help-43

[serene blade]

is that wrong?

but since the top absolute value is multiplied by -1 would i take that -24 and make it positive?

[eternal pulsar]

That is wrong

[serene blade]

i see that

[eternal pulsar]

Did you simplify correct

[serene blade]

i just redid the math

probably not

[eternal pulsar]

I get something else

[eternal pulsar]

I see that you guys went over that |x^2-10x|=-(x^2-10x) for this particular limit

[eternal pulsar]

That’s what I got

[serene blade]

says its wrong

[eternal pulsar]

This might be a system error

I see absolutely nothing wrong with what you have

[serene blade]

ptrobably something wrong with the numerator one

[eternal pulsar]

,w when is (60-6x-|x^2-10x|)/(|x^2-100|-64)=(60+4x-x^2)/(36-x^2)

[boreal girder]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by exiled_hype

[eternal pulsar]

Ohhhh

Fuuuuuck

[serene blade]

?

i think i know

let me cook

it was the negative before the absolute value

@eternal pulsar

i just cooked

.close

[compact pewter]

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[slim crystal]

am i right with D here?

[subtle zealot]

please show your work

[slim crystal]

i would but its on paper

[stone jackal]

✅

[slim crystal]

THANK YOUUU

[stone jackal]

If it's correct, why must they show their work?

[subtle zealot]

to confirm they did it right

[slim crystal]

sorry dude i dont got a phone on me rn

im on pc

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[sudden carbon]

hi can someone please explain what “coordinate of the Cartesian vector” actually means? Like the coordinate is the change in x and y, so [2,-5] but what exactly does the 2 and -5 mean? Im new to Cartesian vectors so im a bit confused by the notations

[junior vault]

when split into x and y components

could also be considered displacement from point A if you want

[sudden carbon]

OHHHHHHHHHHHHHHHHHHHHHHH

TYSMMMM

[sudden carbon]

so then in order to convert a cartesian vector into coordinates, we always do tail - tip for both x and ys?

[junior vault]

well you need an origin to start from since it isn't defined by the vector directly

but pretty much

[sudden carbon]

what does this mean btw

also if you're just given the coordinates [2,-5] is there an infinite amount of equivalent vectors that could be drawn?

[junior vault]

like you don't know from the vector (2, -5) that it starts at A

[sudden carbon]

aa i see

[junior vault]

technically yes, since a vector is just direction based no matter where you draw it it will always have the same direction

but typically they are drawn at the origin

0,0

[sudden carbon]

ohh okay thanks

makes sense

also i have another question

[junior vault]

sure

[sudden carbon]

how do i convert a vector into a unit vector?

like

for example

given vector v = [5,-1], how do i express it as 2 unit vectors that are collinear with v

[junior vault]

have you seen the formula $\hat{v} = \frac{1}{|v|} \cdot v$

[boreal girder]

BuilderDolphin

[sudden carbon]

yup, tho i dont really understand what it truly means behind the symbols if that makes sense 😹

lke what does it mean exactly when we express a vector as a unit vefctor

[junior vault]

a unit vector is a vector with magnitude 1

sort of related to the unit circle

r = 1

[sudden carbon]

yea that makles sense, but what does it mean if we rewrite v as a unit vector

[junior vault]

it means we divide v by its magnitude, resulting in unit vector v with magnitude 1

$\hat{v}\cdot |v| = v$

[boreal girder]

BuilderDolphin

[junior vault]

unit vector v is colinear with v

[sudden carbon]

huhhhhhhhhhhhhhhhhhhhhh

[junior vault]

i should probably clarify |v| means magnitude

if that was confusing

$\hat{v}$ is just saying the unit vector v

[boreal girder]

BuilderDolphin

[junior vault]

what seems to be the confusion

i can reexplain if needed

[sudden carbon]

thank you

okay so

[sudden carbon]

why does it give you v tho

[junior vault]

(mightve gone too fast)

[sudden carbon]

OH

wait is it because

v with funy hat is magnitude 1

so that times magnitude v = v

[junior vault]

yep

it's colinear to v

so if you multiply by the magnitude of v

you just get v

[sudden carbon]

so we're saying v is colinear to v?

lol

[junior vault]

v hat is colinear

but yes

[sudden carbon]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

uhh

so like

they got some very funny thing

they got

[5/sqrt26, -1/sqrt26] and [-5/sqrt26, 1/sqrt26]

[junior vault]

are those unit vectors

looks like it

[sudden carbon]

yea

huh

[junior vault]

uh what was the problem with those

[sudden carbon]

so i dont get like

how you get them

and im kinda confused again on what it means

[junior vault]

well you have this equation

[junior vault]

can be rearranged to this

which is more practical to use

[sudden carbon]

so basically each time we just do the 1/magnitude * x , 1/magnitude * y?

magnitude like length right

[junior vault]

correct

[wanton inlet]

You can alternatively say x/magnitude and y/magnitude

||Just to avoid any notational confusion||

[sudden carbon]

does this always work?

