help-10

[timid silo]

Yes

2^4x?

no

SORRY LOL

3(something) + (something) = ???(something)

what is ???

4

ok

let something= 2^(2x)

what is this now

2^2x + 2^2x?

where are ur coefficients disappearing into though

like there is a 3 there

Oh i thought

But how do I have it inside

what does that mean

How do I have it inside the equation

similar to how I did it here

3*2^(2x) I'm not sure what you are looking for

8^2x

no

Im not sure

Oh

[timid silo]

you already got this

3(something) + something = 4(something)

you dint have to do anything else

just

replace something with 2^(2x)

and that's it

no more no less

4(2^2x)

yes!

Yay!

Aaah i see!

I got it:)

Thank you!

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[timid silo]

u can further simplify this if u want

2^(2x) = 4^x

so like 4*4^x is Like

Liike this?

yeah that works

you can also make it a 4

2^(2x+2) = 4^(x+1)

all good either way

Oh wow

I dont know how I could find the +1 in the second equation

Other then by knowing the first

u can

,, 2^{2x+2} = 2^{2\8{x+1}} = \8{2^2}^{x+1} = 4^{x+1}

[timid silo]

Aah yes!

But that’s what I mean

Unless that is what you mean by further simplifying

Then that is another perspective!

[timid silo]

it's usually all factoring but yeah

it's all equivalent

Yeah!

How do I become better at math over the summer lol

that's a big ask haha, anyways just pick up some decent looking precalculus book and go through it on your own and solve as many problems as you could

that's what university people do and it's been working well for me

[timid silo]

Okay! Lol im taking calculus next year lets see how it goes

your typical calculus 1 course is much easier if you have good precalculus knowledge

so yeah definitely worth working on that

Yeah lol

Even now im starting to have trouble with the next problem

[obtuse pebble]

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[timid silo]

same principle

[dim temple]

what do you need help with

oh nvm

[timid silo]

Just make ax^-3 - bx

?

well there is an extra step here

so like

,, a^{m-n} = \4{a^m}{a^n}

[timid silo]

use this

[timid silo]

yea

so like

what is 2^3

?

[timid silo]

8

[timid silo]

ok i mean don't just summon it from nirvana into the second coefficient somehow

[timid silo]

LMAO

Lets see

its [
\4782^x-2^x
]
like you had originally

[timid silo]

Yes

[timid silo]

ok now same procedure

7/8(something) - (something) = ???(something)

don't let the fraction scare u its the same thing

but anyways what's ???

-1/8?

ye

so what is ur thing now

-(1/8)*2x

Iii thinkk

no

you kinda just

ok i mean you could've typo'd 2^x for 2x so im guessing its that

but yeah -(1/8)2^x

Oh yeah I typoed that lol

but yes thats correct

but we can simplify

remember how we said 2^3 = 8

can you figure out a way to use that such that u have something of the form -(???^???)

1/8 = 2^-3

Yeee

So u just

-2^x-3

? :OO

yeppie

Yeeiiipiieieiie

Now now 😈

Its time to go to bed

lmao Alright

goodnight and good luck

You too! Thank you for all the help!

Goodnight

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[graceful marten]

Im a bit confused

This I understand (Orange)

But how did they get

3/2 kT

Because from what I understand is that v^2 is the same as Crms^2

[frank monolith]

they replaced m(Crms)^2 with 3kT as mentioned at the top of the page

[graceful marten]

In the context of finding Ek of one particle and finding Ek of multiple particles

[graceful marten]

Oh

Ohhhh

I seee

Thank you

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[gritty drum]

Why is ‘xy=1’ not a linear equation?

[gaunt walrus]

Do you know wut is linear equation?

[gritty drum]

The highest power in equation is 1

[gaunt walrus]

tell me, what is an equation

[gritty drum]

Expression = Expression

Just assume that xy=0 here btw

[gritty drum]

.

[gaunt walrus]

oh ys

hi

umm

technically, yes

but x = 0 and y = 0

[gritty drum]

Ok so

Let’s make it =1

xy=1

[gaunt walrus]

Ok, that's not

[gaunt walrus]

No

with the term xy, that isn't linear equation

[gritty drum]

Why? (Literally my main question)

x and y both have a power of 1, also xy has a power of 1

The highest power is still 1

[gaunt walrus]

Linear Equation Definition: A linear equation is an algebraic equation where each term has an exponent of 1 and when this equation is graphed, it always results in a straight line. This is the reason why it is named as a 'linear' equation.

[gritty drum]

Ok, xy doesn’t give a line

[gaunt walrus]

it gives a line

but not straight line

[gritty drum]

But I want a more algebraic, intuitive reason

[gaunt walrus]

umm

that I cannot help you, sorry

maybe you ask others

[gritty drum]

Alright

<@&286206848099549185>

[valid trench]

Cuz it’s not an equation

[gritty drum]

Edited

[valid trench]

Oh

Then it’s a linear equation with 2 variables

[gritty drum]

It’s not tho…

It’s in 2 variables, but not a linear equation

[valid trench]

How?

