help-19

[glass vault]

$\frac{2^{(9^x)}}{2^{3^x+1}}=2^{9^x-3^x-1}$

[clever fjord]

ImOakley

[slow cradle]

wait

is the +1 meant to be for the x?

like

2^3^x+1

cos u have 3^x x 3

[glass vault]

Yeah

[slow cradle]

ok

from that step

[glass vault]

If you have the same base multiplied together you add the exponents

[slow cradle]

i did like 9^x= (3^x)^2

[glass vault]

Divided you subtract

[slow cradle]

so the

image is wrong

[glass vault]

No

[slow cradle]

how

[glass vault]

Oh i didnt see the bottom bit

[slow cradle]

(2^3)(3^x)

[glass vault]

The image simplifies it differently

But it uses the same rules

It cross multiplies and then gets the 4 in base 2

And adds the exponents

[slow cradle]

bro

this is insane

bro lowk its so over

so

these type questions

make same base

so u can use

law of indices

[glass vault]

Yeah

Are you like having trouble memorising them

[slow cradle]

no

i just

never thought of it

in this

im a bit rusty

its been like

a month

since ive done math

and ive kinda

just jumped into the really hard end

[glass vault]

Honestly indicies and logarithms isnt too hard bc its really just knowing the rules and applying them, theres not too much of a curve

Indicies doesnt get any harder than that question except maybe with function problems but that wouldnt be much harder if at all

[slow cradle]

for this question

is this always the case

when finding

shortest distance

between two graphs

[orchid torrent]

Yes - if the curves are smooth and non-intersecting, then the distance is minimized somewhere where the tangents are parallel

Although this seems like a lot of work

[slow cradle]

can i even use it

in circles

[orchid torrent]

Just take the point on the parabola to be (t, t^2+4) and you need to minimize |t^2-2t+6|

[slow cradle]

wait nvm

thats useless

cos id need to know

the gradient of the poiints

of the circle

[orchid torrent]

tangent is perpendicular to radius

[slow cradle]

well yhh

but like

in the

closest point of circles

this cant apply

cos

they never tell use

the gradient

of radius

or anything

[orchid torrent]

meh

[odd edge]

@slow cradle Has your question been resolved?

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[celest acorn]

Question: "for every natural n>1 show that there exists 3 naturals a,b,c, such that 4/n = 1/a + 1/b + 1/c"
My progress:
"so i know if I show this for prime n I am done, and I have shown it for prime n not 4k + 1 for some k,
Now for n = 4k+1 and prime I have shown this is the same as
4q+3/(4kq+q+3k+1) = 1/r + 1/s
For some q r and s. where integer q>=0 and r and s are natural
for odd k I can just take q=0 and it is easy"

[celest acorn]

Or in other words there exists some t such that

4q+3-t|4kq+q+3k+1
And
t|4kq+q+3k+1

Or there exist a and b such that

a|bk+ (a+b+1)/4
And
b|ak + (a+b+1)/4

[sturdy cape]

Isn't this an open problem?

https://en.wikipedia.org/wiki/Erdős–Straus_conjecture

[celest acorn]

It was given as a challenge problem in a problem set

[sturdy cape]

Right, but it's a challenge problem because it's unsolved

[celest acorn]

Oh

[sturdy cape]

The same problem set won't have the answer for that challenge

[celest acorn]

Oh

[sturdy cape]

There has been some work to show that in certain cases solutions do exist: https://oeis.org/wiki/Erdős–Straus_conjecture

[celest acorn]

Hm, thanks

[wooden python]

guh

what level are you at again

[celest acorn]

Level?

[wooden python]

level of education

hs, undergrad, postgrad

[celest acorn]

First sem undergrad

[wooden python]

and they give you a "challenge problem" thats actually a fucking open problem

[fluid tundra]

to be fair sometimes people do that as a joke but they make it very obvious it is not a typical 'challenge' problem

[south plume]

are you sure you rewrote it correctly? because if we allow negative values for a,b,c then solutions are known

[sturdy cape]

Yeah, typically if a challenge like that is unsolved they'll at least SAY it's an unsolved problem

[celest acorn]

I can send link to pdf

[south plume]

and aren't hard to find

[celest acorn]

https://bikramhalder.github.io/BMath-Sem1-2021-2022/Analysis-1/Assignments/Assignment-2.pdf

[sturdy cape]

