help-10

[austere hamlet]

Mhm

[timid silo]

so in the diagram, you are given 2 points that line p runs through right

[austere hamlet]

Yes

[timid silo]

(0,2), and (6,4)

so do you see anything special about the first point

(0,2)

[austere hamlet]

Hmmm

[timid silo]

do you see that it intercepts with an axis

how the point is on the y axis

[austere hamlet]

Well yes

[timid silo]

that means that the point is the y intercept right

since it intercepts with the y axis

[austere hamlet]

I mean yeah

[timid silo]

so in the equation y=mx+b

where b is the y intercept

do you see how you can use that point to fill out b

[austere hamlet]

Wait, I just noticed this was a graph💀

Okay so

We are going to give y a number

Right

It's either 0 or 2?

Assuming it's 2 tho

[timid silo]

yess

so b is 2 right

[austere hamlet]

Ye

[timid silo]

since the y intercept is 2

so the formula becomes

y=mx+2

[austere hamlet]

Yes

[timid silo]

2 replacing b

so you have a second point right

( 6,4)

[austere hamlet]

Yeah

Do you mean

That x is 6

?

[timid silo]

yea, for one ponit

when x is 6, y is 4

[austere hamlet]

Which we don't want

Right

[timid silo]

so you can plug in the coordinates (6,4)

into the equation

y=mx+2

6 is x, 4 is y

[austere hamlet]

Wouldn't that be y = m6+4

[timid silo]

waittt

but the equation was y=mx+2

[austere hamlet]

Yeah

[timid silo]

you are replacing y and x in the equation

so it becomes

(4)=m(6)+2

[austere hamlet]

Oh

[timid silo]

does that make sense

[austere hamlet]

SO LIKE THAT

lol

[timid silo]

lol

so 4=6m+2

and you can solve for m

[austere hamlet]

M = 1?

So is that like the final answer?

[timid silo]

you can solve for m

by subtracting both sides by 2

[austere hamlet]

Omg

[timid silo]

rich becomes 2=6m right

[austere hamlet]

My bad

M = 1/3

[timid silo]

yess

so you know m and b in the equation y=mx+b

so you just have to replace m and b

with 1/3 and 2

[austere hamlet]

Y = 1/3+2

Right

?

[timid silo]

yess

wait

1/3x+2

not 1/3+2

[austere hamlet]

Oh

Yeah

Wait do I fill the x?

[timid silo]

so the final function becomes y=1/3x+2

[austere hamlet]

That's like the answer?

[timid silo]

no, the general form of a line has both y and x

[timid silo]

ye

[austere hamlet]

Woah, that sure was insane

But I should have noticed that it was a graph

[timid silo]

so basically to sum up, because the y intercept was given, you can find b in the equation y=mx+b, and you can plug in a second point to find m

[austere hamlet]

Yeah

I understand well now

Thank you

Anyways there is one more

17. Express the deviation of the line p from the 15th problem from the coordinate axis x with a relation

[austere hamlet]

@timid silo

[timid silo]

<@&286206848099549185>

yo

[austere hamlet]

Oof

@tacit spade since we are here, can I get help for one more question

I need to get the area of the rectangle

So here is what I think

a = ?
b = 50

a = b/sin(ß)

a = 50/sin(50°)
a = 70,71

c² = a²+b²
c = 86.60

S = a•b
S = 2598.08

[vernal yarrow]

but you can solve this using sine law

where a/sin A = b/sin B

a/sin 100 = 50/sin50

[austere hamlet]

ohhh

i could have used that too

[vernal yarrow]

a = 70.71 x 0.9848

[austere hamlet]

damn so my calculation was wrong?

[austere hamlet]

couldnt this work?

[vernal yarrow]

so a=69.635, 69.635 x 30 = 2089 around

if i did my calculations correctly

<@&286206848099549185> can you check my work

[austere hamlet]

well technically your method is correct, but mine should be too... or not

[vernal yarrow]

you cannot use the pythagorean theorem

it's not a right triangle

it's a 50:30:100 degree triangle

<@&286206848099549185> hopefully someone may help check

[austere hamlet]

okay but we know that one side is 50 and now that we found out that the other one is 70,71 wouldnt it be possible to use pythagorean theorem?

[austere hamlet]

from what did you get 0,9848

since we are here

i need more help

11. Specify the scope of the function

12. The graph of the quadratic function for x € R is a parabola, determine the coordinates of its vertex

13. The graph of the quadratic function for x € R is a parabola, determine its range of values

[obtuse pebble]

@austere hamlet Has your question been resolved?

[austere hamlet]

Pls

[austere hamlet]

<@&286206848099549185> uhh

[austere hamlet]

.close

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[fervent nacelle]

Linear Algebra
Hey guys, my linear is pretty rusty and I'm a little confused:
I have a symmetric bilinear form, which I've converted into a matrix;
Now, I've supposedly found a base where the matrix is diagonalized:
The base being the rows of the left-hand matrix, and the diagonalized matrix being the result at the end.
My friend has reached a different result.
Does this make sense? Or can both of us be right?

[mighty geyser]

how did you find the eigenvectors?

[fervent nacelle]

Pardon?

You mean the diagonal matrix?

[mighty geyser]

hold on, no eigenvectors, let me check

[fervent nacelle]

"Matrix congruence" is what I used

[mighty geyser]

did your friend get the same values on the diagonal?