[gritty drum]

For some reason multiplying products means no linear equation

[gritty drum]

Idk literally why I opened this post

I’ve got 3 sources tho, my book, Gemini, and DaWhirs

Well yea, cuz it’s not linear

It’s curving

Also I want an algebraic reason

The graphical one is not satisfying me

[wild swallow]

the term xy isn't "linear"

it has degree 2

you'd say that it's a quadratic expression

the degree of a term is the combined degree of each variable

to see the quadratic nature more clearly you can make a substitution x = u + v, y = u - v

[vestal burrow]

If you have xy=1 and solve for y, what do you get?

[wild swallow]

then the equation turns into u^2 - v^2 = 1

[obtuse pebble]

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[ember frost]

the following graph is a third degree polynomial

[ember frost]

calculate the amount of real roots $f(f(x))=0$ has

[warm shale]

FunguDeu

[ember frost]

my solution is 8, apparently its wrong and i dont know where i trip

or rather 8 is where im stuck

[ember frost]

so i differentiate this and got f'(x)f'(f(x))

f'(x) has 2 roots, f(x) has 3 roots, which maps to a total of 6 roots when plugged in f'(f(x))

so the total amount of roots this has is 8

but from this i cant find the amount of roots f(f(x)) has

[dusk widget]

why did you differentiate?

[ember frost]

find the amount of extrema, to see if variation table works

[wild swallow]

if there's 8 extrema then there's 9 roots no?

[ember frost]

who knows if the graph is whackass like this

8 extrema but only 1 root

[dusk widget]

it's a cubic

[wild swallow]

yeah but well

[dusk widget]

this can't happen lmao

[wild swallow]

you can just check the number of roots

like you did for f'(f(x))

[ruby path]

all three roots of f(x) will also be roots of f(f(x)) right

not exact value

[ember frost]

oh

[ruby path]

what I mean is since f has 3 roots

so will fof (at least)

[ember frost]

wait im a dumb

yeah f(x) has 3 roots

so when plugged to become f(f(x)), each root maps to another 3 roots

which total to 9 roots

[dusk widget]

yeah

[wild swallow]

degree 9 polynomials usually have 9 roots

[ember frost]

maximum 9 roots?

[wild swallow]

well most of the time there's no multiplicity

[ember frost]

again who knows if its decides to be funny and has only 1 or 2 roots

but yeah you got a poiny

[wild swallow]

(most of the time you work in an algebraically closed field too)

[ember frost]

so yeah, that should be it

thanks yall!

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[little coral]

I’m having trouble understanding what the professor is doing in this clip regarding the numerator. Why does it go from all 1’s to x0 - (x0+ delta x). He kinda skips explaining it but I don’t know how he got there

[red ice]

Remember the definition of the derivative

That's $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ and $h$ is the same thing as $\Delta x$

[warm shale]

south

[red ice]

So if you substitute in f(x) = 1/x

You get the same as what the prof has

Well, $f(x)$ evaluated at $x = x_0$

[warm shale]

south

[grizzled shore]

How would you add $\frac{1}{a} + \frac{1}{b}$

[warm shale]

Frosst

[inland matrix]

After factoring it becomes $\frac{1}{\Delta x}\left(\frac{1}{x+\Delta x}-\frac{1}{x}\right)$

And now just simplify the thing within the brackets

[grizzled shore]

Use \left( and \right) to make the brackets auto size

[red ice]

Ah right I didn't explain the step you wanted

[warm shale]

Sulphur

[inland matrix]

Nice

[grizzled shore]

It reduces to this problem

You should know this by heart at the level of doing calculus

If you’re rusty on your fractions do go back and revise on it

Calculus will assume that you can do all sorts of algebraic tricks with ease

[little coral]

yeah I am rusty. I haven't done true math in a while. I can keep up w most of the basic calculus but some of this algebra slips past me and i need the reminder

[inland matrix]

Wait I thought you opened this channel

Mb

[little coral]

no i did

[grizzled shore]

Yeah, it’s a bit rough if you aren’t on top of the algebra, nothing wrong with taking a look back at old topics to refresh your memory

[little coral]

I just kinda let solidworks do it for me when i design something

but i see your point. I'm kinda jsut relearning it as I go, but once its there i remember it pretty well

wait a sec I already understood this. The part i didn't understand was how multiplying it together gives you x0-(x0+delta x)

like the factoring out makes sense. Even multiplying the denominators together makes sense, I'm guessing that's where I'm forgetting something about the numerators but still

<@&286206848099549185>

[obtuse pebble]

@little coral Has your question been resolved?

[obtuse pebble]

@little coral Has your question been resolved?

[little coral]

<@&286206848099549185>

:(

<@&286206848099549185>

[little coral]

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[ember frost]

Given complex number $z$ such that $|z-1-i|=1$, complex number $w$ such that $|\bar{w}-2-3i|=2$. Calculate the minimum value of $|z-w|$

[warm shale]

ngDe

[ember frost]

using the information i can construct the locus of z and w, which is

respectively

but i struggle to see what |z-w| would be

[unkempt vault]

The minimal value of that is the distance between the closest points on the loci

[ember frost]

ah i see what you mean

alright thanks

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[drifting sandal]

how i do this one?

