help-10

[uneven flicker]

sin(2x) = sin(210 degrees) and sin(330 degrees)

so 2x = either 210 or 330 + 2pik

we dont add pik, but 2pik, since if we add pik, we fall under the quadrants where sin(x) is positive

x = 105 +pik, x = 165 + pik

i intentionally wrote 185 wrong, to see whether you were paying attention

[haughty cliff]

no but the anwsers in my book are diffrent

[uneven flicker]

you can group that with this for a single line solution with x

[uneven flicker]

yeah but we didnt group with the cosx

[haughty cliff]

oh ok

[uneven flicker]

here i dont think there is any sol

for this case scenario

so cos(2x) = 1

2x = pik

x = pik/2

oh nvm

[uneven flicker]

this is wrong

,w cos(pi)

[warm shale]

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[uneven flicker]

cos(2x) = 1 only at 2pik

so x = pik

,w cos(2x) + 2sin(2x)cos(2x) = 1 + 2sin(2x)

[warm shale]

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[uneven flicker]

ok yeah pik is from the cos(2x) = 1 case and the other 2 from cos(2x) = 0 case

i just dont have pen and paper to bring to a single solution case, but i think you can do it

its simple

[haughty cliff]

I have tried doing it again on my own from the beginning. Can you check if I did it correctly?

[uneven flicker]

cant see shit

bad picture qual

[haughty cliff]

hold on

now?

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[tiny basin]

Show that $\frac{1}{n} \times Y^{2}_i$ is an unbiased estimator of $s^2$

[warm shale]

catnwaffles

[tiny basin]

I have gotten to $\frac{1}{n} \times n \times E[Y^2]$. Do I just take the second moment of my distribution and simplify?

[warm shale]

catnwaffles

[frosty spoke]

what is the definition of s, and Y_i, and how are they related to each other?

s is the standard deviation of each Y_i?

[tiny basin]

s^2 = population variance. Y_i are just observations

[frosty spoke]

population variance of what?

Y_i?

[tiny basin]

X ~ N(0, s^2). Y = |X|(half normal)

Variance of X.

[frosty spoke]

that information would've been very helpful if you included it in the beginning

then what's Y_i?

independent identically distributed samples?

what's little n

[tiny basin]

IID from half normal

[tiny basin]

size

[frosty spoke]

and little n is how many you took, yeah?

and I assume you actually want to show that $\frac{1}{n}\sum\limits_{i=1}^n Y_i^2$ is an unbiased estimator for $s^2$?

[warm shale]

Saccharine

[tiny basin]

Yeah, for population variance

[frosty spoke]

the variance of X_i

well

[tiny basin]

The question is very confusing sorry

"s^2 is the variance of X, it is not the variance of Y"

[frosty spoke]

well, in a much more rigorous manner, you'd say X_i are independently and identically distributed random variables

with X_i having distribution N(0, s^2)

and suppose you have Y_i = |X_i|

then show that $\frac{1}{n}\sum\limits_{i=1}^n Y_i^2$ is an unbiased estimator for $s^2$

[warm shale]

Saccharine

[frosty spoke]

right?

[tiny basin]

Yes

[frosty spoke]

first, do you see how this is radically different from what you first asked?

not the least that you don't even have a sum in your original question

[tiny basin]

I didn't know how to latex a sum

[frosty spoke]

you can always just write it out in text

something is better than nothing

so you took the expectation of this estimator

and using that Y_i are identically distributed

[tiny basin]

$\int _0^{inf}:x^2\cdot \frac{\sqrt{2}}{s\sqrt{pi}}e^{\left(-\frac{x^2}{2s^2}\right)}$

[warm shale]

catnwaffles

[frosty spoke]

you concluded that this was $\frac{1}{n} \times n \times E[Y_i^2]$ yeah?

[warm shale]

Saccharine

[tiny basin]

Yes

[frosty spoke]

can you see why $Y_i^2 = X_i^2$?

[warm shale]

Saccharine

[tiny basin]

No idea

[frosty spoke]

okay, why don't you try writing Y_i in terms of X_i?

[tiny basin]

Well isn't Y = |X|

[frosty spoke]

it is

Y_i = |X_i| by definition

[tiny basin]

I see why Y^2 = X^2

[frosty spoke]

okay

so then does it make sense that you're computing $\frac{1}{n} \times n \times E[X_i^2]$ then?

[warm shale]

Saccharine

[tiny basin]

Would it matter which one I compute if they're both equivalent?

[frosty spoke]

well, not in the sense that you'd get a different answer

but definitely in the sense that one is much easier to compute than the other

you probably don't even need to compute it with an integral

just stating it as a fact about normal distributions is enough

[tiny basin]

So squaring a half-normal results in a normal dist, so I can just use that fact to substitute X^2 in for my expectation, and use a normal pdf?

[frosty spoke]

squaring a half-normal does not result in a normal distribution

it results in a chi-square distribution, but that's irrelevant

the point is that |X|^2 = X^2, all the same

so computing E[|X|^2] is literally the same thing as computing E[X^2]

[tiny basin]

So how can we go back to the normal distribution here?

[frosty spoke]

wdym by that?

[tiny basin]

I agree that if Y = |X| that Y^2 = X^2

How does that tie into X ~ N and Y ~ half normal

[frosty spoke]

okay, so we've so far proven that the expectation of that estimator is equal to E[X^2], yeah?