[fervent nacelle]

Nu uh

Diag(1, 0.25, -0.25) is what they got

I thought a given matrix only has a single diagonal form (well not for, the values can be in different order), then again intuitively this feels wrong

[mighty geyser]

say you multiply by diag(2, 1, 1)

[fervent nacelle]

Yeah?

[mighty geyser]

left and right multiply your result by diag(2, 1, 1)

it should result in a congruent matrix

[fervent nacelle]

I'm not sure I follow

I'm not looking for a congruent matrix, the matrix in the middle of the equation and the resulting matrix are congruent already

I simply don't know if the rows of the matrix on the left of the equation are indeed the base I'm looking for
I don't think it's possible we got different results, but I don't really know

[mighty geyser]

but if you left multiply by a matrix and right multiply by the transpose, you get a new matrix which is congruent to the original matrix right?

[fervent nacelle]

Seems so, yeah

As in, I've found a a matrix M which indeed fulfills:
M^t * A * M = diag(4,1,-1/4)

Oh and Det(M) != 0

[mighty geyser]

yeah, invertible

[fervent nacelle]

Yeah that

[mighty geyser]

So let's do it again with diag(2, 1, 1)

that should get another matrix congruent to the original matrix right?

[fervent nacelle]

As in, M = diag(2,1,1)?

[mighty geyser]

Okay this might be easier, instead of M, use 2M now

[fervent nacelle]

Huh

Alright

[mighty geyser]

see what result you get

[fervent nacelle]

(16,4,-1), as one might expect

This is just 4(M^t * A * M) innit

[mighty geyser]

yeah, but now you have different numbers

[fervent nacelle]

Oh

[mighty geyser]

so it is possible to get different numbers

[fervent nacelle]

Yeah but like, the ratios are the same

[mighty geyser]

now take M*diag(2, 1, 1)

[fervent nacelle]

Oh alright

Oh or did you mean M and diag alone

[mighty geyser]

instead of using M on the original matrix, use M*diag(2, 1, 1)

yeah

[fervent nacelle]

Oh

As in M = M * diag(2,1,1) so to speak?

[mighty geyser]

not sure, what you mean

just try out i guess

[fervent nacelle]

Like M * diag(2,1,1) * A * (M * diag(2,1,1))^t?

[fervent nacelle]

I'm just not sure what you mean by this, basically

It's what I wrote just now, isn't it?

[mighty geyser]

(M * diag(2,1,1))^t * A * M * diag(2,1,1)?

thought you put the M on the right side

[fervent nacelle]

Yeah you do, it doesn't erally matter though

This is the convention, I just defined M to be (what is traditionally) M^t

I swapped the two's roles, basically.

Uh, it's a little hard to do with a calculator

Could you just tell me if it's possible that we got different results?

We were simply supposed to find a base for A to be diagonal in, there are (quite evidently, as you've just shown me) multiple bases - it's just that the ratios between our two matrices aren't the same. That's the weird part

As in Diag(4,1,-0.25) and Diag(1,0.25,-0.25). The value in the middle on the first matrix is "-4 times bigger" than the last value. His isn't though

[mighty geyser]

imagine scaling one of the basis vectors

[fervent nacelle]

Oh

I think I understand,

[fervent nacelle]

Like this??

Very cool
I apologize, my intuition on the subject matter might as well be nonexistent

[mighty geyser]

yeah, i'm just giving concrete calculations to give some examples

[fervent nacelle]

So essentially, there are many different forms and bases in order to diagonalize* some matrices?

[mighty geyser]

Actually one can prove something regarding the forms

It's known as Sylvester's Law of Inertia

but I think it would be more educational to try to discover it yourself

Can you make it into the identity matrix? (Is A congruent to the identity matrix?)

If not, how close can you get?

[fervent nacelle]

I probably could, but I'm afraid I've already spent 3 hours on this silly exercise

Wouldn't you reckon that this is something I'll encounter during the course?

Within the context of subspaces even

Oh, and the diagonalized matrix of eigenvalues is specifically for Q^-1 * A * Q

I see now.

Which is why there's only a single "variant" of an eigenvalue matrix, but many (infinite?) forms of those M matrices.
That makes a lot of sense in all honesty

I appreciate the help

Seems like I've lost you
Sorry for any language barrier frustrations, have a nice day 🙂

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[silk shell]

I've tried so many different ways to get x, but i genuinely don't understand how to do this question.

[timid silo]

i'd assume you use the property

where if a quadrilateral is inside a circle

the opposite angles of the quadrilateral are supplementary

so i think how you'd go about doing that is

[silk shell]

thats what i thought too

[timid silo]

basically, angle ABF is 180-x right

[silk shell]

but when i started inputting the values nothin made sense

[silk shell]

yeah

[timid silo]

then you can figure out the value of angle AFB right

in terms of x

since angle baf is 54, angle abf is 180-x

angle afb is 180-54-180+x

since it's 180 - (54) - (180-x)

[silk shell]

that makes sense

[timid silo]

so this simplifies to x-54 right

[silk shell]

yeah

[timid silo]

using vertical angles

you can know angle DFE right

since vertical angles are congruent

[silk shell]

cant we say that x = 54?

