help-0

[pliant cedar]

do u know the chain rule

[past current]

what's that

[pliant cedar]

oh okay so u dont

its a rule u can use to differentiate

then maybe u shud do it wthout calculus

[limpid turret]

In short, @past current, there are multiple ways to solve this. You can do it algebraically, geometrically, trigonometrically, or with calculus.

[pliant cedar]

or maybe u do and u dont remember?

[limpid turret]

We just trying to find the solution that suits what class you are in

what's the cover of your book say?

[past current]

more likely than not, I know the concept but not the name for it

[pliant cedar]

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

[ocean seal]

SilverSoldier

[pliant cedar]

u know this?

[past current]

and this is the practice paper name

[past current]

no I don't

[pliant cedar]

oh

[limpid turret]

Okay this is introductory algebra

[pliant cedar]

do u know how to find $\frac{d}{dx}f(g(x))$ in any way at all?

[ocean seal]

SilverSoldier

[limpid turret]

no calculus should be used here

[pliant cedar]

oh

okay then do it geometrically

[limpid turret]

nah

I see what you need to do now

[past current]

this is all I know in terms of differentiation, so, probably not

[limpid turret]

Solve your circle equation in terms of y

[remote heron]

toki kijeta

[limpid turret]

Then set your line and circle equations equal to each other

[past current]

toki!

[limpid turret]

This will give you a quadratic equation

For the circle and line to just touch, the quadratic must have only one solution

[past current]

I've been trying to find b^2-4ac = 0 somehow

[limpid turret]

That is, its determinant must be 0

[limpid turret]

Do it by how I just told you

[past current]

I'm not sure I'm getting anywhere😅

[mellow stag]

will give 5$ if u help

[limpid turret]

@mellow stag open your own help channel and receive help for free

[limpid turret]

Solve for y in the equation of your circle

Do that first

[mellow stag]

how

[lone heart]

Please read #❓how-to-get-help

[past current]

hang on, just realized, what I've done is solving for m & x 😭😅

[limpid turret]

What you have now

Is great

You ended up where I wanted you

But faster

[past current]

oh right ok

[limpid turret]

Now simplify it into standard quadratic form

[past current]

what do I do w the m

[limpid turret]

leave it

Just put the whole thing in standard quadratic form

[past current]

can I pull like terms together further?

[limpid turret]

yes

[past current]

without interfering w/ m

[limpid turret]

Ax²+Bx+C=0

Put it in that form

You have x² and (mx)²

Group them

Also group 6x and -12mx

[past current]

I doubt this is correct?

[limpid turret]

x²+m²x² is not 2mx²

[past current]

2m²x² ?

[limpid turret]

You may wish to review how to add variables

but it's (1+m²)x²

Distribution law of algebra: ab+ac=a(b+c)

[past current]

oh wait I see how it gets there

[limpid turret]

good

[past current]

in a way, it's like factorising out x²

[limpid turret]

yes exactly

On that note, fix 6x-12mx

[limpid turret]

huzzah

You have your quadratic

[past current]

so I just find two m values where b²-4ac = 0?

[limpid turret]

Now the line will "touch" the circle when there is only one solution to x

That is, only when the determinant is 0

That is, only when b²-4ac=0 (as you said)

So solve for m (as you said)

@past current how's it going?

[past current]

i must have got something wrong somewhere

it's not solvable

[limpid turret]

show your work

because you are correct up to this point

[limpid turret]

Oh I see

The sign of -16 is wrong

Should be +16

Find where that happened

Then you are done

[past current]

ohhh nvm I found it

-8 * -2 = 16, not -16

all the way at line 1

[limpid turret]

yeah I missed that too

[strange sleet]

Take a walk down the Stern-Brocot tree, along the red bold line up to $\frac{2}{3}$, then walk down that entire green infinite subtree and collect everything you find. That is the meaning of $ Q \cap [\frac{1}{2},1] $

[ocean seal]

agilepotato

[past current]

I think I've done it, thank you soo much for your help and patience, @limpid turret ! <3

[limpid turret]

Looks good to me

[lone heart]

@past current Has your question been resolved?

Channel closed

Closed by @past current

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[stuck cape]

how do i calculate A^k

[tacit arch]

Find a similarity transformation for A

[stuck cape]

?

[livid narwhal]

is X' X^T?

[stuck cape]

yes

[livid narwhal]

have you seen $A = VDV^{-1}$ before?

[ocean seal]

Shell

[livid narwhal]

where D is a diagonal matrix and V is just a matrix?

[stuck cape]

nah

[livid narwhal]

well you need to know about that

[livid narwhal]

if you believe me when I say this is true

[stuck cape]

i think it can also be solved by using Binomial theorem

but its lengthy

[livid narwhal]

then $A^k = (VDV^{-1})(VDV^{-1})...(VDV^{-1})$ k-times so $A^k = VD^kV$

[ocean seal]

Shell

[livid narwhal]

which is useful because if D is diagonal then D^k is just all the entries to the k

[livid narwhal]

when has that ever stopped anyone

[stuck cape]

maybe at some exponent it'll be zero

and if not then time wasted

[naive valley]

there must be some shortcut or this problem is crazy

according to matlab the value of k is not in the single digits

[naive valley]

just to add to the fun, -1 doesn't have full geometric multiplicity

[livid narwhal]

yeah i just noticed that

so its not even diagonalizable

[stuck cape]

[naive valley]

by brute force computation, the answer is k=10, perhaps that can give a clue as to how one might solve this without brute force

1 10
2 9
3 4
4 15
5 -2
6 21
7 -8
8 27
9 -14
10 33

left number = k, right number = x' A^k x

[stuck cape]

this is the solution

the person who made this problem is pro
but who solved is ultra pro max

thanks

[naive valley]

oh they only looked at even values of k

the odd ones don't give that simple form

[stuck cape]

tnx for your help

.close

[lone heart]

Channel closed

Closed by @stuck cape

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[lilac thicket]

hi everyone !
My question is: Do homomorphisms from smaller structures to bigger structures generally exist ?

