help-36

[mighty scaffold]

When we split the term

Why is it that there is only 1 3 that applies to a variable

U= 3x, V=Y, why is it not U=3x and V=3y?

That was my question

I’m fine with differentiating

But that simple foundation of math

Is lacking

From my knowledge

[gaunt scroll]

so u saying 3x*3y=3xy

[mighty scaffold]

Oh damn lol

HAHAHAHAHAA

[gaunt scroll]

that doesnt make sense at all

[mighty scaffold]

Oh my fucking god

I’m an idiot

Yo

You’re a good man for that

Loool I’m just high on THC

[gaunt scroll]

its 3rd lvl maths bruh

damn

[mighty scaffold]

When I’m sober I’m a cooker

But I can’t cook whilst intoxicated

[gaunt scroll]

😶

[mighty scaffold]

Me & you. Battle till death.

[cold gorge]

You didn't particularly answer my question though.

I hope you are, though it hardly seems like it.

[mighty scaffold]

I’m in the dusty trap

[gaunt scroll]

what question?

[mighty scaffold]

If my day was good

.close

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[daring pendant]

Can I get some idea of what to do here?

I thought to take integral but then I immediately realized I don’t know how to do that with multivariable yet

I don’t think I can do 3 variable elimination because theres a quadratic

So I’m kind of at a loss

[robust sedge]

Just integrate f_x with respect to x and integrate f_y with respect to y and compare to try and combine them to make a function that differentiates partially to get f_x and f_y.

[daring pendant]

We haven’t done ANY integration in class so there must be another way

[robust sedge]

Note. When you differentiate f(x, y) with respect to x, any functions within which only depend on y will get eliminated. So when you integrate f_x with respect to x, you get f(x, y) = x^3 +xy + c(y), where c(y) is a function of y, it can just be a constant too. A similar treatment can be done for f_y.

[robust sedge]

If you've been given this that would seem extremely unusual.

You've never done integration but are given partial derivatives?

[daring pendant]

Integration with multi variable functions

I know how to like u sub and trig sub

[robust sedge]

You essentially treat y as a constant (the same way as you would with a regular number) because it is fixed. For each y value of the 3D graph you're essentially looking at each curve as a function of x. If you considered the graph above y = 5, you'd have a curve in only x as you would substitute 5 into it. Just imagine that y is simply a number here.

Integrating f_x with respect to x thereby gives f(x, y) = x^3 + xy + c(y). Again, c(y) is there because if we differentiate with respect to x, assuming y is a fixed number, c(y) would be eliminated. This is done to avoid "destroying" information.

[final saddle]

@daring pendant Has your question been resolved?

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[merry lagoon]

The piece of cardboard is folded into a silhouette of a small sailing ship as shown in figure 2. The folding is done along a line CD so that B is brought over into E
Determine DC

[ashen tree]

One thing you could try is to express your answers in terms of trigonometric functions.

i.e. Try to find right triangles

[merry lagoon]

I forgot to mention one thing

I also have this information

@ashen tree how would i do it i am clueless

[ashen tree]

Do you find any right triangles?

[merry lagoon]

Why would it help to find right triangles?

[ashen tree]

Because it lets you use trigonometric functions

[merry lagoon]

Well, AEC is a right triangle

Nevermind, i was too fast with the coordinates

[ashen tree]

So, it doesn't have to be between named points.

[ashen tree]

seems reasonable.

Did you calculate 26,24

by yourself?

[merry lagoon]

Yep

I can't see a right angled triangle in there tho

Only ABC in the first figure

[ashen tree]

If F is the intersection of EC and AD, then AFC is right as indicated by the picture

And by introducing F, there also appears other right triangles.

[merry lagoon]

I don't know how it will help me but i will just follow your instructions i guess

So right angled triangles: FDC, FED, FAC

[ashen tree]

Yes. Which of these do you know any angles of?

[merry lagoon]

E is 41

rIGHT?

[ashen tree]

hmm

isn't <ECA = 41?

[merry lagoon]

It says B has become E and b was 41 before so e must become 41 too

[ashen tree]

hmmm, if the angles at B and A are 41 each, then the angles in ABC sum to 41 + 41 + 82 < 180

[merry lagoon]

Yeah, a and b are 41

[ashen tree]

I'm not convinced

[ashen tree]

But 41+41+82 = 164 which is not 180

[merry lagoon]

Wait, sorry, the angle is 49

I messed it up

[ashen tree]

no worries

[ashen tree]

Anyways, That is one angle. I recommend you try finding <FCD

[merry lagoon]

FCD = 24,5

[ashen tree]

Why?

[merry lagoon]

Because if you look at this it's one fourth

[ashen tree]

Great!

However, what is 82/4 ?

[merry lagoon]

20.5

[ashen tree]

Okay, so we have one angle in triangle CDF. Since it is a right triangle, you can find the ratio of any two of its sidelengths. (using trigonometry)

The same is true for the other right triangles.

[merry lagoon]

Wdym exactly?

[ashen tree]

You now have all the angles you need. With these you can find lengths.

