help-33

[hot cloud]

ye

[willow owl]

so you may not always get only one solution

[hot cloud]

we’re trying to find the vertices

to maximize profit of the objective function

[willow owl]

most of the time you will not only get one solution

ah wait

there are mistakes here

for example in the first inequality you should have 7x+4y=<168 instead of 7x+46=<168

[hot cloud]

yeah sorry that’s what i meant

typo

[willow owl]

same mistake in the second inequality of the first system and the first inequality of the second system

[hot cloud]

yeah i meant to put 4y

[willow owl]

they are probably typos

[hot cloud]

yeah

[willow owl]

np

other than that you are fine

if you are to find the vertices

then you need to solve the analogous systems of equations of these

not the systems of inequalities

[hot cloud]

wth is that

[willow owl]

7x+4y=168
14x+2y=210

7x+4y=168
14x+5y=252

14x+2y=210
14x+5y=252

these systems

[hot cloud]

hollup

lemme fix the ti pod

typos

7x+4y<=168
14x+2y<=210

7x+4y<=168
14x+5y<=252

14x+2y<=210
14x+5y<=252

[willow owl]

so you need to solve these systems of equations

[willow owl]

instead of these systems of inequalities

[hot cloud]

they’re the same thing tho

oh wait

the equal signs

yeah i changed them when i solved it i forgot to type it

[willow owl]

so basically what you did here was that you solved the systems of equations

not the systems of inequalities

which is the correct thing to do

[hot cloud]

ight bet

so i got the vertices correct right

wait but in the question it said write a system of linesr inequalities

[willow owl]

i checked the solutions of the systems and yes they are correct

[willow owl]

idk what the restrictions are but isnt this what you did

and then isnt there a follow-up question asking for the vertices ?

[hot cloud]

ye

[hot cloud]

yeah

there aren’t anymore vertices right?

just those 3?

[willow owl]

ok thats what you did

what do you think

thats a question for real

[hot cloud]

wouldn’t there be (24,0) (15,0) (18,0) (0,42) (0,50.4) and (0,105)?

bc that’s when they meet the x and y axis

[willow owl]

no

they form vertices when they intersect with each other

[hot cloud]

but x>=0 and y>=0 are apart of the system too

[willow owl]

that wasnt mentioned before

but ok if these are included too then yes

[hot cloud]

wait

how do ik if they r included or not

[willow owl]

because you will have more than these systems here

[hot cloud]

i’m not sure if it is included

how do we know when to include x>=0 and y>=0

[willow owl]

i am not pretty sure too tbh

you better wait for someone to give you a sure answer for this

[hot cloud]

alr

in that case can i ping the helpers role again

[willow owl]

i am not sure if the one(s) who will respond to it will help or just tell you to not ping it and go

if you want wait for 15 mins and then ping

[hot cloud]

alr i’ll wait 15 mins

[willow owl]

good luck

[marsh citrus]

@hot cloud Has your question been resolved?

[desert dirge]

and they were never heard from again

[tulip idol]

This is indeed correct.

[hot cloud]

oh shoot

wait so that means

[tulip idol]

Should've written what you're supposed to maximize as well

[hot cloud]

oh yeah that was in a previous question

[tulip idol]

To maximize: 28x + 8y

[hot cloud]

the vertices are (8,28) (12,21) (13,14) (24,0) (15,0) (18,0) (0,42) (0,50.4) and (0,105)?

[hot cloud]

yep

[hot cloud]

is this right?

[hot cloud]

i lwk gave up after there were no helpers helping lol

[tulip idol]

Idts there'd be this many coordinates. Note that the inequality 14x + 2y ≤ 28 automatically implies the two other 7x + 4y ≤ 168 and 14x + 5y ≤ 252

So you're only required to check the inequalities x ≥ 0, y ≥ 0, 14x + 2y ≤ 28, to maximize 28x + 8y

[hot cloud]

wait why only 14x+27<=28

i mean 2y

not 27

[tulip idol]

Check the graph

14x + 2y ≤ 28 or rather, 7x + y ≤ 14 clocks within the triangle (0,0), (2,0), (0, 14) which is well within the other two inequalities

[hot cloud]

oh yeah i see it forms a triangle

but so does the other ones alone

wait

[tulip idol]

Yes. But the triangle formed by 7x + y ≤ 14 is enclosed within the other triangles and so, satisfying this inequality alone is sufficient

[hot cloud]

ohh ok

so then vertices are (0,0) (2,0) and (0,14)?

and (0,14) maximizes profit

[marsh citrus]

Channel closed

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[hot cloud]

.reopen

[marsh citrus]

✅

[hot cloud]

?

[tulip idol]

Yes. Btw the profit function is Selling Price - Cost Price = (28x + 8y) - (14x + 5y) = (14x + 3y)

And yes, if you're evaluating on the boundary condition, i.e., 7x + y = 14, then you have

Profit = 14x + 3y =2(7x + y) + y = 28 + y, function of y

So you maximize y, i.e., y = 14 ✓ so (0, 14)

@hot cloud

[hot cloud]

mb

ight bet

ty bro

[marsh citrus]

@hot cloud Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[hot cloud]

.reopen

[marsh citrus]

✅

[long temple]

What’s ur question

[marsh citrus]

@hot cloud Has your question been resolved?

[hot cloud]

this is the system and the vertices r (0,0) (0,2) and (0,14) right

[marsh citrus]

@hot cloud Has your question been resolved?

[tulip idol]

Yes

(0, 14) maximizes

[marsh citrus]

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[marsh citrus]

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[jagged gale]

how do i find the domain and range of f(x)=x^2-x ?

[ivory sundial]

Domain is the set of possible inputs, so to find the domain just look at what values of x "won't work" (usually get this means looking for square roots of negatives and divisions by 0)

Range are the possible values you can obtain, best way is to plot it.