[glacial obsidian]

You have to find $r_1(s)$ such that $|r_1'(s)|=1$ for all $s$

[warm shale]

d

[drifting sandal]

why

[glacial obsidian]

Because that is equivalent to the definition of arc length parametrization

[drifting sandal]

ive never heard of that

[glacial obsidian]

It is easy to prove

You can try to prove it yourself

[drifting sandal]

y = 5x = <t, 5t>

then what

[glacial obsidian]

Transform that

so that the derivative is a unit vector

Or maybe write (s(t), 5s(t))

Where s is a function that depends on t

The derivative is (s'(t), 5s'(t))

And its norm is $\sqrt{1^2+5^2}|s'(t)|$

[warm shale]

d

[drifting sandal]

i dont understand

[glacial obsidian]

A parametrization is a change of variables

t ---> s

So it can be written as a function

s = s(t)

And you want that $|r'(s)|=1$

[warm shale]

d

[drifting sandal]

yes

[glacial obsidian]

That is, |r'(s(t))=1|

[glacial obsidian]

Since r(t) equals this, you now write r(s(t))

[glacial obsidian]

like this

And do what I wrote above

To obtain |r'(s(t))|=sqrt(26) |s'(t)|

[drifting sandal]

r(s(t))

how do you go from r(s(t) to sqrt(1^2+5^2)s'(t)

[glacial obsidian]

This is the derivative

[glacial obsidian]

And this is the norm of the derivative

| (s'(t),5s'(t)) | = | (1,5) | times |s'(t)|

[drifting sandal]

the derivative is 5...

where do you get sqrt(1^2+5^2)

[glacial obsidian]

The derivative is a vector

The norm of a vector v is sqrt(v1^2+v2^2)

[drifting sandal]

idk what a norm is

and how is the derivative a vector

[glacial obsidian]

Wait

[glacial obsidian]

Maybe I did not understand this correctly

[glacial obsidian]

What I understand is that r is a curve which depends on s and whose image is the line y=5x

So that r(s) is a vector for each s

Maybe the question is refering to something different

[drifting sandal]

idk

ive never heard of a arc length parameterization

[glacial obsidian]

Maybe you could provide some more context

[glacial obsidian]

Then why are you trying to do an exercise about arc length parameterizations?

[drifting sandal]

because its on the homework

[glacial obsidian]

Then you must have studied it

[drifting sandal]

no

f 𝐫(𝑡)
is any parametrization such that 𝐫′(𝑡)≠0
for all 𝑡
, then

𝐫1(𝑠)=𝐫(𝑔−1(𝑠))

is an arc length parametrization, where 𝑡=𝑔−1(𝑠)
is the inverse of the arc length function 𝑠=𝑔(𝑡).

Recall that the arc length function is

𝑠=𝑔(𝑡)=∫𝑡0‖𝐫′(𝑢)‖𝑑𝑢

i found this

[glacial obsidian]

Do you understand this text?

[drifting sandal]

no

[glacial obsidian]

Well, then you need to study and understand this before you can do exercises about it

[drifting sandal]

thats what i am actively trying to do

[glacial obsidian]

It can be a bit long to explain everything about differentiable curves, and now I can't do that

Perhaps some other helpers could

[drifting sandal]

i need to integrate r'

but i dont have r'

and i also need a t value which i was not given

[glacial obsidian]

As I said before, you don't have to

There is a different method

But in this case you can integrate as well

[glacial obsidian]

Yes, you do have r'

[drifting sandal]

where

[glacial obsidian]

If r=(t,5t), then r'=(1,5)

So |r'|=sqrt(26)

[glacial obsidian]

It is the length of the vector

[glacial obsidian]

Because r is a vector as well

[drifting sandal]

the length of the vector. What vector

[glacial obsidian]

r

[drifting sandal]

i dont have r

if i had r the question would be done

[glacial obsidian]

r = (t,5t)

You said it before

[drifting sandal]

ok r = (t, 5t) r' = (1, 5) length = sqrt(1^2 + 5^2) then i integrate that from 0 to t but i dont have t

[glacial obsidian]

Then you obtain s as a function of t

If you integrate, you obtain sqrt(26) times t

So s = sqrt(26) t

t = s / sqrt(26)

[glacial obsidian]

And then you substitute that here

[drifting sandal]

thank you

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[timid silo]

I need help

[timid silo]

I don't understand how I'm supposed to get g(x)

Like am I supposed to translate on the line or

[raw torrent]

What translation do u see

I mean

It looks like graph got shifted down

Or I’m tripping

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Yes it's a translation

It says it in the question

[obtuse pebble]

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[forest tapir]

Could someone help me with sin cos tan

[forest tapir]

