help-17

[half jetty]

oh okay

sorry for the late response i was working on smth else

okay hold on

I just replace theta with 2x and 2alpha respectively yeah?

Kinda stuck here

[vocal sleet]

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[carmine edge]

Any idea how i would solve a lights out puzzle like this?
Black = off, red = on, I need to turn it all on
If theres a lights out calcuator that doesnt require a perfect square of tiles, that would be really helpful

[dark kiln]

the only thing i know about this is it's not always solvable

[carmine edge]

people have solved this puzzle before

[dark kiln]

oh good

[carmine edge]

but its in a different state cause i was messing around with it for a bit

[dark kiln]

i can try making a calculator

[carmine edge]

its on minecraft 💀 so people have solved it before fs

[dark kiln]

i give up though, you can't brute force it

[carmine edge]

i mean you can ig you just gotta get really lucky

[hybrid flicker]

code is done

very brute force

[hybrid flicker]

I'm gonna need the exact look of the grid

to try the code on

[carmine edge]

^

oh

[carmine edge]

i just got reset and theres a whole puzzle to get to the lights out thing, but i believe this is what it still is

it shouldnt have changed

ill be back at that area to check in a bit

[hybrid flicker]

it looks like some tiles like this one are not usable

[carmine edge]

yeah

its not a perfect square

[hybrid flicker]

not but like there's no light coming from them

[carmine edge]

wdym

this image is the exact layout

theres no other missing tiles

[hybrid flicker]

no but look at the mc screenshot, it looks like there are some grey tiles that are neither turned on nor turned off

[carmine edge]

anything thats in the white area from this screenshot is just asthetic stuff

[hybrid flicker]

if I look at this grid it would correspond to the tile on 2nd row 3rd column (counting all columns)

[hybrid flicker]

look at the screenshot, start from the tile that's lit up at the bottom

and look at the tile on its right

[carmine edge]

that screenshot is from a different angle

like the screenshot is tilted 90 degrees from the diagram or something

[hybrid flicker]

but it doesn't really matter, that's one of the tiles that you put in black anyway

when there's no light switch in them

[carmine edge]

OH

youre right

my bad

[hybrid flicker]

anyways I'll base myself off of this

code is running

should arguably take a long time since it's bad

[vocal sleet]

@carmine edge Has your question been resolved?

[vocal sleet]

@carmine edge Has your question been resolved?

[carmine edge]

should i say yeah or what

to the resolved ting

i assume not

[dark kiln]

wait i didn't even check

[carmine edge]

what does that mean

[dark kiln]

nvm

@carmine edge yeah it works

it puts the lights out

[carmine edge]

this?

[dark kiln]

yes

[carmine edge]

1 is the tiles you would activate right

[dark kiln]

yeah

[carmine edge]

oh wait

[dark kiln]

maybe the perimeter stuff is wrong

[carmine edge]

this grid was wrong

[dark kiln]

oh ok

[carmine edge]

there was 2 spaces missing

its alr i cant be bothered to try all that anymore anyway 😭

thanks

.close

[vocal sleet]

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[spiral turtle]

@untold onyx how well do you know your prime factorizations?

I would solve it by reducing everything into prime factors and combining them

Well, not really solve, but simplify and expression

[vocal sleet]

@untold onyx Has your question been resolved?

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[fair ibex]

Im having trouble with this problem. due to the helium shortage helium prices are prediccted to rise as high as $84.00 per thousand cubic feet in the neaer future. How much would it cost to fill a party balloon whose volume is 4.00 liters

[glacial osprey]

how many cubic feet are in 1 liter

[fair ibex]

28.3168

oops wait

0.0353147 cubic feet

[fair ibex]

my book says the answer is $0.00297

i cant get it to make sense tho

[vocal sleet]

@fair ibex Has your question been resolved?

[fair ibex]

<@&286206848099549185> sorry for the ping, i just need help quickly

[fair ibex]

^

[gaunt willow]

this was my thought process

[fair ibex]

yeah kinda what i was thinking too, but the book says the answer is $0.00297

[gaunt willow]

hm

well thats 1/4 my answer

so maybe they meant only 1 liter?

[fair ibex]

maybe but i cant believe a textbook like this would make an error like this

[spiral turtle]

,w $84 / (1000 ft^3) in ( $/(liter))

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by omnipotententity

[spiral turtle]

Yeah, I'm getting the same result as you guys

@fair ibex it's certainly a textbook error. They happen.

[vocal sleet]

@fair ibex Has your question been resolved?

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[fair ibex]

Okay, thank you guys

:)

[vocal sleet]

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[next burrow]

Does this look correct

[next burrow]

the tree diagram specifically

[vocal sleet]

@next burrow Has your question been resolved?

[fallen hound]

Idk what a tree diagram is but your answer looks correct

[vocal sleet]

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[sly sierra]

!status

[vocal sleet]

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

[vast shale]

1

[sly sierra]

do you think it's true or not?

[vast shale]

I "feel" it is not true

[sly sierra]

your feeling is correct

there is a standard counterexample

it's not super likely that you would discover it on your own

you can look for "non-analytic smooth function" to find it

[desert hornet]

$$f(x)=\begin{cases}
0 &: x\le 0\
e^{-\frac1{x^2}} &: x>0\
\end{cases}$$

[twin meteor]

kheerii

[desert hornet]

I believe it’s this

[sly sierra]

yep

-1/x exponent also works

[desert hornet]

All its derivatives at 0 are equal to 0, however it isn’t identically 0

[desert hornet]

Are you sure?

Ah yeah i guess it does

[vast shale]

How do you find derivative of this function

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[vast shale]

!status

[vocal sleet]

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

[vast shale]

1

Anyone ?

<@&286206848099549185>

[hybrid flicker]

Do you think it's true?