[wanton inlet]

Also, these two values represent the vertical component (the one parallel to the y-axis) and the horizontal component (the one parallel to the x axis) of the unit vector, the unit vector together with its components, form a right angle triangle where the hypotenuse’s length is always 1

[sudden carbon]

ohh

i still dont get

how the factors work

😭

like the x/mgnitude

what does that even meannnnnnnnnnnnnnnnnnn

how does that even relate to the vector itself

[junior vault]

$\hat{v}\cdot |v| = v$

[boreal girder]

BuilderDolphin

[wanton inlet]

The original vector is <-1, 5> right?

[sudden carbon]

yea

[wanton inlet]

So

[sudden carbon]

oh

this makes sense

but ike

5/sqrt26...

OH

WAIT

YOU GET

5

if you multiplty 5/sqrt26 by magnitude (sqrt 26)?

🙀

[wanton inlet]

Yes

[sudden carbon]

woah

woaaaaaaaaaahhhhhhh

[wanton inlet]

yk, i have no idea what clicked for you but nice

[sudden carbon]

this kinda helped i guess

i did not understand the x/magnitude

and how that unit vector relates to the vector itself

[wanton inlet]

Ok

[junior vault]

but the equation literally shows it 🙏

[sudden carbon]

yeah i didnt see it before

but now i do

[wanton inlet]

The vector is a multiple of the unit vector

[sudden carbon]

the dots connected 💯

took some time but

[junior vault]

oh

[sudden carbon]

ty both

[junior vault]

wait you said did

i thought you said do

nvm

[sudden carbon]

huh

[wanton inlet]

Fire stuff

[sudden carbon]

what did

[junior vault]

i thought this was "i do not"

instead of did

but it whatever

[sudden carbon]

oh

i get it now

‼️

[wanton inlet]

Well, we’re glad we helped a cat learn math

[junior vault]

[sudden carbon]

btw i jusy started cartesian vectors, is the unit itself gonna be hard? cuz i knwo wwe're gonna do dot product rule and do you guys have any tips for actually understanding the concepts cuz i think it's like pretty different than regular math

well it's a new way of seeing things i guess

[sudden carbon]

🙇

[wanton inlet]

Well, unit vectors aren’t be hard, and as a tip, i find right angle triangles as a good way to visualize things

Especially when you’re computing an angle using the dot or cross product, you literally can see the angle and connect the dots more easily

[junior vault]

vector projection is cool

(helps visualize dot product)

[sudden carbon]

thanks

so break it up into x and ys?

[sudden carbon]

thanks even tho no idea what that is just yet

[junior vault]

splitting into components is always good when it comes to adding/subtracting

[wanton inlet]

Yeah if you find yourself struggling to understand, it helps break things up

[junior vault]

you'll learn it along with dot product

[sudden carbon]

okayy

thank you both

[wanton inlet]

Any time

Like literally, i wanna procrastinate more, so pay a visit whenever you feel like it

[sudden carbon]

haha

lock in 🔥 but ty

also what are you in math undergrad

[wanton inlet]

Will do, ty

[sudden carbon]

what do you plan to do after

[wanton inlet]

Im not even math major

[sudden carbon]

huuuuhhhhh

what major are you

[wanton inlet]

Those math peeps confuse me alot

Im just an engineering undergrad

[sudden carbon]

OG

OMG

what enggg

[wanton inlet]

Mech

[sudden carbon]

wowwwwwwwwwwwwwwwwww

is the math for eng rlly hard

[wanton inlet]

Not really

We focus more on formulas and computing

[sudden carbon]

do you do proofs?

[wanton inlet]

But calculus is a bitch everywhere

[wanton inlet]

Rarely

[sudden carbon]

ong

what year are you in

[sudden carbon]

thank god

proofs are NOT it

😭

[wanton inlet]

Well i like em actually but yeah, they get over my head sometimes

[sudden carbon]

you must be super good at math...

[wanton inlet]

Half way through the degree that’s all imma say

[wanton inlet]

Well if i failed calculus 2 times in a row, i better be

[sudden carbon]

omfg 😭

[wanton inlet]

Btw ya got any more questions?💀

If not .close and let’s switch to a general channel

[sudden carbon]

nope, thanks for reminding

.close

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[wanton inlet]

Anytime

[compact pewter]

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[slim crystal]

so this is my question

its part A

just finished it

and wanted to ask if i did everything correctly

sup brodie

😭

@native shard worry bout these mfs

[native shard]

💀

dawg

[slim crystal]

ggs

[native shard]

wild

[slim crystal]

https://tenor.com/view/kaicenat-spider-man-kai-gif-26805446

[native shard]

how old are you bro

[slim crystal]

.close

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[copper sierra]

,preban 1290993603937570857

[boreal girder]

[Ticket #350](#〖senior-console〗 message): Prebanned kqllin2k.

[compact pewter]

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[red forge]

can someone help with this question, im kinda lost on the process to go through with this question

[red forge]

like ik u can peform the ratio test to etst for convergence with alternating series but, after that??

[red forge]

and why does this even matter

[compact pewter]

@red forge Has your question been resolved?

[compact pewter]

@red forge Has your question been resolved?