[rocky grove]

Topology

[lilac thicket]

I am sorry if I am not exact on some terminology , what I mean is, if I have a graph for example with 5 nodes and 6 edges, could there be a homomorphism to a graph with 10 nodes and 9 edges

[lilac thicket]

`:D?

because from what I have seen so far the homomorphisms were always from a graph with more nodes than the graph it was mapped on.

but just from the informal definition from google I'd say such homomorphisms should be possible as the main requirement for a homomorphism is to preserve the structure

[vale wigeon]

vertex and edge counts alone cannot tell us whether such a homomorphism could exist (in this case, at least)

maybe it could, maybe it couldn't. it depends on the particular graphs you're looking at

[lilac thicket]

I just wanted to provide a rough example of what I meant

by bigger, and smaller

we could also say from a smaller set of elements to a larger set of elements

[vale wigeon]

i mean, yes, there DO exist graphs G and H such that G has less vertices than H and yet a homomorphism from G to H exists.

[lilac thicket]

Can I imagine it like this :
For every node in G where |V(G)| < |V(H)| there simply is a cycle in H ?

[vale wigeon]

|V(G)| < |V(G)|???

[lilac thicket]

I am sorry

[vale wigeon]

anyway, i think you are overthinking it.

[lilac thicket]

could be :D

[vale wigeon]

a homomorphism is just a mapping of vertices to vertices and edges to edges, nothing more.

[lilac thicket]

yes but I just dont fully get when it is correct to build a homomorphism from a smaller to a larger structure

dont really have a great sense for homomorphism yet

so I am trying to get some examples into my brain

[lilac thicket]

for these two I could find h: G -> H right ?

[lone heart]

@lilac thicket Has your question been resolved?

[ruby current]

the trivial homomorphism always exists between any two structures

[vale wigeon]

wdym by "correct"?

[lone heart]

@lilac thicket Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[alpine sable]

If a sequence has one element say 5
Does that mean it converges to 5?

Please confirm

[limpid turret]

yes

passes cauchy test iirc

[alpine sable]

Thank you!

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[alpine sable]

Please help
Prove that the converse of part c is false by finding the sequence {xn} such that {rn} converges but xn doesn't not

[alpine sable]

Part a is: prove if xn -> L then rn -> L

For a given sequence xn we define its sequence of geometric mean to be rn where rn = (x1.x2.x3...xn)^1/n

[naive valley]

probably you mean the exponent to be 1/n instead of n?

[alpine sable]

Yes

Sorry

[naive valley]

np

[alpine sable]

Fixed

[naive valley]

is there any constraint like the x_n's have to be strictly positive?

if they can be zero then it's easy to find a counterexample

[alpine sable]

Ooh yess
There are no constraints

So it can be 0 or any other number

The ans will converge to 0

Makes sense

[naive valley]

seems like you could just let x1 = 0 and then x2... can be any divergent sequence

[alpine sable]

Fair enough

And then xn won't converge cause we will have 2 cluster points right?

[naive valley]

well the product will always be zero

so the geo mean will always be zero

but the sequence will do whatever you make it do

for example: 0,1,2,3,4,...

diverges to +infinity

[alpine sable]

ah I see

Thank you so muchh!

[naive valley]

sure

[alpine sable]

.close

[lone heart]

Channel closed

Closed by @chrome path

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[dusky kraken]

I understand how to get b c d but I do not know how to figure out what a is

[vale wigeon]

f(0)=4

a * 1 * (-1) * (-4) = 4

[rancid hazel]

exactly

and then work from there

[dusky kraken]

Ok interesting so i would assume that a is 1?

[rancid hazel]

yes

[echo socket]

assume

[vale wigeon]

no need to assume

a*4 = 4, you can solve for a

[dusky kraken]

So I understand that the numbers in the brackets multiplied gives the y-int but i just didn’t understand what to do with the a

[echo socket]

Solve for it

And plug it into the final expression

[alpine sable]

https://www.desmos.com/calculator/95ir8qe0yi

Note that the roots are not quite enough to completely define a polynomial, since we can stretch it by some factor and still preserve the points where it crosses the x axis

[lone heart]

@dusky kraken Has your question been resolved?