[desert hornet]

Let $0\le f_{\text{min}}\le f_{\text{max}}\le 1$

[twin meteor]

kheerii

[desert hornet]

A justification like this should work

[hybrid flicker]

Though if ||f = 1/2...||

[vast shale]

I have only one example in mind and it is true for that function

[hybrid flicker]

Maybe look for more

[vast shale]

f(x) = 1

[desert hornet]

Yeah i meant that working like this can give us examples where it doesn’t work

Does $f^2(t)=f(f(t))$ or $(f(t))^2$?

[twin meteor]

kheerii

[vast shale]

Second one

[boreal remnant]

does [0,1] -> [0,1] mean EVERY value in the codomain is acheived?

[vast shale]

No

[hybrid flicker]

Definitely not, this is the regular notation for any function, and "any" function is not always onto

[boreal remnant]

ok then we have:
f >= f(t)^2

wait

isn't this just IVT

[vast shale]

If we consider f(x)= 1/2 we are done right ?

This statement does not hold for this function

[boreal remnant]

there exists a such that:
f^2(a) = int_0^1 f^2(t) dt
f^2(a) <= f(a)
if f(a) is achieved, and f is continuous and monotonically increasing, then f^2(a) must also be achieved

emphasis on the monotonically increasing part

i think this holds for monotonic functions

[vast shale]

It is a True/False question

[boreal remnant]

yeah no i agree the statement is false

but just saying that you need a stronger condition for it to be true

[vast shale]

.close

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

@vast shale Has your question been resolved?

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[elder scaffold]

$T_n = \int_0^\pi sin^n(x) dx, show that T_n= \frac{n-1}{n}T_{n-2}$

[twin meteor]

Galaxy

[elder scaffold]

I forgot how recurrence relationships work

I know that im after an integration by parts and that I should pull a sin^2 from sin^n

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@elder scaffold Has your question been resolved?

[vocal sleet]

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[paper swallow]

Can someone help me solve this simultanerous equations

[vocal star]

solve for one of the variables for one equation and plug it into the other

[lethal plank]

you had the right idea. After plugging in y, you just get a normal quadratic equation for x

solve that using normal methods for roots of quadratic equations

either factoring if it's obvious or the quadratic equation if it's not

then plug in the x solutions into the equation for y in terms of x

[vocal sleet]

@paper swallow Has your question been resolved?

[paper swallow]

I cant solve for answer

[lethal plank]

in what sense

you can just plug it into the quadratic formula after expanding every term

and grouping the terms for each power of x

this is just computation

@paper swallow

[paper swallow]

Ok

Tq

[vocal sleet]

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[toxic oracle]

Hey folks, I'm looking for an algorithm that eventually gives me all combinations of length k from the elements of a set but maintains an approximate equal distribution over the elements and "subcombinations" in in the combinations. I'm not really sure how to make this more specific, but as an example:

Let S = {1, 2, 3, 4, 5, 6, 7, 8}, k=3
The algorithm generates combinations c_i:

c_0 := {1, 2, 3} // Or any other set of 3
c_1 := {4, 5, 6} // Any 3 elements except {1, 2, 3}
c_2 := {7, 8, 1} // 7 and 8 forced, 1 is arbitrary
c_3 := {2, 4, 7} // Not 1, since it already appeared twice. Not 3 since it was in a set together with 2. Not 5 or 6 since they were in a set with 4.
Other options for c_3 are {2, 4, 8}, {2, 5, 7}, {2, 5, 8}, {3, 4, 7}, {3, 4, 8}, {3, 5, 7} and {3, 5, 8}

Is there a well-known algorithm that will give me "equally distributed" combinations like that?

[toxic oracle]

Eh, too specific

.close

[vocal sleet]

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[paper swallow]

Do you get the x=3 straight from this step or you need to continue working to get this answer

[vocal sleet]

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[hollow sleet]

Hello, In a Cartesian coordinate system, I have a circle with radius r and center (0,0). I want to move any point with coordinates (x,y) which belongs to the circle with a displacement vector of norm u. How can I express u_x and u_y in terms of r, u, x and y? Thanks !

[hollow sleet]

Maybe it's more clear with that

I think there is the mains equations

[vocal sleet]

@hollow sleet Has your question been resolved?

[hollow sleet]

<@&286206848099549185>

[dire lake]

youre correct now subtitute in eqn 1 to get both values of y

[solar barn]

maybe rotation matrices can help here. The idea is to take $(x,y)$ and rotate it counterclockwise by the angle $\theta$, which can be decided by $r$ and $u$, then $(x,y)$ gets translated to $(x+u_x,y+u_y)$, so the whole thing would be
$$\begin{pmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos \theta\end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} x+u_x \ y+u_y \end{pmatrix} $$
then solve for $\begin{pmatrix} u_x \ u_y \end{pmatrix}$ as
$$\begin{pmatrix} u_x \ u_y \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} - \begin{pmatrix} x \ y \end{pmatrix}$$

[twin meteor]

Crystopher

[solar barn]

the only problem I see with the approach is that deciding $\theta$ can be ambiguous sometimes, or rather $\sin\theta$

[twin meteor]

Crystopher

[hollow sleet]

yes that's the main problem I have here, I don't know how to define theta

[solar barn]

you can have $\cos\theta$ using cosine's law
$$r^2+r^2-2r\cdot r\cdot\cos\theta = u^2 $$
$$2r^2-2r^2\cos\theta = u^2$$
$$1-\cos\theta = \frac{u^2}{2r^2}$$
$$\cos\theta = \frac{2r^2-u^2}{2r^2}$$

[twin meteor]

Crystopher

[hollow sleet]

I see yes, and by solving the system I will get ux et uy

[solar barn]

the problem here is $\sin\theta$,
$$\cos^2\theta + \sin^2\theta = 1$$
$$\cos^2\theta = \frac{4r^4-4r^2u^2+u^4}{4r^4}$$
$$\sin^2\theta =1-\cos^2\theta$$
$$\sin\theta = \pm\sqrt{1-\cos^2\theta}$$
$$\sin\theta = \pm\sqrt{1-\frac{4r^4-4r^2u^2+u^4}{4r^4}}$$
$$\sin\theta = \pm\sqrt{\frac{4r^2u^2-u^4}{4r^4}}$$
$$\sin\theta = \pm \frac{u}{2r^2} \sqrt{4r^2-u^2}$$

[twin meteor]

Crystopher

[solar barn]

depending on $\theta$ it is either $\frac{u}{2r^2} \sqrt{4r^2-u^2}$ or $-\frac{u}{2r^2} \sqrt{4r^2-u^2}$

[twin meteor]

Crystopher

[hollow sleet]

I am wrong in saying that if I want to turn in the trigonometric direction it is +?