[timid silo]

but the angle isn't 0

[silk shell]

ohh true

[timid silo]

but since we know the value of angle AFB

using vertical angle definition angle DFE is congruent and also x-54

[silk shell]

yeah

[timid silo]

so then you can know the value of angle FDE

which is 180 - (x-54)-(32)

since a triangle adds up to 180

[silk shell]

yeah

[timid silo]

and angle FDE is 202-x then right

so we can figure out angle FDC

since FDC and FDE form a line

so FDC is 180 - (FDE)

which is 180 - ( 202 -x )

which simplifies to x-22

and coming back to

[timid silo]

this

x-22 and x are opposite angles of a quadrilateral in a circle

meaning they add up to 180

so x-22+x=180

2x=202

x=101

[silk shell]

oh my god

thanks a ton

that makes so much sense but i just doubt i would ever think of that in an exam

[timid silo]

sometimes just playing around with angles work

[silk shell]

let me show you what i done

[timid silo]

tbh there was a more efficient way to do this i realized half way

[silk shell]

you're going to be shocked

[timid silo]

but even if ways are slightly inefficient

just playing around with numbers and angles sometimes work

[silk shell]

i mean if you get the same answer, its fien with me tbh

sorry this might be easier to look at

[timid silo]

ah wow

this is very complicated lol

[silk shell]

i got it wrong because i was assuming what an angle would be

[mighty geyser]

yeah the 90 degrees is off

[timid silo]

yh it's better never to assume an angle

unless you are extra extra extra extra desperate

[silk shell]

i assumed that because i can form 2 semi-circles with a line (that i drew, meaning i doubt its through the diameter) both angle suspended from that would be 90 degrees

which is where i made the BIG mistake

[timid silo]

ah i see

[silk shell]

i didnt even think of making an equation

[timid silo]

a method that seems to work for me and most people is filling out blank angles using all the knowledge you have

[mighty geyser]

you can label all the angles and write down all the relations you can think of between them

[timid silo]

it often helps you come up with equations and things like that

[timid silo]

yh exactly

anyways idk when your exam is but gl :)

[silk shell]

its in a month, but revision now is better than never

anyways, thanks a lot

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[hardy swan]

Why do I do 6/35

To solve for the answer

nvm

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[frank crag]

What is the inverse function of ln(3x) ?

[frank crag]

3*e^x ?

or e^x^3

or even e^x/3 xd

[trail cloak]

$e^{\ln(x)} = x$

[warm shale]

VulcanOne

[trail cloak]

And

$\ln(e^x) = x\ln(e) = x$

[warm shale]

VulcanOne

[trail cloak]

So when you want to find the inverse of y = ln(3x)

You do the usual

Replace x with y

$x=\ln(3y)$

[warm shale]

VulcanOne

[frank crag]

hmmmmmmmm

[trail cloak]

You then take $e^{x} = e^{\ln(3y)}$

[warm shale]

VulcanOne

[frank crag]

b) f(x) = e^3x ; h(x) = ln = (3x)

f^-1 (x) = 1/3 ln (x)
so the term h(x) is not the inverse function of f(x). It would be right if f(x) = ?what?

This is the task

[frank crag]

this is right?

[trail cloak]

Yep

[frank crag]

hm

[trail cloak]

You use this trick when you want to cancel out the natural log

[trail cloak]

Because this works

[frank crag]

okay

thanks ❤️

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[frank crag]

now... topic: Definition of the derivatives

task: determine f´(0), f´(2) and f´(3)

f´(0) means the x value = 0

and there should be no slope there

but the solution says 2

[sage geode]

I'd try deriving the equation of the graph by assuming that it's a parabola

[sage geode]

Why?

You can still draw the tangent line to the graph at x = 0

[frank crag]

f´(number) = slope ?

or did I misunderstand that part

[sage geode]

Yes f'(x) is the slope of the tangent line to the graph at the point x = 0

[frank crag]

OOOOOOOOHh

so if I drew the tangent line from x=0

it would reach y=2, x=1

2/1 = 2

[sage geode]

Yeah but just for the safety I'd not do such task only visually

[frank crag]

yeah

makes sense 😄

thanks!

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[valid lodge]

Hi can someone tell me where my teacher gets - √ 2 from?

[valid lodge]

I understand how x < √ 2 is found

but not -√ 2 < x

[distant moth]

Well square rooting both sides, would give you |x| less than equals sqrt(2)

That gives x belongs to [- sqrt(2), sqrt(2)]

[valid lodge]

I dont see where the - sqrt(2) is coming from

[distant moth]

do you know the graph of abs(x)

draw that, and then draw y= sqrt 2

then see for which x, the inequality is satisfied

[valid lodge]

i understand more clearly now but in which cases does this apply?

im guessing that its with symmetrical graphs, but when can I know it is symmetrical and isnt then

[unreal musk]

What may make it become more apparent, would be to graph 2 - x^2, find where it intercepts the x-axis, and consider where the graph is >=0

[distant moth]

Drawing graphs to check for inequalities? Always as long as you can draw it

[distant moth]

It doesnt have to be symmetrical

For inequalities tho, for stuff like this, drawing graphs would take too much time, google wave curve algorithm, there should be videos of that on yt

[valid lodge]

no i mean in which cases are both the zeros of the function the same like -x and x

[distant moth]

What

[valid lodge]

here -sqrt(2) and sqrt(2)

[distant moth]

sqrt(2) and - sqrt(2) are not the same

But i think what youre asking about refers to even functions

[valid lodge]

just negative and positive

alright thx for the help

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[sour granite]

Im doing numerical integration with simpson's rule, and the order of the error should be 4 (the slope of ln|error| against N, the number of intervals). However, for some functions, the slope is smaller (around 3.4), could anyone help explain why?