[tiny basin]

agree

[frosty spoke]

we know that E[X] = 0 and E[X^2] - E[X]^2 = s^2, since we really said that 0 is the mean of X and s^2 is the variance of X, no?

[tiny basin]

Ah

Yes since E[X^2] = s^2 + mu^2 but mu = 0

[frosty spoke]

so then E[X^2] = s^2

[tiny basin]

thank you

do you have any recommended resources for math stats? My class is kind of a drag

[frosty spoke]

hmm not sure; it's been a long time since I did it

I would say that it helps tremendously to get a solid grasp of probability beforehand

like knowing what random variables are, all of the weird things that happen with it

then if you represent statistics as just reducing real-world problems to those involving questions about random variables

it's much easier than the sorta ad-hoc way statisticians use probability

[tiny basin]

Alright thanks again 🙂

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[coral sable]

Can I integrate like this?

[grizzled shore]

Yes

But only for indefinite integrals

[coral sable]

K thanks

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[grizzled shore]

Definite integrals will have its limits change

[obtuse pebble]

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[balmy steeple]

how do i go about determing if there exists an inner product such that a matrix is self-adjoint with respect to this inner product. And then from there how do i find said inner product

[balmy steeple]

i know that if a matrix is self adjoint then its eigenvalues are real

so my first thought is to compute said eigenvalues ; if they are not real then an inner product does not exist, but if they are then there may exist an inner product

[kind hawk]

see eg https://mathoverflow.net/questions/35618/matrices-self-adjoint-with-respect-to-some-inner-product

[balmy steeple]

okay thanks

[obtuse pebble]

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[mellow yoke]

hi can someone please explain or direct me how to do this modulo operation?

[mellow yoke]

say theres field Z = 7 and i want to know 3^-1 in this field

how do i do that? I can't find anything on it. hopefully i've asked theq uestion clearly.

[haughty coyote]

Find the a solution to the Bezout equation 3a + 7b = 1
Then b = 3^-1 in Z_7

Unless you just use the naive method of trial and error of course

[mellow yoke]

thanks. so the answer would be 4?
and how do I use this equation in other examples? say Z=5 and i want 5^-1
would it be 5a+5b=1 where b=5^-1? lol

ima be so embarassed if it isnt 4

[haughty coyote]

Check it

Also pls don't write Z=n, wtf is this notation

[haughty coyote]

My bad

[mellow yoke]

what's the proper notation?

[haughty coyote]

Obviously a = 3^-1

[mellow yoke]

we use a weird Z for fields

oh so b is 0

[haughty coyote]

$\mathbb{Z} / n \mathbb{Z}$

$\mathbb{F}_{p^n}$ is also possible for fields with $p^n$ elements

[mellow yoke]

that wasnt obvious to me lol

[haughty coyote]

And that is why things are proven in mathematics, not asserted

[warm shale]

mateo713

[mellow yoke]

so another example,
say Z=5 and i want 5^-1
would it be 5a+5b=1 where a=5^-1? lol

[haughty coyote]

If you have such a solution, it should be clear that 3a = 1 mod n

[mellow yoke]

sorry about the notation lmao

[haughty coyote]

Think about what 5 is in Z_5

[mellow yoke]

it's 0, right?

[haughty coyote]

Also it's a=5

[haughty coyote]

Is 0 invertible ?

[mellow yoke]

no.

?

lol

[mellow yoke]

what did you mean by this? I feel like it went over my head. when i think of 0 in Z_5 i know Z_5 is 01234 so 5=0, no?

[haughty coyote]

Yes

So you should absolutely not be asking what its inverse is

[mellow yoke]

and 0 is invertible, yes? so what should i gain from that that i'm not gaining lmao

ohhhhhhhh

right.

[haughty coyote]

I would recommend learning basic group theory and notice that 0 is not invertible

That would probably do you goid cause you look very shaky right now

[mellow yoke]

gotcha. do you have any recommendations maybe youtube wise? i haven't heard anything about group theory yet. this is all coming from my first year linear algebra course.

[haughty coyote]

What is this doing in linalg

This is not linalg whatsoever

[mellow yoke]

a lot of it pops up when solving systems of equations in Z_n

they're so much fun

lmao

;-;

[haughty coyote]

Would be silly to solve systems in Zn before actually properly studying Zn itself

[mellow yoke]

they told us nothing about doing modulo arithmetic(?). i've been drawing clocks and doing calculations by using a clock.

lmaooooooo

so it hasn't worked for inverses

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[pseudo thunder]

I had a question

[pseudo thunder]

About how to solve

A physics question

Question 2

[obtuse pebble]

@pseudo thunder Has your question been resolved?

[clear condor]

!status

[obtuse pebble]

What step are you on?
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2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
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6. None of the above

[pseudo thunder]

I just don’t know what to plug into what solution

<@&286206848099549185>

I just need help to like

Plug in the coefficient of KE

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[crude dawn]

its me again with similar problem

[crude dawn]

i was trying put arctgx as t and arcsinx as a t but got stuck

is it good way..

[glacial obsidian]

You could try writing arctg x / arcsin x as (arctg x / x) · (x / arcsin x)

[crude dawn]

omg it orks

got rid of arcsinus

amazing

thanks

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[north anchor]

in this equation how i can indentify if is antiderivate

[obtuse pebble]

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[hot sonnet]

Hello, integrate twice the differential equation y'' = 2x+5

don't forget the constants of integration

then use the initial conditions given y(0)=1 and y'(0)=1 to find the constants

[north anchor]

A dale pa thx

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[wanton oar]

I have a question

[wanton oar]

why are the following two functions not equal to eachother?