[solar barn]

you mean turning counterclockwise?, in such case $\theta$ gives a rotation of the point $\theta$ degrees/rads counterclockwise with respect to origin

[twin meteor]

Crystopher

[hollow sleet]

Yes I see, ok

I will try with all of that, thank you so much !

[fringe flicker]

Confused on this homework question

[solar barn]

what's confusing? what have you done so far?

[spiral turtle]

! occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fringe flicker]

I tried it with replacing a=1

[spiral turtle]

@fringe flicker ^

[fringe flicker]

Ok

[hollow sleet]

.close

[vocal sleet]

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[uneven pagoda]

is 2^12:2^10 = 2^2 ?

[hidden kelp]

Yes

[uneven pagoda]

1^100 = 1?

[hidden kelp]

1 to any power is 1

[uneven pagoda]

what about 1000^0 ?

[hidden kelp]

Anything (except 0) to the 0th power is 1

[uneven pagoda]

thanks

[hidden kelp]

No worries

[uneven pagoda]

5^2-3^2 = 2^2?

[mellow oyster]

you can't always conclude that (a^n - b^n) = (a-b)^n

[uneven pagoda]

:))

i tried

[hidden kelp]

Yeah when in doubt, plug numbers and calculate to see if something works or not

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

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[vocal sleet]

@oak sand Has your question been resolved?

[oak sand]

<@&286206848099549185>

[whole oasis]

Let's give it a shot

[oak sand]

👍

it said ping after 15 minutes

sorry?

[whole oasis]

Have you tried doing a sketch and finding the missing angles?

[oak sand]

npnp

[dire lake]

find MQ using SOHCAHTOA

[whole oasis]

you'll see some nice looking properties

[oak sand]

what is SOHCAHTOA?

[whole oasis]

e.g. that MQR is 30, 60, 90 and PMQ is isosceles

[oak sand]

ok let me retry that

[dire lake]

this is going to to be hard to explain, mind if i send a vn?

[oak sand]

please

would be much appreciated

[oak sand]

ok

[dire lake]

sin=opp/hyp, cos=adj/hyp,tan=opp/adj

[oak sand]

ahh, i think currently its expected we solve without trig

as we have not been taught that yet

[dire lake]

what else do they expect you to solve it with?\

im genuinely confused, what topic is this supposed to be under?

[oak sand]

geometry 1

[whole oasis]

isosceles triangles, 30, 60, 90 or 45, 45, 90 triangles

[dire lake]

we are asked to find the length, right?

[oak sand]

yes

[whole oasis]

What does it change?

It's about their specific properties which let us solve it without trig

[dire lake]

ok so how is it done?

[oak sand]

MQ is easy to find

i dont know how to determine MR however

[oak sand]

i have all the angles, i dont know how to draw side lengths from it

nevermind, i just figured it out

thank you very much modus and sext

.close

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[muted ember]

If you apply integration by substitution and then back-substitute, do you have to adjust the limits of the integral again?

[plain minnow]

If you back-substitute, you don't really have to mind the limits at all. You can just keep them as before doing any substitution

[muted ember]

So if i would now back-substitute the limits still would be a and 0?

[plain minnow]

As an example: If you had $\int_{0}^{1} 2xe^{x^{2}} ; \dd x$ and substituted $u = x^{2}$ to get\

$\int 2xe^{x^{2}} ; \dd x = \int e^{u} ; \dd u = e^{u}$\

And the back-substitute $u = x^{2}$ to get the antiderivative $e^{x^{2}}$ you just have to evaluate the antiderivative at the original limits $x = 0$ to $x = 1$.

[twin meteor]

Mikkel Angelo

[muted ember]

What if a limit is a fixed but arbitrary variable, you cannot and do not want to calculate the integral, then the limits of the integral remain the same after the back substitution, even in the example, can you simply remove the sign of the a by substituting and back substituting?

Or did you mean that i have to put The limits back to the value they had before the Substitution?

So the limits would be again -a and 0?

[vocal sleet]

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[waxen ruin]

Hello is this correct?

[sweet flower]

A, B, C are correct, D looks wrong

[waxen ruin]

Mmmm

is it 3?

Since there 3 bumps

[oak magnet]

More ? No ?

[waxen ruin]

Where there is 2 relative extreme because one is at the top

And the other is at the bottom

No?

Ohhhh nvm it should be 4

[sweet flower]

There are no global/absolute extrema on this graph

[waxen ruin]

O

[oak magnet]

The relatives ones are asked

[waxen ruin]

So it is 4

?

Thanks

[sweet flower]

And you see why?

[waxen ruin]

Yes

The 4 lil bumps

Each is min max min max

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[fickle mason]

Guys how do I solve this

[calm light]

4 could be written as 2^2

and 2^2x = (2^x)^2

[fickle mason]

Yeah ik

[calm light]

wait this is much simpler than i thought

[fickle mason]

This is what I did

[flint idol]

this can be rewritten as a quadratic

[fickle mason]

Rly how

[flint idol]

using this fact

if we let u=2^x

then we’ll have the equation 4u^2=8u-4

now just solve for u and then use the initial substitution to get back to x

[fickle mason]

OHHH

That is genius

I see

Thanks

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[queen chasm]

Please help me figure out how find the ordered pairs of x^3+6x^2+5x-12

[pallid zenith]

what do you mean "the ordered pairs"

[queen chasm]

the points on the graph of the equation

[kind light]

so the roots?