Channel closed

Closed by @dusky kraken

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[daring citrus]

Currently unsure as to where to go from here

.close

[lone heart]

Channel closed

Closed by @daring citrus

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[plain sundial]

hey

[plain sundial]

where did i go wrong here

<@&286206848099549185> 🥲

[wanton pebble]

don't ping helpers

[plain sundial]

been 3 days

if i don't ping no one's gonna come lol

[plain sundial]

anyways the answer sheet says that the p=2 and q=16

[wanton pebble]

okie

see u got p

[plain sundial]

i subbed that in too i don't really see what's wrong here

[frigid stream]

I would argue that it is more likely that they come if you don't ping them right away... 😄

[wanton pebble]

and x=4 is a solution

so put x=4

[plain sundial]

provlem solved we're not arguing no more

[plain sundial]

thats what i did

[wanton pebble]

okie

see its minimum value is 0 at x =4

can you make a quadratic equations whose both zeros are 4

i will be like (x-4)^2

so compare these 2

maybe it will help

[plain sundial]

so

the thing is

do you know

where did i go wrong

did i do the question wrongly

did i do it right

[wanton pebble]

your second step was worng

[plain sundial]

this part?

[wanton pebble]

expand (x-p/q)^2

yeah

[plain sundial]

if i do it again that way

i mean the way i did in the photo

but correctly

the answer's gonna be correct no?

oh

crap

i just realized

it's complete the square i'm supposed to write q/2p

THANKS fantastic mr. fox

.close

[lone heart]

Channel closed

Closed by @plain sundial

Use .reopen if this was a mistake.

[wanton pebble]

i don't know man... honestly I don't even know what did you do here

[plain sundial]

i was doing the complete the square thingy

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[plain sundial]

i forgot the b part was supposed to be divided by 2

[wanton pebble]

i mean as a whole solution

cause you are solving a very simple question with such difficult way

[lone heart]

@plain sundial Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[dim matrix]

It is fine if we get two different triangular matrices from a given matrix?

[wanton pebble]

what?

[feral turtle]

prolly meant "is it"

[mortal trellis]

If you chose a different order of the eigenvalues then the triangular matrices will look different

[dim matrix]

I solved this here

But when I check online calculators, I am getting a bit different answer

[mortal trellis]

Oh you mean rref

[worn fox]

rref is unique iirc

so you've gone wrong somewhere

[mortal trellis]

Well those aren't reduced

ref is not unique. Only rref is

[worn fox]

ah yes

[mortal trellis]

If you subtract the third row from the second and then flip the sign of the second row in the pic, you get your result

Bit weird that you added the third row to the second instead of the second to the third

Not wrong of course but you need to switch the rows after that

[dim matrix]

But is it necessary to solve it further once I get a triangular?

[dim matrix]

I'm new to math 🙂

[mortal trellis]

Wdym. My message just shows that the screenshot and your result are equivalent

[dim matrix]

Always ran from it but ended up here again

[dim matrix]

Yeah I understood this. Just asking if I should solve it further to match the result with the online calculator, or if I don't need to solve it further.

[mortal trellis]

Well what is the original question

[dim matrix]

This one

[mortal trellis]

Well then either is fine

From both you get the same rank

[dim matrix]

okay, many thanks @mortal trellis 👍

.close

[lone heart]

Channel closed

Closed by @dim matrix

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[tranquil anchor]

hey

[wanton pebble]

hy

[tranquil anchor]

i would like some help with a if possible

[gray isle]

what have you tried?

[tranquil anchor]

what does it mean size like area?

[gray isle]

size of the angle

(in degrees)

[tranquil anchor]

ohhhhhhhhhh

[wanton pebble]

like can u use trigonometry?

[royal plank]

ngl size does sound like area

[tranquil anchor]

i didnt see the the ^ above x

i am blind

[royal plank]

ah

[tranquil anchor]

yea it does xd

[tranquil anchor]

on my screen its practically invisible

ok if its angle i got this

lemme do it and see if i got it correct xd

[tranquil anchor]

i got 50 degrees

i used sin90/13 = sinx/10 and put it in my calculator it gave me 50.3

[wanton pebble]

what?

[gray isle]

that gets you <WXZ

which isn't quite what the question wants

[tranquil anchor]

wait i add 50.3 to 105? bc it wants to find the entire thing

or am i wrong

[wanton pebble]

yeah you are right

[tranquil anchor]

aight tyvm

for this one

where it says find the angle of EMD

does it mean the whole angle bc that would be 180 or the angle on the triangle on the right?

[tranquil anchor]

oh

[tranquil anchor]

70 right?

cuz the total must be 180

[wanton pebble]

how

[tranquil anchor]

like total of both angles of M must be 180 since its a straight line

and since M2 is 110 because that would make M1 70

how do i do B here

i am questioning my life rn

.close

[lone heart]

Channel closed

Closed by @tranquil anchor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[stoic folio]

Are radians more popular or prevalent to use over degrees?

[stoic folio]

I'm very surprised to learn that Microsoft chose to use radians in their [System.Math] class.

[marsh rapids]

degrees are for geometry and children. Radians are for everything else

As my teachers say, "you aren't children anymore, in radians"

Except in optics in practice because it's easier to work with very small values when they're in degrees

Because no one wants 1.3e-5 rad as their angle change

But that's basically just geometry then

[stoic folio]

You're answer is very interesting and timely considering the article I just stumbled onto yesterday. https://www.theverge.com/tldr/2018/3/14/17119388/pi-day-pie-math-tau-circle-constant-mathematics-circumference-diameter-radius-holiday-truth

[vale wigeon]

actually, snipers use a unit called angular mils

which is an approximation of 0.001 radian

it is more convenient this way to adjust your aim by small increments

[stoic folio]

.close

[lone heart]

Channel closed

Closed by @stoic folio

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[thick lynx]

Is $\lim \sum ab = \sum \lim a \lim b$?

[vale wigeon]

gonna need some clarification on that notation.