[north anchor]

this is inverse function

[wanton oar]

so basically it means cot(x)?

[north anchor]

in this is a another expression algebragically

[nocturne minnow]

Not that kind of inverse

[wanton oar]

oh

I see

so say that instead of -1 the power was 2

would the functions be the same

[nocturne minnow]

Think of it like this, the opposite operation of addition is subtraction, and the opposite of multiplication is division, the opposite of tan is $tan^{-1}(x)$

[warm shale]

dldh06

[wanton oar]

right right

[nocturne minnow]

Whereas $(tan x)^{-1}$ is the reciprocal, 1/tan

[warm shale]

dldh06

[wanton oar]

I see

So how would you take the derivitive of that?

what rule would you use?

[nocturne minnow]

Of which one?

[wanton oar]

inverse tan

[wanton oar]

is this just a rule you have to memorize

or can you get this using the basic rules

[nocturne minnow]

Pretty much or your teacher would give you th

[wanton oar]

i get it now

thanks a lot

[obtuse pebble]

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[mortal sun]

For a), when we find the error, do we subtract the trapezoidal rule approximation from the exact value?

[mortal sun]

this is how I did it

[obtuse pebble]

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[mortal sun]

<@&286206848099549185>

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[vale trout]

for this function, when I wrap it around to find the 4 answers how much and I supposed to add by for each wrap

the reference angle is 33.7

oh I think it is by 180 (or pi) but im not 100% sure

[iron eagle]

first put the 3 on the other side

then since tangent is negative the angle is either in quadrant QII or QIV

[vale trout]

yes I simplified it to tan^-1(2/3)=2 theta

im currently at the angles 146.3 and 326.3

but from there what do I add by to wrap it?

[iron eagle]

$QII: 180^\circ - \theta$\
$QIV: 360^\circ - \theta$

[vale trout]

ah I see

[warm shale]

Tangerine

[iron eagle]

this is a better picture

[vale trout]

yes I get it now

wowza

I am slight bozo

thx

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[strange sphinx]

Can someone please help me?

[leaden snow]

!status

[obtuse pebble]

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[modern palm]

is (m+1)! <(m+1)^(m+1) the same as (m+1)*m! < (m+1) * m^mwhere m! is factorial of m.

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[sleek mural]

Hi, do my answers seem correct?

[sleek mural]

These r the questions ^

,rotate

[sleek mural]

And here r my answers ^

<@&286206848099549185>

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[foggy forge]

I dont get it..?

What is the question ?

[sleek mural]

This right here

@foggy forge

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[obtuse shoal]

Why is it that when calculating the probability of the union of events A and B we subtract the probability of A*B? Seems to me that P(A*B) would satisfy the condition?

[royal basin]

A*B?

do you mean A&B?

[obtuse shoal]

yeah

[royal basin]

the reason we subtract it, i.e. why P(A ∪ B) = P(A) + P(B) - P(A&B), is that P(A)+P(B) counts the intersection twice.

[obtuse shoal]

well what i read online said P(AB) so I assumed it was multiplication

[obtuse shoal]

ah that makes sense thank you

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[south rivet]

Draw a picture

[obtuse pebble]

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[tiny isle]

hi there

can someone show me the correct approach to this question?

this is the answer, but i do not know how to attain them

like,

i know it can be discovered using binomial distribution, but where exactly does the question tell you to use it

[plain stag]

It doesn't, and it shouldn't have to

It presents you with a situation where the binomial distribution is appropriate to model the situation, and counts on you recognizing it and then applying it

[tiny isle]

ok

what about this one

[plain stag]

What about this one?

[tiny isle]

so i assumed that it should be solved in the same way

but it's not

[plain stag]

What are the requirements for a binomial distribution to be used?

[tiny isle]

only 2 outcomes

fixed chances of success

independency

i think this model takes all 3

[plain stag]

Are you sure that's all of them?

[tiny isle]

ahh

i can't remember more of it

there might be one about

having a fixed number of observations or something?

[plain stag]

Fixed number of trials, yes

[tiny isle]

ok

so in this case 5 spins

hence it is fixed

[plain stag]

Not at all

If your first spin is green, you stop

[tiny isle]

oh

[plain stag]

That's only 1 spin

[tiny isle]

i see

so instead of using cumulative

i should simply find out the possibility of each individual outcome?

[plain stag]

That's what the distribution tells you, yes

[tiny isle]

wait

so how do actually do that

find out the powers of 4/7 ?

that's not right though

perhaps the calculator will help

[plain stag]

Consider thinking out the scenario that reflects each value of the random variable X

What does it mean if X=1?

[tiny isle]

no matter

i got it

[plain stag]

Alright

[tiny isle]

basically i misinterpreted the question

i thought it means that the number of times G hits

it is actually about the number of attempts prior to hitting G

[tiny isle]

thank you for the help though

[plain stag]

Np

[tiny isle]

.close

[obtuse pebble]

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[frosty valve]

The first 3 terms of the expansion (3-1/9x)⁵ can be written as a + b/x + c/x²

[neat ingot]

sorry

[drifting wraith]

np

[plain stag]

Dog pfp...

