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[sacred socket]

Proof for two linearly independent solutions of a second order ODE
I do not understand where those intitial conditions come from?

[edgy willow]

I assume the theorem lets you choose whatever initial conditions you want

so they chose the simplest ones for the matrix

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@sacred socket Has your question been resolved?

[sacred socket]

If this was the case, could they not have chosen:
0 0
0 0
And then get a W(x) = 0 thus disproving the theorem

[edgy willow]

uh what's this section trying to say

it's probably just that it's possible to find two independent solutions

[sacred socket]

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[sour talon]

for u = cos x

i got thi s

$-\int_{0}^{1}\frac{du}{\left(a^{2}-b^{2}\right)u^{2}+b^{2}}$

[solid kiln]

RizzlerFly

[sour talon]

and now i'm stuck

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@sour talon Has your question been resolved?

[austere cedar]

From here you can complete the square and cast this as an arctan integral

[clear flower]

Hint: You may apply another substitution and use this formula.

(expect that your next substitution is related to u = ... tan k)

@sour talon

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[left oriole]

ok so you want the line to go in the direction (3,2,0)

and pass through (-1,7,4) yes?

[woven nova]

yes, that does work. since the dot product is 0, that vector is parallel to the line

and generally, we can set up a line like this:
line = anchor point + t * direction

like this

you can visually think of this like this:
we go to the anchor point, and from there on, we go along the direction vector

if that direction vector is the vector you found, then this line is also parallel to the plane

and if our anchor point is (-1,7,4), then the line passes through that at t=0

yep

But yes, you can get infinitely many solutions, since there are infinitely many parallels directions to the plane

and these all can have different representations

what i mean with representations:
(0,0)+t(1,1) and (1,1)+t(2,2) have the same shape

yes

here i tested it

on the picture you can't see it, but when i rotate it around, it is clearly parallel to the plane and does pass through the green point

nah, i use https://mathcha.io/editor

it is LaTeX but for lazy people like me

it has a graphing tool an you can write latex by searching for things

instead of knowing the latex command \Leftrightarrow, i can just search for Arrow and search what i need and learn the correct words that way

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[still stirrup]

what are the points on the graph of the function f(x)=x^4 where the tangent line has a slope of 1/2?

[balmy spear]

How can you work out the slope of f(x)?

[real pebble]

when you hear tangent line you think differentiation

[still stirrup]

i m stuck in how to solve 4x^3=1/2

[real pebble]

try isolating x

[still stirrup]

x=^3√(1/8)

like this?

[heady lava]

yes

any particular number you can think of that is cbrt(1/8)?

[still stirrup]

idk how to solve...

only know i can be 1/2

are there 3 solutions?

[heady lava]

1/2 is right

there are 2 other complex solutions that you most probably dont need to find

[still stirrup]

so only with this is ok?

[heady lava]

yes

[still stirrup]

ok, then the points are (1/2, 1/16) right?

[heady lava]

yes

[still stirrup]

tysm

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[dusky ocean]

can someone do a quick sketch for this

[cunning gate]

We cannot provide answers for you, we can help you get to the answer tho

[dusky ocean]

are you allowed to confirm or deny

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[steel herald]

hello, how do you find the values of y & r from csc 240 degrees?

[steel herald]

would you convert the degrees into radians?

oh

i realized my mistake

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[quartz nymph]

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[slate basin]

help? how do i do this?

[slate basin]

just please guide me

[real pebble]

k so

multiply the number of merits by its corresponding number of students

and the mean number of merits will be based on that

[slate basin]

so like 1 x 10?

[real pebble]

the reason being that the number of students is simply a frequency

[real pebble]

yep

[slate basin]

give me a sec ill go do the calculations

[real pebble]

don't forget to take the 8 students that got 0 merits into account later on

[slate basin]

i multiplied all the numbers now what?

[real pebble]

take the mean of those numbers

[slate basin]

im really bad at mean how do i do that.. 😭

[real pebble]

uh

the numerator will be all the numbers you had summed up

and the denominator will be the total number of students

[slate basin]

wait wdym?

[real pebble]

like yk how you multiplied the numbers

[slate basin]

yea

[real pebble]

so you got 10, 32, 54 ec

[slate basin]

mhmhmh

[real pebble]

add them up for the numerator

and divide by tootal students

total students being 8 + 10 + 16 +...

[slate basin]

ok i got 2.8

lemme do that with the second

i got 3.01 😭 i feel like thats wrong

[real pebble]

sounds right

if you look at the data

what it means is

on average the students got 3 merits

and the most students got 2-4 merits

so that's probably correct

and that just means that the year 11 students did better on average

[slate basin]

okay so how do i do who did better? wouldnt be year 11?