[marsh rapids]

If the sum is finite yes. If it's a series it only holds under certain conditions

[vale wigeon]

i'm not sure what the notation even maens.

means*

[marsh rapids]

So clarifying your notations is necessary

[thick lynx]

I have $\lim_{n \to \infty} \sum_{x = 0}^n \frac{n!}{(n - x)!n^x} \frac{1}{x!}$

[vale wigeon]

ouch. x as summation index

[thick lynx]

Now it would be really nice if I could pull that limit in

Since that would prove that this is e

[thick lynx]

Should've picked k, lol

[vale wigeon]

$\lim_{n \to \infty} \sum_{k=0}^n \frac{n!}{k!(n-k)! n^k}$

[ocean seal]

Ann

[vale wigeon]

hm wait hold on

that sum is a binomial expansion of (1 + 1/n)^n isn't it

[thick lynx]

Yes

[vale wigeon]

i'm not sure how your interchange of limit and summation would work symbolically though

$\lim_{n \to \infty} \sum_{k=1}^n a_k b_k \overset?= \sum_{k=1}^{\infty} ....???$

[ocean seal]

Ann

[marsh rapids]

Fortunately ak and bk have limits as n goes to inf so it actually is writable

[vale wigeon]

oh then they are doubly-indexed sequences?

yeah i guess they depend on both n and k

[marsh rapids]

Yes

[thick lynx]

$\lim_{n \to \infty} \sum_{x = 0}^n \frac{n!}{(n - x)!n^x} \frac{1}{x!} = \sum_{x = 0}^n \lim_{n \to \infty} \frac{n!}{(n - x)!n^x} \lim_{n \to \infty} \frac{1}{x!} = \sum_{x = 0}^n \frac{1}{x!} = e$ if this is valid

[lime swallow]

12m 🧒

[vale wigeon]

$\lim_{n \to \infty} \sum_{k=1}^n a_{n,k} \overset?= \sum_{k=1}^{\infty} \lim_{n \to \infty} a_{n,k}$

[ocean seal]

Ann

[vale wigeon]

it's this, by the looks of it...?

the product thing was just a distraction

[thick lynx]

Well the limit of a product is the product of the limits

[thick lynx]

so we just need to prove or confirm that, yes

[marsh rapids]

e^(n ln(1+1/n)) is linearly approximated as n -> inf by e^(n * 1/n) = e

[vale wigeon]

well, now i at least understand what kind of manipulation was meant

at least from a symbolic standpoint

but i am not sure whether it's always true, or what conditions could be placed on a_n,k to guarantee its truth.

[marsh rapids]

You can see it as a sequence of series

[vale wigeon]

can you...?

[marsh rapids]

A finite sum is a series over an almost-null sequence

Not saying it's valid though

That manipulation

[vale wigeon]

well, its validity is what i am doubting.

[alpine sable]

i think there must be an easy counterexample

a_n,k = k/n should break it

[thick lynx]

I just found a question about this on Math StackExchange
https://math.stackexchange.com/questions/1937534/limit-of-infinite-sum-equal-infinite-sum-of-limits

So it does not follow in general

Thank you

.close

[lone heart]

Channel closed

Closed by @thick lynx

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[stuck cape]

is this correct question

yes it is

[lilac thicket]

where exactly are you struggling ?

[stuck cape]

CHOOSE ONE OF THEM BELOW


1. did not know to find median
2. don't know statstics
3. don't know math
4. don't know your name
5. i'm dumb
6. all of above

[lilac thicket]

bruh

[hard patio]

Woah woah woah 💀

[stuck cape]

PULL REQUEST MERGED

type your problem in chat gpt it'll solve for you

[hard patio]

Imagine making up that option only to end up choosing one from 1 to 5

[lone heart]

Channel closed

Closed by @sterile pagoda

Use .reopen if this was a mistake.

[hollow sparrow]

#❓how-to-get-help

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[mortal trellis]

How old are you

[alpine sable]

are you trolling

either way tbh it's bannable so 🤷

[stuck cape]

CHOOSE ONE OF THEM BELOW


1. don't know to add, subtract and multiply
2. don't know english
3. don't know math
4. don't know your name
5. i'm dumb
6. all of the above
7. none of these
8. write custom range between 1 to 7

[alpine sable]

<@&268886789983436800> underaged user

tbh seems like you are either underaged or heavily trolling

so

tbh i doubt a 4 year old is on discord

[mortal trellis]

Obviously trolling

[alpine sable]

but

[gray gorge]

.close

[lone heart]

Channel closed

Closed by @gray gorge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[vagrant meadow]

Can anyone explain me in an intuitive way that why should gradient of f is parallel with gradient of g in langrage multiplier where
f is original function
g is constraint function

[last ether]

Yeah so a function has an infinite number of level sets

The intuition is that you're maximizing/minimizing the level set values, and that's only the points where the constant is tangent to a level set.

Because a function has an infinite number of level sets, you're guaranteed that the constraint is tangent to at least one level set.

Gradients are orthogonal to level sets, and by extension, the constant (where they are tangent, ofc), thus their gradients are parallel

Here's a shitty drawing

The very far-reaching "parts" of the constraint is guaranteed to be tangent to some level set

Long story short, what LaGrange multipliers guarantee is that the constraint, which has to be in the same dimension as the level sets, is tangent to at least one level set, which can be a max or min.

@vagrant meadow hopes this helps

[lone heart]

@vagrant meadow Has your question been resolved?