[red forge]

<@&286206848099549185>

Ive gottin this far

[kind crane]

you have a sign error, but other than that, it's right up to the last limit. your conclusion after that is incorrect though

you have to use the given range of t from the problem

[red forge]

better conclusion i wrote here

[red forge]

what do you mean by sign error?

like just wrong + or -

[kind crane]

the - after the equals sign is wrong

[red forge]

i thought since the -1^n+2
-1^n+1

i.e when n = 5

-1^7 and -1^6

so -1/1

so -1

hence move outside?

[red forge]

and would this be a correct approach :)

[kind crane]

you ignored the absolute value sign

[red forge]

oh

yea i see

oops

mmm

[red forge]

so i assume this would be allg, except not done
because we know this convereges absolutely for values of t > 0, but the only option left is when t = 0
so you can sub in t = 0 and then test for convergenece in other ways, (alternaating series probably) and hence you have yor interval of converegence?

is there anything here that is to do with the radius of converegnce, becuase ik its something related here but i have no idea how, not that it matters i dont htink if i can just find the interval?

[kind crane]

youre not asked for radius

can you write t>0 as an interval

[red forge]

yes?

t E (0 , inf)

[kind crane]

then do it

[red forge]

do we not have to test for convergence when t = 0

since ratio test is inconclusive

[kind crane]

oh yes you do have to

[red forge]

also yea im just asking like in regards to this question, what is the radius of convergence, comapred to the interval, i undersand the interval is literally the interval of values of t where T(t) is convergent to a value hence well defined, but in this case what does radius mean

im aware it doesnt ask for it

or is it impossible to excplain in this regards

[red forge]

would this be correct please and thanks

.close

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[lone citrus]

Hey I just finished AP Calc AB and I'm currently studying Calc 2. Thing is the BC curriculum covers parametric equations and polar coordinates before sequences and series, meanwhile my college does the opposite. Which one is better to learn first? Is there even a correlation between the two?

[subtle zealot]

theyre relatively unrelated, so it should not matter which you do first

[solemn wasp]

they're pretty interchangeable, yeah

[stone jackal]

I would recommend starting with parametric functions because they are a better continuation of what you did in AP Calculus AB.

[lone citrus]

Okay, thank you guys

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[fresh ginkgo]

how do i prove for p(x)=(x-1)*f(x) that if x=1 is a root of p`(x) then f(x) is product of at least 2 (x-1)?

[hushed magnet]

set g(x)=(x-1)^d h(x) where h has no powers of (x-1)

then differentiate with product rule

[fresh ginkgo]

also does p have to be a polynomial? or can the powers of x-lambda_i be rational/irrational

[hushed magnet]

well then generally it wouldnt be differentiable at eg x=1

[compact pewter]

@fresh ginkgo Has your question been resolved?

[compact pewter]

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[graceful fractal]

Hi, I'm a bit stuck on these graphs regarding in my statistics course. I'm not sure how to interpret them. What about these graphed power functions shows that one type of error is greater than the other?

[graceful fractal]

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[split granite]

Can someone guide me to simplification which happens in the very last step? Where root x by 2 goes into power of e?

[compact pewter]

Please don't occupy multiple help channels.

[spice saffron]

Indice law,

x^(n/m) = mth root of x^n

[split granite]

isn’t root of root x smthn like x^(1/4), and the other thing being e^(1/2)

[split granite]

Yeahhh

[cloud granite]

$\sqrt{e^{\sqrt{x}}} = (e^{\sqrt{x}})^{\frac12}$

[boreal girder]

SELVATOR

[spice saffron]

[split granite]

But

root x isn’t in power

root x is in multiplication w e

[cloud granite]

Look twice

Look at second sentence

[spice saffron]

Look at the first line.

[split granite]

Ohhhhhh

[spice saffron]

The guide has made a typo, so yeah

[split granite]

Yeahhh

Okay thankyouu

🙏

[spice saffron]

No worries

[split granite]

.close

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[slim crystal]

i'm on Part D here

this is what i've done so far

i'm just wondering which quadrant my sum vector would fall in?

its a positive x and y so does that mean its quadrant 1?

<@&286206848099549185>

[brazen quiver]

Yes exactly

[slim crystal]

okay thank you sm

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[drowsy tusk]

this solution has an error yeah, you need to multiply the error term by the 1/2

question is just find the firrst order Taylor series of arctan(x) around x=1, including the error term

[kind crane]

yup

[drowsy tusk]

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[forest token]

status : 1

[unkempt sail]

I see

Convert it into telescopic

Not exactly a telescopic

!status

[compact pewter]

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

[forest token]

;/

[unkempt sail]

[forest token]

technically 2

i put n=1000

[unkempt sail]

How did y9u begin

[forest token]

and n=1

[unkempt sail]

Yesbthats the way

[forest token]

and got f(2) in terms of f(1)

[unkempt sail]

Do the same for all n lying in between 1 and 1000

[forest token]

then i tried to get f(2) in terms of f(999)

[forest token]

huh i didnt get that

try to get all those n in terms of f(1)?