I don’t have a question I just want to know how you use it

For context I’m having a National test in math on 27th April on sin cos tan and I wanted to learn it early

[thick torrent]

the usual way theyre introduced is with a right triangle like this

there are a lot of other interesting properties about them but im not sure how much you're expected to know if you dont have a question on hand

[forest tapir]

Quick quest what is like hypotenuse and the adjacent

[thick torrent]

the hyp is the longest side of a right triangle

[forest tapir]

It’s like you want to learn a new language you just want to learn how to talk it

[thick torrent]

the adjacent is the adjacent side to the angle $\theta$

[warm shale]

jack

[thick torrent]

sure but youd ask questions like 'how do i say...'

i cant know what 'basics' you refer to unless you have a question

'basics' may mean different things for ppl of varying lvls of education

do you have examples from schoolwork

try using the khan academy series then

https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func

do you find it hard?

so you're getting tested on things you havent been taught?

uhhh

its been forever since i was your age and i probably had a different schooling system too

https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-trig-ratios-intro/e/trigonometry_1

you asked about sin cos tan my guy

i'm not an (american?) schoolteacher idk what you're supposed to learn in the sixth grade

you asked for questions involving sin cos tan so thats a pretty basic type

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-geometry-topic

isn't this what you're learning? ^

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[brittle swan]

Could someone help me find my error? I've been stuck for so very long
(slight mistake with ± at the last term but it's not the main issue)

[brittle swan]

I found a complex quadratic in zw, then tried to generate a formula for |zw|², but arrived at a non-integer answer

(I know the answer is an integer but I don't know what it is)

[obtuse pebble]

@brittle swan Has your question been resolved?

[brittle swan]

<@&286206848099549185>

.closs

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[gleaming karma]

hello, could someone check if i am right? i am totally not sure about this

[tranquil geyser]

lol Maybe a cheat like ploting the graph and ordering them according to steepness

[tardy moat]

generally speaking, factorial always wins, any polynomial beats any power of a logarithm, and exponential beats polynomial

[gleaming karma]

alright thank u

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[tardy moat]

@gleaming karma I have some links to some proofs of these facts if needed

[gleaming karma]

yes i would be rl grateful

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[hoary idol]

in a garden there is apple and orange trees. bob picked 300kg from one apple tree andd 800kg from orange tree . from one tree he picked 600kg average what percent of trees are the apple trees

[hoary idol]

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[obtuse pebble]

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[torpid shuttle]

.open

the point C(2,6) is translated 2 units right. What are the coordinates of the resulting point, C'?

[torpid shuttle]

.close

[obtuse pebble]

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[torpid shuttle]

.reopen

[obtuse pebble]

✅

[gilded needle]

what is your doubt?

[torpid shuttle]

the point C(2,6) is translated 2 units right. What are the coordinates of the resulting point, C'?

i dont know what to do

[torpid shuttle]

to move it 2 times right

i dont think its it though

[gilded needle]

what did you get when you tried that?

[torpid shuttle]

(4,6)

[gilded needle]

that is correct, assuming "right" means +x

[torpid shuttle]

oh

[gilded needle]

which is the usual convention

[torpid shuttle]

then yeah

thanks

baaka

jk

[gilded needle]

বাকা

jk

[obtuse pebble]

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[obtuse pebble]

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[proven cape]

I don't know the equation for 11

[dense imp]

think about, what's v'(t) ?

[proven cape]

v'(t)=s(t)

[dense imp]

s'(t) = v(t)

velocity is the derivative of the position, and the derivative of velocity is what?

[proven cape]

a(t)

[dense imp]

right, so how can you use that?

a(t) is given right?

[proven cape]

It is 32?

[dense imp]

well what's the most logical way to orient your axes? what's your negative and positive directions?

[proven cape]

velocity is 90 feet per second

for upwards

[dense imp]

and gravity is accelerating your projectile in what direction?

[proven cape]

down

[dense imp]

so a(t) = ?

[proven cape]

-32 because it is gravity?

[dense imp]

right

a(t) = v'(t) = -32

so v(t) = ?

[proven cape]

32

-32t?

[dense imp]

almost, don't forget the constant of integration

v(t) = -32t + C so now we can use the initial condition (the velocity given) to solve for C

you are in calculus \ know integration right? i forgot that sometimes people learn this stuff with just given crutch motion equations before learning calculus

[proven cape]

the integral from 0 to 3?

[dense imp]

we're just doing an indefinite integral to get the equation

we know v(0) = 90 right?

so v(0) = -32t + C and that allows you to find C

[proven cape]

C=186

Yes?

[dense imp]

?

v(0) = 90 = -32(0) + C

C is just 90

[proven cape]

ok

v(3)=-32(3)+90 then

sso it is -6

-6 for velocity

so 11b

Set it to 0?

[dense imp]

right

set s(t) = 0 that is

[proven cape]

t=2.8125

So 11c

net distance?

How do I calculate that?

You here?

[dense imp]

integrate the s function between the times

[proven cape]

0 and 2.8125

[dense imp]

did you just learn integration?

