]

pnoןɔ

[ionic pendant]

so you can find the other functions with SOH CAH TOA

[wraith hinge]

idk if i’m just being blind

[ionic pendant]

you have it that $\theta = \cos^{-1} \ab(\frac 1x)$, so $\sin \ab ( \cos^{-1} \ab(\frac 1x)) = \sin \theta$ which you can find from the triangle

[solid kiln]

pnoןɔ

[wraith hinge]

yeah

i got it

thankyou

idk why i was being completely blind

thanks for ur help!

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[jagged wharf]

The above is an extract from my measure notes, on the Lebesgue Stieltjes measure mu_F. Isn't the Cantor function a function from [0,1] to [0,1]? Also I don't understand the last sentence; what is the author trying to get across?

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[fair meteor]

tan is opp/adj right? like mentioned earlier

and cot is 1/tan

so what do you think cot is in terms of opp adj and hypothenuse

[obsidian cedar]

basically just find tan(y) and take the reciprocal

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[molten perch]

How come u substitution doesnt work here?

[wraith arch]

it does work

[molten perch]

Is my work wrong?

[wraith arch]

do you remember the log rule?

[raven pagoda]

Also, it's ln(Abs(x))

[wraith arch]

where $ln(ab)=ln(a)+ln(b)$

[solid kiln]

y0shi

[wraith arch]

the two answers are just off by a constant

[fair meteor]

its just integral of x times 1/2 wso really its just integral of x which is ln(abs(x))

[molten perch]

and the constant goes into the +c? so it doesnt matter?

[fair meteor]

you overthought

[wraith arch]

yeah basically

[molten perch]

how about this one

[wraith arch]

the C's for the one with substitution and the one without substitution are different

[raven pagoda]

Your "u" is the equation inside the square root here

[molten perch]

this is the key

[molten perch]

but is it not possible the other way? im just trying things to get used to integrals

[raven pagoda]

Yes, this is how you do it

[raven pagoda]

No. You can't do that

[molten perch]

why

[wraith arch]

both answers up there are the same actually

$2\sqrt{2}=2^{\frac{3}{2}}$

[solid kiln]

y0shi

[wraith arch]

and we can put that back into the 2-t bit

[wraith arch]

and still get this

[molten perch]

would i get deducted points for doing it an unconventional way?

or do most people not care

[wraith arch]

they are both fine in my opinion

really depends on your teacher tho

counting which is more simplified

but really just u sub the equation inside

much easier and less work overall

[molten perch]

thx

.closae

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[full tusk]

for each of the following im supposed to draw the nullclines and phase plane, is this how i do it?

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[full tusk]

forgot to post this

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[dense nebula]

!help

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[dense nebula]

@sunnyz

[raven pagoda]

Get your own channel

[dense nebula]

@opal inlet

[sour plinth]

<@&268886789983436800>

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[wraith hinge]

(I opened a help channel yesterday about this but I abruptly had to leave)
I want to expand on what's being talked about here

[wraith hinge]

as far as I can tell, this is saying functional linear dependence is stronger than pointwise linear dependence

but why doesnt functional linear independence also imply pointwise linear independence?

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[ionic pendant]

to take the 1D case, the functions f(t) = t and g(t) = t^2 are linearly independent because there are no nontrivial constants a, b such that
a f(t) + b g(t) = 0
for all t in a given interval. but for any given point t=T in the interval, we can choose some a,b such that
a f(T) + b g(T) = 0
(in the 1D case we are guaranteed pointwise convergence, since 2 real numbers can't be linearly independent)

[wraith hinge]

yeah its like we are flipping between the \forall \exists and \exists \forall

and apparently \exists \forall is generally stronger which I dont quite get

[ionic pendant]

well if we instead picked f(t) = t and g(t) = 5t, then clearly we could pick a, b such that
a f(t) + b g(t) = 0
for all t in a given interval. but since it works for all t, clearly the same coefficients work for any given particular T as well

[frosty slate]

I have a question, what kind of math is this?