[real pebble]

since 3.01 > 2.8

[slate basin]

now i have to pray i didnt f anything up 😭

HEYY I WAS RIGH

thank you dude !!

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[slate basin]

.reopen

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[slate basin]

ahem im confused again

is this just 1 ?? 😭

[worldly wing]

the people who visited 0 times also count as less than twice

[slate basin]

ah

i forgot ab that

okie now im done

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[lament plaza]

open open open

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[small meadow]

Hey, i'm actually studying for exams and one question rlly bother me atm.

translation: when two integer are chosen as answer the inequation: x^2 - 2x - 15 < 0. which is the probability to be natural numbers ?

why is the answer 10/21?

[zinc dove]

Do you know the integer solutions to the problem

[small meadow]

i find smgh like 5/7 cause x1= -3 and x2 = 5

[zinc dove]

Well -3 and 5 don’t actually work themselves right

[small meadow]

yup

[zinc dove]

It’s just everything between them

[small meadow]

so there's 7 integer for the solution

[zinc dove]

Yea

[small meadow]

and 2 are negztive

[zinc dove]

Oh how are you defining natural numbers

I know the French do it weird

In USA natural numbers are 1,2,3

[small meadow]

natural number; N

integer; Z

[zinc dove]

Haha I know

But what is the least natural number for you

[small meadow]

oh

0

[zinc dove]

Usually for us it’s 1

[small meadow]

damn sounds weird lmao

so for you 0 is in Z ?

[zinc dove]

Ofc

[small meadow]

oh okay

but i think i missed smgh about the "when TWO integer"

[zinc dove]

anyway I think if you assume the problem means the integers have to be distinct

[small meadow]

why they're talking about 2 ther's only x

no y or smgh

[zinc dove]

Then it’s 4/7 * 3/6 /2

Wait how is the answer 10/21 are you sure

[small meadow]

yup

[zinc dove]

Hmm

I don’t even see how you can get to a denominator of 21

Besides 7*6

[small meadow]

yh but reefering with others answer there's definitly smgh with 7 at the denominator and idk what to do with that

may be enumeration but i'm quite lost

[zinc dove]

Hmm maybe the two numbers can be the same

So then it would be uh

(7*6)/2 + 7 possibilities total

And then (5*4)/2 +5 possibilities for Ns

[small meadow]

oh damn

found it !

[zinc dove]

Nice

[small meadow]

There it is !

Thx !

i shloud tried that lmao x)

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[zinc dove]

Oh it was this

It’s just you guys include 0

So 5/7 * 4/6 /2

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[robust wagon]

hi! im having a little trouble with proving the third sequence's infimum and supremum

[digital bison]

what have you tried?

[robust wagon]

so for the previous problems ive been utilizing the archimedean property but for the 2 previous sequences it was a little more clear on what the max and min of the sequences were

[kindred pier]

First, do you know the limit of third sequence as n approaches infinity?

[robust wagon]

i believe it would be 0

[digital bison]

since every term in that sequence is positive, and the limit is 0.
do you have an idea of the infimum?

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[robust wagon]

so the infimum must be 0

[robust wagon]

how could i algebraically prove this? my professor wants us to avoid using calculus

[robust wagon]

<@&286206848099549185>

[boreal siren]

what

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[robust wagon]

？

[undone solstice]

The sequence is always decreasing from some point so if you choose a positive epsilon youre bound to find a term less than epsilon so you can prove that way that the limit is 0

[robust wagon]

i see okay. how would i approach the supremum?

[undone solstice]

The sequence would be decreasing from n = 1 ( and strictly decreasing from 2 onward)

And if you evaluate the sequence at those points youd get two which is the supremum/maximum

[robust wagon]

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[empty nova]

?

huh

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[west smelt]

Could someone help me out, not sure where I went wrong.

[rugged latch]

id recommend log differentiation if u were gonna do it again but lemme check

i dont think u power and chain ruled properly on the first step

@west smelt

[rugged latch]

yea that seems right

oh hm

[west smelt]

The derivative seems correct, although in the answer in the numerator the 169 is meant to be negative, why is this?

[rugged latch]

can u show the answer key

did they put the minus sign outside the fraction

[rugged latch]

yes they put it outside the fraction

when they do that, it means the minus sign distributes to every term in the numerator

u wrote the minus sign in the numerator, but they didnt

ur answer isnt wrong

[west smelt]

My answer isn’t wrong?

[rugged latch]

.

[west smelt]

Wait so if the negative sign is in the numerator we don’t distribute it to every term?