[queen chasm]

no they should be ( , )

something

[kind light]

what

!original

[vocal sleet]

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.

[queen chasm]

I am trying to graph it but need to find the zeros

[pallid zenith]

wwhat tools are you allowed to use

[queen chasm]

a scientific calculator

[pallid zenith]

my scientific calculator has a root finder on it

[kind light]

uh

you can find all the roots with rational root theorem right?

[queen chasm]

I'm not looking for the roots

[kind light]

1+6+5-12=0 so x=1 works

[kind light]

huh

[pallid zenith]

the zeros are the roots

[queen chasm]

oh

[pallid zenith]

its the same word

although sometimes 'zeros' also means y-intercept

but im guessing you dont want that

[queen chasm]

y intercepts

[pallid zenith]

theres only 1

[queen chasm]

?

[pallid zenith]

y intercept

a function only has a one y intercept

[queen chasm]

Im so confused

[pallid zenith]

you probably want the x intercepts

[queen chasm]

sure

[pallid zenith]

or roots, or zeros

[queen chasm]

at this point I just need to learn something

XD

[kind light]

do you want to find the x'es such that x^3+6x^2+5x-12=0?

[queen chasm]

yes

[kind light]

thats called an x intercept

[pallid zenith]

have you tried factoring this? the roots look nice

[queen chasm]

The way my professor showed us we take all the coefficents and multiply them by factors of the leading coefficent or a and the denominator factors of the constant

I dont know how

I lost as hell

I did this no problem in class

[kind light]

could you show an example problem?

[queen chasm]

but I am taking 5 classes in this summer semester and I literally have no time for anything

So I am probably gonna fail

I also have two finals this week XD

I am passing all my other classes with an A

but algebra is killing me

I havent done it in 10 years

[kind light]

just try and factor this

[queen chasm]

I dont know how

[kind light]

do you kniw rational root theorem?

[queen chasm]

nope

[kind light]

oh

probably the easiest way is jusy try to guess a root

if you want try to search up "rational roots theorem", it could soeed up the process

[queen chasm]

Thanks for the help guys but I am literally gonna fail

.close

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[kind light]

bruh ok then

[queen chasm]

no for real test tomorrow its 10:17 I have 4 more modules to complete

[pallid zenith]

this is actually not a super straightforward problem

i wouldnt feel too bad

[queen chasm]

I cant spend anymore time on this problem

[pallid zenith]

good luck

[kind light]

its 1 3 4 btw

[queen chasm]

Thanks

[kind light]

the roots

[queen chasm]

I dont know how to use that to graph it

[pallid zenith]

,w solve x^3+6x^2+5x-12 =0

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[pallid zenith]

you need to know these

then you need the leading coefficient

so if its x^3 or -x^3

[kind light]

i think you can use calculus to find local maxima and minima (i think its what its called)

[pallid zenith]

it makes the shapes like this

[pallid zenith]

so the graph goes through these points

[pallid zenith]

and looks like the left here

[kind light]

3x^2+12x+5=0
critical points

[pallid zenith]

i dont think they will understand this

[kind light]

fair

[vocal sleet]

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[kind light]

i dont even know where to start with this lmao

let the diagonal be x

ABED=AE×2×x/2×1/2=xAE/2
BEDFC=EF×x/2×1/2+EC×x/2×1/2
DFC=FC×x/2×1/2

2AE:EF+EC:FC=1:3:2

oh

AE=EF

AE:FC=1:4

so asqrt2=6EF
cool

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[somber holly]

can anyone let me know what is wrong with this solution?

Laplace transform

[vocal sleet]

@somber holly Has your question been resolved?

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[somber holly]

sneeze

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[vast shale]

Can someone recommend me video content at most 6 hours long to properly cover these topics:

1. Increasing decreasing functions
2. Maxima minima
3. Differentiation (techniques and methods)

[vast shale]

I was studying from a YouTube channel earlier but I am not getting any better

And as ashamed as i am to admit it, I have an exam in a day so I’m running on a bit of a tight leash

So any help would be appreciated

[vocal sleet]

@vast shale Has your question been resolved?

[kind light]

maybe the organic chemistry tutor for 3?

[flint idol]

3blue1brown's essence of calculus series could help too

[pale pendant]

For Q2 why resultant velocity is not taken to calculate time and distance? Without resultant velocity answer comes correct else it comes wrong

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[next bay]

Can I let α=1 now?

[verbal slate]

dx?

[verbal slate]

i'm 99% sure that you integrate it with respect to x

then evaluate at $\alpha = 1$

[twin meteor]

TheOracle

[next bay]

Oh yeah I forgot it its dx

Just want to know if you can let α=1 before finishing the integral

[sudden compass]

Feynman actually makes this tougher iirc

[next bay]

Ok thanks

[sudden compass]

Not really, because u would have to integrate a constant

[next bay]

Oki

Thank you

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[dense gust]

How could I know the domain and range of the function f(0)=√ x-1?

[dense gust]

I dont get how I would write it out

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[scenic ravine]

if gcd(a,b)=1, then show gcd(a^2,b^2)=1

so this implies 1|a; 1|b

[scenic ravine]

nvm

got it

but how would I do this without bézout's lemma

[torn timber]

Cube it

[scenic ravine]

That would involve bezout's lemma

[torn timber]

ax+by=1

[scenic ravine]

yeah, I get that bit

[torn timber]

Oh you mean without that mb

[scenic ravine]

yea

[grim lotus]

prime factorization

[grim lotus]

this is smart tho

[grim lotus]

i mean this

[scenic ravine]

I don't know Prime factorisation though(In NT)

It hasn't been defined yet

[torn timber]

It's snows idea

[grim lotus]

it's just a product of primes

you should know it

[scenic ravine]

I know it, but my book hasn't covered it yet, so I can't use it

[torn timber]

Well (a^2,b)=(a,b^2)

[scenic ravine]

I have to prove that first then

[torn timber]

It's easy

[scenic ravine]

I guess I could use bézout's lemma

to prove that

wait

that won't give us the same answer would it

hmm

xa^2+by=d_1

nvm, that won't work here

[scenic ravine]

hint please

oh, hi kheeri

[desert hornet]

Assume $a^2=dx, b^2=dy$ where $d>1$ then try getting some sort of contradiction

[twin meteor]

kheerii

[desert hornet]

Maybe it works idrk

[desert hornet]

Hey wai

[scenic ravine]

hmm

[desert hornet]

try arguing that d must be a perfect square

[scenic ravine]

d?