[wraith hinge]

linear algebra

[wraith hinge]

ohhhhhhhh

alright that makes more sense

ok i think i get this thanks

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[lavish sail]

can someone help

[lavish sail]

genuinly nothin

cause i was absejnt at class

and i dont understand any math videos

[orchid wagon]

have you tried drawing the diagram

[lavish sail]

ye

da traingle

[orchid wagon]

ok, so do you know the information the question gave?

[lavish sail]

no

[orchid wagon]

well, have you learned similar triangles yet?

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[zealous nova]

hello. i need some help with excel and stats

[zealous nova]

im trying to make a graph like this

the red is the control and the blue are the different sets of data i have

how do i make it such that it each data point is compared to the control vs something like this

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[turbid ravine]

can someone explain how the answer is C?
I worked it out and I got -cos(x)-2 in contrast to the answer 2-cos(x)

[orchid wagon]

show your work

[turbid ravine]

Start off: Cos(3x)-2
Reflected in the x-axis: -Cos(3x)-2
Dilation of factor 3 from the y -axis : -Cos(x)-2

nvm found out why

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[wraith hinge]

So I square rooted the equation then solved but that didn't wrk

I don't understand what the first line is

[tribal jetty]

the first line the derivatives of r=(9t^2,18t)

[strange ice]

the 1st one is deriving it

[wraith hinge]

Why must we derive the points?

[strange ice]

derivatives give the tangent of the curve

at the relative x value

[wraith hinge]

You mean the gradient of it?

[strange ice]

yea

they are the same thing

well have the same gradient

[wraith hinge]

Why did they only derive the y value

[strange ice]

they did both

[wraith hinge]

Oh yeah I see that

Then they subbed it into the equation for finding a line

[strange ice]

yea

[wraith hinge]

Ok I get that

But why is square rooting y^2 = 36x then differentiating it wrong?

[strange ice]

idk, i think it's cuz u need to incorporate t into the equation too

[wraith hinge]

Could it be because I'm losing solutions?

[strange ice]

idk really, i haven't learned derivatives with parametric equations (looks like parametric equatinos)

[wraith hinge]

Ok thanks though

[strange ice]

cuz if u jus sqrt its only y and x

[tribal jetty]

(9t^2,18t) means x=9t^2 and y=18t

[wraith hinge]

Oh yeahhhhhhhhhhhhhhhh

Yeah I get that

[tribal jetty]

y^2=18^2 t^2

y^2 = 18^2 (x/9)

y^2 = 36x

[wraith hinge]

Gotchu

Thanks for the help

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[tribal jetty]

but shouldn't be wrong

if you take y=sqrt(36x) you are missing y=-sqrt(36x)

prob is that

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[storm mist]

help

[storm mist]

how is 1-phi1.285 = 0.0994

phi 1.285 = 0.8997 + 0.09 no?

then take 1- that

nvm

0.0009

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[storm mist]

solved it myself thx

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[cold vapor]

Find (a+b)/(a-b) If i found the value of b/(a-b) can i multiply b and value by 2 and make it like this? Seems logical but not sure 1+[(2b)/(a-b)]

[lean spear]

?

if you know b/(a-b) split the fraction into a/(a-b) + b/(a-b)

[cold vapor]

Ight ty

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[crisp storm]

Confused about the ((A and B) <= B): Why is that implication flipped around? Can i just do this? (B => (A and B))

[pseudo grotto]

can you provide context

o wait

[crisp storm]

im supposed to simplify this logic formula

[pseudo grotto]

you're asking if you can just rewrite it

[crisp storm]

Yes i am

[pseudo grotto]

yes you can

[crisp storm]

So A => B == B <= A ?

[pseudo grotto]

no, if you just rewrite it the statment becomes $(A\implies B)\lor (B\implies(A\land B))$

[solid kiln]

M8 of 48

[crisp storm]

Yeah that i understand, i was just wondering about the general implications not considering this specific case

but that answers my question, thank you very much!

[pseudo grotto]

oh wait

[pseudo grotto]

yes i misread this

that is correct

the two are equivalent

[crisp storm]

Thank you :) Have a nice day, you rock!