But if it’s outside the fraction we do

[rugged latch]

yea i mean, as u have it, the minus sign is only on the 2x^2

why would it distribute

[rugged latch]

yes

[west smelt]

Oh I see, so your saying that in these two steps I should’ve left the minus sign outside of the fraction, when combing the numerator and denominator terms

[rugged latch]

i mean u can if u want to

it doesnt rlly matter

Just convention

[west smelt]

But I got the wrong answer, when I did it my way.

The numerator being -2x^2 + 169

[rugged latch]

u did not get the wrong answer

dude i told u thats correct lol

[west smelt]

Oh nvm

I get it

Okay, I’m still getting it wrong.

Here is the full question

Basically I'm tring to find the rate of change of area with respect to time (dA/dt)

I know dx/dt = -7.5 which I had previously worked out, and I know dy/dt = 3 which is given.

Here is my working

I got dA/dt = -35.7ft^2/min

However the correct answer is +35.7ft^2/min

Any idea where I went wrong?

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[somber onyx]

If the ladder is sliding down then dy/dt should be -3

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[languid mauve]

is this the correct way of finding the kernel of a lin transpo

[languid mauve]

<@&286206848099549185>

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[edgy willow]

yea everything's right except f. sort of because you can just put {[1 0]}

[languid mauve]

aklright ty

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[wraith hinge]

Hi sorry if i need to retake the screenshot! but can anyone tell me what i did wrong here? i think i need help with the reason for ABD congruent to CBD that might be wrong

[jade berry]

do you know all the congruencies

can you name them

[wraith hinge]

Yeah for the one of the bottom?

SSS, SAS, ASA, AAS, SSA, AA, HL

[jade berry]

and there's a slight hint for you that will help you to arrive at a correct answer 90^{\circ} is present in the triangle

[wraith hinge]

Im unsure what {\circ} is

[jade berry]

$$90^{\circ}$$

[solid kiln]

ℕaive

[wraith hinge]

Yeah okayyy i see

yeah so i know its SA

but past that im unsure if theres another side or angle

[jade berry]

try to see whats given, the details and choose the congruency relation based on it

[wraith hinge]

Ah i got it

ITS HL

[jade berry]

i don't what that means, RHS right?

[wraith hinge]

hypotenuse leg

[jade berry]

fair enough

you did well

[wraith hinge]

but i figured it out so ill close it here:D

Ty!

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[viscid cairn]

Reading a book on tikhonov regularization and I am wondering what regularity means with derivatives

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[green locust]

Don't know if you guys really deal with "mechanical" maths but...here is THIS

I know the question isn't really that hard...what I want to know is which given "length" do I have to use the length of the stroke (s) which I'd have to convert to metres then divide by 2

Or do I have to use the connecting rod length

[green locust]

There's also these statements if they are of any help

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[green locust]

<@&286206848099549185>

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[dusk loom]

Could someone please provide a solution. I need an answer and I’ve tried multiple things

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[undone solstice]

@dusk loom for the first sum, the term is equivalent to 1/2n^2 which is a riemann sum which is convergent

So first sum must be convergent

For the second if you compute 1/n*u_n youd get zero which means the general term of that sum dominates 1/n which means the sum is divergent by comparison

[dusk loom]

Thank you

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[jade flint]

Is this math valid?

[split chasm]

yes

[jade flint]

thx

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[lone mauve]

before I click Submit, does this look correct?
it's my last attempt, and I'm not sure about radians or degrees on this

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[burnt mulch]

Looks right to me

[lone mauve]

ty

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[static cloud]

Is a transformation being onto have something to do with pivots and/or free variables?

[ionic pendant]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/05%3A_Linear_Transformations/5.05%3A_One-to-One_and_Onto_Transformations

the rank (number of pivots) of a matrix (representing a linear transformations) does determine if it's onto (see the link for details)

[static cloud]

so if the transformation is consistent then it's onto?

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[static cloud]

no

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[umbral maple]

The problem is the angles

I get tan -1 (1/2) = 26.5

but that’s wrong

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[fringe marsh]

can someone explain the bound for theta

[fringe marsh]

I lowkey am sort of confused

i did it from -pi/2 to pi/2

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[copper flame]

Working on my Trig homework as a preparation for an exam, would like some 1 on 1 help

[eager summit]

!da2a

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[copper flame]

<@&286206848099549185>

[analog arrow]

What exactly do u need help with

[copper flame]

how do i approach this kind of problem

[analog arrow]

Hmmn

Well from what I see the question is asking for the exact value of sin(θ/2) given that sin(θ) = 1/4 and θ lies in quadrant I.