If you can't plot it, just know that minimums and maximums occur at the following places:

• Stationary points
• Boundaries

So find all the stationary points and all the boundaries (yes infinity is a boundary)

[jagged gale]

sorry i dont really understand any of that TT

i understand domain and range but idk stationary points or boundaries

[odd orchid]

what do you understand by domain and range

[jagged gale]

its all possible x and y values of that function

[odd orchid]

now look at your function

are there any x values

that arent valid

[jagged gale]

no?

[odd orchid]

so what is the domain of the function

[jagged gale]

all real numbers?

[odd orchid]

now the y values

are there numbers that cannot be outputs of the function

[jagged gale]

im not sure

[odd orchid]

the range of a function is the domain of its inverse

thats one way you can find the range of a function

[jagged gale]

when i graph the parabola stops at around -0.25

[crystal lagoon]

you're not supposed to graph it since that will only give an approximation

well since you've drawn it now, can you see that there's a minimum point?

[odd orchid]

okay a graph will tell you the range

[crystal lagoon]

try and find that coordinate

[odd orchid]

but you can still find the minimum without

[jagged gale]

(0.5,-0.25)

i dont really understand how im supposed to find the domain and range without graphing

[crystal lagoon]

have you learnt completing the square

[jagged gale]

i dont think so

just learned what domain and range is today

[crystal lagoon]

Vertex form? $(x-a)^2+b$

[elfin berry]

AkitoLite

[crystal lagoon]

@jagged gale does that look familiar to you

[jagged gale]

i probably did it like 3 years ago idk

i didnt pay attention

[crystal lagoon]

definitely not 3 years ago😭

[jagged gale]

?

[crystal lagoon]

you can find the minimum from that expression

Wait

If $y=x^2+bx+c$, then
$y=(x+\frac{b}{2})^2+c-(\frac{b}{2})^2$

man what😭

[elfin berry]

AkitoLite

[crystal lagoon]

oof I need to put glasses or something

[jagged gale]

i have no clue whats going on TT

[crystal lagoon]

the eqn on the right is completing the square

It's a formula but I can show you the step process on how you get there

[jagged gale]

ah ok

yeah im really confused cause this is stuff i should have done 3 years ago if i didnt slack off TT

[crystal lagoon]

have you done differentiation? Since you said you did that 3 years ago🙏🏻

[jagged gale]

no clue

i never did anything in class so i just remember hearing random things

i need to find the domain and range of f(x)=-3x+2 and f(x)=-x^2-x+6

[marsh citrus]

@jagged gale Has your question been resolved?

[marsh citrus]

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[severe owl]

[(sinx+3)/x]/[(2x+1)/x]

So we get 0/2

= 0

[fierce phoenix]

Oh thank you

[marsh citrus]

@fierce phoenix Has your question been resolved?

[severe owl]

.close

[marsh citrus]

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[rotund temple]

i wanna know how to do it step by step like im not sure what they r doing

i get the first few parts

[crystal lagoon]

Which part do you not understand?

The 2I part?

[rotund temple]

everything after the cos x = sin x

oh and the step after it

[rotund temple]

basically yeah

[desert socket]

do you know about the king's rule

or do you wanna do it the substitution way

[crystal lagoon]

So the bounds part? You can switch the bounds of the integral by dividing by -1 since cost is an even function

an odd function is bassically just f(-x)=f(x)

[rotund temple]

iI have to do it the substitution way

that’s what the question asked

[rotund temple]

yep i know that

but why do we do the I and 2I

[crystal lagoon]

because we have 2 expression that mean the same thing, so we are able to do that.
It is done because as you can see, when you put them into the same integrand, and same fraction, they cancel out.

[rotund temple]

what’s an integrand?

so what doe s the I represent

[crystal lagoon]

Err bassically the name of the function we are integrating

[rotund temple]

ahh okay

[crystal lagoon]

I? It represents your original expression

[rotund temple]

and why do we multiply by 2?

[crystal lagoon]

You use the original integral, then add the modified one which is the one you obtained

[rotund temple]

also why do we need to modify it

[crystal lagoon]

so that you can add both the original and the new one together

[crystal lagoon]

because they cancel out 🤷🏻‍♂️
There's a reason the substitution is given because it is not so obvious at least imo

[rotund temple]

wait so

why can’t we just do the original equation

without modifying it

[crystal lagoon]

you can, just harder to do

[rotund temple]

i see

Is there any like

videos i can watch

to fully understand this

[crystal lagoon]

idk sorry. But I can give you another example when you can use this

[rotund temple]

sure

what is this called specifically

integration by sub and what else?

[crystal lagoon]

idk either 😭😭

[rotund temple]

😭

alr alr

[crystal lagoon]

for example if you want to integrate sinxcosx only using IBP, you can keep doing by parts until you reach your original equation

[rotund temple]

ibp?

[crystal lagoon]

Integration by parts

[rotund temple]

oh yea

i know that

[rotund temple]

Is that the same?

[crystal lagoon]

It's not IBP but the same last step

Wait

Actually here's a better one
$\int e^xsinx = e^xsinx- \int e^xcosx$ \
Take the new integral aside, \int e^xcosx = e^xcosx - \int e^xsinx$

[elfin berry]

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

[crystal lagoon]

Notice how integrating e^xcosx gives the equation we are trying to solve

Therefore, take $\int e^xsinx = I$ \
$\int e^xsinx = e^xsinx - e^xcosx - \int e^xsinx$

[elfin berry]

AkitoLite

[crystal lagoon]

Then $I = e^xsinx - e^xcosx - I$ \
This gives $2I = e^x(sinx-cosx)$ \ \
ie $I = \frac{e^x(sinx-cosx)}{2}$

[elfin berry]

AkitoLite

[marsh citrus]

@rotund temple Has your question been resolved?