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[last falcon]

can someone help me review my polynomial test. I took it yesterday and have some wrong questions and i wana get over these

[last falcon]

firstly

listen I know long divison okay i did tons of practice but i couldnt do this one

this is my work

it didnt work and it didnt make sense

[ornate sleet]

im not to sure but ur supposed to bring down 3rd term i think

ye

[shrewd vessel]

^

[last falcon]

but like shouldnt i be going in order?

[ornate sleet]

after each division ur supposed to bring down successive terms i think

[regal bough]

why didnt the -x^3 cancel?

[last falcon]

idk

i tried every way

[shrewd vessel]

this is an issue that arised because you're dividing by 2x^2-1, which doesn't have the term x

my tip is

try writng 2x^2 -1 as 2x^2 +0x -1

[last falcon]

like here x^3 is not that the thing u show

if i would try to pull it down

i thought havign 3 terms wouldnt be great

3 numbers or

i know it'd neutralize

[shrewd vessel]

no, it's fine

[last falcon]

but idk

[last falcon]

oh

i always saw

2

so thought it wouldnt work

also got these

[regal bough]

hole is like

if u have a function f(x)=x/x

wait bad example

f(x) = (x+3)(x+2)/(x+2)

u can simply cancel out x+2 right?

to get f(x)=x+3

[last falcon]

yeah

[regal bough]

but there is a hole

when x=-2

u r essentially doing 0/0

the function's domain excludes that

[last falcon]

so 0/0 is a hole?

[regal bough]

places where u get 0/0 are holes

if u draw a graph

[last falcon]

oh

but not 0's right

[regal bough]

u just place a hole to represent discontinuity at that point

[last falcon]

so w is a hole

bc upper is 0 when x=-3

[regal bough]

like vertical asymptotes?

vertical asymptotes arent holes

[regal bough]

roots arent holes either

[last falcon]

no like anyways

[last falcon]

but

x=-3

va= -3

and root =-3

[regal bough]

r u referring to this function

[last falcon]

and when x=-3 f(x)= 0/0

[last falcon]

no

w(x)

[regal bough]

which one

oh

[last falcon]

on my page

[regal bough]

lemme see

here right

[last falcon]

yep

[regal bough]

when x=-3

thats not a hole

its a VA

wait

lemme factor

[last falcon]

but upper is 0 too

alr

-27+54-27

no need to factor just plug it in

[regal bough]

yeah u r right

its a hole

[last falcon]

alr

uhm one more

question

u see c)

where it says no vertical astmpote

i did the x^2 ones

but they added w(x) as answer too

why's that is it because it's a hole?

[regal bough]

yeah

theres no vertical asymptote since the denominator cancels out

[last falcon]

okay

mhm

alright thank u

im closing the chat

[regal bough]

np

[last falcon]

have a great day

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[lament plaza]

open you dumbfuck

[lament plaza]

yeah right

so

im thinking conditional identities

reaching out to you @regal bough
since no one helping

RIP

<@&286206848099549185>

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[lament plaza]

.reopen

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✅

@lament plaza Has your question been resolved?

[lament plaza]

.close

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[lament plaza]

here @silk sequoia this the ques

.reopen

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✅

[silk sequoia]

Oh yeah I'm not doing that. Gl though lmao

Just be patient. Someone will come around eventually.

In the US people are in class. In Europe people are getting home and eating etc.

Helpers will be in en masse in the next couple hours

[lament plaza]

[trim jolt]

@lament plaza Has your question been resolved?

[trim jolt]

@lament plaza Has your question been resolved?

[tribal jetty]

i want to see the solution for this problem

[regal bough]

!status

[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[lament plaza]

🤓

[lament plaza]

1

[regal bough]

im on the same step

[raven pagoda]

Explain me your problem @lament plaza

[lament plaza]

bro thinks she me

i have no idea what to do

[raven pagoda]

Can you transfer it to LaTeX? Your writing makes it unclear tbh (sorry)

[lament plaza]

@regal bough
hey there buddy
my good friend "you're super cool
the greatest person ever

(mind helping me with the latex )

[raven pagoda]

Yes

[lament plaza]

its basically

tan nd argument is binomial coefficient

in form of nCr

[raven pagoda]

$\tan\left(2024_{C_{1000}}\right)$ ?