[copper flame]

oh wait i was still reading through that

[analog arrow]

My bad

[solid kiln]

tori

LaTeX source sent via direct message.
```Compilation error:```! LaTeX Error: Unicode character θ (U+03B8)
               not set up for use with LaTeX.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 ...on is asking for the exact value of sin(θ
                                                  /2) given that sin(θ) = 1...

You may provide a definition with```

[analog arrow]

Is ur problem solved?

[copper flame]

yes, could I post more problems?

[analog arrow]

Sure

[copper flame]

i figured how to do the product but not the difference

i have a feeling that Im over complicating this

@analog arrow

<@&286206848099549185>

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[sick flame]

Is it correct

[mystic surge]

no

gamma have to be in both sides dividing

[sick flame]

Ohh actually the original Equation was something like this :

@mystic surge

[mystic surge]

complex numbers are my weak

[sick flame]

Ohhh I have been trying to get help since morning but couldn't get 🤧

@mystic surge

<@&286206848099549185>

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[cold vapor]

Find x is the question tried a few things didnt work

[cold vapor]

,lcw

[buoyant bronze]

lets see

whats the point name thats out of frame?

D i guess

[cold vapor]

Yeh

[buoyant bronze]

alr and x is EDC angle?

[cold vapor]

X is bax

Bac

[buoyant bronze]

oh okk

1 sec

[main sigil]

I got it

[buoyant bronze]

me too

[cold vapor]

Nice

[buoyant bronze]

you extend ba to the right

[main sigil]

okay that sounds more interesting

i just worked with the given angles

[buoyant bronze]

oh wait not right left

[buoyant bronze]

me too basically. shall i explain or do you want to?

[main sigil]

explain

ill watch

[buoyant bronze]

alr

[buoyant bronze]

so you can extend BA to the left

[cold vapor]

Ight

[buoyant bronze]

then the angle that is created like cb something will be equal to edc because ab//de

[cold vapor]

Oh using the Z rule

[buoyant bronze]

yeah yeah

[cold vapor]

Then what did u do

[buoyant bronze]

can you express the dce angle in terms of x?

[cold vapor]

90-x?

[buoyant bronze]

yes.

now can you express angle edc in terms of x? (DCE is equilateral)

[cold vapor]

180-90-x/2 90+x/2?

Right one

[buoyant bronze]

(90+x)/2 yes

alr and that will be equal to the cb something angle we said in the beginning right?

[buoyant bronze]

here

[cold vapor]

Yeh

Still cant find cba

[buoyant bronze]

that angle and cba form 180o right?

[cold vapor]

Nvm i can

[cold vapor]

Yes

[buoyant bronze]

and what would it be

[cold vapor]

270-x/2

270+x/2 not the one above

[buoyant bronze]

it would be 180-(90+x)/2 right?

[cold vapor]

Ah this forgot there was a paranthesis in division

[buoyant bronze]

great now take the sum of the angles in the cba triangle

and solve for

x

[cold vapor]

Think i did smthn wrong found 210

[buoyant bronze]

lets see

we have

[cold vapor]

But this should work

[buoyant bronze]

(270-x)/2 +x +x =180 right?

[cold vapor]

Yeh

Oh 270

[buoyant bronze]

(270-x)/2=135-x/2 right?

[cold vapor]

Thought that was -270

Should be 30

[buoyant bronze]

great

what grade question is this btw

[cold vapor]

10th

[buoyant bronze]

cool

[cold vapor]

Not 10th

9th forgot i was solving 9th class geometry for a sec

Anyways ty

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[buoyant bronze]

np

[cold vapor]

.reopen

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✅

[cold vapor]

Btw was there any other solutions to this

[buoyant bronze]

probably yeah cause mine was kinda hard

[cold vapor]

Maybe something with ba de being parallel actually

[buoyant bronze]

@main sigil did you have an easier solution?

[main sigil]

I had a very similar one

[cold vapor]

Can u post it

Remember u sending a ss before deleting it

[main sigil]

First i founs out that CDE is just 2x, thats because CBA = 180 - 2x = 180 - CDE

[cold vapor]

İght

[main sigil]

Then, because of ACE + ECD + ACB = 180

(180 - 4x) + 90 + x = 180

90 = 3x

X = 30

[cold vapor]

Why is 180-CDE 180-2x

[main sigil]

CBA is 180 - CDE

[cold vapor]

Oh right

Lol ty

[main sigil]

Its similar solution to skycrew

[cold vapor]

Yeh also thought about doing something with b+d=180 but couldnt do anything

Anyways ty

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[fringe marsh]

can someone explain how to find the bounds for this

[fringe marsh]

iwas confused by the example tbh

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[frosty lake]

you have to use two integrals on the function, like so:

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[wanton wyvern]

Hey, could anyone help me transfrom this riemann sum to a definite integral?
I know the general formula, but the j-1 is throwing me off..