[desert hornet]

Yeah

[scenic ravine]

how are you so good at NT

[desert hornet]

Olympiad prep lmao

[scenic ravine]

I don't see how that helps

[scenic ravine]

this

[torn timber]

Wai? Is that your name physicsrocks

[desert hornet]

If d is a perfect square then x and y are also perfect squares

Then take the square root on both sides

You’ll get that gcd(a, b)=sqrt(d), which implies d = 1

[scenic ravine]

yea, don't follow this bit

[scenic ravine]

an alt name

[torn timber]

Wai hmm what does it mean?
Sorry I'm going off topic

[scenic ravine]

why am I here

[desert hornet]

$a^2=dx, b^2=dy$

[twin meteor]

kheerii

[scenic ravine]

yes

[desert hornet]

if you can prove $d=n^2$ where n is a natural number

[twin meteor]

kheerii

[desert hornet]

then x and y must also be perfect squares

[atomic forum]

Oh you’re still doing this problem

[scenic ravine]

yeah

[desert hornet]

oh, I guess that conclusion comes from factorisation

[scenic ravine]

I'm really bad at NT

[atomic forum]

It’s ok

I’d still recommend the route with Euclid’s lemma…

Idk to me it feels the simplest

[desert hornet]

euclid's lemma?

[scenic ravine]

I'm not sure of how to apply it

[atomic forum]

Want to try again?

[scenic ravine]

sure

[desert hornet]

oh factorisation?

[atomic forum]

So, Euclid’s lemma states that

[desert hornet]

$p|ab, p$ is prime $\implies p|a$ or $p|b$

[twin meteor]

kheerii

[scenic ravine]

yeah

[atomic forum]

Let $d, a, b$ be integers such that $d | ab$. Then if $\text{gcd}(d, a) = 1$ then $d | b$

[twin meteor]

Pseudonium

[atomic forum]

Yes, so you can derive this from the version I posted

[atomic forum]

So, remember the universal property

We need to show that if $d | a^2$ and $d | b^2$ then $d | 1$

[twin meteor]

Pseudonium

[atomic forum]

We split into cases

Either d = 1 and we’re done

Or d > 1

If d > 1, you can use the hint I suggested

So there’s some prime $p$ which divides d

[twin meteor]

Pseudonium

[atomic forum]

And then…?

[scenic ravine]

then d is a multiple of p

[desert hornet]

what can you say about a then

[scenic ravine]

as is a

a is a muliple of p too

[atomic forum]

How?

[desert hornet]

actually we said gcd(a,d)=1

I meant what can you say about b

[atomic forum]

You can derive the version for primes (the one you posted) from what I posted

We’re using the version you posted for this, since p is prime

[desert hornet]

I'm a bit confused

by what you said

[atomic forum]

What’s confusing?

[desert hornet]

what version of what theorem are we using lmao

[atomic forum]

The one you posted

[torn timber]

Mmh my head is not getting it now imma give up rn

[atomic forum]

The special case for primes

[desert hornet]

I haven't really gotten into this side of NT, the proving every result side so idrk

[desert hornet]

ah okay

isn't that a corollary of the fundamental theorem of arithmetic though

[scenic ravine]

this is my first introduction to proofs too, is that. a bad idea?

[atomic forum]

Usually you use Euclid’s lemma to help prove the fundamental theorem for arithmetic

[scenic ravine]

using NT to understand proof writing

[atomic forum]

So I’d say using FTA isn’t allowed

[desert hornet]

ah yeah true

[atomic forum]

It’s not a bad idea

[desert hornet]

uhh, you just need to know which results you're assuming to be true and which you aren't

otherwise it should be fine

[atomic forum]

So, we have that p is a factor of d, right? We can write this as $p | d$

[twin meteor]

Pseudonium

[atomic forum]

Moreover, we assumed that $d | a^2$ and $d | b^2$

[scenic ravine]

yes

[twin meteor]

Pseudonium

[atomic forum]

From this, what can you deduce?

[scenic ravine]

not too sure

[atomic forum]

Ah

Ok, so there’s a neat result about divisibility that’s useful here

Namely, that it’s transitive

[scenic ravine]

oh

[atomic forum]

So if a, b, c are integers, and $a | b$ and $b | c$, then $a | c$

[twin meteor]

Pseudonium

[atomic forum]

Can you prove this?

[scenic ravine]

yeah

let ne try

[desert hornet]

where do we go from here?

[scenic ravine]

$b = k_1a$

[atomic forum]

(So, one can form a category whose objects are natural numbers and a morphism n -> m means n | m)

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

$c=bk_2$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

$but b=k_1 a$

[desert hornet]

yeah this works

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

so $c/(k_1 k_2_=a$

[twin meteor]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[desert hornet]

yes indeed

[scenic ravine]

so $\frac{c}{k_1k_2}=a$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

oh

[desert hornet]

uhh, sure, if you want to write it like this

$a|b, b|c\implies a|c$

[scenic ravine]

so a|c

[twin meteor]

kheerii

[atomic forum]

So…

There’s a few things I’d fix with this proof

First, you should say explicitly that $k_1$ and $k_2$ are integers

[twin meteor]

Pseudonium

[desert hornet]

yeah this is important

I feel like a backseat teacher lmfao

[atomic forum]

Also, it’s not guaranteed that $\frac{c}{k_1 k_2}$ makes sense

[twin meteor]

Pseudonium

[atomic forum]

It could be the case that $k_1 k_2 = 0$

[twin meteor]

Pseudonium

[scenic ravine]

ah

right

[atomic forum]

However, it will always be true that $c = k_1 k_2 a$

[twin meteor]

Pseudonium

[atomic forum]

This just follows from substitution

[desert hornet]

hmm, this would imply that $k|0\forall k\in\mathbb{Z}$

[twin meteor]

kheerii

[desert hornet]

this is true?