[rotund temple]

ok tyty

[marsh citrus]

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[dark rampart]

I am unsure how to get the primite function to 5^(-x)

[dark rampart]

I initially did (5^(-x))/(ln(5)) + C but that was incorrect

it was supposed to be a "-" before

[dark rampart]

like this: -(5^(-x))/(ln(5)) + C

[arctic adder]

Is there any calls I can join to talk about math with people.

[vernal forge]

indeed

[vernal forge]

if it helps, you can do a u-sub with u=-x, du = -dx

if you know what u-sub is

[marsh citrus]

@dark rampart Has your question been resolved?

[marsh citrus]

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[hot cloud]

.reopen

[floral swan]

Don't know how to proceed

[floral swan]

i know the projection formula but

i reach that <t3,q> * 2/7 * t3 = 1+t

what do I do from here?

[marsh citrus]

@floral swan Has your question been resolved?

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[still temple]

(x^2 -8x+7):(x-7)

[patent sandal]

!status

[marsh citrus]

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

[rotund pike]

You can write exponents like x^2.

[still temple]

1💀

[crystal lagoon]

try factoring the numerator or long/synthetic division

[still temple]

it starts with (x^2

[still temple]

How?

photomath says x(x-1)* 7(1-x)/(x-7)

[patent sandal]

Do you know how to do long division?

[still temple]

but thats weird

[still temple]

not really

[patent sandal]

ok

its kind of like normal division

but each time you try to multiply the divisor with a term so that it eliminates the first term in the

dividend

do you know how to factorise polynomials?

[crystal lagoon]

Have you not learnt factoring?

[patent sandal]

Yeah honestly it would be best if you brush up on long division or factorisation before attempting the problem

[still temple]

i did

[crystal lagoon]

you just factorise it

[still temple]

factorizing is like “n!” right

[patent sandal]

no thats factorial

[still temple]

ohh

sorry my english is weak

so whats factorizing

[patent sandal]

its where you express a polynomial as a product of two "simpler" polynomials

for example x^2 + 5x + 6

= x^2 + 2x + 3x + 6 = x(x+2) + 3(x+2) = (x+2)(x+3)

so the bigger polynomial x^2 + 5x + 6 can be expressed as the product of two linear polynomails (x+2)(x+3)

[still temple]

yeah got it but how do you know you should split 5x into 2x and 3x and not 4x and 1x

[patent sandal]

that i did on intuition

but there are some tricks

for example u have to split the middle term into two terms

[crystal lagoon]

vietas formula

[patent sandal]

such that their product is the product of the first and last term

[crystal lagoon]

$x_1 + x_2 = \frac{-b}{a}$

$x_1x_2 = \frac{c}{a}$

[elfin berry]

AkitoLite

AkitoLite

[patent sandal]

imo this is not required for simple factorisation

[crystal lagoon]

that's bassically the principal

Like that's bassically what you're thinking when factorizing

[patent sandal]

i mean yeah

[still temple]

im in high school and not specialized in math

[still temple]

It’s kinda hard to understand this

or why it’s needed here

[crystal lagoon]

$x_1$ is the first root of the quadratic, and $x_2$ is the 2nd root. in other words, they are the answer to $x$ when $y=0$

[elfin berry]

AkitoLite

[crystal lagoon]

ok so to obtain the formula we gotta work backwards

let's start off with $(x-a)(x-b)$

[elfin berry]

AkitoLite

[crystal lagoon]

When expanding, we get $x^2-bx-ax+ab$

[elfin berry]

AkitoLite

[patent sandal]

[crystal lagoon]

I think I'm overcomplicating it a bit yeah lol😭🙏🏻

but like that's the proof

[patent sandal]

this is like bringing guns to a fist fight

@still temple

remember this principle to factorise:

[crystal lagoon]

Can't use formulas without knowing how to prove them

[patent sandal]

u have to split the middle term into two terms such that their product is the product of the first and last term

[still temple]

i tried continuing from here

where did I mess up

[crystal lagoon]

Photomath is wrong

[patent sandal]

photomath was wrong

jinx

[patent sandal]

try this first

[still temple]

oh okay

[patent sandal]

did you know that mathematicians used to study calculus, derivatives and apply them for a long time without a proper definition for limits?

[still temple]

can someone explain by simply solving it?

[patent sandal]

ok ill walk you through

[still temple]

then if i dont understand i ask questions

[still temple]

thank you

[crystal lagoon]

I mean yeah but like it's not like they didn't even know ANYTHING about limits

[patent sandal]

x^2 - 8x + 7 = x^2 - 7x -x + 7

try to continue from here

[still temple]

Why is there a =

OH

[obtuse flint]

hi

[still temple]

my bad

[patent sandal]

lol

[still temple]

thought it turned into an equation lol

[still temple]

x(x-7)-7(x-7)

why is this wrong

[patent sandal]

it doesnt work

if u expand it

u get x^2 - 14x + 49

which is not the original equation

[patent sandal]

Id recommend watching the video "This Is the Calculus They Won't Teach You" by "A Well-Rested Dog". (Skip to 24:14)

[still temple]

So what should I do

from there

[patent sandal]

from here?

[still temple]

yes

The task says “do the division” if it changes anything but i guess not

[patent sandal]

ok

[still temple]

so we have x^2-7x-x+7

[patent sandal]

then u seperate the common factors

so from x^2 - 7x -x + 7

= x(x-7) - (x-7)

= (x-1)(x-7)

putting this back in the original question

we can cancel the x-7

so that we are left with x-1

which is our answer

id highly recommend you to learn polynomial division

[still temple]

So wait how did 7 turn into a -7?

[patent sandal]

cus i took out a -1

by factoring

so -1 * -7 = 7

[still temple]

took it out from -1x?