[sour granite]

All i can see is that the function is relatively flat in that interval

Here's the plot when integrating one function

And this is for the "flatter" function

[kind hawk]

intuitively I would say that for flatter functions getting more intervals doesn't change the error as much because you can't approximate the function that much better. extreme cases would be constant functions I guess

also maybe you are just generally running into accuracy issues?

[sour granite]

Hm ok yeah sounds fair

Thanks!

[obtuse pebble]

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[untold yarrow]

how to calculate the force to compress an object to a given dimension<

[sharp pecan]

thats physics innit?

what type of material do you have

[untold yarrow]

I want to do it for any material

[sharp pecan]

the formula depends on the type of material, temperature and state of matter

[untold yarrow]

ok can zou show me the formula<

?

[sharp pecan]

shouldn't you have one from where ever the task is from

I don't think its that simple to have a universal one

for gasses you can use compression-density formula

[untold yarrow]

ok let's say i have a wooden cube with a density of 70 cm3

[sharp pecan]

okay thats insane

basically you need to look up what pressure what material needs to compress by 10 or 100 percent, and wood has a ton of different ones

whats your actual task?

[sharp pecan]

even worse, wood has a crushing strenght limit, meaning at some point it will splinter

whats the actual task you have to doi

[untold yarrow]

i'm just asking im been interested with pressure machines

[sharp pecan]

why are you asking in a math discord though

[alpine raven]

you apply a pressure on the material which formula is the force divided by the surface where the force is applied

[obtuse pebble]

@untold yarrow Has your question been resolved?

[obtuse pebble]

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[silver plover]

say u have f(x)

[silver plover]

if u find f(kx)

will it have same minimum value as before

my lect says it does

but it doesnt look like it

cuz clearly the x coordinate will be divided by k

so surely the minimum value cant be the same

[cold thistle]

I think it depends on the actual function

[silver plover]

but is it always the same

it isnt right?

[cold thistle]

The function f(x) = ax^2 preserves its critical value for all a in reals

But ax^2 + ax for example doesn't

I think critical points would have the same x-value but not necessarily the same y

[obtuse pebble]

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[fossil garden]

i need help on how to do the long division version of it (somehow my teacher didn’t teach us how so 😭) also may i ask for an example on paper 😭

translation : “divide the following polynomials”

[teal turret]

¿Así necesitas hacer P(x)/Q(x)?

Un ejemplo^

[fossil garden]

Si
idrk spanish much but i understand

[teal turret]

Isn’t ur problem in Spanish?

[fossil garden]

yea but like my parents put me into a spanish school (cause i moved to spain) and i dont understand anything thats why i need help ;-; long story it sucks

[teal turret]

Oh dang rip, Aight

So first set it up like regular division problem

[fossil garden]

but yes like that i have a problem doing long division with it cause it has 3 terms i think its called- like its “2x2^ + x + 4” instead of “2x + 3”

[fossil garden]

okay okay

[fossil garden]

how do we do that ;-;

[teal turret]

So if we had something like 56x/4, how would we set it up in a long division problem

[teal turret]

Or just refer to this as an example

[nocturne minnow]

https://www.youtube.com/watch?v=_FSXJmESFmQ&ab_channel=TheOrganicChemistryTutor

Or that

[fossil garden]

okay thank you so much 😭

.close

[obtuse pebble]

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[timid silo]

Can anyone explain Euler's Identity?

[rigid lintel]

i dont think just discord text can do eulers identity justice

theres great videos on youtube

that explain it better than anyone here

[trail cloak]

Try out 3B1B or Blackpenredpen

[wooden cipher]

It also may help if youre familiar with taylor series

[obtuse pebble]

@timid silo Has your question been resolved?

[hollow harness]

can you help me ?

[trail cloak]

You can get your own help channel and there we can help you :)

You can pick #help-1

[obtuse pebble]

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[narrow fiber]

Is there another way to solve this without graphing?
"Find the equation of the circle that touches the two axes and touches the line y=6"

[loud tangle]

graphing would help but you can consider the general form of the circle equation

if it touches the line at y = 6 that means it has a radius of 3

[narrow fiber]

But how did that go

What do I write in an exam

[loud tangle]

idrk proof wise since some countries have it different

[cold thistle]

I assume by "touches" it means touches exactly once without intercepting, otherwise you could just make a circle about the origin with radius >= 6

[loud tangle]

btw it said it touches both axies too

[narrow fiber]

Yeah that's the inteded meaning

[loud tangle]

meaning its not at the origin

you can have 2 cases

one where its origin is in the upper right quadrant and one where its in the upper left one

at exactly (3, 3) and (-3, 3)

since you said it is touching both axis

[narrow fiber]

I feel like this relies on graphing

[loud tangle]

visualisation is basically your way of intuition but you can prove it without a graoh

graph

[narrow fiber]

You imagined the line y=6 and you got
2r = 6 => r = 3
And then the rest is simple

that's the part that I want to know if there's another way to do it

[loud tangle]

you can deduce from.the fact tbat its touching the line ar y = 6 that the function's maximum is 6

and since its tangent minimum is 0

then work with that

and since its a circle equation you know the distance between the maximum and minimum is 2r so you get r = 3

then as u said ez

[narrow fiber]

This captured my intuition well

Btw

How would we do it if it gave any line instead of y=6 ?