[atomic forum]

Yes!

[desert hornet]

ah

[atomic forum]

Everything divides zero

[desert hornet]

hmm, that does make sense I guess

[atomic forum]

(0 is a final/terminal object)

[desert hornet]

k*0=0

[atomic forum]

So, the way I would structure the proof is as follows

Let a and b be integers

We say that $a | b \iff \exists k \in \mathbb{Z} \text{ such that } b = k a$

[twin meteor]

Pseudonium

[atomic forum]

This is a definition of what divisibility means

Then, suppose a, b, c are integers such that a | b and b | c

This means that there exist integers $k_1, k_2 \in \mathbb{Z}$ such that $b = k_1 a$ and $c = k_2 b$

[twin meteor]

Pseudonium

[atomic forum]

Just by unpacking the definitions

Then, by substitution, $c = k_1 k_2 a$

[twin meteor]

Pseudonium

[atomic forum]

So if we let $k = k_1 k_2 \in \mathbb{Z}$, we have $c = ka$

[twin meteor]

Pseudonium

[atomic forum]

Thus, by the definition of divisibility, $a | c$

[twin meteor]

Pseudonium

[scenic ravine]

yeah, can't believe I forgot that

sorry

[atomic forum]

That’s ok

So now we know that divisibility is transitive, right?

[scenic ravine]

yes

[atomic forum]

Let’s go back to what we were doing before

We need to prove that if $d | a^2$ and $d | b^2$ then $d | 1$

[twin meteor]

Pseudonium

[atomic forum]

To do this, we split into cases

If d = 1 we’re done

If d > 1, then we know that d has some prime factor p

I.e. that $p | d$

[twin meteor]

Pseudonium

[scenic ravine]

yes

[atomic forum]

Then, what can you deduce?

[scenic ravine]

$p|a^2$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[atomic forum]

Yep!

[scenic ravine]

and similarly $p|b^2$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[atomic forum]

Indeed

[atomic forum]

Then, you want to use this

[scenic ravine]

so $p|(a^2b^2)$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[atomic forum]

No

Well, that’s true

But it’s not helpful

[desert hornet]

(Hint: $p|a\cdot a$)

[twin meteor]

kheerii

[scenic ravine]

ah

[desert hornet]

how do we know this?

[scenic ravine]

$p|a^2$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[atomic forum]

You can use induction

[torn timber]

FTA?

[desert hornet]

we said we weren't using FTA

[atomic forum]

This fact usually gets used in proving FTA, so I don’t think FTA is allowed

[desert hornet]

oh, hmm

that kinda makes sense

[torn timber]

Well this will prove the FTA tho

[desert hornet]

yeah, but we need to prove this first

[atomic forum]

Mhm otherwise it’s a circular argument

[desert hornet]

how would we induct?

[atomic forum]

So have you met strong induction?

[scenic ravine]

how?

[atomic forum]

$p | a \times a$ and p is prime

[twin meteor]

Pseudonium

[atomic forum]

So what does Euclid’s lemma tell you?

[scenic ravine]

yeah

[desert hornet]

I have

[scenic ravine]

p|a

[desert hornet]

indeed

[atomic forum]

Ok so - assume the statement holds for m < d, and then we need to show it holds for d. Either d is prime in which case we can use p = d, or d is composite, so d = n*k for some integers n, k both bigger than 1. Then 1 < n < d so by our inductive hypothesis, n has a prime factor p. And then since p | n and n | d, we have p | d, so d has a prime factor

[desert hornet]

and can you say the same thing for b?

[desert hornet]

ahhhh yeah that makes sense

assuming d is composite implies d=nk

[atomic forum]

Indeed

[desert hornet]

that's the step i was missing

[atomic forum]

It’s basically what “composite” means

[desert hornet]

yeah okay

nice

hope wai caught all that lmao

[scenic ravine]

yes

[atomic forum]

Yes, this is correct

[desert hornet]

which implies?

[scenic ravine]

p|ab

[desert hornet]

uh, well, yeah

but

[torn timber]

Not really but yeah

[desert hornet]

you already arrived at a contradiction

[atomic forum]

Now you want to use the assumption on a and b

[desert hornet]

$p|a, p|b\implies ???$

[twin meteor]

kheerii

[atomic forum]

Namely that gcd(a, b) = 1

[scenic ravine]

hmm

but that's given

right

[desert hornet]

what does THIS tell you about gcd(a, b)?

remember that all of this relied on an assumption, i.e d>1

[scenic ravine]

I'm really not sure, sorry

[torn timber]

P=(a,b) but that is equal to 1 thus

[atomic forum]

You should use the universal property here

[desert hornet]

if we can show that this assumption causes a contradiction we are done

[atomic forum]

Have you met the universal property of gcd?

[torn timber]

D=1

[scenic ravine]

not by that name

[desert hornet]

$d|a, d|b\implies d|(a,b)$

[twin meteor]

kheerii

[scenic ravine]

that I have

yes

[desert hornet]

so use that

[scenic ravine]

so p|(a,b)

[desert hornet]

here

[desert hornet]

yeah

do you agree that that is a contradiction?

[atomic forum]

It says the following. Given integers m and n, the integer gcd(m, n) is the unique integer satisfying $\forall d \in \mathbb{Z}, d | m \text{ and } d | n \iff d | \text{gcd}(m, n)$

[twin meteor]

Pseudonium

[desert hornet]

yeah Pseudonium said it way better than me lmao

[scenic ravine]

ok, yeah, that works

[scenic ravine]

so this

[torn timber]

Well he is an experienced proofwrtiter

[atomic forum]

(Um please ask my pronouns first)

[atomic forum]

And what’s (a, b)?