[unkempt sail]

No no no

Write the equations for all n in between them

Write them one below the other

To get a clear picture

[forest token]

okay

i kinda see what ur getting at

[unkempt sail]

Good

[unkempt sail]

Yes

Now do you see what you have to do

[forest token]

add everything?

[unkempt sail]

Yes

[forest token]

i got it

love u bro

.close

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[compact pewter]

Please don't occupy multiple help channels.

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[unkempt sail]

[compact pewter]

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[uneven siren]

.reopen

😭

[uneven siren]

.close

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[cursive vector]

Can someone help me understand this result? How do they go from Cij = TijklmnEklaEmnb)VaVb to the second expression? Is TijklmnEklaEmnb isotropic because it is the multiploe if isotropic tensors?

I am asking about part (d)

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@cursive vector Has your question been resolved?

[compact pewter]

@cursive vector Has your question been resolved?

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[slim lodge]

I understand how to get the second deribvative...

how do i find the bounds for when t is suppose to be c.up?

t-1=0

so t cant be 1

and t cant be 0

but what determines the < or > signs?

like uhh

idk how t explain

let me rephrase

t<0 i understand, but why t>1? shouldnt it just be t cant equal 1

idk what im saying

[rigid perch]

we are interested in where is the second derivative positive. there are two things which can switch signs:
t - 1 and 4t^3

so you have to account for the sign changes of both of them

[slim lodge]

ok

but it would still be positive for all values greater than 1 right

[rigid perch]

yes, that's what they conclude

[slim lodge]

im confused about the t<0 now, am sorry

can yoou explain the t<0 condition to me?

[rigid perch]

if t < 0 then both t - 1 and 4t^3 are negative, so it cancels out and the whole thing is positive

[slim lodge]

i see

thank you

🙏

.close

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[slim lodge]

.reopen

[compact pewter]

✅

[slim lodge]

horizontal tangent is when

dy/dx = 0 right?

where do i find vertical tangent?

[eternal pulsar]

Yes

[eternal pulsar]

Specifically here when dx/dt=0 for when dy/dt is not 0

[stone jackal]

dx/dt = 0
dy/dt =/= 0

[stone jackal]

Take caution when using this, as you might forget about holes within the function.

[slim lodge]

this sounds hard to do

[stone jackal]

Instead, consider finding where dy/dt = 0 and dx/dt =/= 0.

[slim lodge]

how do i find when dx/dt = 0 but dy/dt cant be 0

dont u just set dx/dt to 0

thats it

[eternal pulsar]

It really isn't

Just do when dx/dt=0, if dy/dt=0 at the same time t we end up with 0/0

[stone jackal]

Once you find those t values, you must plug them into dy/dt to test dy/dt =/= 0.

[eternal pulsar]

Which is a big uh oh

(Undefined and lots of bullshit occurs with 0/0, so we prefer dy/dt to not be 0)

[slim lodge]

Just to clear up on this real quick, Im doing problem 23, i have dy/dx = -3sin(3theta) / -sin(theta)

would i set the top to 0

individually

and bottom

[eternal pulsar]

Yes

There's some intersecting solutions, so be careful

The whole point of the secondary condition is to just not run into 0/0

[slim lodge]

oh i see

[eternal pulsar]

Hence

[stone jackal]

Also, to prove [this statement](#help-43 message), observe what happens at theta = 0.

[carmine garden]

Is this, stewart ?

[slim lodge]

yes

[slim lodge]

for the top

i have pi/3 and 2pi/3

for the bottom its pi, 2pi

what about intersecting stuffdo i need to look for?

because i dont see it

[stone jackal]

Are you restricting $\theta \in [0, 2\pi]$?

[boreal girder]

@stone jackal

[slim lodge]

yes

[eternal pulsar]

Just see when the solutions intersect (if both theta values produce 0 in numerator and denominator)

[stone jackal]

$$\left. \dv{y}{x} \right|_{\theta = 0} = \frac{-3\sin(0)}{-\sin(0)} = \frac00$$

[boreal girder]

@stone jackal

[slim lodge]

ah i see

it doesnt produce a 0 here

so

[stone jackal]

Oh, I misinterpreted your question, sorry.

[slim lodge]

my horizontal tangents would be = pi/3 and 2pi/3, and pi and 2pi?

[stone jackal]

How did you get 2pi/3?

[stone jackal]

Also, what do pi and 2pi represent?

[compact pewter]

@slim lodge Has your question been resolved?

[slim lodge]

sorry

pi and 2pi coem fromt he denominator

sin(0) = pi,2pi

-3sin(3theta) =

pi/3 and 2pi/3

comes from numerator

[stone jackal]

Oh, I understand now.

[slim lodge]

are these my 4 points

?

[stone jackal]

4 points?

[slim lodge]

i mean

points where this is horizontal tangent

[stone jackal]

Your vertical tangents are wrong. Your horizontal tangents are correct.

[stone jackal]

Sorry for the long delay; It took me a while to understand what you were saying.

[slim lodge]

oh i mean i didnt solve for vertical tangents yet

i dont think

[stone jackal]

Oh, are you trying to prove that denominator =/= 0?