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[cold vapor]

Adc is folded to ad and the c point is brought to c1 thingie find acb

[cold vapor]

Dont understand how to do these folding

,rcw

[cold vapor]

Questions and where to start

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[wraith hinge]

Hello \

[wraith hinge]

is there a way to know how to Separate sums ?
or just by luck

[wraith hinge]

ik its a weird question
but solving irreducible polynomial can be done only this way

[bitter hearth]

Factoring isn't about luck—it's about spotting patterns. You split -24x^4 into -8x^4 and -16x^4 to group terms with common factors. Group 10x^2 with 16x^6, and -8x^4 with -16x^4. Factor out the common terms from each group, and you'll see a common (2x^2 - 1) emerge. Factor that out, and you've cracked it! It's like a puzzle, just find the pieces that fit together.

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[wraith hinge]

thank you

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[hasty idol]

Does a group of order 35 contain an element of order 5? of order 7? If G is cyclic it just x^5 and x^7, but otherwise I don't know how to prove it

[nova spire]

You can use Sylow's theorem, but that could be overkill

basically :

|G| = 5*7, 5 being prime, and 7 does not divide 5

thus we can apply this theorem

(the general case is |G| = p^n * q)

and so G has a subgroup of order 5...

[hasty idol]

Same logic for 7

Ok i think I got it, I'm going to read more about that theorem

Thanks

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[rotund oyster]

Hello, I am making a tycoon game and have a question

How can i possibly figure out how much things should cost if i want players to play for a certain average time when there are so many paths they can take that could altar it.

They can do multiple things to earn money, and they can spend money in multiple ways. I also want progression to feel linear, like they are earning more and spending more, really getting further in the game

I dont really know what to do, anything helps

[slender shard]

I don't think linear is what you want. You want it somewhat exponential. The more a player spends, the more they are rewarded.

[rotund oyster]

Right, i dont want the average play time to be 60 minutes and have 60 thing to buy and calculate that, it would be the same the whole time, not exponential

[slender shard]

Based on my experience, there should be a locality bonus to encourage players in the same time zones to clan up and get bonuses. All too often I see clans fail because of too many clan members in far too many different time zones unable to play with other clan members.

[rotund oyster]

Clans? Time zones? Im just trying to figure out how I can get an average play time when there are so many paths they can take that could altar how long they play.. What if they buy and do all things that give money and only buy the things that dont give money after

[slender shard]

Oh wait, you're talking about a different type of game.

[rotund oyster]

Yeah tycoon style

[slender shard]

So like Idle Miner, Idle Title. and whatnot.

Or farming games.

[rotund oyster]

Yeah like a farming game, not everything you buy produces more money, you can go multiple paths of placing nice things or focussing fully on earning more

I want the average play time to be the same whichever path

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[edgy ridge]

what can i let the non-homogenous part of teh equation to be? i tried letting c(n) = A*B^n

[edgy ridge]

so its like 4*6^n = 12c(n-1) - 36c(n-2) - c(n)

and then subbing in c(n-1) = AB^(n-1) and c(n-2) = AB^(n-2) and c(n) = AB^n

so 4*6^n = 12AB^(n-1) - 36AB^(n-2) - AB^n

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[gaunt wave]

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[delicate lance]

why does it turn into this ?

[ionic pendant]

,tex \logrules

[solid kiln]

pnoןɔ

[ionic pendant]

product rule for logarithms

[delicate lance]

ohh

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[latent axle]

Antiderivitive 1/x^5

[latent axle]

I think it’s some extension of the derivative 1/x

But idk how

[devout drift]

$\f1{x^5} = x^{-5}$

[solid kiln]

ColdTe²

[latent axle]

Yep I got it from here

Thanks

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[latent axle]

Antiderivitive of this, I’m lost

[unique stratus]

first

An antiderivative of a function f(x) is a function whose derivative is equal to f(x). That is, if F′(x)=f(x), then F(x) is an antiderivative of f(x). x33,x33+1,x33−42,x33+π.

orgainc chme

did

a good video

it helped me tons

chek it out

[latent axle]

Ik what antiderivites are idk how to find the one here

[split chasm]

split into two fractions, then power rule

[frosty slate]

Or this

[jolly basin]