(or rather deduce lengths from other lengths)

I am sorry, I gotta go now

[merry lagoon]

Bruh

Ok

Sendme the answer after

[final saddle]

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[tranquil pine]

how do i check if {e1,e2,v}

[tranquil pine]

to dertermine whether the subspace spanned by each of the following sets ia line, a plane or r^3

[smoky plover]

what is e1 and e2?

(x,y,z)?

[tranquil pine]

see idk

it just says that

so idk what to do

[smoky plover]

if e1, e2 and v are not in R^3 then it's not a subspace of R^3

[tranquil pine]

that isnt an option

the idea is you take your three vectors

[tough tiger]

What is the actual question?

[tranquil pine]

avector + bvector2 + c*vector3

reduce it with to row form

see if there are any free varialbes

[smoky plover]

so {e1,e2,v} is a subspace of R^3 u just have to find out if it generates a line, plane (R^2) or all of R^3 ?

[tranquil pine]

yh

[smoky plover]

first condition is that the vectors must be in R^3 if in the question they specify that {e1,e2,v} is a subspace of R^3 then u don't have to prove it

[tranquil pine]

its based of this figure

3.6

[rotund citrus]

are you asking if v is in {e1, e2}?

[tranquil pine]

no figure 3.6

e1,e2 and v

[rotund citrus]

spans R^3

[tranquil pine]

@rotund citrus

yes

[rotund citrus]

it does.

[smoky plover]

it does span R^3 because the vectors are not linearly dependent

[tranquil pine]

what does that mean precisely?

[rotund citrus]

v and w both occupy e3, it would span r^3

[smoky plover]

the vectors are not coplanar

[tranquil pine]

havent heard of that

but i assume it means

[smoky plover]

it just means that the vectors are not on the same plane

[tranquil pine]

yh ok

[smoky plover]

in order to span R^3 with 3 vectors they must not be in the same plane

[tranquil pine]

and by same plane

is to be on top of one another?

[rotund citrus]

3 vectors must be linearly independent.

Ax = 0 must only contain the trivial solution

Ax = b for every b, etc

[smoky plover]

top of one another? u mean in the same plane ?

[tranquil pine]

e4 is on top of e2

[smoky plover]

u mean collinear vectors.

[rotund citrus]

e2 = (0, 1, 0)

[tranquil pine]

also how much do you differ to span a different plane

do all these lines span different planes?

[smoky plover]

to span a plane u just need 2 non collinear vectors, if that's what u're asking about

[rotund citrus]

if they are flat along e1, e2 then no

they would span the e1, e2 plane

[tranquil pine]

yes i mean that the are exact similar to e1

other than they point in another direction

does v and w span different planes?

they are only different in the x-axis right?

[smoky plover]

yes

[tranquil pine]

why

their y and z are the same

(deltax,0,0)

if you were to make it in change

[smoky plover]

oh sorry i didn't notice that

v and w span a line

that's why {e1,v,w} span a plane even though the set contains 3 vectors

[tranquil pine]

a line?

[smoky plover]

cause v and w are linearly dependent

[tranquil pine]

r^2

doesnt w span a line

but v is in r^3

or am i mistaken

[smoky plover]

it is

[tranquil pine]

w = (0,y,z)

[smoky plover]

in the other photo it says {e1,v,w} spans a plane

[tranquil pine]

but v = (x,y,z)

[smoky plover]

do u have coordinates of the vectors ?

[tranquil pine]

they are in the picture

are they not?

or am i totaslly lost

[smoky plover]

haha i mean exact coordinates

like numbers

[tranquil pine]

then no

[smoky plover]

u can express e1 in terms of v and w so e1,v,w are linearly dependent which is why {e1,v,w} spans a plane when they are 3 vectors

[tranquil pine]

ok i am a little new to this so lets start of with the very basics

[smoky plover]

but {v,w} spans a plane too because the vectors are not linearly dependent

[tranquil pine]

spanning (x,y,z) just means you have a non-zero in all coordinates?

[smoky plover]

spanning means generating

how many and which vectors do u need to generate a line, plane or space (which is R^3)

[tranquil pine]

is this a question?

[smoky plover]

no xd

just an intuition to what span means

to generate a line u just need ANY vector

[tranquil pine]

yes

[smoky plover]

to generate a plane u need AT LEAST 2 non colinear (parallel) vectors

to generate space (R^3) u need AT LEAST a plane (2 non colinear vectors) and another vector that is NOT IN that plane

i hope this is clear

[tranquil pine]

i thought space was just 3 dimensions

damn

[smoky plover]

yeah 3 dimensions so 3 vectors or more

a plane is 2 vectors remember

[tranquil pine]

yh i can imagine

[smoky plover]

so if ask u if {r,t} spans R^3 ur immediate response should be no because 2 vectors are not enough to span (generate) R^3

[tranquil pine]

yes that makes senes

sense

but that isnt the quesiton

[smoky plover]

ur questio is if {e1,e2,v} spans R^3?

[tranquil pine]

yes

or like show me how

not just yes or no

[smoky plover]

ok to span R^3 what do u need ?