[plain stag]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[frosty valve]

lemme show the working for mine hold on

@plain stag is this right

[plain stag]

Assuming you've done the multiplication right, your logic seems fine

You can simplify the 2nd and 3rd terms, but otherwise it looks good

[frosty valve]

oh i have to simplify?

shit i left the answer is a=243 b=-405 c=270

💀

there goes my 4 marks

[timid silo]

💀

[frosty valve]

i didnt know we could simplify this

[plain stag]

,calc 405/9

[warm shale]

Result:

45

[plain stag]

,calc 270/9

[warm shale]

Result:

30

[plain stag]

b = -45 and c = 10/3 it seems

[frosty valve]

ok🤡

thx😭

[obtuse pebble]

@frosty valve Has your question been resolved?

[obtuse pebble]

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[uneven night]

. The three sides of a rectangle are called length, width and height. What values ​​can its volume
suppose the height + width = 5 dm and the height + length = 8 dm?

[uneven night]

b(15+2b-b^2)

i don't know how to move on

i expressed everything in terms of b which is the base

so I used the formula V=bxhxl and expressed everything in terms of b

how can i proceed

should i take the derivative

<@&286206848099549185>

[median dome]

do you mean a cuboid?

[uneven night]

yes

[median dome]

but what do you mean by base

what is b

[uneven night]

V=bhl

right

[median dome]

so you mean the width

[uneven night]

yes

[median dome]

according to the language of the questin

[uneven night]

yes

[median dome]

how did you get that

[uneven night]

i expressed height and length in terms of width

so i only could get one variable to work with

<@&286206848099549185>

[obtuse pebble]

@uneven night Has your question been resolved?

[obtuse pebble]

@uneven night Has your question been resolved?

[obtuse pebble]

@uneven night Has your question been resolved?

[obtuse pebble]

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[wind obsidian]

how do i solve this

[wind obsidian]

<@&286206848099549185>

[void compass]

search on chat gpt

i cant help you

@wind obsidian

[obtuse pebble]

@wind obsidian Has your question been resolved?

[obtuse pebble]

@wind obsidian Has your question been resolved?

[obtuse pebble]

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[timid silo]

I understand how to do this with functions

.close

[obtuse pebble]

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[umbral nest]

identity functions and linear functions are the same right

[jagged nimbus]

How do we get (-1) after solving (w^2+w)? I am confused

[ashen dove]

w is root of x²+x+1

[jagged nimbus]

I am sorry could not understand since I am not good with maths. Can you explain further? Thanks for the reply btw.

[rich plume]

Consider the equation x^2+x+1=0
Thus x^2+x=-1 and this is why it is substituted

[jagged nimbus]

Oh, I got it. Thank you so much 🙂

[obtuse pebble]

@umbral nest Has your question been resolved?

[sour jacinth]

nope

[umbral nest]

how are they different

also how is this a isosceles triangle

@sour jacinth idk if i should ping but i figured i replied after a while so might as well

[sour jacinth]

linear functions have degree 1

[obtuse pebble]

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[sour jacinth]

but identity function is a specific function y=f(x)=x

where as linear function is as follows y=f(x)=ax+b

[obtuse pebble]

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[umbral nest]

sorry yes i understand

[sour jacinth]

its isosceles right triangle

[sour jacinth]

??

[umbral nest]

this one i figured it out

so makes sense!

[sour jacinth]

you see slope is nothing but the tan theta

of that line

[umbral nest]

oooo i see makes sense thanks

[sour jacinth]

welcome

so you are learning function??

[umbral nest]

ahh thank you

dam

[umbral nest]

yea kinda lmao i wanted to learn trigonometry but turns out i forgot functions

[sour jacinth]

oo

[umbral nest]

.close

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[round flame]

Can I send my problem here?

[raven condor]

yea

[stoic salmon]

dont ask to ask

send it

[round flame]

Alright

I don't get the last part of the problem...

Does it mean...I have to put the values of x and y =1 in the first step and then find out the value of C and in the next step putting the value of y=4 and C (the value achieved from the previous equation) proceeding to find out the value of X?

[obtuse pebble]

@round flame Has your question been resolved?

[obtuse pebble]

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[soft elm]

Can someone explain this to me, the answers are a = 3 and k = -2 but i got a = 4 (difference in x co ords) and k = -1 as its flipped

[timid silo]

when you flip it

0 stays 0

add 3 then you got 0;3

[high lily]

do you mean k=-1?

[soft elm]

yh

[high lily]

your k value is incorrect

[soft elm]

why is k not -1?

[high lily]

note that you didn't consider/observe the vertical stretch

[timid silo]

i dont see it

[soft elm]

oh yeah

[timid silo]

theres no stretch wdym

am i blind

[soft elm]

theres a +2 difference at the top line rather than the +1 diff on the bottom (from the y-int to the y co-ords)

[timid silo]

oh i see it

[soft elm]

very subtle though

[timid silo]

im blind yay

[soft elm]

so is the why k =-2 then as its flipped n streched

and the the +3 for a is just cause all points are up by 3

but the stretch throws that off no

[high lily]

yeh

[soft elm]

okay thank you

sneaky question just need to practice more

added to my work-on list

!done

.donme

.done

.finished

.close

[obtuse pebble]

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[soft elm]

got there in the end

[timid silo]

frfr

[obtuse pebble]

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[fair urchin]

Thé answer is supposedly 1/4

[mellow glacier]

huh?