[proven cape]

Yeah I used Math+9 on my TI-84 and got 126.5625

[dense imp]

what did you have for your s function

[proven cape]

-32t+90

[dense imp]

.... that was our v function

when you solved part b) you found the s(t) function right? what'd you get for that?

[proven cape]

okay so -32t^2+90t?

[dense imp]

when you integrate the -32t you it becomes -16t^2, you forgot to divide by 2

s(t) = -16t^2 + 90t + C but since s(t) = 0 we see that C = 0

[proven cape]

yeah

~237.305

[dense imp]

so you need to redo the calculation from b now if you used the wrong function

to find the t when it hits the ground

[proven cape]

-32t+90=0

t=2.8125

[dense imp]

"when it hits the ground"

which function is 0 when it hits the ground?

[proven cape]

oh it is the s(t)?

[dense imp]

right, the s(t) function is our position function right?

in this particular problem we are measuring vertical distance from the ground

so it would stand to reason we can interpret "when it hits the ground" to mean when s(t) = 0

[dense imp]

^

there would be no logical reason why v(t) is necessarily 0 when it hits the ground, in fact that wouldn't make sense, why would something not be moving as it's impacting the ground

so learn to think of how to interpret s, v, and t logically

s(t) is like your odometer on a car, v(t) is like your reading of the speedometer at any given moment

[proven cape]

t=2.3717082 or 2.372

[dense imp]

how are you getting this

we have s(t) = -16t^2 + 90t correct?

[proven cape]

Yeah

-16t^2+90t=0

[dense imp]

so you can factor s(t) = t (16t - 90) and set it equal to 0
t (16t - 90) = 0
can immediately see the t = 0 is one of the roots (That's when the projectile first started at t = 0 and was at the ground) and the other root is?
16 t - 90 = 0

[dense imp]

90t not 90

[proven cape]

oh yeah

[dense imp]

youre overusing your calculator, this problem hasnt really had any numbers that need it

[proven cape]

0 and 5.625

[dense imp]

right

anyway i gotta go but i suggest you review the basics of integration and re-read the chapter or whatever you have on s(t) / v(t) / a(t) type stuff

[proven cape]

I wasn't there on Tuesday because I was sick

.close

[obtuse pebble]

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[late flare]

i chose B

but apparentl am wrong

my question is how as lim approaches 0 from the positive side, it goes to infinity

[alpine verge]

well x=0 is a vertical asymptote

and the definition of a vertical asymptote is that $\lim_{x \to a} f(x)=\pm \infty$, where x=a is the vertical asymptote

[warm shale]

y0shi

[alpine verge]

ty

[late flare]

but am trying to imagine it graphically

i tried to do it graphically, i came up with B

[white sandal]

its C

since it is positive

[late flare]

can you explain

graphically

i tried to do it

ona. graph

[white sandal]

u know how the function f(x) is positive

then it would reach a value higher than 0 as x approaches infinity

and since x > 0, if the lim as x -> 0 from the right

it would equal infinity

alr hope this helps

[latent walrus]

stop spamming people

[white sandal]

buddy

i only said it twice

[obtuse pebble]

@late flare Has your question been resolved?

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[pallid light]

hey, is there someone that talks french or understands it enough to explain me on a call the pdf ?

[obtuse pebble]

@pallid light Has your question been resolved?

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@pallid light Has your question been resolved?

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@pallid light Has your question been resolved?

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[haughty creek]

I have no idea what the second half of this question means

[haughty creek]

Sigma is assumed to be 1? And also if sigma is a variable how in the world am i supposed to be able to plot this as a quantile quantile graph

This next question builds off this one and i dont understand

[obtuse pebble]

@haughty creek Has your question been resolved?

[haughty creek]

<@&286206848099549185>

[obtuse pebble]

@haughty creek Has your question been resolved?

[obtuse pebble]

@haughty creek Has your question been resolved?

[obtuse pebble]

@haughty creek Has your question been resolved?

[plain tulip]

sigma?

https://tenor.com/view/pout-christian-bale-american-psycho-kissy-face-nod-gif-4860124

[obtuse pebble]

@haughty creek Has your question been resolved?

[red ice]

The idea here is that you take many samples from a normal distribution with mean 0 and variance 1

Just as how the sample mean is almost never going to be exactly 0; it's going to be close to 0 of course

In fact the sample mean for a normal distribution will follow another normal distribution

So the same thing can be said for the variance: the sample variance won't be equal to the true variance which is 1

If you plot the distribution of the sample variance, you should notice that it will compare really well to a normal distribution with mean sigma^2 and variance 2 sigma^4 / (N - 1)

That's the theoretical result and there are proofs online

[red ice]

So yeah if you have a 95% confidence interval for the sample variance, it's just the mean plus/minus (z-score for 97.5%) * (the standard deviation), which you can get from the variance 2 sigma^4 / (N - 1)

And yes sigma^2 = 1 so sigma^4 will also be 1

So you have 2/(N - 1) and you should be able to notice what happens to the variance of the sample variance as N gets larger and larger

[visual rain]

Pls help me solve the 17 no question

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185> pls

[timid silo]

I honestly don't know how to do it

[timid silo]

Just don't spam lol

[visual rain]

bruh its urgent

[timid silo]

That looks like a quadratic formula

[visual rain]

yes

but how the hell do i solve it

[timid silo]

Lemme think

[cold thorn]

log?