[tranquil pine]

3 vectors

that arent

the term you used

linear independent i believe it was

[smoky plover]

that's too abstract for a beginner

[smoky plover]

let's stick to this definition for now

colinear means parallel vectors btw

[tranquil pine]

alright

yes but when you say parrellel

parallel*

do they have to be the exact same

exepct one is "longer"

[smoky plover]

no we don't care about their magnitude

colinear or parallel vectors means that they have the same direction

the same "slope"

[tranquil pine]

isnt that what i said

[smoky plover]

u were talking about magnitude

[smoky plover]

the exact same what

[tranquil pine]

arent all these parallel

[smoky plover]

correct

so back to ur question, to generate R^3 we need a plane (2 non parallel vectors) and a vector that is not in that plane

check if {e1,e2,v} verifies that condition

[tranquil pine]

well

two non parallel

that would be e1 and e2

[smoky plover]

yup

and is v in that plane

(xy) plane

[tranquil pine]

i thought it would not be in the plane

[smoky plover]

yes and that's because it has a z coordinate

[tranquil pine]

so it would span r^3

[smoky plover]

yup

[tranquil pine]

isnt this th plane that they span

[smoky plover]

exactly

which is (oxy) plane

[tranquil pine]

how is w in this plane?

[smoky plover]

it's not

[tranquil pine]

so how can you tell me its not in r^3

[smoky plover]

i didn't say that

[tranquil pine]

then e1,e2 would also span the r^3

and v

not just those two

[smoky plover]

2 vectors can not span R^3

[tranquil pine]

i just said that

read what i said right below it

[smoky plover]

u said e1,e2 would also span the r^3

[tranquil pine]

no

[smoky plover]

haha ok

[tranquil pine]

i love paint xd

at any rate

how about e2,e2 ,w

they would not span r^3

correct

[smoky plover]

u mean {e1,e2 ,w}?

[tranquil pine]

yes

no

e2 e3 w

[smoky plover]

yup they can't span R^3

[tranquil pine]

they all have no x

[smoky plover]

because u can express w in terms of e2 and e3 which means that w is in the plane generated by {e2,e3}

which means that the vectors are linearly dependent

[tranquil pine]

how do you that

i can see how you could

exprexx w by e3 and e2

but what equation

[smoky plover]

$w=a.e2+b.e3 \ , \ (a,b)\in \mathbb{R}^2$

[tranquil pine]

so two constants

[soft zealot]

Adam Chebil

[smoky plover]

yes

[tranquil pine]

last question

i need to show the implicaiton, hypothesis and conclusion of

if x = y then |x| = |y|

[smoky plover]

what does this mean??

[tranquil pine]

you know this

logic

p -> q

biconditional and that sorta thng

[smoky plover]

i study math in french

[smoky plover]

u just need to prove this ??

[tranquil pine]

yes xd

but its seems a bit too easy

[smoky plover]

x and y real numbers ?

[tranquil pine]

since its true by definition

yes

if x and y is a positive number then |x| = x and y = |y|

[smoky plover]

i think Proof by contradiction would work here

try it

[tranquil pine]

ok

thanks for help

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[primal juniper]

What am I supposed to do here, I am confused here

[primal juniper]

or how to prove it rather

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@primal juniper Has your question been resolved?

[primal juniper]

<@&286206848099549185>

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@primal juniper Has your question been resolved?

[primal juniper]

<@&286206848099549185>

[flat iris]

so im guessing you dont understand the problem？

[primal juniper]

Yea more so what to do

[flat iris]

i mean it says to use induction

or u start proving

n=1

[primal juniper]

I never done induction through word problems

but is it the same way, base case, hypothesis then prove K + 1?

[flat iris]

sure

yea

[primal juniper]

ill do them and then when I'm done I will show u

[primal juniper]

is n = 1 easier?

[flat iris]

you start with that

and then continue with your n=k+1

[primal juniper]

swap the 2nd and third, but how is this?

[primal juniper]

yea idk if I did it right

[flat iris]

👍

W

[primal juniper]

thx

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[twin adder]

First I just solved for x and y, so we have x = s - t and y = t. Then I found the Jacobian and just got 1

How do I find the bounds?

if I plug in the definitions of x as functions of s and t I get the ellipse bounded by s^2 + 3t^2 <= 1

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[woeful raptor]

could you explain how this equation was simplified from step 1 to step 2?

[royal gust]

Looks like some ys were cancelled

[warm ether]

expanded the right bracket, got a y^-1, so multiplied everything by y

and pulled out a 3

[royal gust]

Well, let's look at the numerator alone:
3x²y³ - 3x³y²(-x³/y³)

Can you add those fractions and put everything above y³?

[woeful raptor]

i'll try writing it out

okay i got 3x²y³ +((3x^6)y²/y³)

which then i can simplify to 3x²y³ +((3x^6)/y)... right?

[royal gust]

Yeah you can do that

Still, gotta add those fractions together and put everything above y

[woeful raptor]

ohhh

okay!

now i have (3x²y^4)/y +((3x^6)/y)

so now i can bring the y down to the denominator, making it y^7

and then factor out the 3 in the numerator!

that makes so much sense!

thanks so much

.close

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[teal aspen]

how do I divide the numerator by x^2 ?

[ashen tree]

Perhaps it helps if you instead of dividing by $x^2$ first divide by $x$, then by $\sqrt x$ and then lastly by $\sqrt x$ again.