[vagrant meadow]

Now i have two questions hw do u know that constraint is tangent to atleast one set, what if there exist a case where no tangent exist
And why tangent, why not any other point

??

Ok forget the second qe, answer for the first one alone

<@&286206848099549185>

[cerulean glen]

how can I study the parity of a function?

[west sentinel]

bruh this doesnt work like this u cant just ask in middle of someone dbt switch over to #discussion or smt and delete these msg @cerulean glen

[lone heart]

@vagrant meadow Has your question been resolved?

[vagrant meadow]

Now consider this case where 4 1 2 are the value of f level curve, now the one intersecting the curve has maximum value whereas the one touching it has neither maxima nor minima, now here won't the logic of lagrange gets defeated?

.close

[lone heart]

Channel closed

Closed by @vagrant meadow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[lilac iron]

help pls

[late mango]

equilateral triangle

[lilac iron]

oh

[late mango]

that's isoceles

[lilac iron]

shi

right

.close

[lone heart]

Channel closed

Closed by @lilac iron

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[idle sand]

Can anybody tell me how am ment to solve this comulative distrabution function

[plain flame]

solve it?

that is the cdf

you have the cdf

[idle sand]

yea thats why i don't understand

what the question wants from me

it already gave me the cdf of all the values of x

[idle sand]

this is the textbook answer if that helps

[lone heart]

@idle sand Has your question been resolved?

[desert zephyr]

So you want the derivative of the CDF? Which is the PDF

@idle sand

[idle sand]

yea

[desert zephyr]

Well it's a piecewise function so all you have to do is take the derivative of each piece

[idle sand]

oh ur right

so it would be f(0)= 0.02 and f(4)= 0.04(4)

[lone heart]

@idle sand Has your question been resolved?

Channel closed

Closed by @idle sand

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[queen salmon]

Having 5 boxes numbered 1-5 and having 5 papers numbered 1-5 in the boxes, how many possibilities are there that no paper is in the box with the same number?

[queen salmon]

its an easy one, but i cant get the solution and im guessing the solution is wrong

but just wanted to make sure

[chilly storm]

sretno

so the way it'd be good to start is seeing how many events are possible to happen

if you can get the number 1,1,3,2,2 in the box, that means those are permutations with repetition

in your case we have 5 different number on each paper, and we can mark that as n = the number of possible outcomes for each event

then we have r = the size of each permutation, which is the number of boxes you have

the formula being n to the power of r, gives you 5 to the power of five = 3125 (the number of permutations with repetition)

[queen salmon]

yeah thats what i got

[chilly storm]

and the rest i wouldn't know (how to calculate what is the possibility of having zero papers with the same number as the one on the box)

[lone heart]

@queen salmon Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[buoyant sandal]

How do you get the turning points of a quartic/cubic/quadratic equation?

[mortal trellis]

well the same way you would do it for other functions

[last ether]

turning points as in maxes and mins or inflections?

[lone heart]

@buoyant sandal Has your question been resolved?

[buoyant sandal]

How?

[buoyant sandal]

Yes

[mortal trellis]

or?
yes

...

[last ether]

first derivative test\

[buoyant sandal]

The turning point for a quadratic is the minimum/maximum value with Y

How do I do so

<@&286206848099549185>

[weak bison]

you can find local min and max values of a function by taking its derivative and solving for when it = 0

[buoyant sandal]

What is the derivative

is it M

m for gradient?

[weak bison]

derivatives is from calculus

[buoyant sandal]

Wait

Not gradient

[weak bison]

what level math have you taken so far

[buoyant sandal]

just a few weeks of A level pure maths

1st year or so

[weak bison]

ok so no calculus then

for quadratic functions you will have it be of the form ax^2 + bx + c
you can find the min/max by evaluating the function at -b/(2a)

[buoyant sandal]

Oh discriminant

Nvm I read it too fast

[weak bison]

yea it comes from part of the quadratic formula not the discriminant tho

[buoyant sandal]

-b/2a but you remove the +- square root of the discriminant

seems good

ty

.close

[lone heart]

Channel closed

Closed by @buoyant sandal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[glossy matrix]

@blissful whale can you help?

[blissful whale]

Do you want to compute the limit?

[glossy matrix]

Yes

[blissful whale]

I don't think it exists

[glossy matrix]

How?

I mean

How do I prove it?

[blissful whale]

you can find two sequences of points that converge to different values for x/sqrt(y) but both approach 0

try to make x or y very large in comparison to the other for that

[glossy matrix]

I don't know if that's a valid method. Here's one method from my textbook. Can I just use x = √y curve instead of x=y line, like in this example?

[blissful whale]

it's valid

but if you insist you can also pick an entire curve isntead of just some points on it

[glossy matrix]

Oh sorry, bad wording. I meant like, it might not be valid from exam point of view. Teacher might not give me marks if the method is not in the textbook

Although I'm not really sure

[blissful whale]

ah

[glossy matrix]

Just don't wanna risk it

[blissful whale]

[glossy matrix]

Ah, alright

[blissful whale]

c: t -> (sqrt(t), t) curve seems to work indeed

[glossy matrix]

So how would you write it?

The proof?

In proper mathematical terms

The textbook way is way too wordy

[blissful whale]

true imo 😄

well you just define two curves and say they cross 0

but they do not converge to the same trhing

when put in the function

I would simply do that and only write the conclusion from it.