[timid silo]

frfr

[fair urchin]

Images are still loading

[mellow glacier]

french revolution french revolution?

[timid silo]

yes

[fair urchin]

I have one bar…

[timid silo]

lmao

[fair urchin]

For a) the answer is 1/4 but I’m not sure how to get that answer

[obtuse pebble]

@fair urchin Has your question been resolved?

[fair urchin]

<@&286206848099549185>

[fossil crane]

WAIT 15 MINS

[obtuse pebble]

@fair urchin Has your question been resolved?

[obtuse pebble]

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[broken hare]

.

[median dome]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[broken hare]

im not sure how to start this

[median dome]

draw a reference triangle

[broken hare]

so find r first?

or hypotenuse

[median dome]

what?

[broken hare]

ok drew triangle

??

[median dome]

show

[median dome]

say, a=tan^-1(-2/3)

mark one of the angles as a

and using that, see what you can figure out about the ratio of the two sides

[broken hare]

i dont understand

can u elaborate on that

[median dome]

just look up a video on reference triangles

I can't really explain it over texts

[broken hare]

uhh ok

if anyone else can help that would be appreciated

[drifting badger]

Using a geometric interpreration of the values (i.e. by drawing a triangle like here) it can first be useful to use that arctan is an odd function (that is arctan(-x) = -arctan(x)) to rewrite arctan(-2/3)

Are you following this far?

[broken hare]

kinda, so how does that help

[drifting badger]

This does so we are working with -arctan(2/3) instead of arctan(-2/3), where the expression arctan(2/3) gives us some useful information for our drawn triangle.

[broken hare]

i guess thats what i dont understand is why it's useful

[drifting badger]

If we label the down-left-cornered angle α, what is a way to express this angle in terms of arctan, if we know the values of the triangle's sides?

[broken hare]

would the answer be (2 * sqrt(13)) / 13

[drifting badger]

Yes. That is correct

[broken hare]

okay, so i can use pythagorean to find the missing side and then just use the values that would apply to sin in this case

[drifting badger]

Yeah

[broken hare]

oh ok

but would it be negative?

since the y would have to be negative in this case

since inverse tan functio spans pi/2 to -pi/2

and all x values would have to be positive in that case

so the 2 would be negative

[drifting badger]

Here we're using that the trigonometric values are the same for similar triangles, so we could construct a right triangle with two sides, one with length 2 and the other with length 3. This allows us the express one of the angles in the triangle as α = arctan(2/3)

[drifting badger]

Yes. We first use that arctan is odd and then that sin also is odd

So sin(arctan(-2/3)) = sin(-arctan(2/3)) = -sin(arctan(2/3))

And sin(arctan(2/3)) has a geometric interpretation

And then we are interested in the negative of this value

[broken hare]

okay it makes sense now thank you

[drifting badger]

[broken hare]

.close

[obtuse pebble]

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[broken hare]

.reopen

[obtuse pebble]

✅

[broken hare]

@drifting badger can i ask for one more

[drifting badger]

Sure

[broken hare]

this one is a little tougher

[median dome]

no

should be negative

[broken hare]

read further

[median dome]

oh

[drifting badger]

What exactly is the problem presented here

[broken hare]

i guess im confused cause now theres a variable for the y value so i can't use pythag

[drifting badger]

The value of y will vary depending on the value of x, so I'm not really sure what the problem is. Unless there is some missing information?

[broken hare]

just says to solve for y without a calc

[drifting badger]

Hmmm. I suppose the task should be interpreted as trying to find the value of y expressed in terms of x but without the use of trigonometric functions

Don't know why that confused me

Sorry, just had a brain fart

[broken hare]

none of my notes even have a problem like this

[drifting badger]

Well, first off: recall that $\csc (x) = \frac{1}{\sin (x)}$. This means that $y = \csc (\arctan (\frac{x}{\sqrt{3}})) = \frac{1}{\sin (\arctan (\frac{x}{\sqrt{3}}))}$

[warm shale]

Mikkel

[broken hare]

ok

[drifting badger]

From this, we can then take the reciprocal on each side, to get $\frac{1}{y} = \sin (\arctan (\frac{x}{\sqrt{3}}))$

[warm shale]

Mikkel

[drifting badger]

This should hopefully look familiar

[broken hare]

i dont remember doing this

[drifting badger]

.

[broken hare]

yea, but the taking reciprocal of both sides

go on, i might just be confused

[drifting badger]

This just means that we go from $a = b$ to $\frac{1}{a} = \frac{1}{b}$

[warm shale]

Mikkel

[broken hare]

ok yea im still with u then

[drifting badger]

Then you can use your knowledge from question c. to get an expression for 1/y

[broken hare]

i solved c by finding hypotenuse and then just plugging in the numbers that would apply to sin, i dont see how thats possible in this case

[drifting badger]

What was your method for finding the hypotenuse in c.

[broken hare]

a^2 + b^2 = c^2

[drifting badger]

Using which values for a and b?

[broken hare]

2 and 3

[drifting badger]

So if we now construct a triangle in a similar way from $\sin (\arctan (\frac{x}{\sqrt{3}}))$ what would be the values of this triangles sidelengths?