[soft zealot]

Ivar Ängquist

[ashen tree]

Assuming x>0, this is the same as dividing by x^2

[ashen tree]

Also, remember that $\frac{\sqrt A}{\sqrt B} = \sqrt{\frac AB}$

[soft zealot]

Ivar Ängquist

[teal aspen]

ah I think I get it, thanks!!

[ashen tree]

np

[teal aspen]

.close

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[nocturne pawn]

yo

[nocturne pawn]

will someone jus like do my shit for me

i do not feel like doin alla this

[junior token]

no, sorry, that’s not what this server is for, but if you ever want to learn how to do a problem yourself, then you are welcome to ask here

[nocturne pawn]

uhh ok then

[junior token]

what have you tried?

[nocturne pawn]

i do not know how to do any of this

i am

fuckin brain dead

bro i jus went off of copy pasting but it aint givin me shit for answers so i have to learn the problem

[nocturne pawn]

jus slide some help

for ya boy

[junior token]

I’m trying to help, so I’m asking what you understand first

[junior token]

This is related to the vertex of a parabola

Do you know how to complete the square?

[nocturne pawn]

an imma tell u absolutely none of it

im a fresh baby OUT DA WOMB when it comes to dis

imma take a guess alright if i get it right tell me fr

I STILL

wanna know how to do it

[nocturne pawn]

but let me guess first

12.25

GUESSING

[junior token]

has your teacher ever mentioned completing the square?

[nocturne pawn]

i do online school so ive just been teaching myself

i dont do online school bc i want to

i do it bc i did an accidental illegal

so im in online school

i dont have any teachers bc this program is nuckin futs

[junior token]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:vertex-form/v/vertex-form-intro this video might help

[nocturne pawn]

bro u want me to do alla that

[junior token]

it’s a video

[nocturne pawn]

nvm my shit jus closed on me an it got locked

on god

i aint tryna get outta this

im on opera gx an i pressed x an my shit closed

[junior token]

this channel is not closed yet

[nocturne pawn]

no i mean like

the problem i was on

alright

we should be goo

now

@junior token

im on a new problem

bc my last one yknow said fuck u

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[novel schooner]

What does the subscripts mean?

[novel schooner]

I think thats what their called

[tired walrus]

subscripts do not have any fixed meaning

but here $b_1$ and $b_2$ refer to the two bases of the trapezoid

[soft zealot]

AnnGhost

[novel schooner]

Is that universal? With all math problems?

[tired walrus]

no

nowhere near

subscripted letters are just an extra set of variable names

[novel schooner]

Ok, i understand.

My last question is: Are there any steps i would have to do that are not listed in the formula?

I would solve it
((7+10)/2) + 5

[tired walrus]

times 5.

not plus 5.

[novel schooner]

Ah ok. The space means multiplication. What about simplifying?

Ive had a few where simplification is needed or the answer would be wrong

Actualy never mind on that

I think that all my questions

.close

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[tranquil pine]

how did she go from 2^5/2 --> (2^5)^1/2?

[unique aspen]

$a^{bc} = {\left(a^b \right)}^c$ always

[soft zealot]

Kaisheng21

[tranquil pine]

i hope you have a blessed day and your pillow is warm

[rich tide]

,tex .exp rules

[soft zealot]

B-eard

[tranquil pine]

thank you

.close

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[sturdy adder]

Why these 2 functions are different?

[sturdy adder]

If you equal P(x) to 0 and do some algebra you get h(x)

But when graphing it's different

why

[spring haven]

how?

[sturdy adder]

-6/2 then + 169/16

that is 121/16

[spring haven]

that doesn't mean that P(x)=h(x)

[sturdy adder]

what does it mean then

[spring haven]

it just means that they have the same roots

[sturdy adder]

I am just very confused

I always asumed that if you did the same operations on both sides the equation does not change

[oak vortex]

if you multiply an equation the roots dont change but the graph will stretch or shrink

[sturdy adder]

yea

but it's literally changing the function

[spring haven]

also, the two functions that you have written do not even have the same roots

how did you get h(x)?

[sturdy adder]

factoring g(x) by completing the square

no wait

that's p(x)

[spring haven]

wait my bad

it's correct

[sturdy adder]

h is another way of factoring by completing the square

but it outputs a different function

[spring haven]

$x=0 \implies 2x=0$, but the graphs of y=x and y=2x are different

[sturdy adder]

well tbh it's the same stuff just doing algebra to get rid of the 2 and the extra terms

[soft zealot]

kheerii

[sturdy adder]

but it modifies the function

[spring haven]

the way you have done it gives you $p(x)=2\cdot h(x)$, so they will not have the same graphs, even though they have the same roots

[soft zealot]

kheerii

[spring haven]

however, if the two functions had the same roots AND the same leading coefficient, then the two graphs will coincide

for example, the graphs for g(x) and P(x) in your case will coincide

[sturdy adder]

right

if I *2 all of h(x) I get the same thing

that is kinda weird I gotta say and non intuitive

[spring haven]

well you are changing the function by dividing by two

[sturdy adder]

I don't think I've ever had anyone in my classes that actually knew this

[spring haven]

there is a difference between a function and an equation

[sturdy adder]

but if I also divide by 2 the other side why I am changing it?