[torn timber]

Sorry my bad

So what are your pronounce

[desert hornet]

we know that $(a, b)=1$. let $d=(a^2,b^2)$. if $d=1$ we are done. if $d>1$ then d must have a prime factor, say $p$. $p|d$, hence $p|a^2, p|b^2 \implies p|a, p|b$ from Euclid's Lemma. so, $p|(a,b)$ which is a contradiction as $(a, b)=1$. hence, $d=1$. q.e.d.

[twin meteor]

kheerii

[scenic ravine]

oh

I see

[desert hornet]

sorry I thought the question was done lol I just wanted to write the proof in short

[scenic ravine]

I'm going to def want to do this problem again

[desert hornet]

of course this is missing a lot of details but this is the gist of the proof

[scenic ravine]

yeah, I was lost

[atomic forum]

she/her today

[scenic ravine]

I need to probably revise this unit again, seemed to have missed out the nittie -gritties

[atomic forum]

The contradiction comes from the fact that no primes divide 1

[scenic ravine]

sorry for wasting your time, and thanks for the help

[torn timber]

Well she is an experienced proofwrtiter then

[atomic forum]

It’s ok

[scenic ravine]

go t it

Thanks!

[desert hornet]

@atomic forum does this look good

[torn timber]

Thanks pseudonium

[atomic forum]

Yope

[scenic ravine]

Wanted to ask, is it normal to striggle with problems like this if I'm new to Proof writing and NT?

[atomic forum]

Oh yeah totally

I did

[desert hornet]

is that meant to be an in between of yep and nope lmao

[torn timber]

Yes :(

[atomic forum]

Oh sorry it’s a meme among my friends

Your proof is fine

[desert hornet]

oh cool

[scenic ravine]

thanks everyone!

[desert hornet]

i haven't really gotten into this side of number theory, like, ever

so it's pretty cool

[atomic forum]

Huh..

You do Olympiad stuff right?

I never did much Olympiad stuff

I thought they would’ve covered this

[desert hornet]

I used to

[torn timber]

Nah you're better off than me i remember for proving a simple question I was stuck for 3 hrs

[desert hornet]

they did cover things like this but it wasn't about proof writing or making sure everything is proven or anything like that

it was more about higher level theorems and applications

[atomic forum]

I see

[scenic ravine]

can I close this now?

[atomic forum]

Sure

[scenic ravine]

thanks

[atomic forum]

If you’re satisfied

[torn timber]

Yes

[torn timber]

Exactly what she said

[scenic ravine]

I am, wish I could star some of these messages for later reference though, will probably have to take screenshots instead

.close

[vocal sleet]

Channel closed

Closed by @scenic ravine

Use .reopen if this was a mistake.

[worthy citrus]

make your own server and paste links to messages in it

[scenic ravine]

will do

thanks

[vocal sleet]

Available help channel!

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[placid osprey]

Problem 15. Consider the function $f$ given by $f(x, y)=\arctan \frac{y}{x}$.
a) Determine $f(1, \sqrt{3}), \nabla f$ and $\nabla f(1, \sqrt{3})$
b) In which direction from the point $(1, \sqrt{3})$ is the directional derivative of $f$ maximal? What is the maximum directional derivative at that point?
c) Determine the directional derivative at the point $(1, \sqrt{3})$ in the direction given by the vector $(1,1)$.

[twin meteor]

dghf

[placid osprey]

Need help with b) and c)

[atomic forum]

Damn why’ve I never thought of this

[twin meteor]

dghf

[placid osprey]

But I don’t know what a “maximum directional derivative” means, and how the gradient vector, when normalized, give us just that.

I know the partial derivatives fx, fy, and fz are necessary in the formula for Du*f(x, y, z)

But this whole “maximum directional derivative” stuff is vague to me

The gradient vector I reckon is the direction a function grows in?

[harsh canopy]

I believe the maximum directional derivative is when the direction is grad f

[bitter pilot]

https://openstax.org/books/calculus-volume-3/pages/4-6-directional-derivatives-and-the-gradient#:~:text=The maximum value of the,point in the same direction.

[harsh canopy]

You should be able to see it since $D_{\hat{\mathbf{u}}}f = \nabla f \cdot \hat{\mathbf{u}} = |\nabla f| \cos \theta$

[twin meteor]

eugene_krabs_has_cake

[placid osprey]

We don’t have the directional derivative so how can we derive the unit vector, since u = v/||v||

[harsh canopy]

Where theta is the angle between grad f and u

Duf is maximised when theta is 0

[placid osprey]

We have no information about theta in the problem, no?

[harsh canopy]

We are trying to find the vector u

Such that the directional derivative is maximised

And in that expression it happens when cos(theta) = 1

[placid osprey]

I see

In what order would you solve the problem now

Now that we have the gradient of f atleast

According to my good friend Claude:

$\frac{\nabla f(1, \sqrt{3})}{|\nabla f(1, \sqrt{3})|} = \frac{(-\frac{\sqrt{3}}{4}, \frac{1}{4})}{\sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2}} = \frac{(-\sqrt{3}, 1)}{\sqrt{\frac{3}{16} + \frac{1}{16}}} = \frac{(-\sqrt{3}, 1)}{\frac{\sqrt{4}}{4}} = (-\sqrt{3}, 1)$

[harsh canopy]

Well we have grad f since we know the point

[twin meteor]

dghf

[harsh canopy]

What’s with the last 2 equalities

Also where did you multiply the top by 4 in the first equation for?

[placid osprey]

Oh must be a latex formatting issue

But the denominator shouldn’t vanish lmao

[placid osprey]

Is the overarching idea correct though

I.e. the gradient vector evaluated at that point divided by the gradient vector’s magnitude at that point

[harsh canopy]

Yes

[placid osprey]

And then we get the unit vector

So we get the angle by doing tan^-1 on the x and y then, right?