[solid kiln]

Tittom_123

[raven pagoda]

@lament plaza

[lament plaza]

yes like this

next term is
m times tan (2023C1000)

[raven pagoda]

Can you quickly explain me nCr? I am ok to help, but I don't know that much about this yet

[lament plaza]

then - tan(2023C999)

[lament plaza]

its

where r is k

[raven pagoda]

$\tan\left(2024_{C_{1000}}\right)+m\tan\left(2023_{C_{1000}}\right) - \tan\left(2023_{C_{999}}\right)$

[solid kiln]

Tittom_123

[lament plaza]

yes we are equating it to the product of same kind almost

same tan nd all
just extra n

[raven pagoda]

$\tan\left(2024_{C_{1000}}\right)+m\tan\left(2023_{C_{1000}}\right) - \tan\left(2023_{C_{999}}\right) = n\tan\left(2024_{C_{1000}}\right) \cdot \tan\left(2023_{C_{1000}}\right) \cdot \tan\left(2023_{C_{999}}\right)$

[lament plaza]

no its product in the right hand side

[regal bough]

what course is this

[lament plaza]

Algebra

higschool

[raven pagoda]

?

[lament plaza]

prolly

[regal bough]

no like what exam r u preparing for

[lament plaza]

its ntan(arg) times other tan times other

[solid kiln]

Tittom_123

[lament plaza]

yes this

[raven pagoda]

Wouldn't it be better to just switch nCr to it's fractional form?

[lament plaza]

its the entrance to the best mathematics institutes in the country or equivalent

[regal bough]

no

[lament plaza]

idts if it will simplify things

looks ugly

[regal bough]

is this a multiple choice?

[lament plaza]

integer type

need to find m^2 -n

[regal bough]

whats the range of the integer possible

if they give u that

[lament plaza]

i have no idea

could be numerical

or could be 1 to 9

more probability for former

[raven pagoda]

Isolate m and n in the equation?

[lament plaza]

can you do that?

[raven pagoda]

Probably, if they mutiply a term

[lament plaza]

😭
i have no idea how to approach it

[raven pagoda]

Isn't nCr something in probability?

[lament plaza]

no 😭

its just a factorial nnotation
or saying number of ways to choose r things from n things

[regal bough]

combinatorics

the most important part of studying math is to know when to give up

nows the time

[lament plaza]

so i just fail

like dat

[regal bough]

yes

go to a foreign university

its gonna be easier than whatever this shit is

[raven pagoda]

Using (n k) could be better than the C notation

[regal bough]

ncr is best notation when typing or writing

n k is better in binomial theorem

[lament plaza]

i applied

[raven pagoda]

Never give up. That's not a good mindset

[lament plaza]

i dont got enough aid to attend NYU

[regal bough]

yeah spend ur whole life on one problem

[raven pagoda]

Yes

[regal bough]

u could apply to germany

its free tuition

or china

[lament plaza]

too late now

i'd think of transfer

i wanted to apply for tshingua

[regal bough]

the applications close in may or something

[lament plaza]

too late to make up my mind now

[raven pagoda]

@lament plaza you know that tan takes an angle in radians or degrees, right? First of all, in your problem, is it in radians or degrees?

[regal bough]

radians

[lament plaza]

radians ofc

ig

[lament plaza]

why are you soo...

nerd

[regal bough]

...

[lament plaza]

why do you know everything

why

[raven pagoda]

Then find the smallest angle value in the trig circle?

[lament plaza]

idk how that helps

[raven pagoda]

Idk too, but we gotta try

[lament plaza]

im guessing there gotta be a manipulation here

[regal bough]

do the tan approximation

tanx=x

[lament plaza]

did you try it yourself ?

[regal bough]

no because it wont work

[lament plaza]

axxy axxy

you're so literally me

😭

[regal bough]

🤓

[raven pagoda]

Maybe you could try using logarithms?