[patent sandal]

try to expand what i wrote

you will get back the original expression

-(x-7) = -x + 7

[still temple]

ok gimme a sec

[patent sandal]

nonononon

[still temple]

in the original it’s +7 at the end

[patent sandal]

not -7

what is -1 * -7 ?

negative 1 times negative 7?

[still temple]

-7 but where is “ -1* “here?

[patent sandal]

-1 is contracted to -

its infront of the (x-7)

[still temple]

yes

im slow

[patent sandal]

its fine

[still temple]

so

[still temple]

u sure this is correct?

[patent sandal]

yes

[still temple]

i think im lost

[still temple]

can we get to this with smaller steps from the start?

if you don’t mind

[patent sandal]

yeah

x^2 - 7x - x + 7

= (x) . (x) + (x) . (-7) + (-1) . (x) + (-1) . (-7)

= (x)((x) + (-7)) + (-1)((x) + (-7))

= x(x-7) -(x-7)

= (x-7)(x-1)

the . is the multiplication symbol btw

[still temple]

(x^2-8x+7)

[: (x-7)👈🏻we put this part aside]

x^2-8x+7
x^2-7x-x+7 i understand til here
x(x-7)-1(x-7) then we get this and I don’t know why +7 turns into -7 without first doing the multiplication with -1 first and why x doesnt turn into -x on the right side

and i dont know why the x and 1 we took out disappears in the last step:
(x-1)(x-7)

[still temple]

sorry i didnt read this yet maybe it will clarify gimme a sec

[still temple]

On the right side (-1)*(x+(-7))=
(-x+(+7))

Oh you took out (X-7)?

at the end

[marsh citrus]

@still temple Has your question been resolved?

[brave knoll]

What is going on ?

What is your current question?

[still temple]

(x2-8x+7) : (x-7)

[brave knoll]

yes, what do you want to do with this expression?

[still temple]

Task just asks to do the division

[brave knoll]

Alright
First you notice that 7 is a solution of x² - ix + 7
So the rest will be 0

I suggest you to study x² - 8x + 7 (find its roots)

then if C and 7 are roots, x² - 8x + 7 = (x-C)(x-7)
The solution directly follows

Unless you want to do euclidian division of polynomial, (which is not done in high school in my country)

Do you get it ?

[still temple]

i dont get this first step

7 is a solution?

[brave knoll]

It is

But if you are in high school, step by step you should \

1. Find roots of $x^2 - 8x + 7 = a x^2 + bx + c$ \
2. Write this polynomial as $a(x-x_1)(x-x_2)$ \
3. Then look at the division $a(x-x_1)(x-x_2) : (x-7)$

[elfin berry]

Jy

[brave knoll]

And "magically" (because you are a non Maths major at high school), everything will perfectly simplify

[still temple]

so first step

a=1 b=-8 and c=+7

?

[brave knoll]

yes but you have to find the roots of this polynomial

ever heard of ∆=b² - 4ac ?

[still temple]

i will revise polynoms then

[still temple]

i think i did but can’t remember clearly

[brave knoll]

then x = (-b ± sqrt(∆))/(2a) if ∆≥0

[still temple]

Whats the triangle?

[brave knoll]

the delta

[still temple]

looks like rocket science to me

[still temple]

ahhh

[brave knoll]

It is called the discriminant

[still temple]

Yea I remember something like that

[brave knoll]

$\Delta$

[still temple]

yea

[elfin berry]

Jy

[still temple]

i have to revise

[brave knoll]

I don't know what you study, but if you are willing to pursue something with a decent level of maths, second degree polynomial solving is a must

[still temple]

i was missing a ton of classes recently and couldnt revise with the rest of the class

This one seems hard

idk if it actually is

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[ancient copper]

@still temple so $\frac{x^2-8x+7}{x-7}$?

[elfin berry]

;(

[ancient copper]

[ancient copper]

okay, take it like this; $\frac{x^2-7x-x+7}{x-7}=\frac{x^2-7x}{x-7}-\frac{x-7}{x-7}$

[elfin berry]

;(

[ancient copper]

now factor out the x in the first terms, and simplify the second and first once you have done that

[marsh citrus]

@still temple Has your question been resolved?

[weak wind]

i think bro disappeared

[marsh citrus]

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[untold wyvern]

For a collection $C$ of open sets of real numbers. There is a countable subcollection ${O_i}$ of $C$ such that $\cup {O\in C} O=\cup{i=1}^\infty {O_i}$.

[elfin berry]

somethingwrong

[untold wyvern]

Im having trouble proving this, I know that the union is itself an open set and hence the union of a countable collection of disjoint open intervals

but im not sure whether this helps

[untold wyvern]

i guess i could try showing that each disjoint open intervals is subsets of the union of some countable subcollection of C but not im not sure if this is true

[marsh citrus]

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[untold wyvern]

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[sharp quarry]

I want to draw those graphs in my calculator... with demos it works but i coudlnt figure it out how it works with my calculator

[sharp quarry]

this is the one i am using

<@&286206848099549185>

cus if i define the funktion it doesnt do anything

how should i define them so i can draw them?

[safe shuttle]

Maybe use the web version of Desmos?

[sharp quarry]

i cant use it on the exam

nvm i figured it out

-close

.cloe

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[still temple]

Hi guys, does anybody know how to get knowledge specifically on compound renewal processes?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

I know it's something advanced and there isn't much people interested in stochastic processes, but I haven't found much information on this topic and it's something I need

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

Well, it seems getting help here is hard

[brave knoll]

I don't know what that is, but I suggest you to check books online / at a library

[still temple]

I already did that

I looked at more than 30 books I think

I only got about renewal processes and compound Poison processes

So I decided to ask ChatGPT to see if it could bring some light about the extension of this topic

It gave a lot of things, I've found no information about

[marsh citrus]

@still temple Has your question been resolved?