[slim lodge]

i dont really know what im doing, all i know is that dy/dx = 0 for horizontal tangents

[stone jackal]

$$\dv{y}{x} = 0 \implies \theta = \frac{\pi}{3}, \frac{2\pi}3$$

[boreal girder]

@stone jackal

[stone jackal]

I agree with this part of your statement.

[slim lodge]

not with the pi and 2pi?

[stone jackal]

No, because those theta values cause both the numerator and denominator to be 0.

[slim lodge]

ah

sorry

[stone jackal]

No need to apologize; everyone is here to learn. 🙂

[slim lodge]

as for vertical tangent

can you walk me through per chance

[stone jackal]

$$\dv{y}{\theta} \neq 0, \dv{x}{\theta} = 0$$

[boreal girder]

@stone jackal

[slim lodge]

so set dx/dtheta to 0

set -sin(theta) to 0 then?

but

it does the weird thing

with pi and 2pi

so they cant be vertical tangents

right

so what would be the correct step to appracoh this?

[stone jackal]

,w -sin(t) = 0, 0 <= t <= 2pi

[boreal girder]

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[stone jackal]

That's it. Since you never meet both condiions at any single theta value, you conclude that there are no vertical tangents.

,w plot (x, y) = (cos(t), cos(3t)), 0 <= t <= 2pi

[slim lodge]

i see

[boreal girder]

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[slim lodge]

thank you

[slim lodge]

random question real quick is it possible to have the same points

for horizontal tangent

and vertical tangent

[stone jackal]

This is not possible, because the conditions directly conflict with each other.

[slim lodge]

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[random mica]

Let n be a positive integer. A board of size N = n^2 + 1 is divided into
unit squares with N rows and N columns. The N^2 squares are colored
with one of N colors in such a way that each color was used N times.
Show that, regardless of the coloring, there is a row or a column with at
least n + 1 different colors.

[random mica]

We're supposed to use the pigeon hole principle

[safe leaf]

start small

maybe start with n = 1

[random mica]

yes i have checked N = 2,5

but I'm struggling to get anything definitive

[safe leaf]

maybe try n = 3

perhaps you need more examples to find the pattern

[random mica]

hm ill have a look

is it possible to prove a stronger form where theres at least one column with n+1 different colours?

[unkempt sail]

Wdym by stronger form

[random mica]

i didnt say row or column, i just said column

[unkempt sail]

???

[random mica]

In the original question statement, they asked us to prove that there exists a row or column which has at least n+1 different colours. I'm wondering whether its possible to prove that there exists just a column which has at least n+1 different colours

[unkempt sail]

I mean you can take cases

The question just asks for a general row or column

[random mica]

<@&286206848099549185>

[compact pewter]

@random mica Has your question been resolved?

[compact pewter]

@random mica Has your question been resolved?

[slim osprey]

Ok, assume that a colour was found in r rows and c columns. What can you tell about rc in terms of N?

[random mica]

1 min ill come to this

[random mica]

rc<N^2

[slim osprey]

That’s correct but not what we are looking for

[random mica]

r+c<N ?

no wait

r+c<2N

[slim osprey]

Look for rc first not r+c

[random mica]

but

if say the colour is on every diagonal square

then r=N,c=N and rc=N^2

so im not finding any better bound

do i find a lower bound, an equation??

@slim osprey

[slim osprey]

You find an upper bound for rc

[random mica]

but the best upper bound is N^2

as i explained

at least only in terms of N

[slim osprey]

Ohhh, I’m so sorry you want to find a lower bound. I’m so sorry

Yeah, ofc the upper bound is going to be N^2

[random mica]

right

rc>N

[slim osprey]

Yes! What does this tell us about r+c?

[random mica]

perhaps r+c>2n

[slim osprey]

Alright, this is true, so now what does this tells us about at least how many distinct lines each colour uses?

[random mica]

lines as in rows/columns?

[slim osprey]

Yeah

[random mica]

each colour appears in at least 2n rows+columns

[slim osprey]

Yes. So adding everything up, what is the least number of distinct (colour, line) pairs

[random mica]

N colours * 2n lines gives a lower bound of 2Nn distinct pairs

[slim osprey]

Yes, exactly. Now use pigeonhole principle given that there are 2N lines on the board (aka the rows and columns)

[random mica]

2Nn distinct (colour,line) pairs (pigeons), 2N lines (pigeon holes), means theres a line which has n distinct colours

[slim osprey]

Ok we have a problem, I see what went wrong. But the ideas we have our very close

[random mica]

wait i think r+c >2n + 2

perhaps

in fact how do we even prove that part

[slim osprey]

Instead of 2n did you try 2sqrt(N)

This is what we need, and I’ll show you why

[slim osprey]

You prove it using AM-GM inequality

[random mica]

ohh

(r+c)/2 > sqrt(rc) => r+c > 2sqrt(N)

r+c > 2sqrt(n^2 + 1)

[slim osprey]

Yes, exactly

Now try again with the same steps

[random mica]

N colours * 2sqrt(n^2+1) gives a lower bound of 2Nsqrt(n^2+1) distinct pairs

now we need an integer approximate for 2Nsqrt(n^2+1)

and it has to be something like 2Nn+1

[slim osprey]

We don’t need to. By pigeonhole, we can say that one of the 2N rows must have at least sqrt(n^2+1) different colours. Then we can take the ceiling of this which is just n+1

[random mica]

oh

ok this was a pretty cool proof!

i guess the central idea was the distinct (colour,line) pigeons and the line holes

[slim osprey]

Thanks, sorry if I was confusing

[random mica]

then we started finding inequalities

[slim osprey]

Yes

[random mica]

no you werent at all

[slim osprey]

Exactly

[slim osprey]

Okay that’s good to hear

[random mica]

ill keep this in mind for next time thanks so much!