[lament plaza]

imma try

[raven pagoda]

Like this

[lament plaza]

i see how does that help tho

[raven pagoda]

Uhm

[raven pagoda]

Maybe this could help

[lament plaza]

this makes sense
but now

this seems trivial but how do we manage the equations is smth i cant figure out

im trying

[raven pagoda]

There might be formulas to express (n-1 k-1)

[lament plaza]

@raven pagoda @regal bough

i 've done it

omg

you all its so easy

im so silly
why didnt i think of this earlier

😭

like bro wtf

😭

listen

[raven pagoda]

Ok

[lament plaza]

DO NOT OK

LISTEN TO MY SOLUTION

so we observe

Let A = 2024 C 1000

B = 2023 C 1000
C = 2023 C 999

we observe that

B+C = A

(binomial property)

take tan both sides

nd compare

gives m = -1
n = 1

so final answer 0

nooo wayyyy
this was so easy

axxy im better than you

they both not appreciating my solution

[raven pagoda]

Good job

[regal bough]

cool

now close the channel so others can use it!

[slim saffron]

.

[trim jolt]

@lament plaza Has your question been resolved?

[lament plaza]

@tribal jetty

[lament plaza]

jealous ass

.close (she prolly gonna bully me)

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[burnt knot]

For a finite polynomial ring $\mathcal{R}=\mathbb{Z}q[\text{X}]/\text{X}^N+1$, we define a subset of the ring, $\mathcal{T}={a\in\mathcal{R} : a = \sum{i=0}^{N-1}a_i\text{X}^i, a_i\in{0,1,q-1}}$ and denote the polynomial $a$'s multiplicative inverse is $a^{-1}$ (of course in \mathcal{R}). The question is that if I uniformly sample an element from $\mathcal{R}$, what's the possibility to sample an inevitable polynomial?

For a finite polynomial ring $\mathcal{R}=\mathbb{Z}q[\text{X}]/\text{X}^N+1$, we define a subset of the ring, $\mathcal{T}={a\in\mathcal{R} : a = \sum{i=0}^{N-1}a_i\text{X}^i, a_i\in{0,1,q-1}} and denote the polynomial $a$'s multiplicative inverse is $a^{-1}$ (of course in $\mathcal{R}$). The question is that if I uniformly sample an element from $\mathcal{R}$, what's the possibility to sample an inevitable polynomial?

For a finite polynomial ring $\mathcal{R}=\mathbb{Z}q[\text{X}]/\text{X}^N+1$, we define a subset of the ring, $\mathcal{T}={a\in\mathcal{R} : a = \sum{i=0}^{N-1}a_i\text{X}^i, a_i\in{0,1,q-1}}$ and denote the polynomial $a$'s multiplicative inverse is $a^{-1}$ (of course in $\mathcal{R}$). The question is that if I uniformly sample an element from $\mathcal{R}$, what's the possibility to sample an inevitable polynomial?

[solid kiln]

[RRFA]DukeAble

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[rapid dock]

my answer was one first

then i did the math again and got 0

[quasi wasp]

show your process and what you worked so far

[rapid dock]

well i changed sin x to 1- cos x

[marsh forum]

you can't do that

$sin(x) \neq 1-cos(x)$

[quasi wasp]

1-cos^x = sin^2

[solid kiln]

ƒ(Why am. I here)=I don't know

[rapid dock]

ok i tried to do smt similar to a problem in my lesson

[solid kiln]

Talent Unlimited

[rapid dock]

but what would i change it to

[quasi wasp]

multiply numerator and denominator by 1+cos(x)

and then simplify

[rapid dock]

so sin x is 1 + cos x

[quasi wasp]

no

basically what you want to do is make the denominator in the form of 1-cos^2(x)

[rapid dock]

hmm ok

[quasi wasp]

to do that you need to #help-38 message

[rapid dock]

i have sin x (1 + cos x)/ 1-cos x ( 1 + cos x

)

[quasi wasp]

yes

now simplify the denominator

[rapid dock]

2

[quasi wasp]

what would you get?

huh?