[loud tangle]

wdym any line

[narrow fiber]

How do we solve it in the general case

[loud tangle]

you deduce its maximum value is the y value of the line

same goes for minimum

since y value of OX = 0

[narrow fiber]

You can't use that if it gave the line x-y=2

Because then which y value do you choose?

[loud tangle]

oh

well write it in y = mx + b form

[narrow fiber]

y=1x-2

[loud tangle]

i dont think you have a choice but to draw it out to help visualise there

intuitiokn has to come from graph

[narrow fiber]

So in conclusion:

I'll write "just look at it bro" on my paper

[loud tangle]

write it somewhere else

then do the intuition once yk how to do it with the drawinf

[narrow fiber]

Yeah I know i'm just kidding

I'll write the question

Then do the drawing with 2 circles and the line

And then first line of text is 2r = 6

[loud tangle]

look man i kinda gtg

[narrow fiber]

.close

[obtuse pebble]

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[obtuse pebble]

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[spice chasm]

is this right

[forest sinew]

you can use desmos, iyw

it looks alright, tho

[lethal egret]

facts

[obtuse pebble]

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[rapid relic]

so I was wondering do 2 mutually exclusive events always have to equal 1

[rapid relic]

I had a question here

and I know 1-0.43 =0.57 is one of the solutions

right

but i'm not sure why that's the case

it's x+0.57

[civic zealot]

no

[rapid relic]

can someone please explain that to me?

How does it work then?

[civic zealot]

they can sum to 1, but they do not have to sum to 1

[rapid relic]

oh ok

But then why in this case do I do 1-0.43

I thought P(A)+P(A') =1

and I use that formula

Why do I use b here?

[civic zealot]

that's true, because A U A' is the whole set, which has probability 1

But just because A and B are mutually exclusive does not mean that B = A'

[rapid relic]

then what are we suppose to do over here?

A little confused on how to solve it

[civic zealot]

since P(B') = .43, the P(B) = .57 by the property you gave

so P(A U B) = P(A) + P(B)

because they are mutually exclusive.

[rapid relic]

Ohhhh ok

ty that actually makes sense to me now

That's a lot simpler than I thought

tyty I appreciate the help

.close

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[storm temple]

are either of my solutions correct? 😭

[fierce lagoon]

,calc Sin(17 Pi/12)

[warm shale]

The following error occured while calculating:
Error: Undefined function Sin

[fierce lagoon]

Bruh

,w Sin[17 Pi/12]

[warm shale]

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[storm temple]

Oop

[fierce lagoon]

Bottom one

You're fine

[storm temple]

My bottom solution is correct?

[fierce lagoon]

Ye

[storm temple]

Yayyyyyy thanks lol

[fierce lagoon]

You have the mathematical equivalent

[storm temple]

.close

[obtuse pebble]

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[timid silo]

first picture is the question, second picture is my work, third picture is the answer key

I know my work is hard to read but basically what I did was set u=e^x whilst in the answer key its u=1+e^x

why this and not that?

(u-substitution)

[warm shale]

Stephen

[teal turret]

Those two aren’t equal if the != wasn’t clear

[timid silo]

wait so I just made an algebra mistake?

[teal turret]

Yea, from that step yes, I didn’t look into the rest of ur work

[timid silo]

oh fuck I'm an idiot I split it up the denominator as if it was the numerator lmaoo

man fuck finals week

well, thanks

let me see if this fixes the problem

[teal turret]

Np, I think u coulda done that u = e^x sub

[timid silo]

that's what I was worried about

good then

I missed class when we went over that 😭

[teal turret]

Rip

Yea u coulda used u = e^x but u would have to use another sub after that

Using u = e^x + 1, u skip that step

[timid silo]

do you have any advice for how to pick the most efficient u

[teal turret]

Hmm

Good question

Honestly it just comes with experience, lots of practice with integration techniques and such. I’ve just searched on google and saw a video (I haven’t watched) but I’ll link because it looks promising. Additionally, I’ve also seen something that says u-sub is just chain rule in reverse, so if you’re able to think about it like that, that might be helpful. But overall I’d just recommend practice with different types of problems, and when practicing, try different u-sub’s in the same problem to see if they work well, u may notice some trends which may help u to determine the most efficient paths. Here’s the link: https://m.youtube.com/watch?v=evdYUQnHrBo

[timid silo]

damn thanks

I'll definitely be practicing tonight

[teal turret]

Good luck

[obtuse pebble]

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[ancient stag]

can someone explain to me why it's important to take care of the fact that the cos function will oscillate between positive and negative?

[ancient stag]

so my lecturer has decided to put a modulus around the expression

and we're considering the limit as it approaches 0

and im confused because if the h term is tending towards 0 then the whole function would tend to 0?

so why would it matter if it oscillates between +/- as it gets close to 0

[obtuse pebble]

@ancient stag Has your question been resolved?

[ancient stag]

<@&286206848099549185>

[nimble dew]

hes just being super careful about the proof using the squeeze theorem

[ancient stag]

.close

[obtuse pebble]

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[distant rampart]

The complex number 𝑢 is given by the following equality: 𝑢 = 16𝑖. Solve the equation 𝑧⁴ = 𝑢, and represent its solutions in a complex number plane.