[split vale]

Is this right <@&286206848099549185>. The question is what are the chances of a ball, 2 balls, and 3 balls landing all into one of the 6 cards

[warm geyser]

What particular thing you want know?

For this you need a particular compound or something then you can find it, in general "compound renewal process" for many things it is not specifying any particular thing @still temple

[marsh citrus]

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[marsh citrus]

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[spring tree]

How to prove the Leibniz test works?

[spring tree]

If $\left{a_{n}(x) \right}$ is a monotonically decreasing sequence of nonnegative reals s.t. $\lim_{n}a_{n}(x)=0$ uniformly in $x$, then $\sum_{n=1}^{\infty} (-1)^{n}a_{n}(x)$ converges uniformly.

[elfin berry]

babario

[spring tree]

Let $S_{n}(x)=\sum_{k=1}^{n} (-1)^{k}a_{k}(x)$. Then $n \ge m$ implies $|S_{n}(x)-S_{m}(x)| \le |a_{m}(x)|$ since $\left{a_{n}(x) \right}$ is monotonically decreasing and alternating terms cancel. Given $\varepsilon>0$, choose $M$ s.t. $m \ge M$ implies $\sup_{x}|a_{m}(x)|<\varepsilon$. Then $n \ge m \ge M$ implies $\sup_{x}|S_{n}(x)-S_{m}(x)|<\varepsilon$. Since $\varepsilon$ is arbitrary, $\left{S_{n}(x) \right}$ is uniformly Cauchy and uniformly convergent by uniform Cauchy criterion.
is my proof correct?

[elfin berry]

babario

[marsh citrus]

@spring tree Has your question been resolved?

[spring tree]

<@&286206848099549185>

[sterile lynx]

Yo

Need help on this ?

[spring tree]

yes

just to verify the proof real quick

[sterile lynx]

@maiden zealot

Come here

Nah this mf my stand

But yeah i don’t fw that shit

But Astro sure does

Im more with algebraic

And maybe a bit of linear algebra

[spring tree]

LOL

[sterile lynx]

These shit easy

You’re on more text than math

And I don’t like literature

God bless you we’re not in complex analysis

[spring tree]

tiz just real 😭

[sterile lynx]

lol

[spring tree]

<@&286206848099549185>

[sterile lynx]

@maiden zealot aint you helping my man ?

[maiden zealot]

?

what da problem

@spring tree

[spring tree]

just to verify this proof i wrote to prove the alternating series test is correct

[maiden zealot]

seems right to me

[spring tree]

nice

thanks then

[sterile lynx]

You gotta explain to me where do you situate yourself

[maiden zealot]

?

[sterile lynx]

Like mathematicallt

At what level

Undergrad

Grad

Phd

[maiden zealot]

idk

prob on border between undergrad and grad

[sterile lynx]

You’re certainly not undergrad level anymore

I see

Yeah you fine asf

All that in khan ?

[maiden zealot]

na

[sterile lynx]

Didn’t you used other ressources ?

[sterile lynx]

Oh cool

Where ?

[maiden zealot]

i have asked questions in math class that my teacher doesn't know the answer to because its so advanced (he has a masters in mathematics)

[sterile lynx]

My prof has a doctorate but he’s shit in differential geometry

He didn’t liked it so he didn’t done them during his cursus

But he’s good with algebra

https://theses.fr/2009NAN10081

What do you think of his thesis ?

(It’s in French but you can translate )

[spring tree]

.close

[marsh citrus]

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[pallid terrace]

For recurrence relations/reduction formulae, how do you decide between whether you should go down to I_1 or I_0?

[devout mauve]

as you always go back two steps, you need a start value for the odd values and one for the even values

[pallid terrace]

Sorry I don't really understand what you mean by that
What if the question doesn't say anything about odd or even values?

I have a pdf for a velocity distribution and I need to integrate it

(I can't send the question as I'm being assessed formally on it but it's an open book exam so I can ask for general help)

[devout mauve]

here you have I_n expressed in terms of I_(n-2)

so if you for example wanted to compute I_7, then you would need I_5, I_3 and I_1

[pallid terrace]

Why?

[devout mauve]

but if you wanted to compute I_8 then you would need I_6, I_4, I_2 and I_0

[pallid terrace]

Interesting I had no idea, I must have missed that, thank you

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[slim quartz]

im not sure how to do this

[severe owl]

IBP

[slim quartz]

how do i know what to choose tho

like for u or dv

[severe owl]

u = 1

v = arctan(4x)

[severe owl]

Ig that's intuition and practice

[tulip idol]

1. Are you able to integrate arctan -> Yes? Then go ahead and put bounds write answer
   No? Then you must wish to differentiate arctan. Are there other functions you can integrate? Is "1" a function? Then integrate 1, differentiate arctan

[marsh citrus]

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[fossil tide]

can someone who speaks German or knows this theorem explain it to me i think it is really phrased in a weird manner by my professor

[fossil tide]

it is the Fischer-Riesz theorem

this is an analysis 3 lecture

[main idol]

Translate to English and can you be more specific what you don't understand

[fossil tide]

$$ \text{let } 1 \leq p \leq \infty , f_n \text{ p integrable and let } \tilde{f}_n \text{ be a cauchy sequence.} $$

$$\text{ then a subequence converges almost everywhere to a p integrable function f and } $$

$${\lVert \tilde{f}n - \tilde{f} \rVert}{L^{p}} \to 0 \text{ , in particular } L^{p}(\mu) \text{ is a banach space}$$

this is the translated version

@main idol

[elfin berry]

ابو حيدر

[main idol]

And which part of that is confusing to you

[fossil tide]

is f_n a function or a sequence of funtions

[main idol]