[warm shale]

Mikkel

[broken hare]

x would be opposite and sqrt3 adjacent

[drifting badger]

Exactly

Then you can find the hypotenuse here, expressed in terms of the variable x

[broken hare]

but then i would two variables for the hypotenuse

or wait

so hypotenuse would be sqrt 3 + x^2

[drifting badger]

sqrt(3 + x²) just to make it clear that both terms are inside the square-root, but yes

[broken hare]

so final answer (x * sqrt(3 + x^2)) / (3 + x^2)

[drifting badger]

First off we can simplify this to x/sqrt(3 + x²)

And second, we need to remember that we took the reciprocal earlier

So the information we have now is actually 1/y = x/sqrt(3 + x²)

And we want an expression for the value of y alone

[broken hare]

if im just using pythag why do i need to take reciprocal

[drifting badger]

.

[broken hare]

wouldn't i just solve for hypotenuse and then use those values for csc

yea but my final answer is r/y

[drifting badger]

We first rewrote our equation so it was in terms of sin instead of csc, since that gave us the geometric interpetration we wanted

[broken hare]

i dont understand why thats necessary

[drifting badger]

I'm not really familiar with any geometric interpetrations of csc(x) = 1/sin(x), so I just flipped the fraction (took the reciprocal) so we could draw out a solution from our right triangle

And to keep the equality of our original equation, we also had to take the reciprocal of y which is 1/y

[broken hare]

if csc is just sin flipped can't i just get all the sides of the triangle and use those without manipulating the equation

[drifting badger]

Then we arrived at

1/y = x/sqrt(3 + x²)

where we can take the reciprocal again on each side (this flips the fraction) and leaves us with

y = sqrt(3 + x²)/x

[broken hare]

im not understanding why y needs to be flipped

my answer is still in terms of csc

[drifting badger]

We didn't have to. It's just much easier (in my opinion) to interpratete sin(tan(x/sqrt(3))) geometrically than 1/sin(tan(x/sqrt(3)))

[broken hare]

so is my answer right in terms of csc?

im just trying to solve this based on how we learned it

[drifting badger]

The final line of this should be your answer

[broken hare]

or how i think the teacher wants it since i haven't come across a problem like this

ok so how do i get there

[broken hare]

did i do something wrong here?

[drifting badger]

Okay. First off we have that

y = csc(arctan(x/sqrt(3)))

Which we can rewrite as

y = 1/sin(arctan(x/sqrt(3)))

This can be a bit hard to interpretate geometrically (and therefore hard to rewrite). What we do know though, is how to interpretate

sin(arctan(x/sqrt(3)))

Like we did similarly in c. with the triangle. Luckily we can rewrite 1/sin(arctan(x/sqrt(3))) as sin(arctan(x/sqrt(3))) by flipping the fraction, but for us to do that with

y = 1/sin(arctan(x/sqrt(3)))

We also need to flip the left-hand-side giving us

1/y = sin(arctan(x/sqrt(3)))

This flipping is just a temporary step for us actually being able to visualize the right-hand-side. When we have rewritten that side, we can flip each side right back again

So flipping it around is an 'extra step' per say. But it's a step that helps us actually visualizing how the right-hand-side can be rewritten

[broken hare]

right, so doing that is just a temp manipulation but we are getting different answers which is where im confused

[drifting badger]

You forgot to flip back

[broken hare]

oh ur right

lol

sorry

[drifting badger]

It's alright lol

[broken hare]

thanks for ur time

.close

[obtuse pebble]

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[drifting badger]

[obtuse pebble]

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[tawdry vortex]

Hello

[tawdry vortex]

How would this be graphed?

thanks also sry im not sure how to flip

the pic

[tardy epoch]

,rotate

[tawdry vortex]

can it be a rough sketch as llong as it looks like it is not touching

like

ill draw one and see

[obtuse pebble]

@tawdry vortex Has your question been resolved?

[obtuse pebble]

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[cerulean plaza]

trying to find basis for eigenspace, where did i mess up? im kinda confused where I made a mistake, the answer is listed as [-1/2 t -s, t, s] in the textbook

[cerulean plaza]

wait nvm

.close

[obtuse pebble]

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[kindred lichen]

the way to solve this is to get the transpose then make the fourth row with the given calories fat and carbs then solve the system normally?

[kindred lichen]

this is how the answer was on matlab

is that correct?

[timid silo]

looks fine

if you want to check, multiply each cereal with the corresponding answer entry, then sum

[kindred lichen]

oh yeah lol, dont know why i forgot to do that

it is indeed correct

thank you good sir

.close

[obtuse pebble]

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[primal flicker]

Hi I am confused by how to calculate the mean of this

[primal flicker]

What I have done so far is:

E[X_t] = E[W_t] + E[(-1)^(t-1)*W_(t-1)]
E[X_t] = 0 + E[(-1)^(t-1)*W_(t-1)]

but I am confused by how to evaluate this expectation:

E[(-1)^(t-1)*W_(t-1)]

[tardy epoch]

pull the constant out

[primal flicker]

do I have to write it as an integral?

its the expectation over t right?

[tardy epoch]

pain

[primal flicker]

D:

[tardy epoch]

do you have to split it in terms of odd and even t

if t = 2n, then E(-W_{2n-1}) and similarly for t = 2n + 1

[primal flicker]

Oh wait that’s smart

The mean would be 0 for both cases?

[obtuse pebble]

@primal flicker Has your question been resolved?

[obtuse pebble]

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[restive solstice]

For number one how do i answer it in words it is functions btw

[blissful bane]

@restive solstice

[restive solstice]

Yea

[blissful bane]

What do you know about domain of a function?