[distant rampart]

i've got r, which is 16

but i dont know how to get 𝛉

and i dont know how to get the - angles inside cos and sin

[tardy epoch]

Imagine instead of 16i, you're just looking for z^4=i

Do you know how to write i as a complex exponential?

[distant rampart]

The complex number $i$ is defined as the square root of $-1$. It can be written as a complex exponential using Euler's formula, which states that for any real number $x$:

$$e^{ix} = \cos(x) + i \sin(x)$$

Therefore, we can write $i$ as a complex exponential by setting $x = \frac{\pi}{2}$:

$$i = e^{i \frac{\pi}{2}} = \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) = 0 + i 1 = i$$

So $i$ can be written as a complex exponential as $e^{i \frac{\pi}{2}}$. Note that this representation of $i$ as a complex exponential is only one possible representation; there are many other ways to write $i$ using complex exponentials.

[warm shale]

dgh

[distant rampart]

@tardy epoch This is what ChatGPT gave me.

Is it right?

In the complex plane, the complex number $i$ is represented by a point on the vertical axis, above the origin. This means that $i$ is pointing directly upwards on the complex plane. To find the angle of the complex number $u = 16i$, we can use the fact that $i$ is pointing upwards on the complex plane. The angle of $u$ is then simply the angle that $u$ makes with the positive real axis, which in this case is $\frac{\pi}{2}$ radians or $90^\circ$.

[warm shale]

dgh

[distant rampart]

How do we get the - angles, or why do we write it like that?

[daring rock]

Lol don't ask an AI to do your homework

[tardy epoch]

.

[narrow vault]

from what i've seen chatgpt is pretty good at homework as long as the user has enough awareness to extract the useful and correct parts of it

[tardy epoch]

If you understand it, then go for it. I'm not reading all that

[daring rock]

I've asked it quite a few questions about math. Sometimes it's surprisingly accurate, and sometimes it's complete nonsense

[narrow vault]

ye it's hit and miss and you have to figure out when it's missing lol

[distant rampart]

I did, but this one is more simpler #help-10 message

[distant rampart]

Can someone explain this?

[tardy epoch]

I'm still not reading all that nonsense

[distant rampart]

Yeah I understood that the first time

[daring rock]

Do you recognize that 9pi/8 for example is coterminal to -7pi/8

in the case of z_3 there

[distant rampart]

Hmm, I know that in the polar form of a complex number, the angle 𝛉 can be any value between -π and π.

But I've lost memory of the reason why we prefer the negative angle.

Is it because we subtract 2π?

[daring rock]

Yes, if you add or subtract 2pi, you get a coterminal angle

[distant rampart]

I see!

[daring rock]

As for why we prefer angles between -pi and pi, I think it's mostly convention, It plays nicely with the arctan function

[distant rampart]

So if we have z^4, we want two plus angles, and two minus angles?

i.e. distributing it evenly?

[daring rock]

That will generally be true for even exponents, yes

but just make sure they're on (-pi, pi]

[daring rock]

you don't really need to memorize a rule like this

[distant rampart]

Is this true?

"if we used an angle outside of the range -π to π, the arctan function might not return a value that is consistent with the angle of the complex number in the complex plane. For example, if we used the angle 5π/4 to represent the complex number -1 + i, the arctan function would return a different value:

angle = atan(imaginary part / real part) = atan(1 / -1) = 3π/4

This result is not consistent with the angle of the complex number in the complex plane, because it is not between -π and π."

[daring rock]

It should say -5pi/4, not 5pi/4

-5pi/4 and 3pi/4 are coterminal but -5pi/4 is not in the interval (-pi,pi]

[distant rampart]

I see, I have a exam in 3 hours, so I'm cramming lol

So for coterminal angles we only consider the angles in the range (-π, π]

[daring rock]

for complex numbers, you should typically only express them by angles in that range

[distant rampart]

Ok ok

[daring rock]

but yeah I think that's basically what you meant

[distant rampart]

Another thing, on the notation ( and ]

The notation (-π, π) is often used to indicate the range of values that are not included in the range of coterminal angles, because coterminal angles are defined as angles that differ by a multiple of 2π and that are within the range (-π, π]. In other words, coterminal angles are not in the range (-π, π).

This is wrong right? Because π is a coterminal of -π

Oh nvm, it's not lmao

[daring rock]

no you're right, pi and -pi are coterminal

[distant rampart]

Or is it 🤔

Yeah lol

I have not slept all night

[daring rock]

that's exactly why we only include one of them

positive pi, by convention

[distant rampart]

But why did it say "In other words, coterminal angles are not in the range (-π, π)."

[daring rock]

was this from the ai?

[distant rampart]

Yea

[daring rock]

that's why

[distant rampart]

Okay lol

[daring rock]

it is not really a reliable source

[mighty geyser]

i think it means something

if x, y are coterminal and in the range (-pi, pi), then x=y

i.e. x cannot be different from y

although it's not very clearly stated

[daring rock]

isn't this also true for (-pi, pi] ?

[mighty geyser]

yes, it is also true for that

[distant rampart]

btw, how do i best memorize the trig values up to π/2?

for cos, sin, and tan

i don't have the best memory of that

i remember there was a triangle you could look at

[daring rock]

I think this is probably better than the special triangles

@distant rampart

[mighty geyser]

draw equilateral triangle and square

the special triangles are hiding inside

[daring rock]

Btw, if you needed a reason not to use chat GPT for math help, here's one

[lost tree]

x^4 = x?