Sequence

n=1,2,3,... Usually

[fossil tide]

and \tilde{f}_n is an equivalence class of these functions i presume

[main idol]

Unclear from what you've provided so far

Could be a subsequence of f_n

Could be entirely different

[fossil tide]

we have defined it a few pages prior to be the equivalence class of some integrable function

so if i have a function f i define an equivalence class namely two functions are equivalent if

then f tilde is the equivalence class containg all functions equivalent to the representant f

[main idol]

I see so yes this is correct based on your notes

[fossil tide]

i dont understand this part

[main idol]

Do you know what L^p norm is

[main idol]

Yes plug that in

Treat p<inf and p=inf separately

[fossil tide]

okay i will try it out and see if it makes sense

[main idol]

Then take the limit as n goes to inf

[fossil tide]

do you know what the applications of such a theorem are

[main idol]

Yes

[marsh citrus]

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[twilit siren]

does anybody know how to solve these
i know how to do it when theres no square in the sinus or cosinus but with squares i dont know how

[twilit siren]

and its a unit cirkel

[summer grove]

the first one you can take the root

and thus get $\pm 1$

[elfin berry]

Bonk

[twilit siren]

2x - 1/3pi = 1/2pi ?

[summer grove]

almost

sin is periodic, right?

[red nimbus]

we sin periodically

[twilit siren]

ye

[tulip idol]

Find solutions to $\cos \left(2x - \frac{\pi}{3}\right) = 0$

[elfin berry]

Arya

[tulip idol]

That's Q1.

[twilit siren]

is this good?

[summer grove]

yes

[twilit siren]

what do i do with the sin square 2

[tulip idol]

Where?

[twilit siren]

here

[tulip idol]

$1-\sin^2 x = \cos^2 x$

[elfin berry]

Arya

[tulip idol]

Send the left sin²x to the right and you'll have basically

[twilit siren]

ah ok

[tulip idol]

Understood?

[twilit siren]

yes but do you know the second one>?

[tulip idol]

Similarly for Q2. You can simplify to sin²x . cos²x = 0

How?

[twilit siren]

is this good so far?

[tulip idol]

Yes, and cos² x - 1 = -sin² x, so yeah

You get to, either cos x = 0, or sin x = 0, both satisfy the equation

Clearly that implies x = npi/2, right? cause every even multiples of pi/2, sin x = 0, and at every odd multiples, cos x = 0

[twilit siren]

is that so

[tulip idol]

welp what do you think

what's the sol for cos x = 0?

[twilit siren]

1/2pi

[tulip idol]

uhh, general solution

[twilit siren]

uh idk

[tulip idol]

.

[tulip idol]

Also, this is not true

[twilit siren]

wat why?

[tulip idol]

The general solution for cos x = 0, is x = (2n + 1)pi/2

Because, cos pi/2 = cos 3pi/2 = cos 5pi/2 = cos 7pi/2 = 0

Cos every odd multiple of pi/2 = 0, so the soln to cos x = 0 is this

Similarly, sine of every integer multiple of pi is 0

[twilit siren]

bro my test tmrw might fk me :/

[tulip idol]

so sin x = 0, assumes the solution x = npi

well, atb

At least if this question comes, I hope you get this right

[marsh citrus]

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[frank mesa]

would this be considered a proper subspace of V?

W = {f in V | f(1) = 0}

closure under addition:

f, g in W.
(f+g)(1) = f(1) + g(1) = 0 in W

closure under scalar multiplication:

f in W, c in R
(cf)(1) = cf(1) = c * 0 = 0 in W

[devout mauve]

there is a third thing to show for subspaces

nonemptiness

[frank mesa]

mhm

let me go over that real quick mb

[frank mesa]

subbing [f(n) = 0, for all n in N] into the recurrence relation:

f(n) = 0 + 0 = 0. f in V, W

so W isnt empty?

[devout mauve]

yes

[frank mesa]

ty

[devout mauve]

the zero vector is the one you usually show is in a subspace

but like you hopefully noticed, its actually not that easy here to give a nonzero vector in W

like, its not hard. but not trivial

this is a nice example that this condition really shouldnt be forgotten

[frank mesa]

I see

tysm

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[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

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

[elfin berry]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

x1 -2x4 = 0

x2 - x3 - 4x4 = 0

x1 = 2x4

x2 = x3 + 4x4

(x1,x2,x3,x4)=(2x4, x3 + 4x4, x3 , x4)

x4(2,4,0,1) + x3(0,1,1,0)

T = <(2,4,0,1),(0,1,1,0)>

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

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[buoyant jetty]

.reopen

[marsh citrus]

✅

[marsh citrus]

@buoyant jetty Has your question been resolved?

[buoyant jetty]

can I get some help

[serene briar]

Maybe, by synthesis analysis ?

[buoyant jetty]

elaborate

[slim surge]

I would try this. Compute T perp n H, then pick enough linearly independent vectors in H but not in the intersection to span S so you get the equality.