[sturdy adder]

yes, and I just learnt this

[spring haven]

it's true that you can divide both sides in an equation, but that is changing the functions on both sides

$f(x)=g(x) \implies \frac{f(x)}{2}=\frac{g(x)}{2}$, but the graphs of $y=f(x)$ and $y=\frac{f(x)}{2}$ will not coincide (unless f(x) is identically equal to 0)

[soft zealot]

kheerii

[sturdy adder]

thx

idk how many years of math classes and not a single teacher/professor ever bothered teaching this properly

that is the problem when excercises are robotic and either make you graph without factoring/finding roots or simply find roots and be done with it

if I've ever had excercises where I had to do both and graph the final function I would have learned this before

thank you

[final saddle]

@sturdy adder Has your question been resolved?

[final saddle]

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[molten mesa]

Hi

[molten mesa]

gimme a sec to write the question

So if we are converting the parent function $\frac{6}{x-3} + 5$ to the child function $\frac{6}{-3x-3} +5$ . IS THIS reflection over the y axis or the x axis. And is it a vertical dilation of 3? Or is kt a horizontal dilation by a factor of 1/3? Or is it horizontaly dilated by 3?

[soft zealot]

ahaan

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@molten mesa Has your question been resolved?

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[lofty kindle]

maybe im missing something basic here but how did it turn negative on the third part 😭

[abstract walrus]

y-x = -(x-y)

[spring haven]

$\frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = -1$

[soft zealot]

kheerii

[lofty kindle]

oh fuck im dumb

i was thinking the y-x was x-y and divided it out

thanks for clearing that up, i legitimately did think that was a x-y on top

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[tranquil pine]

meaning of pi

[tranquil pine]

Pi is the ratio of the circumference of any circle to the diameter of that circle, and is written as the Greek letter for p, or π

[molten mesa]

Ugh

[tranquil pine]

?

[molten mesa]

this

just nvm

[tranquil pine]

oh i see

[molten mesa]

HOW ARE THERE NO HELP CHANNELS LEFT

[tranquil pine]

💀

[tranquil pine]

thanks

np?

you could have searched it

.closerequest

.close

[tranquil pine]

can you close it

[tranquil pine]

uh how

type .close

.close

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[vast coral]

so

[vast coral]

I have this line

in R3 (line given by a point and vector)

[steel roost]

ok

[vast coral]

and I want all a,b,c,d from R where a plane with equation ax + by + cz = d includes line p

[steel roost]

a plane is defined by two vectors, right?

[vast coral]

yea

[steel roost]

so thats all you need

one vector you can get from p

[vast coral]

yeh obv i will get vectgor u that is perpendicular to v(v being the -1,1,0 vector)

and it includes the points A

thats 1 plane

[steel roost]

so you already found the perpendicular vector?

[vast coral]

it is necessary to describe all a,b,c,d of the given condition.

[steel roost]

now you need a vector along line p

[vast coral]

there is unlimited amount of them

u=(1,1,0)

[steel roost]

yes

[vast coral]

for example

[steel roost]

now you must find a vector along p, right?

[vast coral]

what does "along p" mean?

[steel roost]

along line p

[vast coral]

parallel?

[steel roost]

yes

[vast coral]

well that can be the one given

-1 1 0

[steel roost]

thats in the direction of p?

[vast coral]

no, thats a point

the t(asfdasdfa)

[steel roost]

yes

[vast coral]

after t, thats the vector

[steel roost]

but p is the first part plus t(asdasda)

[vast coral]

yes its a line, given by a point and a vector

righbt?

[steel roost]

yes

i made a mistake

p is a vector that describes a line

so the vector in p's direction would just be p

[vast coral]

yeh well, i understand, it must be 2 vectors that are linearly independent

[steel roost]

yes

[vast coral]

thats what i said

[vast coral]

yeh i have the geometric imagination about it, its just how to write it

so can anyone help me please?

https://cdn.discordapp.com/attachments/908076393198419988/1167187276221124772/image.png?ex=654d36db&is=653ac1db&hm=96163777e3fb56e13005eb5efd8fdb90d9b07025bd922dea68249f8e5e049d94&

and I want all a,b,c,d from R where a plane with equation ax + by + cz = d includes line p

[steel roost]

you need a point on line p and your normal vector

you already have the normal vector

if you let t=0, then you get the point (3,1,2)

a plane is defined as

where r0 is the vector given by your point <3,1,2>

and r is any vector <x,y,z>

thus

take the dot product:

<-1,1,0>*<x-3,y-1,z-2>=0

and you will obtain your equation

[vast coral]

so that answers the question for what a,b,c,d € R does the plane given by general equation ax+by+cz=d contain the line p?

cuz there is unlimited amount of those planes

[vast coral]

. this is normal vecot

vector

@steel roost

why did u put n as the one given?

[final saddle]

@vast coral Has your question been resolved?