Basically the lastz sentance in your solution example

[glossy matrix]

Ahhh, makes sense

Thanks a lot

[lone heart]

@glossy matrix Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[south hamlet]

Hi, can someone help me with derivatives?

[rocky grove]

[south hamlet]

Find the differential dy, given:
Y = -x(x²+3)

[rocky grove]

Okay take the derivative after simplifying

[south hamlet]

You know what, let me ask this instead

[rocky grove]

Uh

[south hamlet]

So this is an answer to one of the problems from my tutor, but I don't know which

[rocky grove]

Handwriting is a bit weird

I can't make anything of it

[south hamlet]

Yeah we were in a hurry

Well okay, then can you explain problem c instead then

Or any one is fine really 💀

[alpine sable]

Is that like $f(x) = \frac{x}{x+1}$

[ocean seal]

♡LexQa♡

[alpine sable]

For c

[south hamlet]

x² + 1

[left mesa]

x^2

[south hamlet]

For the denominator

[alpine sable]

Apply quotient rule

[south hamlet]

Whats the quotient rule?

Sorry, english isnt my first language

[alpine sable]

Like if you have $\dv{x}(\frac{f(x)}{g(x)}) = \frac{f'(x)g(x) -g'(x)f(x)}{(g(x))^2}$

[ocean seal]

♡LexQa♡

[alpine sable]

Lmao I love how I used both Newtonian and Leibniz notation there

Whatever

[rocky grove]

Mood

I use my own notation sometimes xd

[south hamlet]

Took me awhile... 🙃

So would this be correct?

Why does my answer look different

That screenshot is from chegg

Oh never mind, problem solved

[long thorn]

HOW

[south hamlet]

Oh sorry

I forgot

.close

[lone heart]

Channel closed

Closed by @south hamlet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[broken pollen]

Could someone explain what it means by given that one is red whats the probability they are all red?

[oblique trail]

This a test?

[broken pollen]

Nope Hw

[oblique trail]

Ok

[oblique trail]

If you already know that one bead is red

What is the probability both are

Probability of a given b is probability of a intersection b/ probability of b

I think that might be what they want you to use

[broken pollen]

Still dont understand

That would make it 15/63

And that is not the right answer

[oblique trail]

Alright let me see

(5/84)/(1/4)

Wouldn’t it be that

5/21

??

[broken pollen]

Nope

[oblique trail]

Huh

[broken pollen]

Its 5/37

Thats the answer

but idfk how to get it

[oblique trail]

oh my bad

That’s my bad

So you need to know bayes rule I think

If you don’t know that google it I think that will make it a lot easier

[desert fractal]

[broken pollen]

I think I understand now

I get the basic of it

but I still don't fully understand it

[desert fractal]

you figured it out?

oh nvm

[broken pollen]

if anyone can explain please do

[desert fractal]

well i thought this at first too

but notice that it says "at least" one of the beads is red

we can use the formula $P(A|B) = \frac{P(A\cap B)}{P(B)}$ to solve it

[ocean seal]

edwardborn

[desert fractal]

where P(B) is the probability that we draw at least 1 red bead

P(A n B) is the probability that we draw two red beads and at least one red bead

which is the same as P(A)

[broken pollen]

Probability of drawing 1 would be probability of p(r)+p(gr)+p(gr)

Would that make sense ?

[desert fractal]

oh get what youre saying

p(r) is the probability where both beads are red?

[broken pollen]

If that would be the p(b) then we would get the right answer

Nope event when its 1 R

so would be just

16/64

managed to get the right answer with that but I don't understand by own logic behind it lmao

[desert fractal]

do you think there might be a simpler way to find this value?

[broken pollen]

Cant think of any

[desert fractal]

what if i say P(B) + P(not B) = 1

[broken pollen]

since i would say p(r)+p(r' and r) would give us that

[broken pollen]

I still dont get what your trying to say

if we add B and not B obvs it's 1

[desert fractal]

what's equivalent to B not happening?

[broken pollen]

1-p(b)

[desert fractal]

what about drawing no red beads?

[broken pollen]

That would be p(r')

[desert fractal]

yup

and that would be 1 - P(B)

by finding that value we can find P(B)

[broken pollen]

Still lost bro

I think i figured out a easier way to do it

let me write it down

[desert fractal]

do you mean 5/47?

[broken pollen]

nope

[desert fractal]

im dumb

forgot to subtract it by 1

yeah it's 5/37

[broken pollen]

sry about the trashy handwriting at the end

😓 kinda got confused my self

I did it and i think its alright

I get your method a bit but not that well

[desert fractal]

you got a 37/84 there which is a good sign at least

cant be pure coincidence

kinda hard to see but i guess you've calculated it directly

other way to do it is by adding the probability that one is red and that both is red

[desert fractal]

but this is only right if p(r) is the probability that both are red and p(gr) is the probability that one is red

by other words that p(r' n r) you've written on that paper should be multiplied by two

otherwise everything is correct

but yeah i'd still recommend you to find P(not B) instead. Even tho it's harder to notice, it's much easier to calculate

just gotta realize that the opposite of drawing at least one red bead is the same as drawing none at all

then you get the desired P(B) by solving P(B) = 1 - P(not B)

[lone heart]

@broken pollen Has your question been resolved?