[visual rain]

its not logarithm

[timid silo]

According to my scientific calculator
X1 = -1 - √(5)
X2 = -1 + √(5)

[visual rain]

how

[timid silo]

General equation

[icy maple]

are you just solving for x?

[visual rain]

yes

[timid silo]

I used the general quadratic equation

Correct me if I am wrong.
I substituted a = -1, b = -2, and c = 4

According to the R.H.S of the equation

[visual rain]

but there should be a zero in LHS to be a perfect quadratic eq

[icy maple]

and the answer is not -1?

[visual rain]

no -1 inside a sq root is not counted as a solution

<@&286206848099549185>

[timid silo]

I see what I did wrong

[visual rain]

someone pls help me

[haughty creek]

Wtf? You're hijacking my question?

[haughty creek]

Ohh I see

[red ice]

Yes pls move

Also dude why did you ask the same question twice

[haughty creek]

So since sample variance is chi_{n-1}^2 var/(n-1) that just converges to normal with mean var sigma^2 and var 2sigma^4/N-1

[red ice]

Yeah

That's my point exactly

It's not trivial ofc

[haughty creek]

so i should just plot chisquare/N-1 vs normal with mean 1 and variance 2/(N-1)

[haughty creek]

Yes ofc

but i've seen the proof before

[red ice]

Nice

[haughty creek]

The question was just so unclear

[red ice]

Something like that

You shouldn't need chi-square explicitly: just get R or whatever software you're using to randomly choose 10, 50, or 100 points

It's about discovering what happens to the sample variance instead of being too rigorous about it

[haughty creek]

Oh this makes sense

[haughty creek]

Yeah

Okay

Pretty cool

Well too bad i understood it like 1:50 after the p-set was due

but at least i get it now

.close

[obtuse pebble]

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[serene rover]

.reopen

[obtuse pebble]

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[mint oak]

how do yk the derivative of y=2cos3x

[forest sinew]

use the chain rule

do you know the derivative of $\cos x$?

[warm shale]

jan Niku

[mint oak]

-sinx

[forest sinew]

right

[mint oak]

why the chain rule tho?

[forest sinew]

how about $2\cos x$

[warm shale]

jan Niku

[mint oak]

hmmm

Im not sure

[forest sinew]

so in general if the constant is just chilling out front, it doesnt matter, its just along for the ride

$\dv x \qty( 2 f(x) ) = 2 \dv x \qty( f(x) )$

[warm shale]

jan Niku

[mint oak]

so i do

2(-sinx)

-2sinx

[forest sinew]

yup

[mint oak]

ohh ok

[forest sinew]

so almost done

[mint oak]

wait so what does cos3x derivative mean

[forest sinew]

you have one more complication in your problem

$2 \cos (3x)$

oh the bot died lol

[mint oak]

lololol

[forest sinew]

thats not good

$test$

i need it

sad

[warm shale]

jan Niku

jan Niku

[mint oak]

yay

[forest sinew]

woo

okay so

in general

$\dv x \qty( f\qty( g(x)) ) = f'\qty(g(x)) \cdot g'(x)$

[warm shale]

jan Niku

[mint oak]

OH

[forest sinew]

maybe we can drop a lot of the parens and arguments and stuff and write it more simply like

[mint oak]

IS it bc theres no x on the cos so its not product rule

[forest sinew]

$\qty(f(g))' = f'(g) \cdot g'$

[warm shale]

jan Niku

[forest sinew]

the problem is theres another function of x

on the inside

[mint oak]

oh

[forest sinew]

which sometimes isnt a problem like

if you just have cos(x)

then the derivative of x is just 1

and you dont really notice

but here the function on the inside is 3x

since that makes the argument vary more rapidly than if it were just x

you should expect that will impact the derivative some how

[forest sinew]

and this is how

can you identify f and g here?

i mean, whats the outside function, whats the inside function

[mint oak]

is f cos and g 3x

[forest sinew]

yea but you have an extra 3 here

it should be like

$-2\sin(3x) \cdot 3$

[warm shale]

jan Niku

[mint oak]

ohh ok

so when converting cos3x to

f(x) = cosx

g(x) = 3x

so

derivative of cosx

is -sinx

and then you plug 3x in

[forest sinew]

i would start here

you identified the functions perfectly

now remember that $\qty( f(g) )' = f'(g) \cdot g'$

[warm shale]

jan Niku

[mint oak]

yes

[forest sinew]

and plug in the necessary information

[mint oak]

this is also a chain rule?

[forest sinew]

yes, can you tell why?