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[eager stirrup]

i keep getting the wrong answer

[eager stirrup]

i have that using loptials rule i write this as

ln(1-4.5/x)/(1/-7.5x)

and then taking the derivative i have (1/(1-4.5/x))/(1/-7.5x)

which i believe evaluates to 1/0 after taking the limit of the numerator and denominator.

but DNE is wrong as an answer

[tough tiger]

This is not equal to that

What a^b is equal to using exp and ln ?

[eager stirrup]

but its of the form inf^0

wait

[tough tiger]

Sure but you did not write the right formula

[eager stirrup]

im actually not sure

[eager stirrup]

whats the right formula?

[tough tiger]

Well answer this question

To see if you know the right formula

[frosty musk]

Try a change of variable, t=1/x might ease things out

L=lim t->0+ e^(-7.5[ln(1-4.5t)/t])

[eager stirrup]

i have to put it as a power of e?

[eager stirrup]

i forgot what to do from here though

[tough tiger]

Yes exactly

But he gave the answer so now you got it

[eager stirrup]

how do i simplify it though?

[tough tiger]

Do you know the theorem for continuous function ?

[eager stirrup]

i think so but i need a reminder

[tough tiger]

That lim _(x->a) f(x) = f(lim x-> a)

[eager stirrup]

oh right

that one

[tough tiger]

Yes

Try applying this one

[eager stirrup]

so the limit would be e^lim(-7.5[ln(1-4.5t)/t])?

[tough tiger]

Yes exactly

Now I think you’re able to use L’hospital

Did you find it ?

[final saddle]

@eager stirrup Has your question been resolved?

[final saddle]

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[tranquil pine]

Found this interesting question earlier that I want some assistance on.

\vs{4 mm}
Suppose we have two random variables $X,Y$ with corresponding pdfs $\m{f_X}x, \m{f_Y}y$. Suppose $Z = X + Y$. Now the convolution of $\m{f_X}x$ and $\m{f_Y}y$ is [
\m{f_Z}z= (f_X \ast f_Y)(z) = \int_{-\infty}^\infty \m{f_X}x\m{f_Y}{z-x} \dd x
]
The question is, is it possible to prove that the convolution of $Z=X+Y$ is not necessarily the same as that of, say, $W=X-Y$?

[candid hull]

I mean X+Y and X-Y don't even have the same domain most of the time

[opal plinth]

Just find a counterexample?

[candid hull]

how would you expect their distributions to be equal in general ?

@tranquil pine

[tranquil pine]

but would their convolutions not be the same?

[candid hull]

the convolution (of the PDFs of X and Y) is the resulting distribution of X+Y

what special meaning do you assign to convolution here

@tranquil pine

[tranquil pine]

nothing much than what you normally consider the convolution to be

I may have misinterpreted this question

sorry

I'll close the channel then

.close

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[sharp tundra]

hello

[vivid walrus]

👋

[sharp tundra]

so

this is my new algothrim

to define S

[sharp tundra]

does this makes sense

[vivid walrus]

it seems

[sharp tundra]

this math

is mainly different from the rest

bc it creates a infinite illusion of numbers but having a pattern

<@&286206848099549185>

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[weary lark]

When taking the d/dx of what's circled, I am confused as to how it becomes what it is in the next step.

[royal gust]

That's the quotient rule

[weary lark]

Oh okay thank you

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[final saddle]

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[magic ivy]

i cant figure out how to prove it sums to greater than 1/2 for m>2

[gilded schooner]

978??

[magic ivy]

yes

[mint orbit]

well can you bound each term?

which term in the series is the smallest? whats its magnitude

[magic ivy]

@mint orbit the smallest term is 1/(2m). Could you please explain what it means to bound the terms?

[mint orbit]

okay, so the smallest term is 1/(2m)

then $\frac{1}{m} + \frac{1}{m+1} + \dots + \frac{1}{2m} > \frac{1}{2m} + \frac{1}{m+1} + \dots + \frac{1}{2m}$

[soft zealot]

jan Niku

[mint orbit]

ill replace what is probably the largest term in the series with the smallest term

and create something smaller

you can repeat this process

eventually, youll create a very simple series, that is definitely smaller than the original series

or at least as small, if theres onlyone term

because youve replaced every term with the smallest term

make sense?

[magic ivy]

In repeating the process would I next replace 1/m+1 with 1/(2m)?

[mint orbit]

yes, and in doing so youd create a series thats smaller still

because 1/(m+1) is bigger than 1/(2m)

well theres a case where theyre the same, but lets say were not dealing with that

you definitely wont make the series any bigger

you would then repeat this process with every term in the series

$\frac{1}{m} + \frac{1}{m+1} + \dots + \frac{1}{2m} \geq \frac{1}{2m} + \frac{1}{2m} + \dots + \frac{1}{2m}$

[soft zealot]

jan Niku

[magic ivy]

right

[mint orbit]

weve made a lower bound. What is it equal to? is it helpful? is it trash and we need to try to make something better?