[buoyant jetty]

I tried finding the intersection between H and the orthogonal complement of T

[buoyant jetty]

basis for T:

x1 = 2x4
2(2x4)-x2+x3=0
4x4+x3 = x2
(x1,x2,x3,x4)=(2x4, 4x4 + x3, x3, x4)
x4(2,4,0,1)+x3(0,1,1,0)
T = <(2,4,0,1),(0,1,1,0)>

[buoyant jetty]

Orthogonal complement of T
v = (x1,x2,x3,x4)

1. v . (2,4,0,1) = 0
   2x1 + 4x2 + x4 = 0
   x4 = -2x1 -4x2
   x4 = -2x1 -4(-x3)
   x4 = -2x1 + 4x3
2. v . (0,1,1,0) = 0
   x2 + x3 = 0
   x2 = -x3
   (x1,x2,x3,x4)=(x1,-x3,x3,-2x1+4x3)

[buoyant jetty]

x1(1,0,0,-2)+x3(0,-1,1,4)
Orthogonal complement of T = <(1,0,0,-2),(0,-1,1,4)>

[buoyant jetty]

Intersection of the orthogonal complement of T with H:

T : a(1,0,0,-2)+b(0,-1,1,4)
T : (a,0,0,-2a)+(0,-b,b,4b)
T : (a,-b,b,4b-2a)

H : x1 - 2x2 + 2x3 - x4 = 0

a -2(-b) + 2(b)-(4b-2a) = 0

a + 2b + 2b -4b+2a = 0

3a = 0

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[gray laurel]

w

is this thing free

[peak fiber]

nope, look in the available channels category

sending a message there will claim that channel for you

[marsh citrus]

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[regal kite]

I solved c as 87, but according to the answer key it's incorrect

[regal kite]

How I got 87:

I took the most left triangle 32 / 80 and solved the last angle as 68. Then I took the triangle 25 / 68 / C, and did 180 - 25 - 68 and got 87

wait oops

I just realized my mistake lol

^^ and got 43

.close

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[uncut venture]

Im sooo confused

[marsh citrus]

@uncut venture Has your question been resolved?

[robust dragon]

i ran out of space, but i'm gonna try to derive the area equation A and equate the derivative to 0 to find x

i have a really bad feeling about this solution, hold on

no wait

it checks out

10√2 by 10√2 is the max possible area's dimensions

and to get the volume you can just multiply that area by the log height (25)

giving you 200 sq. cm for the max area and 5000 cubic cm for the max volume

[marsh citrus]

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[marsh citrus]

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[honest creek]

i guess this is true but like hmm why is there no need for cases?

if (a,b) is a point on Quadrant 1 then a 180 degree rotation ccw gets you to Q3 and then yeah the mapping would be (-a,-b)

[tight furnace]

Should there be?

[honest creek]

idk

i guess not

like what about quadrant 2 -> quadrant 4

[tight furnace]

Rotating 180° about a point is the same as "flipping" over that point

[honest creek]

well it agrees yeah

[tight furnace]

The point of rotation becomes the midpoint between your original point and your new point

[honest creek]

no no i mean they took (a,b) in Q1 right

and okay that gets mapped to Q3

what about Q2 -> Q4

[crystal lagoon]

It is exactly the same as Q1->Q3 except it's just a reflection

it works because sinx is an odd function ie f(-x)=-f(x)

[honest creek]

i mean yeah it's exactly the same because (-a,b) -> (a ,-b)

but why did they not bother showing it

[crystal lagoon]

I think it's sort of self evident 🤷🏻‍♂️

[tulip idol]

because they supposed, (a, f(a)) on f(x) => (-a, -f(a)) on graph is intuitive for f(a) = -f(-a)

[hushed egret]

(a, b) -> (-a, -b), take a to be -a

[honest creek]

oh yeah i'm pretty dumb

[hushed egret]

i thought you were mmmm7

[honest creek]

😭 whattt i had to look at my name again to confirm

okay wait so

a geometric proof of a 180 degree counterclockwise

wouldn't need me to repeat it for Q2 -> Q4?

like okay if i take a point in Q1 as (a,b)

then i get a congruent right triangle in Q3 right?

the dimensions match and so we get that the point in Q3 is (-a,-b)

[hushed egret]

a point (a, b) can be anywhere

not just Q1

[honest creek]

no but for geometric purposes, would i not fix it to Q1?

[hushed egret]

you shouldn't need to

[honest creek]

because if it is anywhere then how do i get my congruent triangle construction

to prove stuffs

[hushed egret]

uh

you calculate the angle using other means

[honest creek]

I did? 😭

i just used vertical angles

and well both are right triangles

so that too

[hushed egret]

for example, (a, b) and (-a, -b) lie on the same line bx = ay going through the origin

[honest creek]

and then used the definition of a perpendicular bisector

[hushed egret]

and are equidistant from the origin

[honest creek]

correct yeah

y = b/a x

[hushed egret]

thats enough to tell you theyre 180 degrees apart

[crystal lagoon]

I'm confused about what exactly you're confused about 😭

[hushed egret]

i worry about a = 0, but ok

[honest creek]

😭 yeah okay but like this is for Q1 -> Q3 right

[hushed egret]

well

not really

my argument works for any point

except the origin but thats also not really of our concern

[honest creek]

hmmm right what about

my congruent triangle approach

can't that be salvaged into anything meaningful?

for reference, this is what i was going for

[hushed egret]

well it works but now your argument is diagram dependent

and you're forced to think about whether the argument generalises to all quadrants

[honest creek]

so like fixing (a,b) in Q1 then it is obvious that we the same point in Q3 is (-a,-b) but like we haven't showed this to be true in Q2 -> Q4 :sob so like i was confused

[honest creek]

oh so you mean i'd do one for Q2 -> Q4 and then

see if they're the same thing?

[hushed egret]

well

maybe

but you shouldnt have to if you had made an argument that was general enough in the first place

[honest creek]

😭

[hushed egret]

so these diagram dependent arguments are slightly flawed in that sense

[honest creek]

true

okay yeah your argument works for all cases y = (b/a) x and to handle the a = 0 case we just reflect the point across the y axis

so i guess now it works for every a in R?