[broken pollen]

I thought about all the ways you can get 1 r

Then put down how to get 2 r over that

[lone heart]

@broken pollen Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[knotty oracle]

Isn’t it
[ 13 4 ]
[ 12 7 ]

[hollow sparrow]

it's

[knotty oracle]

Thx

Just a random question I forgot how to do on a practice test for admission to high school

.close

[lone heart]

Channel closed

Closed by @knotty oracle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[wispy salmon]

got function investigation question and they asked to draw the graph, i got a question regarding the horizontal asymptotes, can they really "split" like on the left, or the correct way to look at them is like the one on the right?

[last ether]

It really should be |x| -> infinity before splitting

[wispy salmon]

yes but i cant figure out what i has to do with it splitting like that

what i need to calculate to understand how should i split it

[lone heart]

@wispy salmon Has your question been resolved?

[last ether]

|x| -> infinity is like two limits

Can be split into x -> -inf, inf

That's why I said it really should be $\lim_{|x|\to\infty}$ before splitting it

[ocean seal]

Umbraleviathan

[hollow sparrow]

$\inf$ means $+\inf or -\inf$

[ocean seal]

Mehdi_Moulati

[last ether]

That's prob a regional thing

[golden stratus]

u wot

[last ether]

For me, that's just positive infinity

-inf is negative inf,
inf is just positive inf

I've never seen +inf

[hollow sparrow]

a regional thing 100% as you said

[wispy salmon]

why it matters? even if I split i get the same answe, 1 and -1

[last ether]

By definition

[wispy salmon]

well i guess they skipped that part because it gives the same results on this specific case when splitting.
what i still dont got is why the asymptotes split like that, as in on the left side on the graph only -1=asymptotes exsits, and on the right side only the 1=asymptotes exsits while the -1 dont

[last ether]

It just be like that

[wispy salmon]

as in i just need to assume from the overall data i get from the function that there is no asymptote there?

[last ether]

i mean when you find a horizontal asymptote, you need to test both inf and -inf

[lone heart]

@wispy salmon Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[swift terrace]

f(3) = 5 , f'(3) = 1 f'(0) > 1
lets say f(x) is a cubic that has 2 roots
if f(x-f(x)) has 3 roots, would this mean f'(-2) = 0?

[serene junco]

I don't really follow the logic, can you explain how you get f'(-2) = 0?

Also can you clarify what you mean by a cubic with 2 roots? Do you mean two unique real roots? So one of them has multiplicity 2?

[swift terrace]

yes

[serene junco]

okay, what is the significance of f(x-f(x)) ?

[swift terrace]

lets say f(x)=a(x-b)²(x-c)

since f(x-f(x)) has 3 roots

x-f(x) = b or c

so i set 2 equations

f(x)=x-b, f(x)=x-c

and surprisingly f'(3)=1

[swift terrace]

and those equations in total have 3 roots

[swift terrace]

and this image satisfies all the conditions

@serene junco

[serene junco]

That's not true in general, they could have up to six distinct real roots altogether

[swift terrace]

no but its stated in the question

f(x-f(x)) has 3 roots

[serene junco]

oh i see

[swift terrace]

also isnt it 9?

[serene junco]

9 roots altogether but at least three of them will have multiplicity 2

because (x-b)^2

[swift terrace]

yeah but i didnt solve this algebraically

[serene junco]

yes but if x-f(x) = b, then f(x-f(x)) has a double root

[swift terrace]

whats the problem

[serene junco]

I'm not seeing a path from here to f'(-2) = 0

[swift terrace]

aight so

x-f(x)=b can be changed into f(x)=x-b right

[serene junco]

ok

[swift terrace]

and this means f(x) and x-b intersects somewhere

and in total

f(x)=x-b,x-c has 3 intersecting points

annnd f'(3)=1

which is the slope of x-b

so its tangent right

and f(3)=5

so 3-b=5

b=-2

[serene junco]

ok one sec

Okay

yeah

I agree

that means x=-2 is the double root, which means it's tangent to the x-axis

[lone heart]

@swift terrace Has your question been resolved?

[keen plinth]

f'(0) > 1 means your construction doesnt work

if f has a double root to the left of 3 then the slope will be less than 1 at 0

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[alpine sable]

Help

[alpine sable]

actually nvm-

[last ether]

It do be like that

.close

[lone heart]

Channel closed

Closed by @last ether

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[whole jetty]

Help

[whole jetty]

how do you solve this?

i dont get nothing

[rocky grove]

Angle of straight line = 180

[whole jetty]

How do you find that?

[rocky grove]

Well, it's a property

Of lines

[whole jetty]

So I just write the given numbers then = it to 180?

[rocky grove]

Yeah the givens are all part of 1 big angle

And it is the angle of the straight line

Btw your username reminds me of a colleague I have in my batch. He's like the 3rd now

[whole jetty]

3rd what

[rocky grove]

His last name is super similar to the last bit in your username

[rocky grove]

3rd place out of 300 students

I'm 6th lol

[whole jetty]

wan?

[rocky grove]

lewan

[whole jetty]

ohh

[whole jetty]

woah thats cool

[rocky grove]

Yeah

Hopefully I get 1st place next 2 years

Anyways

Did you solve your question?

[whole jetty]

having a really hard time

[rocky grove]

Here's a hint

[whole jetty]

do you participate in math competitions too?

[rocky grove]

30 + 4x+2 = 180

[rocky grove]

Nope

Wish I could

But I participated in a competition that made me win and go to DC

[whole jetty]

got this one but i dont understand whats next, do you transpose?