[magic ivy]

i am trying to figure the amount of terms on the right side

[mint orbit]

ah thats the key

try doing some cases

say m=2, or 3, or 5

[magic ivy]

It seems to be m+1 terms

[mint orbit]

yea, this sounds about right

[magic ivy]

so the series ≥ (m+1)/2m

[mint orbit]

whats your conclusion

useful lower bound or trash

[magic ivy]

very useful

[mint orbit]

[magic ivy]

Thank you for introducing me to this concept of bounding

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[final tangle]

!patience

[final saddle]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

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[final tangle]

show how you're getting a different result

are you doing work on paper?

anything on which work can be presented

digital screen, black/whiteboard, etc

anything that'll allow you to show others what you're doing

reply pings can be turned off

but otherwise i get seperate results
49 is the expected result

what else are you doing that's leading to something different

yes, as it should

(not an equation)

well get them to show they're work

no justificiation, ignore those claims

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[ruby elk]

no idea what to do or where to start

the fact that i could stare at this all night and it still would make zero sense to me

[tranquil pine]

me too

is hard

[ruby elk]

yeah.

sigh

i really do hate this class

[ruby elk]

i dont get whats going on, but my parents want me to take it because itll look good on a college application

im just worried, that really, im not good at anything

sure, everyhting is an A- or above, but that wont get me anywhere by the time i graduate.

the college scene has become so competitive, i wonder sometimes if ill ever really amount to anything, anywhere

i know it isnt true, but like, my parents constantly tell me it.

i know im wrong, but i just want seem to believe it

...ugh

im so damn tired

[tranquil pine]

me too

[ruby elk]

~by your side, ill be your seasons~

[tranquil pine]

even if i graduated just

its hard

[ruby elk]

im in 9th grade

this shit should be illegal now

[tranquil pine]

cool

[ruby elk]

something about mental health or something

and i think the most aggravating part for me

is that everyone else gets it

everyone else got a little boost

a little help

me?

nope

on my own

ugh.

[tiny star]

can u pls tell me what the sign with the = and wiggly line on top means

[ruby elk]

congruent

[tiny star]

ok ty this makes my life easier

which qn u need help with?

[ruby elk]

literally all of them

in a perfect world id get 1-5 done

6-7 nobody gets

so my black ass sure as hell wont be able to

and then for 1 you make your own given

[tiny star]

sry i cant help u i hate triangles

[ruby elk]

yep, figured

[tiny star]

anybody who likes geometry is a nerd

[ruby elk]

i dont like geometry buddy

i need to pass this class to survive

[tiny star]

i can tell

[ruby elk]

these are actually some disturbingly high stakes in the grand scheme of things when you think about it

one wrong writing stroke and you're out of any major college,

which means youre basically out of any major paying job

anyway

<@&286206848099549185>

[tranquil pine]

is hard maths like

in 1st grade

like

1

i

was

like speedruning

50 ques

sheets

of multipli

cation

and divisin

[ruby elk]

..same

[tranquil pine]

UNTIL *** ECUATIONS AND FUNCTIONS LIKE

f(x) = 2x

like What

how did i continue from

multiplication

[ruby elk]

y= 2x

[tranquil pine]

that

[ruby elk]

y is 2 times x

[tranquil pine]

correct

yes

but

[ruby elk]

...so

[tranquil pine]

you understand?

like

its hard

and i just continued

maths

like

it was easy

but its hard

[ruby elk]

...yep.

[final saddle]

@ruby elk Has your question been resolved?

[ruby elk]

no.

<@&286206848099549185>

[grave steppe]

So what do u understand from the first question?

[ruby elk]

i understand that a perpendicular bisector of a segment

is a line perpendicular to a segment passing through

like

the midpoint

of the segment

but idk how to put it into words

[grave steppe]

Ok what country are u in?

[ruby elk]

the USA

[grave steppe]

Wait hollup i havent taken geometry in a while i messed up

The first 3 arent questions

4 and 5 are the questions

[ruby elk]

they arent?

[grave steppe]

They arent

[ruby elk]

hold on.

brb

[grave steppe]

Theyre more like statements and pieces of info

[grave steppe]

Ight

Ping me when u back

[ruby elk]

back

im 99% positive these are actual things he wants us to prove

but help is help

so

bcz hes had us prove lemmas and corollaries before @grave steppe

[grave steppe]

Fuck bruh im bad at this typa shit😭

I never had to do lemma

[ruby elk]

its fine frl

ill js do 4 + 5

...which im tryna get rn but i dont

[ruby elk]

how do you know this im so confused

im so close to just quitting

its 11:05 PM and i have other work to worry about

.close

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[violet crow]

Hi! I have a question related to Mathematica. I want to plot y = 1/x^3 from x = 1 to x = 2, then revolve the region below it about the line x = -1. How is it possible to revolve around a line? Below commands work for default revolution.

q3a = Plot[{1/x^3, 0}, {x, 1, 2}, 
   Epilog -> {Line[{{1, 0}, {1, 1}}], Line[{{2, 0}, {2, 1/8}}]}, 
   PlotStyle -> Black, Filling -> Axis, 
   PlotRange -> {{0.5, 2.5}, {0, 2}}]
q3b = RevolutionPlot3D[1/x^3, {x, 4, 9}, {\[Theta], 0, 2 \[Pi]}, 
   PlotStyle -> Gray]

[violet crow]

.close

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[final saddle]

@fading pulsar Has your question been resolved?

[final saddle]

@fading pulsar Has your question been resolved?

[fading pulsar]

<@&286206848099549185>

[unreal anvil]

hello Tangerine, what do you need help with?