that's the derivative not the antiderivative

[latent axle]

Is it x-2/sqrt(x)

[jolly basin]

can u explain how you got that

try rewriting the problem as the sum of 2 fractions (hint: same denominator)

then simplify those fractions

[latent axle]

That’s what i dud

[jolly basin]

(x^2+√x)/x^2

x^2/x^2 + √x/x^2

what does that simplify to

[jolly basin]

check the variable you're using

[latent axle]

?

[jolly basin]

otherwise looks fine

[latent axle]

Oh yeah

[jolly basin]

you have x in the answer

[latent axle]

I see

[jolly basin]

what are you actually integrating on

ik it's habitual just be careful abt it

also

u are missing

[latent axle]

sqrt(2) upper
1 lower

[jolly basin]

something VERY important

[latent axle]

That’s the integral

Evaluating sqrt2 would get me what

[jolly basin]

your integral is fine (if you change x to s)

but you're missing something

[latent axle]

I did

What

[jolly basin]

that is very important to indefinite integrals

[latent axle]

?

[jolly basin]

ok well that changes the problem

[latent axle]

That’s what it’s from

[jolly basin]

what you wrote was an indefinite integral

meaning there's an integration constant you didn't write

[latent axle]

What’s it missing

[jolly basin]

if there are bounds then you need to write them

[latent axle]

+C

[jolly basin]

yeah but i didn't realize there are bounds

then just write the bounds on the integral

[latent axle]

There are now

Yeah

But uh

If I plug in sqrt2

I get

Nvm

I got it

Thanks

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[cold vapor]

ADC is folded to AD and the c point is brought to c1 thingie find ACB

[cold vapor]

,rcw

[cold vapor]

Dont understand where to start in these folding type questions

A

[sterile bear]

türkçe yazıyorum bak şimdi katladığın yerde üst üste gelen açılar eşit olucak x diyelim (şekil 2 deki adb açısından bahsediyorum ) geri açınca bdc doğrusu üzerinde 2x +40=180 oluyor x=70 tepe açısı 70+40=110 180-110=70 ikiz kenar olduğu için 70/2=35 derece oluyor

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[cold vapor]

Anladım tşk

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[wraith hinge]

What exactly is direction in vectors and do all, vectors start from origin?

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@wraith hinge Has your question been resolved?

[unique minnow]

A vector is just that : a length and a direction. As for your question, I think the best answer is that it doesn't matter where vectors start.

We like to represent them as starting from the origin, but a vector itself doesn't have a position in space assigned to it.

For instance, you can represent the force exerted on an object from where they are applied, but an equivalent representation could place the vectors at the origin (middle of the object) and you would still get the same situation.

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[cold vapor]

Question is find x I cant find x seperarely no matter what i do

[cold vapor]

,rcw

[trim jolt]

@cold vapor Has your question been resolved?

[dull pilot]

are… three sides equal?

@cold vapor

[cold vapor]

Yeh

[dull pilot]

ah

huh

a sec

pretty frustrating how that is almost a parallelogram

[cold vapor]

Ik

[dull pilot]

do some coordinate magic i guess?

it wouldn’t be pretty but it should work ig

[cold vapor]

Uh

How do i do that

[dull pilot]

uhh let’s see no

god this solution really sucks

[cold vapor]

Uh

Its a solution atleast? Btw this is a 9th grade question if that helps so it can be solved with 9th grade methods

[sacred sapphire]

do you have a reference answer?

[cold vapor]

The answer is 121

[sacred sapphire]

holy shit the answer was so blindingly obvious

and i also missed it

[cold vapor]

Lol

[sacred sapphire]

connect A and C together

[cold vapor]

What were you trying to do

[sacred sapphire]

since its an isoceles triangle, the other two angles must be the same

but the angle is 60

180-60 is 120

120/2 = 60

so actually its an equilateral triangle

[cold vapor]

Thats super obvious lol

[sacred sapphire]

the rest is super simple

[cold vapor]

I also looked at that for 2 minutes and couldnt see it lol

[sacred sapphire]

me too lmao

[cold vapor]

Btw ehat did u try to do

[sacred sapphire]

what i just said

[cold vapor]

Ight ty

.close

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[sacred sapphire]

heres the diagram

rest is super simple

[cold vapor]

Yeh solved it

Also wrote the exact thing didnt see ba at all

[sacred sapphire]

we all are blind sometimes lmao

[cold vapor]

Yeh

[dull pilot]

OMG

lmao.

[cold vapor]

Yo

What were you trying to do

[dull pilot]

I got the coordinates

two lines

😭

[cold vapor]

Lol

Cant see right ahead

Anyways ty

.close

[dull pilot]

yesyes

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[wraith hinge]