[harsh canopy]

No we don’t need to know theta

We just need to know the direction from that

Ie that the direction is grad f

The first part is just asking you for what the direction is

[placid osprey]

Oh

[harsh canopy]

I’m assuming it would want the unit vector

So you do what you did above for that

[placid osprey]

Wait I didn’t understand. Is the direction grad f?

I thought the direction was what the unit vector gives us

Or have I mixed things up

[harsh canopy]

The direction is grad f

[placid osprey]

Okay

[harsh canopy]

But they may want that normalised in the answer

So you would do what you did

Ie divide by the magnitude

[placid osprey]

And we divide by the normalization to get the unit vector, but, what was the point of our unit vector?

To multiply by our grad f?

And this in turn gives us the directional derivative?

I think that’s what they asked for right

[harsh canopy]

I don’t know if people always do this, but i think it’s standard to only use unit vectors when working with directional derivatives

[harsh canopy]

Also the directional derivative would just be the magnitude of grad f by the formula I gave

No need to actually calculate anything I’m pretty sure

[placid osprey]

So what I wrote above is incorrect?

I.e. my questions and assumptions

[placid osprey]

Here.

[harsh canopy]

Yeah you divide by the normalisation to get the unit vector

Maybe the question will just accept the direction being grad f

[placid osprey]

Yes and the directional derivative is not the grad f multiplied by the unit vector?

[harsh canopy]

But if I were doing this question I would normalise grad f

[harsh canopy]

Dotted with the unit vector

[placid osprey]

I can show you how a famous YouTuber solved a problem

But I don’t think we have calculated our Duf yet, right?

[harsh canopy]

Yeah they used the first part of this formula

[placid osprey]

Indeed

[harsh canopy]

So they did find Duf

[placid osprey]

Yes, so now we have our u (if I re-do the calculation), and then I multiply by the grad f, and I’m done?

$\frac{\nabla f(1, \sqrt{3})}{|\nabla f(1, \sqrt{3})|} = \frac{(-\frac{\sqrt{3}}{4}, \frac{1}{4})}{\sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2}}$

[twin meteor]

dghf

[placid osprey]

$= \frac{(-\frac{\sqrt{3}}{4}, \frac{1}{4})}{\sqrt{\frac{3}{16} + \frac{1}{16}}} = \frac{(-\frac{\sqrt{3}}{4}, \frac{1}{4})}{\frac{\sqrt{4}}{4}} = \frac{(-\frac{\sqrt{3}}{4}, \frac{1}{4})}{\frac{1}{2}}$
$= (-\frac{\sqrt{3}}{2}, \frac{1}{2})$

[twin meteor]

dghf

[harsh canopy]

If you dot grad f with that it should give you it

[placid osprey]

Basically multiplying (x, y) by 2 gives the RHS

[harsh canopy]

But with the maximal direction you don’t need that

[harsh canopy]

If you look here it was the magnitude of grad f multiplied by cos(theta)

But cos(theta) is 1

So the directional derivative is just the magnitude of grad f at that point

[placid osprey]

Remind me if I’m correct, is the dot product for when you multiply a constant by coordinates? I.e. distributing the constant to each component in a vector (scaling it)?

[placid osprey]

How do we know it’s 1 in our case?

[harsh canopy]

A dot product is for 2 vectors

[harsh canopy]

Because we found the vector such that cos(theta) is 1

[placid osprey]

We made an assumption that it was 1? Or?

[harsh canopy]

We used cos(theta) = 1 as a criteria

Yeah

[placid osprey]

Wow you can do that? Lol

[harsh canopy]

We don’t find u and then find that it has that property

[harsh canopy]

Well yeah, you can use a vector with any angle theta, we just use theta = 0 so that our Duf equation is maximised

Since the only thing that changes is cos(theta)

The magnitude of grad f is constant

So to find the maximum of Duf is when cos(theta) is maximised

[placid osprey]

Then what’s the point of having it there

The cos

[harsh canopy]

Wdym?

[placid osprey]

Is it necessary to know that cos exists in the formula when a problem gives you a hint that the cos theta is not 1?

I didn’t see the YouTuber take the angle into account for instance

[harsh canopy]

That’s because it gives you the direction

In that problem

In this problem it wants you to find the direction vector

And gives you the fact we want the maximum Duf in order to find that vector

[placid osprey]

Okay. So when the direction vector is given, don’t take cos into account, and when the direction vector is not given, you need to take the cos theta into account?

[harsh canopy]

Yeah

Don’t take cos into account

[placid osprey]

And by taking it into account we simply say “just multiply by 1 to ensure that it’s maximal”?

[harsh canopy]

So in this message only use the first equation, don’t use the second equality when you are given the vector

[harsh canopy]

Yeah you only really use the equation with cos in it when you are either finding the maximal or minimal directional derivative

If it’s maximal you want cos(theta) to be 1

If it’s minimal you want cos(theta) to be -1

[vocal sleet]

@placid osprey Has your question been resolved?

[placid osprey]

I see

I will take a closer look at this problem and ping if I’m stuck

[vocal sleet]

@placid osprey Has your question been resolved?

[vocal sleet]

@placid osprey Has your question been resolved?

Channel closed

Closed by @placid osprey

Use .reopen if this was a mistake.

[vocal sleet]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hollow prairie]

Both of these would be true right?

[hollow prairie]

<@&286206848099549185>

[severe lynx]

Hmm, yes

For 6a, if you let C=2 and n0=2, then it follows.

For 6b, let C=1 and n0=1, then it also follows

Not sure how rigorous they want you to prove both though

If it helps, you can do induction on both and then double existential introduction

But that seems pedantic, how rigorous does your course need us to be? @hollow prairie

[hollow prairie]

Not very rigorous

Brief proof is ok

But yea thx

[vocal sleet]

@hollow prairie Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cunning slate]

i dont need help on a question but

[cunning slate]

i was wondering

hold on

HOW DO i find the area of this shaded region

[urban edge]

are you familiar with how to integrate over polar coordinates?

[cunning slate]

no

ive learned it but like

its kinda hard

[urban edge]

yeah it is