[fading pulsar]

The following integral, you have to do via trig sub.

I got up a form where I had “(sin^2”/(cos^2)^3/2)cos dtheta

[unreal anvil]

ohhh, i was wondering what is this about lol
it's integral, i see

is it a must to use trig sub?

[unreal anvil]

anyways,this is pretty nice, where were you stuck on?

[fading pulsar]

that part, I had to get it back eventually to terms of x and integrate it

[fading pulsar]

Yeah

[unreal anvil]

since you used trig-sub, you can do it in terms of theta first and then get back to x

$\int \frac{\sin²(\theta)}{\cos²(\theta)}\dd \theta$

[soft zealot]

Biscuity

[unreal anvil]

so after simplifying we get this

[unreal anvil]

did you get this? @fading pulsar

[fading pulsar]

no but that would be tan^2thata

Which we can simplify to 1- sec^2theta and integrate

[unreal anvil]

yep

[fading pulsar]

Thank you. Can you show how you got there?

[unreal anvil]

wait

isn't it sec²theta - 1?

[fading pulsar]

Yes

[unreal anvil]

anyways, I'll type

,align
&\frac{\sin²\theta}{(\cos²\theta)^{\frac32}}\cdot\cos\theta\
=&\frac{\sin²\theta}{(\cos\theta)^{\frac32\cdot2}}\cdot\cos\theta\
=&\frac{\sin²\theta}{(\cos\theta)^{3}}\cdot\cos\theta\
=&\frac{\sin²\theta}{(\cos\theta)^{\cancelto{2}{3}}}\cdot\cancelto{1}{\cos\theta}

there we go

[soft zealot]

Biscuity

[unreal anvil]

i think you know what I'm typing 😛

sorry for the typos

[fading pulsar]

Ahh I see thank you

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@fading pulsar Has your question been resolved?

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[viscid sinew]

im totally stuck on how to solve this recursive equation

this is all i came upp with, im very glad if sb could help me!

<@&286206848099549185> ping

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@viscid sinew Has your question been resolved?

[viscid sinew]

<@&286206848099549185> ping

[strange rune]

master theorem?

do you know about this?

https://de.wikipedia.org/wiki/Master-Theorem

its a direct application of it

i think

[viscid sinew]

i have heard oof this but we do not have learned this yet so im not allowed to use it

i need to solve this with induction

[viscid sinew]

<@&286206848099549185> ping

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[dense garnet]

Consider $$f(x,y)=\begin{cases} \sin(x-y)& \text{if } 0<x-y<2\pi \ 0 & \text{else}.\end{cases} $$ It is claimed this function is continuous on $\mathbb R^2$. What are some ways in which I can verify this? I'm not sure where to begin.

[soft zealot]

sunside

[warm ether]

y=0 is continuous,
sin(x-y) should be continuous
so youre really just checking if its continuous at x-y=0 and x-y=2pi

[dense garnet]

yeah, that makes sense 🙂

[dense garnet]

so at x-y=0 and x-y=2pi we have that sin(x-y)=0. Would you say that shows continuity?

[warm ether]

yeah, the limits from each side are equal,
and the limits = the value of f(x,y) at those points also

[final saddle]

@dense garnet Has your question been resolved?

[dull ravine]

@dense garnet what’s still unclear

[dense garnet]

well, I'm still confused about how to write more formally that the limit of sin(x-y) at x=y is 0. After all, x=y is not a point and the same for x-y=2pi.

when we speak about continuity, we speak about continuous at a point, not a line, which x=y represents

I was thinking something along the lines lim_(x,y)->(x,x) sin(x-y)=0, but I don't know if this is something you can write

[dull ravine]

I don’t know what the “correct” notation is

But I think you can still think of the continuity of f as it being continuous at every point

But at every point in a 3 dimensional space, not in a 2 dimensional plane

At every point (x,y,z) to be precise

[dense garnet]

hmm, yeah. Would you say $$\lim_{(x,y)\to (x,x)} \sin(x-y)=0,$$ makes sense?

[soft zealot]

sunside

[dull ravine]

Maybe you can write something like $$\lim_{y\to x^+}$$ and $$\lim_{y\to x^-}$$ to show that $f$ is continuous at every point $(x, x, 0)$

[soft zealot]

FirstNameLastName

[dull ravine]

But make sure to verify this suggestion yourself, cause I’m still in high school and have no clue what the conventional approach here is

[dense garnet]

ok 🙂

[dull ravine]

I do think this should work though

[dense garnet]

.close

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[magic sparrow]

I'm working on a proof that differentiability implies continuity, but I am confused about a certain detail

[magic sparrow]

Essentially it goes as follows

[soft zealot]

Austin

Austin

Austin

[magic sparrow]

But here is where I'm confused. It was elluded in lecture that

we have to do the limit from left and right seperately?

the limit as h->0+ or 0-

or else the proof doesn't work

but I'm not seeing why that is true....

What part of this is false if we do not do that?

@signal shell

[hybrid heath]

Didn't we already prove this?

[magic sparrow]

we did a more complicated version

for multivariable differentiability

and actually

we did the reverse direction

continuous partials => differentiable