[rocky grove]

Well you solve for x

[whole jetty]

u have any tips

[rocky grove]

Well the competition happened because my uni participated in a virtual exchange program

So I guess you'd have to find a college that participates in virtual exchanges hosted by the IIE and Aspen Institute

[whole jetty]

regarding the question

will it become 30+2-180=-4?

[rocky grove]

Hmm

+4x = 180 - 30 - 2

[burnt imp]

where do you find the competitions

questions

[whole jetty]

the 180 should always stay on the right side?

[whole jetty]

i think you can find ones on google

or yt

[rocky grove]

Nah you can have any side you want as long as you keep the equation true

[whole jetty]

how you keep that true

not familiar

[burnt imp]

ok

[rocky grove]

Well

4x = 180 - 30 - 2

Is the same as

180 - 30 - 2 = 4x

Just a matter of organizing your terms in the way you see fit as long as it stays true

[whole jetty]

ah

got it

then just solve it

[rocky grove]

Yep

Don't forget the algebra

And pemdas

:)

[whole jetty]

so

x

x=37

?

[rocky grove]

Yep

[whole jetty]

wow

finally i got it thanks for your help

[rocky grove]

[whole jetty]

.close

[lone heart]

Channel closed

Closed by @whole jetty

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[kindred anchor]

0️⃣

[rocky grove]

Wow

0

[ripe rain]

hey riemann

$\sum !!!!!!!! \int$

[ocean seal]

riemann

$\sum \!\!\!\!\!\!\!\! \int$

[ripe rain]

[rocky grove]

$\sum !!!!!!!!! \int$

[ripe rain]

[keen plinth]

[ripe rain]

[keen plinth]

\ooalign{$\sum$\crcr$\int$}

[tacit arch]

hax

[rocky grove]

Preamble package

[ripe rain]

[keen plinth]

doesnt look right

[ripe rain]

,tex \csmean{ooalign}

[ocean seal]

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

[ripe rain]

oh wait

[tacit arch]

$\sum \hspace{-0.69cm} \int$

[keen plinth]

ooalign is ams?

[ocean seal]

VulcanOne

[keen plinth]

iirc

,tex \csmean{ooalign}

[ripe rain]

uh snow can u dm me my preamble

[ocean seal]

riemann

[keen plinth]

[tacit arch]

0.69 wins

[kindred anchor]

I should be careful to not make help channels Latex playground from next time 😭

$W_1={(a_1,a_2,a_3) \in \mathbb{R}^3 : 2a_1-7a_2+a_3=0} \
W_2= {(a_1,a_2,a_3) \in \mathbb{R}^3 : a_1-4a_2-a_3=0}$

[rocky grove]

Well you began with 0️⃣

[rocky grove]

Wow nice

[kindred anchor]

I want to find W_1 intersection W_2

[rocky grove]

[kindred anchor]

Those two equations are plane in R³, so their intersection must be a line

I tried solving the equations, but I couldn't get to the answer

[rocky grove]

You'll need analytic geometry

[kindred anchor]

I'm only into Introductory Linear algebra

[rocky grove]

Lemme get my analytic geometry book

[keen plinth]

just solve the two equations simultaneously

[kindred anchor]

I get a_2=-3 a_3

a_1 = -11 a_3

[keen plinth]

i think thats right

[kindred anchor]

What is the TeX command for intersection?

[keen plinth]

?

like

$\cap$?

[kindred anchor]

Oh yeah, I forgor

[rocky grove]

1:1/11:3/11 or something I have no clue

[kindred anchor]

So, $t(11,3,-1) \in W_1 \cap W_2$

[kindred anchor]

t is any constant.

[rocky grove]

Yeah

Looks right

Wait I'm getting all positive

[kindred anchor]

What shape in $\mathbb{R}^3$ is something like $5x^2-3y^2+6z^2=0$?

[rocky grove]

That's a hyperbola

[kindred anchor]

,w graph 5x^2-3y^2+6z^2=0

[ocean seal]

Results provided by WolframAlpha

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

Requested by hazysleet#7757

[rocky grove]

Yep

1 negative

1 piece hyperboloid

[kindred anchor]

How do I recognize these shapes?

[rocky grove]

x^2, y^2 and z^2 all positives = ellipsoid

[keen plinth]

at any particular value of y you get an ellipse in the xz-plane

so its going to be a double cone

[rocky grove]

One of them negative = 1 piece hyperboloid

2 of them negative = 2 piece hyperboloid

[kindred anchor]

[rocky grove]

And paraboloid has 2 possibilities

Either elliptic or hyperbolic

For example

x^2 + y^2 = 4z

This is an elliptic paraboloid

But this

[kindred anchor]

I can relate this to conic sections in 2D

[rocky grove]

x^2 - y^2 = 4z is a hyperbolic paraboloid

[rocky grove]

Yep

[kindred anchor]

.close

[lone heart]

Channel closed

Closed by @kindred anchor

Use .reopen if this was a mistake.

[rocky grove]

Also graph things in geogebra to get a fuller picture

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[storm oar]

not sure if this counts as maths since it's physics but it seemed worth asking:

[storm oar]

how do i do the underlined question?

is there a formula?

i just don't know where to start

[placid zinc]

I'm confused by the question. Is the field constant everywhere?

[storm oar]

i think so yeah

i've missed a few lectures but i'm not doing very advanced stuff yet