[hushed egret]

well, i didnt write it as y = (b/a)x

[honest creek]

yeah you wrote it as by = ax

[hushed egret]

i wrote it as bx = ay, which works in all cases

except the origin, in which case the reflection is a bit degenerate anyway

[honest creek]

but when you did that didn't you assume a != 0

😭

[hushed egret]

i did not

[honest creek]

the thing we talked about yday i guess

oh

[hushed egret]

when a = 0, then you get bx = 0

which is just the vertical line

[honest creek]

okay so technically aren't you just testing for a = 0 manually?

if so okay i think yeah that's fair

[hushed egret]

not really

[honest creek]

because isn't it technically written as y = (b/a) x

[hushed egret]

by = ax is always a valid equation of a line as long as (a, b) is not the origin

[honest creek]

oh

u just wrote it in a different form

bleh okay this is weird

[hushed egret]

not just a different form

by = ax is strictly more general than y = (b/a)x

y = (b/a)x cannot represent vertical lines

[main idol]

causing confusion by being too general smh

[honest creek]

what "form" is this 😭

is it close to the "general form"

[main idol]

ax + by + c = 0 is called standard form

[honest creek]

right, blake is using that?

[hushed egret]

well

[honest creek]

okay so i just learnt that standard form > (y = mx + c form)

[hushed egret]

im using the fact that (linear expression in x, y = linear expression in x, y) is always a line

[honest creek]

😭 whattt

[hushed egret]

so all i did was write down some linear equation which both points satisfied

[main idol]

you should figure out how all these representations of lines are equivalent except for a few edge cases

[honest creek]

idk what you meant by the things in parenthesis

[main idol]

by edge case i mean like how b/a can't represent a vertical line

[honest creek]

yeah i guess it does make sense

[hushed egret]

like ax + by + c = dx + ey + f

[honest creek]

no yeah this i understood from what blake said

[honest creek]

i see so tldr

general form works for all cases

even with vertical lines

[hushed egret]

yes

but i don't like this "general form" or whatever speak

[honest creek]

standard form i guess

ax + by + c = 0

okay tbh now that i think about it

[hushed egret]

its just the basic principle that linear expressions give you lines

[honest creek]

all linear equations are expressed in this form

[hushed egret]

so you form any linear expression that works

thinking "ax + by + c = 0" is too narrow

[honest creek]

and i'm guessing it extends to planes and bla bla?

[hushed egret]

yes

[honest creek]

so like linear expression is just $c + \qty(\sum_{i = 1}^n a_i x_i) = 0$

[elfin berry]

mmmm7

[honest creek]

okay yeah i think hmm makes sense

[hushed egret]

sure why not

[honest creek]

fine fine, thank you once again

😭 i feel like i'm not posing my question the right way

but for some reason

u understand my questions

[hushed egret]

[honest creek]

alright thanks again

.close

[marsh citrus]

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[marsh citrus]

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[past flame]

i need help here guys

[marsh citrus]

@past flame Has your question been resolved?

[past flame]

nah

[marsh citrus]

@past flame Has your question been resolved?

[marsh citrus]

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[willow salmon]

How would you make this in NAND Gates

[willow salmon]

NOT(D1) AND D3

[river ocean]

guys can anyone help me with algebric radicals?

[fervent rampart]

you can make a not gate using a nand gate. then it's pretty straightforward

[sleek lake]

yeah

seems easiest to make 2 nots

[willow salmon]

NAND(NAND(D1,D1), D3) then NAND That with itself

[sleek lake]

right

i tried to find a solver but google doesn't know any

[fervent rampart]

wolfram has one

[sleek lake]

i see

[marsh citrus]

@willow salmon Has your question been resolved?

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[sudden steeple]

Struggling a bit to understand this. Ive gotten to fx = -6x + 6y and fy = -6y^2 + 6y +6x

[sudden steeple]

I know your now supposed to make them = to 0 and find values for x and y

but Im not sure how to know if there would be 2 or 4 points

[tight furnace]

There would be 2

[sudden steeple]

is there a trick to make them out?

I mean, -6x + 6y = 0 makes x = y

[tight furnace]

Solving the first equation for x (or y) and plugging into the second gets you two solutions for y (or x)

then, you know x=y

[sudden steeple]

Ye Im stuck

idk what to do with 12y - 6y^2 = 0

can I just quadratic it? ye right

[red nimbus]

6y is a common factor

apply zero product theorem

[sudden steeple]

(0,2) (2,0) ?

wait no

(0,0) (2,2)

Ok I understand now

.close

[marsh citrus]

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[marsh citrus]

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[toxic terrace]

Hey, how do I write the function f(x) = 9xcos(9x) from [0,pi/2] as the limit definition of an integral?

[toxic terrace]

This is what I tried

(the question assumes you're taking the right points so x*_i = x_i)

Not exactly sure where I'm going wrong...

[summer grove]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[summer grove]

whats the function?

[toxic terrace]

(yes I'm aware I forgot the 9 in the cosine)

[summer grove]

ah, i see

mb

[toxic terrace]

f(x) = 9xcos(9x) where x in [0, pi/2]

[summer grove]

i think all thats missing is the 9 in the cos

[toxic terrace]

Yeah I'm checking my x intervals

I think they're fine...

[summer grove]

the x_i=(pi/2*i/n)

[toxic terrace]

Rifht

[summer grove]

so filling that in gives what you have

ohhh

i see that you forgot a i^2

at the start

right?

wait, no

[toxic terrace]

Yeah I did

Okay I got it

Lol I'm kinda stupid

Thanks for the help

[summer grove]

ah yes

so that was the only problem

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[toxic terrace]

If I were to turn this into an integral, it'd be S a=0 b=pi/2, and then the function dx right?

[summer grove]

yup

[toxic terrace]

Alright

Thanks lol

.close

[marsh citrus]

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[marsh citrus]

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[mystic gazelle]

Given function x^2 + y^2 = 9

[mystic gazelle]

Don’t know where to stay

Start

[desert dirge]

knowing the full question would be helpful

you havent given much to go off

[summer grove]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mystic gazelle]

Sorry I was trying to take a photo