[mighty geyser]

yeah - this is why we don't typically let chatgpt posts here

still more efficient to type out ourselves

and ,w for definitions

,w involution

[warm shale]

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[mighty geyser]

,w involution function

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Element118#2437

[mighty geyser]

well that didn't work out

[lost tree]

lol

😝

[daring rock]

I've never seen someone with both the "active" role and the newbie leaf

[lost tree]

lol

well idk lol i keep finding myself returning to discord since i joined this community

[obtuse pebble]

@distant rampart Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[lone viper]

for sine how was it 7/5/2

[lone viper]

and for cosine how was it 3/5/2

[frosty tiger]

they added fractions

1 + (2/5) = 7/5

[lone viper]

oh so it would be 5/5+2/5

right

????

.close

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[timid silo]

Trying to solve this initial value problem:
[
x^2y'' -xy' +2y = 0 \quad y(1) = 5 \quad y'(1) = -1
]

[warm shale]

♡LexQa♡

[wild swallow]

you closed your other channel ig

[timid silo]

i get
[
y = x(5\cos(\ln(x)) - 6\sin(\ln(x))
]
but i am not sure \thonk

[warm shale]

♡LexQa♡

[timid silo]

oh someone actually came and explained it

[wild swallow]

was it before or after approximately 8 hours ago

[timid silo]

it was pretty interesting

[wild swallow]

because i also saw some stuff there

[timid silo]

yes, he posted it immediately after i slept

[wild swallow]

this is one of those DEs which have a name

which im forgetting

but iirc you can guess a power of x

[timid silo]

yeah it was with euler's equations

[wild swallow]

yes

and your solution isn't powers of x which makes me concerned

[timid silo]

i am not good with odes and pdes okay 😵‍💫

[zenith raft]

you cn always check it

[wild swallow]

,w r(r-1) - r + 2 = 0

[warm shale]

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Requested by 夢雪#2250

[wild swallow]

maybe you're right then

the power turned into trig

[timid silo]

i actually inputted that into the system, and it turned to be wrong, and i am just clueless now

[zenith raft]

[wild swallow]

,w A + B = 5, A(1+i) + B(1-i) = -1

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[wild swallow]

[next reef]

[timid silo]

oh wait

i made it work, the system just wanted an extra parenthesis

oopsie

[zenith raft]

lmaooo

it wanted an extra parenthesis or the correct amount of parentheses? 😁

[wild swallow]

[timid silo]

okay...just because you can see through me doesn't mean you can get to expose me like that 😔

[zenith raft]

sowwy 😭

[wild swallow]

[timid silo]

oh yeah snow, i actually have another GF question

it is more of...a question about someone's else explanation

what the hell does this sign mean

[zenith raft]

msg

[wild swallow]

[zenith raft]

i hear that’s bad

[wild swallow]

monosodium glutamate

[zenith raft]

monosodiumsodium glutamate

[wild swallow]

[timid silo]

what

in the hell

[wild swallow]

whats the context

people use symbols for literally everything

[timid silo]

omg im having the biggest brain fart i forgot how to latex

just basically saying that
[ma = \delta \to m\dv[2]{x}{t}]
[\to m\dv[2]{G}{t} = \delta ]

and then

[warm shale]

♡LexQa♡

[zenith raft]

[wild swallow]

looks like physics

[timid silo]

then we defined
[ G = {\color{cyan} \int}\mathscr{G} ={\color{cyan} \int} \frac{1}{m s^2}]

[warm shale]

♡LexQa♡

[timid silo]

G being green's function of course

but

[wild swallow]

metric?

[timid silo]

i am pretty sure this is called...a Bromwich integral?

[timid silo]

yep

wait wdym by that

[wild swallow]

inverse laplace transform

[timid silo]

yep

[wild swallow]

that weird thing

[timid silo]

curious, but where would this kind of stuff go to in this server? kind of want to dig deeper into it

#odes-and-pdes i guess?

[wild swallow]

likely

[timid silo]

okay, im just going to forget about this for now 😵‍💫

thanks snow and layla

.close

[obtuse pebble]

Channel closed

Closed by @native knoll

Use .reopen if this was a mistake.

[wild swallow]

[zenith raft]

i did nothing again 🥰

[timid silo]

its okay ur presence is helpful

[obtuse pebble]

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[zenith raft]

i’m just here to look pretty

[wild swallow]

[timid silo]

goals ✨

[zenith raft]

.close

[obtuse pebble]

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[wild swallow]

lex quick

[timid silo]

wait wot

[wild swallow]

you couldve done .reopen

and yeeted the bot message

[timid silo]

awww

[wild swallow]

or maybe that doesnt work after the channel's been claimed

[timid silo]

i already did it twice and thats what matters 😎

[wild swallow]

[timid silo]

the funniest thing is to send a message and the bot immediately transferring that channel to the hidden section

so the bot has a brain freeze for like 5 seconds and then opens it again lmao

[wild swallow]

[obtuse pebble]

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[frozen quest]

Find the variance of 15,16,17....45?

[frozen quest]

By using consecutive formula that is (n^2 - 1)/12
I can't get the answer

Is there something i am missing

[royal basin]

can we see the problem exactly as stated

[frozen quest]

Wait

[frozen quest]

49th question

[royal basin]

and what value are you using for n?