[mint oak]

because the inside brackets is a function of sin

[forest sinew]

this looks right

[forest sinew]

its because you have another function inside your function

[mint oak]

oh ok

[forest sinew]

when the argument to a function is another function, you expect that to impact how the function changes and what it looks like

chain rule tells you how to account for that

[mint oak]

ohhook

what is this then?

[forest sinew]

how do you mean?

[mint oak]

is that chain rule also?

[forest sinew]

hmm well you'd have to identify if there is a function inside of another function

maybe I can be clear: a function thats not just x

obviously we know what the derivative of sin(x) looks like

[mint oak]

yes yes

[forest sinew]

so, are there extra functions floating around

[mint oak]

yea

the 3x and 4x

[forest sinew]

yup

and how do you account for that

[mint oak]

Ermm

Chain rule

so it becomes

cos(3x)(3)

-(-sin(4x)(4)

[forest sinew]

yea, nothing left to do except to combine it all

[mint oak]

does it multiply in?

[forest sinew]

in?

[mint oak]

cos9x + sin16x

sry was that a stupid question

.close

[obtuse pebble]

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[forest sinew]

#❓how-to-get-help

[obtuse pebble]

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[fading cedar]

I'm so confused on the first and the last

For the 3rd one, they all differ by 5

But I know that rule cannot apply to n =1

And same for the top one, I was thinking of n^2, but that cant apply to n=3

[latent walrus]

when youre constructing something like this
you can take the common difference it 5
so 5n, then to get to the first term its -1, so 5n-1

or 4+5(n-1) as the generic form of an arithmetic term youd normally see

[fading cedar]

Okay this makes sense

the back one they increase in distance

first one*

[latent walrus]

for the first one its a geometric sequence

you multiply by 1/2 each term

1/2, 1/2^2, 1/2^3, 1/2^4,....

any idea when looking at it like that

[fading cedar]

No I'm not getting it

oh

wait

1/2^n

thank god!

[latent walrus]

nice

[fading cedar]

i knew i could get to 8 from 2

but was tweaking

thank you

[obtuse pebble]

@fading cedar Has your question been resolved?

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[vagrant brook]

can someone walk me through this step by step i'm totally lost

[vagrant brook]

i know the steps ur supposed to do in this type of question but im just stuck on what to do

[high lily]

that's a little contradictory

[vagrant brook]

like idk how to apply it to this specific one

[high lily]

describe what steps you think would be required here

[vagrant brook]

so i multiply left by the right denominator

and right by left denominator

and then i expand everything

[high lily]

inefficient

i would recommend first factorising the denominators

[vagrant brook]

sure

factor and then everything i said?

[high lily]

well you wouldn't need to multiply the whole denominator

factorisation helps in identifying common factors

[vagrant brook]

ye

[high lily]

which could lead to potentially using a lesser common multiple

which makes work less tedious

[vagrant brook]

so i multiply left by x-5, and right by x+4?

[high lily]

yeh

well technically (x-5)/(x-5) and (x+4)/(x+4) respectively
multiply to both numerator and denominator

[vagrant brook]

ye

i just finished numerator im gona expand denom rn

ty i appreciate u

[high lily]

don't bother expanding the denom

just leave it factored

[vagrant brook]

my teacher prefers expanding and simplifying everything

i can factor out x from the top i think thats about it

[high lily]

,rotate

[high lily]

4x^2 - 3x^2 isn't 7x^2

[vagrant brook]

oh mb i think i added it

yea

[high lily]

,w expand (x+4)(x+2)(x-5)

[vagrant brook]

is denom wrong

[high lily]

you also seem to have messed up the expansion of the denominator

where's the work of your expansion

[vagrant brook]

i used chatgpt cuz i wanted to give u a quick response

[high lily]

don't use gpt for math

it has nfi what its talking about

[vagrant brook]

i should be good from here its lined up with the answer posted

[vagrant brook]

yea i got bad experiencse w it

[high lily]

if you want a math calculator, use wolfram

[vagrant brook]

ty i think im good from here

[vagrant brook]

alr

i appreciate it

.close

[obtuse pebble]

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[vagrant brook]

,w expand (x+4)(x+2)(x-5)

[obtuse pebble]

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@open spoke Has your question been resolved?

[wispy minnow]

use the factorial definition for nCr

which is n!/k!(n-k)!

i think

[obtuse pebble]

@open spoke Has your question been resolved?

[obtuse pebble]

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[nocturne osprey]

Can someone help me with proving a set is an open set using the definition of an open set?

[gritty vessel]

I can probably help. What's the set?