,align
\gdef\joe#1{\vec{\vb*#1}}
\grad\varphi &= \4{\int \varphi\dd\joe\sigma}{\int\dd\tau} \tss{when} \int\dd \tau \to 0

[wraith hinge]

not sure what's the big idea with this "definition" of the gradient?

tau represents the volume element

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@wraith hinge Has your question been resolved?

[undone solstice]

@wraith hinge ive never seen this before but i think its that you take some volume omega and the surface bounding it sigma then you are essentially looking at the flux going out or in of that volume

[wraith hinge]

but thats not the divergence?

so how is the flux relating at all here

i mean

i understand that it's like

[undone solstice]

Well the gradient gives how much the function changes at any point and in what direction, that integral above is something like scalar flux

[wraith hinge]

hm

why do we minimise the volume though

the integral at the bottom is pretty much the total volume of the thing

[undone solstice]

To get that instantanenous flux i guess

[wraith hinge]

so if its approaching 0 i guess we are considering a very infinitesimal portion of thid

[undone solstice]

The less the volume the more you are approaching a point

[undone solstice]

Which is what the gradient is

[undone solstice]

Yep

[wraith hinge]

right

i wonder whats rhe derivation tho

to make this make sense

[undone solstice]

Same

I would like to know where this is from tbh

[wraith hinge]

i was looking back at my scattered notes for my Electromagnetism course

my professor wrote this somewhere along the lines

I wasn't able to retain the context of it though

[amber python]

wots sigma

[wraith hinge]

area element

so surface integral of the volume thingy

[amber python]

also \vj

[wraith hinge]

yea i just copy pasted it from our server LOL

was too lazy to write it again

[amber python]

isn't this just stokes theorem then

[wraith hinge]

oh uh

whqt was it again

,, \int \8{\curl \vj F} \vd \vc*n \dd \sigma = \int \div \vj F \dd \tau

[wraith hinge]

am i right

i hope im right

[amber python]

,, \int_{\del M} \varphi \dd S = \int_M \Grad\varphi \dd V

[wraith hinge]

[wraith hinge]

oh ok u wrote out the big brain version

[amber python]

so then like

assuming grad phi is continuous

[wraith hinge]

wait

what does dm and m mean again

i keep forgetting

[amber python]

M is a nice enough region

del M is its boundary

the bounding surface

[wraith hinge]

del m is always one dimension lower ig?

[amber python]

ya

[wraith hinge]

aight

[amber python]

so you just have like

[wraith hinge]

hmmm isnt it the divergence of the thing on the right tho

[amber python]

,, \f {\int_M \Grad\varphi \dd V} {\int_M \dd V} \to \Grad\varphi

[amber python]

no

[wraith hinge]

whaaa

oh

[wraith hinge]

i somehow like fucking merged both the divergence and stokes theorems

fuck me ok

[amber python]

i'm talking about discussions greatest moment

wait no

wrong one

i'm talking about literally all of calc 3

[wraith hinge]

wait uh

like

we talked about this right

[wraith hinge]

so what happened here exactly

[wraith hinge]

from here

[amber python]

i missed a vec on S

[wraith hinge]

[amber python]

anyway

,, \f {\int_{\del M} \varphi \dd \vec S} {\int_M \dd V} = \f {\int_M \Grad\varphi \dd V} {\int_M \dd V}

[zealous lark]

Lebesgue differentiation theorem

[amber python]

ye

or just like

assume continuous

life easy

[wraith hinge]

assume smooth

ok sorry

[amber python]

you're lucky you're working with real functions then

there aren't that many smooth complex functions

[wraith hinge]

as the integrator took out his pen...

[amber python]

he realised he was out of ink

[wraith hinge]

yea that sounds like a me moment if imma be fr

anyways

thanks brosephs

until another time

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[queen axle]

onto=surjective
one to one = injective
but is one-one corresondence just another word for injectivity or is it used to denote bijectivity

[queen axle]

ah it means bijective

why do people use these words instead of injective surjective bijective

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[supple copper]

Because they sound very mathy and the engineers might get confused

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[fringe marsh]

Can someone explain this to me

[fringe marsh]

This part specifically confuses me

more so the x = f(y)

[marble wharf]

what about it confuses you

[fringe marsh]

Im not sure under what condition y = f(x) would not be satisfied

or rather the case for using x = f(y)

[marble wharf]

well rather often you would expect y=f(x) not to hold

the condition tells you whether there is an edge between x and y

so for example there is an edge between 3 and 4 if 3=f(4) or 4=f(3)

[fringe marsh]

Honestly, im sorry I still dont get it