[slim osprey]

Sure! No problem

[random mica]

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[turbid surge]

A sub group of $A \subseteq \mathbb{N}^+$ is called a natural line if for every $a,b,c \in \mathbb{N}^+$ such that $a<b<c$ if $a,c \in A$ then $b \in A$. A subgroup $A \in \mathbb{N}^+$ is called a natural line with k holes if A isn't a natural line and there exists k different values $a_1,a_2,...,a_k \in \mathbb{N}^+$ such that $A \cup {a_1,a_2,...,a_k}$ is a natural line. \
Prove/disprove: there exists $m \in \mathbb{N}$ such that every sub group of $\mathbb{N}^+$ is a natural line with at most m holes.

[boreal girder]

prograce

[turbid surge]

I feel like this is wrong because for every m that fits for all the subgroups we can always make a bigger subgroup that needs a bigger m than it

[hushed magnet]

I assume instead of the word subgroup you mean subset?

[hushed magnet]

can we?

[turbid surge]

I think so yeah

[turbid surge]

Yes? if I assume a certain "m" is the utmost number of holes for every subset, then I can simply add to the biggest number "x" the number "x+2" to the biggest subset so now the number of holes is m+1

[hushed magnet]

that assumes the subset has a biggest number

what about {1, 3, 4, 5, 6, ...}

[turbid surge]

A subset can be infinite too by definition?

[hushed magnet]

of course

[turbid surge]

I think statemnt is true then

I can try proof by contradiction?

[azure vault]

if you want to try to prove the statement is true, then idk, maybe

||if the statement is true||

[turbid surge]

Is it not true ?

[azure vault]

well I certainly don't think that it's true

[compact pewter]

@turbid surge Has your question been resolved?

[compact pewter]

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[misty quail]

Given a 7x7 grid, each unit square containing a non-zero real number. Prove that there exists such grids so that the product of numbers in any 3x3 square is equal to the product of numbers in any 4x4 square and that the sum of every numbers in the grid is 2025.

[misty quail]

Sorry my phone died earlier

I have solved the problem but i think you guys would come up with more clever solutions

By the way, i'd like to present my answer but don't know how. How would i create a grid?

[slate sand]

You can try the tabular environment in latex

[eternal pulsar]

Yeah

I guess you can use the matrix environment, but there might be clutter

[short ferry]

idk if something like this would work:
Fill almost everything with 1s, somewhere in the corner place 2 reciprocals, which'd make the total sum 2025

There'd be a total of 47 ones

,calc 2025-47

[boreal girder]

Result:

1978

[short ferry]

a + b = 1978
ab = 1

,w x^2 - 1978x + 1 = 0

[boreal girder]

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[short ferry]

those are the 2 reals you'd place in the corner

stupid solution, perhaps not what they wanted, but works

1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 0.0005 1977.9995

oh and btw in the proof you wouldnt have to compute all that

it would probably suffice to refer to vieta's formula and perhaps present the quadratic

[azure vault]

but then this 3 x 3 square
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 0.0005 1977.9995

doesn't have the same product

[short ferry]

oh wait

yeah

time for plan B then

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[mild heart]

Hey, I don't understand what I did wrong

[mild heart]

The answer is apparently supposed to be
F = 9.0(i-j)N

But I got
F = 9.0(-i+j)N

[prime sigil]

same thing vro

[mild heart]

But my answer is 180° from the correct one

[prime sigil]

Then can't help U

[mild heart]

Ok

[compact pewter]

@mild heart Has your question been resolved?

[mild heart]

It might be because + + repels

.close

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[vale moat]

yo

[vale moat]

how do i do C

[acoustic yarrow]

you would need position of B=position of A

find some position function for B and A

[vale moat]

the position function has to include time

so that we can do simultaneous equations

[acoustic yarrow]

you dont need to.

have you learnt motion in a single dimension

and with constant acceleration

[vale moat]

yeah

[acoustic yarrow]

try modelling car B's position then at time t

[vale moat]

S = 30t

[blazing arrow]

B overtakes A when the total distance covered by B is greater than the total distance covered by A up until that point, try recalling the 3 Newtons laws of motion

[acoustic yarrow]

that is true for Car A

good

now car B

notice car B has an acceleration

[vale moat]

?