[frozen quest]

I am getting first 14 number variance and then first 45 and then subtracting them

[royal basin]

incorrect

variance does not work that way unfortunately

[frozen quest]

Oh

[royal basin]

,calc ((45-15+1)^2 - 1)/12

[warm shale]

Result:

80

[frozen quest]

Last term - first term

Always?

Or is it number of terms

[brave bramble]

Number of terms, yeah

Useful formula! I didn't know that one.

[frozen quest]

And what if there is consecutive gap of 2

How do we get the variance then

[royal basin]

we suffer

the general formula for variance is mean square minus square mean, to put it in a way that isn't symbol soup

[frozen quest]

xD

Thanks for helping

.close

[obtuse pebble]

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[obtuse pebble]

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[mighty crown]

My work

[obtuse pebble]

@mighty crown Has your question been resolved?

[winged aspen]

13+13 +3x = 41
3x = 15

Sorry that it’s shit, trying do it on the phone

[winged aspen]

Your Japanese only should be the people studying both French and Greek. So 2x + x = Japanese only

[mighty crown]

ohhhhhh

that makes sense now thanks Xiszt!

[obtuse pebble]

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[lost radish]

Hello

[lost radish]

Please part 2

Part ii

[timid silo]

,rotate

[lost radish]

Ok?@timid silo

What's else

???

[obtuse pebble]

@lost radish Has your question been resolved?

[acoustic elbow]

Use your answer in part 1.
Let's keep those matrices as A and B. What other (non 0) matrix can you multiply them both by to get the 0 matrix?

[acoustic elbow]

Hint: ||try another matrix that has 0s as 3 entries and a 1||

[lost radish]

.close

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[past solar]

what's the largest possible k where 12^k divides 2022!? need help, can't seem to find a way to use legendre or anything.

[sage geode]

Prime factorisation of 2022 perhaps?

Oh, there's no need though

Cuz 2022 isn't divisible by 4

How'd it be divisible by 12 anyways?

[rigid lintel]

2022!

[sage geode]

Oh

Mb

[wild swallow]

its just legendre

[past solar]

how do you apply cause legendre only works on prime?

[wild swallow]

realise that there will be more factors of 2 in 2022! than 3

[wild swallow]

wait that might not be needed

[past solar]

hmmm.

so find when 2 is used then after that, use 4?

[wild swallow]

well essentially like

if $2022! = 2^a 3^b k$

[wild swallow]

then you can work out like

[wild swallow]

the first inequality might not be true tho

[wild swallow]

dont want to use my brain to figure it out rn

[past solar]

ohhh.

wait how do you find a and b in 2^a3^bk?

[wild swallow]

legendre

[past solar]

so a is 2014.

[wild swallow]

,w nu_2(2022!)

[warm shale]

Results provided by WolframAlpha

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Requested by 夢雪#2250

[wild swallow]

what

[past solar]

and b is 1006.

[wild swallow]

,w sum k=1 to 20 floor(2022/2^k)

[warm shale]

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[wild swallow]

,w sum k=1 to 20 floor(2022/3^k)

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[wild swallow]

guess so

[wild swallow]

[past solar]

the one with the factor with lowest exponent divided 2022! ?

[past solar]

ohhhh, how do you know it don't divide?

[past solar]

OHHHH, thank you so much!

.close

[obtuse pebble]

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[timid silo]

Can i do this?

[timid silo]

Thank u in advance

[rigid lintel]

yes that works

[sage geode]

a = log(e^a) in general

[rigid lintel]

,w sqrt(3) = log(e^sqrt(3))

[warm shale]

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[sage geode]

So why not sqrt(3) = log(e^sqrt(3))?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[sturdy ibex]

Is this graph correct?

[sage geode]

Yes

[sturdy ibex]

alright, thanks

.close

[obtuse pebble]

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[halcyon badger]

.close

[sturdy ibex]

bruh

[obtuse pebble]

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[shell prairie]

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[nocturne minnow]

Don't use the ping multiple times, only once

And just wait patiently

[shell prairie]

a

[obtuse pebble]

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[obtuse pebble]

@shell prairie Has your question been resolved?

[obtuse pebble]

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[white summit]

Hello everybody. I am currently stuck on this integration problem. Namely I am given two points Pinflection(2,2) and Porigin(0,0) and a tangent line of a cubic function at x=2.
I have tried to integrate my function y=-3x+8 (as it should be my second derivative of my main function) and then plug in my two points twice to find the parameters c and d that I receive from integrating. However, I do not get the same result as the solutions.

[white summit]

Here are my calculations so far

<@&286206848099549185>

[obtuse pebble]

@white summit Has your question been resolved?

[white summit]

uhh I hope it's fine to ping again. <@&286206848099549185>

[obtuse pebble]

@white summit Has your question been resolved?

[white summit]

I have done the thinking

.close

[obtuse pebble]

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[woven quartz]

help me with this, please. this was given in an exam a while back, and i got zero on this question, i got zero as the result but our teacher says there's supposed to be two answers

[tardy epoch]

,rotate

[distant moth]

ur missing absolute value

then break accordingly

[woven quartz]

shouldn't that give a zero anyway?

[tardy epoch]

no

integral of a positive function is greater than zero

[rigid lintel]

you cant really integrate implicit functions like youre used to with explicit functions

[tardy epoch]

(over a domain of nonzero length)