[fringe marsh]

f(4) = 2 and f(3) = 1

[marble wharf]

yes

so there is no edge between 3 and 4

[fringe marsh]

whats an example where there is an edge between two nodes

[marble wharf]

between 2 and 4

and between 3 and 1

by your own calculations

[fringe marsh]

Okay that makes sense, but im still confused why both wouldnt have to hold

[marble wharf]

do both f(2)=4 and f(4)=2 hold?

[fringe marsh]

I guess not

okay but due to f(4) = 2 holding, there is an edge between 2 and 4

[marble wharf]

yes

[fringe marsh]

this was really helpful thank you

[marble wharf]

yw

[fringe marsh]

are the edges sequential

like checking for an edge between for instance 0, 1 and then 1, 2 .. 2, 3

[marble wharf]

you have to check all possible combinations

(or you can compute all values of f and then from that get the relevant edges. like I did above)

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[rigid tide]

3rd part is not clear
How do I take such a huge derivative
Is there a way to simplify

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@rigid tide Has your question been resolved?

[limpid heart]

think you might just have to work it out

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[untold yew]

Hi the question is prob rlly easy but i dont get why its wrong 💀

[mystic veldt]

sin theta = 5??

how blud

[untold yew]

Cos tan = sin/cos

Is it not 😭

[mystic veldt]

what's tan equal to?

[untold yew]

5/12

[mystic veldt]

what is cos tan 😭

[mystic veldt]

nah i mean in general

[untold yew]

Cuz tan* 😭

[mystic veldt]

not in this question

[untold yew]

Sin/cos

[whole coral]

It is, but that doesn’t mean that the numerator must be sin and the denominator cos, you can cancel common factors after all

[untold yew]

What?

[whole coral]

Responding to this-

[untold yew]

But if the numerator was cos and the denominator was sin itd be cot

[mystic veldt]

i love your profile ❤️

[untold yew]

Mine?

[whole coral]

Yea, but, like-

For example let’s take a 3-4-5 triangle

[mystic veldt]

yea

[untold yew]

Omg ty 😍

[whole coral]

Then, say, sin(a) is 3/5, cos(a) is 4/5 and tan(a) is 3/4

[mystic veldt]

^

[whole coral]

But clearly sin(a) can’t be 3 and also 3/5 at the same time, similar for cos

[untold yew]

Do i hafta draw a triangle and use pythagorean theorem

[whole coral]

But notice that if you did sin(a)/cos(a) and simplified…

[whole coral]

(And, well, maybe not “have to”, but you really should…)

[untold yew]

Oh okk 😭

TY

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[tribal fractal]

for the first one i made the inequality

[tribal fractal]

0.000005 < 1/n!*x^n

using the lagrange formula

[nova spire]

do you know cos's taylor expansion?

If so, you'll notice you don't need lagrange formula, rather alternating series...

[tribal fractal]

yeah

[tribal fractal]

how would i use that?

@nova spire you there?

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@tribal fractal Has your question been resolved?

[tribal fractal]

anyone

pls help

<@&286206848099549185>

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@tribal fractal Has your question been resolved?

[tribal fractal]

anyone

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@tribal fractal Has your question been resolved?

[limpid dawn]

Wouldn't we approach it that way?

$\ \abs{\cos(1) - T_n(1)} = \abs{R_n(1)} = \frac{f^{n+1}(c)}{(n+1)!}(1-0)^{n+1} \leq 0.000001$

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

The easiest thing would be to write some terms of the cosine Taylor series and see at which n it is for 5 decimals correct

[tribal fractal]

Oh ok

[trim jolt]

@tribal fractal Has your question been resolved?

[tribal fractal]

how would i solve an inequality like this

with a factorial

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@tribal fractal Has your question been resolved?

[limpid dawn]

,, a< c < x \implies 0 < c < 1 \implies 0 < f^{n+1}(c) < 1

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{f^{n+1}(c)}{(n+1)!}(1-0)^{n+1} \leq \frac{1}{(n+1)!} \leq 0.000005 \implies (n+1)! \geq 2 \cdot 10^6

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Try to find the minimum of n so that the last inequality holds

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[dapper swift]

Try drawing a Venn diagram with sets B and C

Where A is the intersection of B and C

Then see what happens if you make the intersection of C and A be B

[rapid horizon]

I feel nothing

[dapper swift]

Yeah so the intersection of C and A is the middle part

Since A is the middle part

[rapid horizon]

Yeah so?

You meant a and b will be equal

[dapper swift]

Exactly

Another way to do this is that A is both in B and in C
So A is a subset of B
And B = C intersect A means B is a subset of A

So A and B must be the same set

[rapid horizon]

Thank you very much