[rapid dock]

mb it sent that by accident

id get sin 2x - cos x

[dapper swift]

You don't need to use the identity 2 sin x cos x = sin 2x here

[rapid dock]

ohh ok

then

[quasi wasp]

just open the brackets

[dapper swift]

So what does (1 - cos x)(1 + cos x) simplify to

This is the important part

[rapid dock]

i dont distribute it then

[dapper swift]

You distribute the numerator still

[rapid dock]

uhhh

[quasi wasp]

(1 - cos x)(1 + cos x) open the brackets in the denominator

distribute

[rapid dock]

-cos^2x - cos x +1

[quasi wasp]

no

[rapid dock]

i dont understand what youre asking then

[quasi wasp]

show your work i'm not sure how you're reaching that answer

[rapid dock]

oh wait nvm

-cos^2x +1

[quasi wasp]

yes

and what would that be?

you should use an identity here

[rapid dock]

isnt that sin^2 x

[quasi wasp]

yes

[rapid dock]

okk

[quasi wasp]

now you can further simplify the fraction

[rapid dock]

so i have sin x( 1 + cos x)/ sin ^2x

[quasi wasp]

yes

[dapper swift]

Yes so you didn't actually need to open the brackets

[quasi wasp]

yeah

[rapid dock]

okk

how would i simplify the numerator

[quasi wasp]

you don't need to

you can directly divide the numerator and denominator by sinx

[rapid dock]

alright

its just sin x then

sin x ( 1 + cos x)

[rapid dock]

what do i do now

[quasi wasp]

.

[rapid dock]

ive done that

[solid kiln]

Talent Unlimited

[rapid dock]

its csc?

csc x

[quasi wasp]

and?

[quasi wasp]

what did you get

[rapid dock]

nvm that i did smt wrong

the only option with csc is cot

but idk how wed get cot

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[wraith hinge]

Does: "Where f(x) <= 0" mean the area of the curve where x <=0

[wraith hinge]

Or is it talking about x^2 +x-6 <= 0

[magic badger]

it's talking about the region below the x axis but above the curve

[shrewd vessel]

yes

wait no

[wraith hinge]

My wording was a bit weird

I understand now though

Thhankyou

[shrewd vessel]

that's not

wait nvm braindead moment ignore me

[wraith hinge]

haha all good

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[devout drift]

What have you tried?

Is that 'a'

[rapid horizon]

Yes

No idea

I got it

.clode

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[steel birch]

can someone pls show the steps for this pls?

[fair meteor]

Can you show which part e specifically

You’re having problem with

Or you don’t know why they did that step

[steel birch]

i dont understand how to calculate the thing i cirlced in red

i understand why they did the step but idk how to expand it from the first step to the solutions

[fair meteor]

You multiply both side by denominator

[steel birch]

oh yeh i mean that one

[fair meteor]

That’s all they did

[steel birch]

wait but how did they get a = 50/81

[fair meteor]

Ah

Try

Organizing them

By x^2, x, and constant

s in your case

[steel birch]

wdym by organizing

[fair meteor]

For example (A + B)x^2

• (C)x

Multiply eveyehting and organize them by order of exponent of s

[shrewd vessel]

organising is not necessary, but it saves you time writing out the long equations every time when you make substitutions to test

[fair meteor]

It’s helpful to say the least

[shrewd vessel]

there are 4 unknowns here, a b c d, and s which you have control over. So make some substitution for s, and get some simultaneous equations to solve

the obvious numbers to try is s = 0 and s = 1, as those give you some zeros

[steel birch]

oh wait

i remember how to do these

[shrewd vessel]

then try two other simple numbers

[steel birch]

form calc bc lmfao

[shrewd vessel]

there is another way actually

[steel birch]

instead of subbing?

[shrewd vessel]

yh

expand all the terms and factorise in terms of s^0, s, s^2 and s^3 (so you have a polynomial with m + ns + ps^2 + qs^3 and you equal this to 5+12s^2-s^3)

and compare coefficients

[steel birch]

um since im currently cramped for time I'll stick with the substitution method for now but I'll take a look into that when after my assignment! 🙂

[shrewd vessel]

yh both methods are good, personally I recommend sub in this case

[steel birch]

wait after doing sub

i get that b is 5/9 and c is -2 and d is 31/81 but how do i easliy get a after that?

wait nvm

i can just sub in a non root value after subbing in all the other constants

and then just eval

alright

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