[acoustic yarrow]

$s = ut + \frac{1}{2} a t^2$

have you used this formula before

[boreal girder]

mia

[vale moat]

S = 20t + 2t^2

i just used the wrong formula

i thought all the formulas give the same results

[acoustic yarrow]

now find t when S_B = S_A

[vale moat]

5

[acoustic yarrow]

yes

[vale moat]

but why specifically that formula

[acoustic yarrow]

this one?

what formula did you initially use

[vale moat]

s = 1/2(U+V)t

[blazing arrow]

thats a formula?

[acoustic yarrow]

yes

in fact you have that

[vale moat]

i didnt learn that yet

[acoustic yarrow]

how would you express V?

you know u = 20 ms^-1

[vale moat]

v = 20 + 4t

so

[acoustic yarrow]

good

now simplify

[vale moat]

s = 1/2(u + (20 + 4t))t

[acoustic yarrow]

mhm

when you expand, you see something really cool

[vale moat]

gimmie sec lemme write this on paper

(u/2)t + 2t^2 + 10t

[blazing arrow]

u havent learnt newtons laws of motion yet ur solving motion in 1D? im sorry what?

[acoustic yarrow]

okay and u is?

[vale moat]

20

so 2t^2 + 20t

[acoustic yarrow]

yes

its the same thing we got before

[vale moat]

nicee

[acoustic yarrow]

if you have done integration, you can see the intuition behind these formulas

[vale moat]

yeah

my course covers it next year

idk why

[vale moat]

they are derived from velocity time graphs right

[acoustic yarrow]

they can be

[vale moat]

okay

thanks

.close

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[hearty grove]

4b)

[short ferry]

have you found A already?

[dense dagger]

if you were able to do 4a, the rest should be easy. You just integrate it directly

[hearty grove]

Yes

A= 4

[short ferry]

then integrate the result of a) instead

integrate 3 + 4/(2x - 3)

[short ferry]

,rcw

[short ferry]

great

now just evaluate it at the given bounds

oh you already did

[dense dagger]

he didn't. should get the correct answer when you add the bounds

keep in mind ln(a) - ln(b) = ln(a/b)

[hearty grove]

Thx

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[uneven viper]

How would one find the partial solution to the ODE

[uneven viper]

I mean I know it's by undetermined coefficients but what do we guess exactly

[versed sigil]

1)write the power series for y^(4) and y''' by using taylors theorem for y and differentiating
2)expand the right side using taylors theorem
3)compare coefficients

[uneven viper]

So using the method of power series?

[versed sigil]

yes. why dont you try it?

[uneven viper]

I don't think that's what the professor wants, we learned that way further into the semester and this ODE problem is from like the start of the semester

Is there any other way to do it?

[versed sigil]

yeah, sure. substitute y''' = u

then you get u' + u = 1-x^2e^{-x}

do you know how to solve equations like these?

[uneven viper]

Yeah, with μ

because it's in the form of u'+p(x)u=q(x)

[versed sigil]

keep in mind i dont know what you learn by the start of the semester, so i might suggest methods that are not the simplest

[uneven viper]

Thanks man, wow haven't thought of that approach

[versed sigil]

no problem 😄

glad to help...

[acoustic yarrow]

im guessing your prof wants you to consider this as a homogeneous equation?

[uneven viper]

But it's not a homogeneous

[acoustic yarrow]

you solve the homogeneous then find the particular solution

[versed sigil]

oh... partial solution.. didnt read that 😄

[uneven viper]

Yes that's what I was doing at first

But I couldn't find the general form of the partial solution

To determine it's coefficients

[uneven viper]

?

[versed sigil]

nothing!

anyways, once you solve for u, you just need to integrate a few times to get y, im sure you know that

[uneven viper]

Yeah

I'm doing that now

Hold on

It's getting extremely large though

[hexed magnet]

wait i might be restarted

but cant u js.

substitute v(x) = y'''(x)

[eternal pulsar]

Already done.

[hexed magnet]

and u get a linear

oh is it that bad?

[eternal pulsar]

^

[hexed magnet]

lemme try

[eternal pulsar]

Yeah...

[hexed magnet]

myb myb

[uneven viper]

Yea it's getting kinda out of control

What to do in this situation.. just patiently integrate?

<@&286206848099549185>

[compact pewter]

@uneven viper Has your question been resolved?

[uneven viper]

@hexed magnet @eternal pulsar @versed sigil sorry to bother guys, but did you find anything?

[hexed magnet]

wait lemme try

oh am i stupid

is it js trivial

[uneven viper]

Wdym

[hexed magnet]

solved for v already

oh u guys got here alr

whats left to the problem after this?

now u js integrate thrice

[versed sigil]

yeah, i dont see whats the problem!

[versed sigil]

the origianl poster seems to be dissatisfied that integrating this function is taking a while

[hexed magnet]

ah

maybe use DI method instead of IBP?