[restive solstice]

Uhh it is the x

I think

[blissful bane]

you can be more precise

[restive solstice]

So i just like say what the x is

[blissful bane]

right, it's those allowed values for x

more like, it's a set of values which can be choose for x

[grizzled shore]

It’s the values of x that do map to something

So the domain of f is the things where f(x) ≠ undefined

[restive solstice]

Its also asking me to use words to describe the intervol notation and set notation

[grizzled shore]

That’s what “allowed values” mean

[restive solstice]

Of the graph

[blissful bane]

If chosen real numbers, do you think all real numbers have mapping ?

for Q1

[restive solstice]

Idk what that means

Man

[blissful bane]

Domain is a "set" which contains values in which f(x) is "defined"

can you comprehend that?

try to get the words; set and defined.

as in, these values belong to {...} and when you plug in to f(x) you should get out a value

[restive solstice]

Ohhh ok thanks

[blissful bane]

That's basically what a domain is; it's a "set" and all values of the domain has the function defined

gotcha?

[restive solstice]

Yea got it

[blissful bane]

Now, all your graphs are drawn in a real plane right?

you have 1, 2, 1.5555, sqrt 2 everything there

Usually when we talk and graph function, we do to with real numbers

[restive solstice]

I dont understand

[blissful bane]

do you see graphs?

do you understand those graphs are drawn with real numbers?

[restive solstice]

Yea

[blissful bane]

what do you know about real numbers?

[restive solstice]

Idkk bro all i need is

How should i describe the number line in words

[blissful bane]

The answer for 1 is nothing but "set of all the real numbers"

in words

[restive solstice]

Oh ok

[blissful bane]

i want you to see why

[restive solstice]

Ok

So is it alwayd gonna start with set of all real mumber

Numbers

[blissful bane]

when dealing with real valued function

we start with set of all real numbers, then check for possible undefined points to avoid then write the domain

[restive solstice]

Ok

[obtuse pebble]

@restive solstice Has your question been resolved?

[restive solstice]

Yes

[obtuse pebble]

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[craggy vault]

<@&286206848099549185>

[obtuse pebble]

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.

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[teal turret]

@craggy vault

[craggy vault]

2

@teal turret

[teal turret]

Show what u have done

[craggy vault]

b/d=6

c7-d=0

a/c=-3

[teal turret]

U mean 7c + d?

How’d u get -d

[craggy vault]

yes

typo

[teal turret]

Ok, there’s one more equation u can make

Look at what the question is asking

How would u get the x-coordinate of the x-intercept if u were given the values of a b c d

@craggy vault

[obtuse pebble]

@craggy vault Has your question been resolved?

[obtuse pebble]

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[formal prairie]

Could somebody explain me the formula of surface integral?

[formal prairie]

I had some questions on how you find the boundaries and why it is a double integral

[clear condor]

[formal prairie]

Well first I wanna know why it is a double integral and not a single integral.

And btw I’m gonna be referring to the formula that helps you find the surface area of a parametrize 3Dobject

[formal prairie]

I know that usually we integrate things in third dimension using double integrals but I was wondering why.

[timid silo]

(you sum up along x and y)

[formal prairie]

Second I know that usually the bounds depend on your parameters and the object your are parametrizing. Let’s say in case of a sphere with x = r cos u sin v
y= r sin u sin v and z= r cos v. How do you know the boundaries of both integrals?

[formal prairie]

Oh so is like (x+y) + z? Kinda that logic?

[timid silo]

what?

for each region $\Delta x \Delta y$ you multiply the height (z-direction) f(x,y)

[warm shale]

hannibal

[timid silo]

base * height volume of vertical columns

[formal prairie]

Yeah I explained my self horribly on that but I get it. You have your base (xy) and then you multiply the third component right?

[timid silo]

right

try performing a double riemann sum by hand

to approximate an integral

you'll understand the process

[formal prairie]

I was wondering how you set boundaries to your parameters. Do you know anything about that?

[timid silo]

can you elaborate?

maybe write it out

[formal prairie]

Sorry. I mean that usually you have a boundaries on your integrals. How do you find the boundaries? More specific how would you find the boundaries of a parametrize sphere where x=r sin u cos v, y= rsin u sin v and z =r cos u.

If you were to solve a problem of finding the surface area of this parametrized sphere. How do you know the boundaries in both integrals?

[timid silo]

you have to find the region D

[timid silo]

this involves spherical coordinates

it's not your typical double integral

[formal prairie]

Btw where do you get this graphics from?

[timid silo]

internet

you could try this

[formal prairie]

Oh thought you input the stuff I was saying into a website and then it graphed it.

[timid silo]

search "spherical coordinates"

no haha

AI isn't that advanced yet or?

[formal prairie]

I haven’t used any that creates images. But I’ve seen people creating some pictures with them.

[timid silo]

x=r sin u cos v, y= rsin u sin v and z =r cos u
this is a triple integral

are you asking about that?

[formal prairie]

Oh but I think I sorta get it. Basically we have a full circle in the center for one parameter, meaning that one parameter does a full cycle, and then the other one is kinda cut in half?

[timid silo]

not exactly

sorry, it's 4am so i can't be asked to write it out

but you learn these with a basis of polar coordinates.

have you done single variable polar coordinates?

these guys

[formal prairie]

No, Im not in multi variable calculus yet, I’m just researching this topic for a homework research paper I have.