[nocturne osprey]

Let set A = {(x, y) in real | 4x^2 + y^2 < 8x} is the set

[gritty vessel]

okay let me think for a second

okay I think the best way to do this is to show that the complement is closed. let $(x_n, y_n)$ be a sequence in $A^c$ and suppose that $(x_n, y_n)$ converges to $(x, y)$. then construct a new sequence $z_n = 4x_n^2 + y_n^2 - 8x_n$ then $z_n$ is at least 0 for all $n$ so $\lim_{n \to \infty} z_n \geq 0$ which you can use to show that $(x, y) \in A^c$

@nocturne osprey

[warm shale]

Awesam

[nocturne osprey]

The worst part is that we are only allowed to use the...what is it with open balls and that being subset of A

[nocturne osprey]

I like this answer tho

[gritty vessel]

really? because the complement being closed is the topological definition technically it's the more "correct" one (even though they're equivalent in a metric space)

[nocturne osprey]

prof just wants us to suffer I guess

[obtuse pebble]

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[gritty vessel]

yeah idk what to do if you have to do it by balls. I'd suggest maybe trying to manipulate the ellipse into the right form and choose an r based on that. perhaps something to do with polar form?

[gritty vessel]

.close

[obtuse pebble]

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[tacit urchin]

Hi, can someone help me with a stochastic integral problem?

[tacit urchin]

This is what I have so far, I think that Bt is a martingale, but I am not sure how to show that e^B_t is also a martingale (if that is the right approach to show this expectation is 1)

[kind hawk]

try one of the advanced channels

[red ice]

#advanced-probability should be the right channel

[tacit urchin]

thanks

[boreal condor]

where did numbers go; hell why do i understand none of this after too many years of calc

[red ice]

"Measure theoretic probability, statistical inference, stochastic processes, sampling, time series, prediction, game theory, causal inference, stochastic differential equations"

Stochastic processes, there you go

[red ice]

Cause it's calculus of random processes

You've probably never seen it before

[tacit urchin]

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[obtuse pebble]

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[boreal condor]

and this is why im not a math major

[red ice]

Yeah this would typically be at a graduate level / upper undergraduate level

Difficulty is relative but yes

[obtuse pebble]

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[proven spindle]

suppose z = x + iy

suppose f(z) = u(x,y) + iv(x,y)

suppose u and v are functions of x and y of respective order at most 1

[proven spindle]

am i right to say that f(z) in terms of z, is at most order 1?

and if so, idk how to prove it

like there is no f(z) = p(a(z), b(z^2), c(z^3)...) such that f(z), when in terms of x, y which are respectively at most order 1, is at most order 1 in terms of z

[obtuse pebble]

@proven spindle Has your question been resolved?

[obtuse pebble]

@proven spindle Has your question been resolved?

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[little helm]

Given x,y are real numbers that satisfy x^2 + 5y^2 - 5xy + 2y + 3 = 0. Find the minimum value of P = xy + 2021

[red ice]

!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 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

[little helm]

i have begun but got stuck midway

[red ice]

Okay then, let's see your working

[little helm]

i tried to i used the above equation to get xy and substitute it into p

which gives (x^2)/5 + y^2 + (2y)/5 + 3/5 + 2021

i was told that i need to complete the square or smth like that

what is that?

[red ice]

They said circle

[little helm]

ok

anyways

[little helm]

how do i proceed from here?

that is if its even in the right direction

[red ice]

Yeah it's this reasoning

Basically if you know the minimum value of 5xy, you know the minimum value of xy and hence P

[red ice]

Complete the square

Of $x^2 + 5y^2 + 2y + 3$, then the minimum value of this is the min of $5xy$

[warm shale]

south

[little helm]

how do i complete the square????

i have never heard of that

[red ice]

This is completing the square

So yeah you need to do 5(y^2 + 2y/5 + 3/5) first

[little helm]

wait

i think im starting to get a hold of it

[red ice]

Ok nice

[little helm]

5((y- 1/5 )^2 + 1/25 + 3/5)

idk if this is correct

5(y- 1/5 )^2 + 16/5

[red ice]

Ah, very nearly

5((y- 1/5 )^2 - 1/25 + 3/5)

You're adding (-1/5)^2 = 1/25 so you need to subtract -1/25 to keep the expression the same

So yeah we have $x^2 + 5(y - 1/5)^2 + \frac{14}{5}$

[warm shale]

south

[red ice]

What's the minimum value of this?

[little helm]

14/5

am i right??

[red ice]

Exactly

So the min value of 5xy is 14/5

min value of xy is 14/25

Yeah and just add 2021 to both sides

[little helm]

so min of p is 2023.8?

[red ice]

It's cause you can make x = 0 and y = 1/5 at the same time, to make the rest 0 + 0 = 0

[red ice]

No

2021 + 14/25

[little helm]

why 14/25??

oh

wait ye i forgot

5xy

[red ice]

Cause at the very beginning we rearranged for 5xy

Yeah I did it this way

I just find it easier to worry about the last few steps later

[little helm]

ok tks mate for helping out

act can u also help me out in 1 more problem?

a^2 + b^2 = 4

find the minimum value of

[red ice]

Ah, so -(a^4 + b^4)

[little helm]

i got it to ab(a+b)^2 -12

[red ice]

= -((a^2 + b^2)^2 - 2a^2 b^2))

okok

[red ice]

I don't think this is very useful unfortunately

[little helm]

ye ur right

we cant tell if ab is positive or not