[uneven viper]

Alright

Just patient integration I guess lmao

Thank you all for your help

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[ivory saddle]

just change it to + C you are allowed

• 2C is the same as + C

ye or just replace + 2C to + C once ur done

[eternal pulsar]

C is arbitrary

[ivory saddle]

its cuz C is some constant, so 2C is also just a constant so u can just call it C for constant

if that makes a sense

its a bit tricky to explain ngl

👍

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[stone jackal]

$$r = 3 + 2 \cos(\theta)$$
$$y = 3\sin(\theta) + \sin(2\theta) \implies \dv{y}{\theta} = 3\cos(\theta) + 2 \cos(2\theta)$$
$$x = 3 \cos(\theta) + 2 \cos^2(\theta) \implies \dv{x}{\theta} = -3\sin(\theta) - 4 \cos(\theta) \sin(\theta)$$
$$\dv{y}{x} = \frac{3\cos(\theta) + 2\cos(2\theta)}{-3 \sin(\theta) -4 \cos(\theta) \sin(\theta)}$$

[boreal girder]

@stone jackal

[stone jackal]

Can someone check my work?

[craggy sluice]

looks right to me

[stone jackal]

Previously, I had been using an ugly formula, but I want to make sure I'm using this new method correctly.

[stone jackal]

Thanks!

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[gilded kayak]

whats the difference?

[gilded kayak]

or, when do i use double integration over single var integration?

[sturdy forge]

well, both the integrals you wrote on the right evaluate to the same thing

so... there is no difference for this particular expression

[gilded kayak]

what about this?

[sturdy forge]

since $\int_{h(x)}^{g(x)}dy=g(x)-h(x)$

[boreal girder]

00100000

[sturdy forge]

oh jesus. my memory is all foggy from my multivariable calc class, so I'm going to refrain from answering further, sorry :^(

[gilded kayak]

o its ok

oh i got it

within double int is z=f(x,y) so im not really just finding areas

finding areas using double int is integrating 1 twice

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[urban glen]

how do I do this on calculator I'm not getting correct one on desmos

[native shard]

so you know dy/dx = 4 at t = 2

because it’s parallel

and dy/dx = dy/dt / dx/dt

[urban glen]

it depends on the value of k

[native shard]

so just plug in t = 2

[native shard]

yes

that’s what we’re solving for

$3ke^{2k} = 4$

[boreal girder]

knief

[urban glen]

whats this for

equating slope?

[native shard]

yes

since they’re parallel

[urban glen]

so i plug in t=2 to find dy/dx and set it equal to 4?

and solve for k

[native shard]

also why did you do g’/f’

y’(t) and x’(t) were already given

no need to differentiate

[urban glen]

oh yea mb 😭

[native shard]

yea using dy/dx = dy/dt / dx/dt

[native shard]

mhm

[urban glen]

i solved it

thank you 🙂

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[tender pasture]

why is c not 17

[tender pasture]

at october 25% is 17

[coral halo]

keyword: greater

[tender pasture]

so is 50-17?

[coral halo]

?

[tender pasture]

how do we do it

[coral halo]

you said its 17 because its the first quartile

[tender pasture]

yh

[coral halo]

but the questions ask for 25% greater than

if you used the first quartile here, it would be 75%

[tender pasture]

is it not like this?

why r we reading it upside down

[coral halo]

follow your scale?

the numbers increase as you go up

not down

[tender pasture]

it says greater then 25%

so wouldnt it be 17

bc 25% is in 17

[coral halo]

25% of them are less than 17

and 75% are greater than 17

[tender pasture]

50% is also greather then 17

bruh

[coral halo]

50% is your median

[tender pasture]

yh

there asking for more then 25% so wht do i do

[coral halo]

use the upper quartile

not the lower one

[tender pasture]

why

it works the ans is correct

but why do we do tht

[tender pasture]

yh

ik this

wait wht

[coral halo]

if you use q1 = 17, you are saying that 75% of them are greater than 17

not 25%

[tender pasture]

is it mot like this

[tender pasture]

oh

i see

[coral halo]

well thats the label on q1 q2 and q3 yeah

[tender pasture]

calm thanks

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[flint wing]

if i am finding the angle between two planes in R3 i used the dot product formula and got cos(theta) = some negative value

[flint wing]

for example cos(theta) = -7/9 or whatever

do i take the positive of this?

like cos(theta)=7/9

or do i use negative

[knotty basin]

waterbeam

[flint wing]

sam do u know]

[sick shard]

Yes you can take positive of it

Taking positive of it will give the acute angle between planes while talking negative will give obtuse

But we usually care about the acute angle so yea it's ok

[flint wing]

ok thanks 👍

.solved

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[deft tangle]

My answer is ABC all looks correct

[kind viper]

"looks"?

do you have proof or is it just vibes

[floral fjord]

@deft tangle

From the above

We can get A U B = A intersection B

And when is that possible

[warped oar]

If there is an element that is not in A ∩ B but in A, then A U B cannot be A ∩ B.

Also, A >= A ∩ B.
So A = A ∩ B.
Similarly, A ∩ B = B.
So A = B.

[deft tangle]

There is no vibe then intersection things mean B=C

And first one means A=B

@warped oar @floral fjord

[floral fjord]

ok

[kind viper]

so $A \cap B = A \cap C$ \textbf{by itself} already implies $B=C$? is that what you claim?

[boreal girder]

Ann

[floral fjord]

what

[kind viper]

(it's wrong)