[timid silo]

oh ok

[formal prairie]

I wanted to understand it a bit more before starting typing.

[timid silo]

coordinate systems are very important

parameterizing often involves changing coordinate system to make the integral easier

you have: rectangular, polar, spherical, cylindrical, etc.

[formal prairie]

.close

[obtuse pebble]

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[idle spade]

hello

any help for b would be good

even just to get the ball rolling

[obtuse pebble]

@idle spade Has your question been resolved?

[idle spade]

.close

[obtuse pebble]

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[wanton hull]

Im pretty sure you know it already
Please don't use help channels for trolling

[trail cloak]

<@&268886789983436800>

[obtuse pebble]

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[tawdry vortex]

how to graph rational function 😭😢

[obtuse pebble]

@tawdry vortex Has your question been resolved?

[fierce lagoon]

If it helps, do polynomial long division to break 2x/(x-1) apart

[tawdry vortex]

Ah ok let me try that one

[obtuse pebble]

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[echo wharf]

idk what to do answer is in the top right corner

[stone spoke]

Maybe try letting u = ln^2(x)

also it looks like you'll have to do IBP more than once, so maybe tabular integration is a better idea

[echo wharf]

whats tabular integration

😭

oh i looked it up

nvm i know what it is

name was unfamiliar

but how can i derive ln^2(x)

dont think ive ever done it directly at least

[stone spoke]

use the chain rule

[echo wharf]

o

so basically (ln x)^2 = 2(ln x) x (ln x) '

[stone spoke]

wait

does x mean times for 2(ln(x)) x ln(x)'

[echo wharf]

ye

[stone spoke]

,w derivative (ln(x))^2

[warm shale]

Results provided by WolframAlpha

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Requested by keto11#8349

[stone spoke]

just had to make sure lmao

[echo wharf]

yeah thats what i got

[stone spoke]

yeah that's right

[echo wharf]

welp gonna have fun going deriving that

give me a sec

@stone spoke

is the answer -ln x/x^2

[stone spoke]

the final answer?

[echo wharf]

yeah

[echo wharf]

when u derive this

ln instead of log ofc

[stone spoke]

oh

,w derivative 2ln(x)/x

[warm shale]

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Requested by keto11#8349

[stone spoke]

did you forget to type the 2?

[echo wharf]

oh yeah i forgot the 2

oops

yeah ig that changes the result

ok close enough

[echo wharf]

doesnt seem to give the right answer i dont think

[stone spoke]

it's not working out?

[echo wharf]

the result would be x(2 ln(x) x 1/x) x ((2-2 ln(x))/x^2)

[stone spoke]

hmm

[echo wharf]

this is supposed to be the final result

[stone spoke]

tbh I've just been doing this in my head so far

lemme get a piece of paper and try it

[echo wharf]

k

[echo wharf]

the goat fr

[stone spoke]

Okay I got the same answer as your answer key

[echo wharf]

damn i suck

can u show me what u did

[stone spoke]

sure

oops

replied to the wrong post

[echo wharf]

all good

[stone spoke]

$\int x\ln^2(x) dx$
$$= \frac{x^2\ln^2(x)}{2} - \int \frac{2\ln(x)}{x} \cdot \frac{x^2}{2}dx$$
$$= \frac{x^2\ln^2(x)}{2} - \int x \ln(x) dx$$
$$= \frac{x^2\ln^2(x)}{2} -(\frac{x^2\ln(x)}{2} - \int \frac{x}{2} dx)$$
$$= \frac{x^2\ln^2(x)}{2} -(\frac{x^2\ln(x)}{2} - \int \frac{x}{2} dx)$$
$$= \frac{x^2\ln^2(x)}{2} -\frac{x^2\ln(x)}{2} + \int \frac{x}{2} dx$$
$$= \boxed{\frac{x^2\ln^2(x)}{2} -\frac{x^2\ln(x)}{2} + \frac{1}{4} x^2 + C}$$

[warm shale]

keto11

[stone spoke]

I think this should be typo free

[echo wharf]

k

lemme check it out

@stone spoke could u explain what u did to get here

[stone spoke]

Just do IBP once on xln^2(x)

[echo wharf]

this?

[stone spoke]

I let u = ln^2(x)

[echo wharf]

o

so wait if u derived ln^2(x)

its (2 ln(X))/x

?

[stone spoke]

yeah that's right

[echo wharf]

so where'd the first part come from

[valid bridge]

Bro I’m so sorry

But

I really need help

[stone spoke]

∫ u dv = uv - ∫ v du, so uv = ln^2(x)*(x^2/2)

[echo wharf]

a^2 + b^2 = c^2

[valid bridge]

I know

But for the second one

The formula isn’t working

[stone spoke]

please make your own server using the help channels

[echo wharf]

you have the area of each square

u can find the sides that way

and since ur looking for one of the sides (we'll say side a) then u need to isolate a in the equation

you'll get a^2 = sqrt(c^2 - b^2)

[echo wharf]

ah ok

makes sense

so instead of taking just ln x

u brought the 2 as well

[echo wharf]

damn bruh that hand writing is heavenly

[stone spoke]

thx

[echo wharf]

alright thanks for the help, means a lot, you mind if i can dm u if i need help with some other ones?

[stone spoke]

yeah fs

I'll be going to sleep in a little while though

[echo wharf]

alr

how long?

[stone spoke]

prob like 45 min to an hour

[echo wharf]

alr bet

ty 🙏

.close