Or at least a screenshot, since jpg and png files are easier to handle

[fluid tundra]

"If you solve this problem, you need not do rest of the Assignment."

yeah that's the part that indicates this is obviously a joke

[sturdy cape]

OH THEY DID THEM FUNNEH

[wooden python]

Analysis [space] -I

bro is taking analysis negative 1

[sturdy cape]

Yh if you COULD answer that, then fok me you defo can do the rest of the paper, the teacher doesn't need proof of that shiet

[celest acorn]

Ok

Thanks

[sturdy cape]

Yeah sorry if you've spent quite a lot of time on this

[south plume]

the teacher watched a little too many "a lecturer described an unsolved problem during class, and a genius undergrad thought it was a homework problem and actually solved it" stories 🥀

[south plume]

i also like how this is challenge problem 2. challenge problem 1 is nonexistent

[celest acorn]

It is there

I solved challenge prob 1

It is in lecture notes

[south plume]

ah

[celest acorn]

So assumed that they were serious

[sturdy cape]

I didn't know Good Will Hunting got a South India adaptation

[celest acorn]

.close

[odd edge]

Channel closed

Closed by @celest acorn

Use .reopen if this was a mistake.

[celest acorn]

oh yeh, by setting q = -1, and b = c

[odd edge]

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.

[acoustic quest]

hi

[acoustic quest]

i wanna know what kinda rule is behind the d/dx

is it just f(x)*g'(x)+f'(x)*g(x)?

or something else?

[trail horizon]

yeah

we assume that $\frac{d^p}{{dx}^p}=(2^px+p\cdot2^{p-1})e^{2x}$

[clever fjord]

DaveyLovesSocks

[trail horizon]

and then use product rule to show that it holds for $\frac{d^{p+1}}{{dx}^{p+1}}$

[clever fjord]

DaveyLovesSocks

[acoustic quest]

Okay thx

.close

[odd edge]

Channel closed

Closed by @acoustic quest

Use .reopen if this was a mistake.

[acoustic quest]

.reopen

[odd edge]

✅

[trail horizon]

?

[acoustic quest]

but btw how can i set which of them are f(x) and which is g(x)?

[trail horizon]

$f(x)=2^pxe^{2x},\qquad g(x)=p\cdot 2^{p-1}\cdot e^{2x}$

[clever fjord]

DaveyLovesSocks

[trail horizon]

pretty sure

[acoustic quest]

uhh okay i see

so before

u start

[round lance]

Is anyone here good at geometry and alg2?

[trail horizon]

!occupied

[odd edge]

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

[acoustic quest]

u put the e^2x in the ()

and then use the product rule right?

[trail horizon]

yes

[acoustic quest]

ahhh okay thx for the help

.close

[odd edge]

Channel closed

Closed by @acoustic quest

Use .reopen if this was a mistake.

[limpid hawk]

$(x-1)^2$

[clever fjord]

green

[acoustic quest]

its closed

start a new one

.close

[odd edge]

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.

[limpid hawk]

what does it mean

[acoustic quest]

now u can ask your question

[limpid hawk]

ohh its turns?

[acoustic quest]

before when u asked i used it and closed it

[limpid hawk]

thank you thank you

[acoustic quest]

yes u see now its occupied by u know

[acoustic quest]

nw

[limpid hawk]

thanks bro

.closed

.close

[odd edge]

Channel closed

Closed by @limpid hawk

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.

[thin whale]

So I'm working through some practice worksheets on discrete mathematics for Relations.

When finding the Cartesian Product of two or more sets, in my case.
A = {x , y}

The question asks what is the cartesian products of
A x A x A
To which I concluded to be
A x A x A = {(x,x,x), (x,x,y), (x, y, y), (y,y,y), (y,y,x), (y,x,x), (x,y,x), (y,x,y)} Which has a cardinality |A| of 8.

But my worksheet answers are treating it like its A x A and saying it only has a cardinality of 4.

Is my worksheet just straight wrong or am I missing something as 3 sets should mean 3 values, as per ordered triples, no?

[low locust]

AxAxA as you wrote it is correct

(tho as some small advice, it would help to write it in a more logical order)

someone fucked up when writing the solution to the worksheet

[wooden python]

Which has a cardinality |A| of 8.
not |A|

[thin whale]

Yeah okay, thank you @low locust I was losing my mind for a bit there.

Thanks for the advise, what do you mean by a more logical order? Is there a specific way that's easier to read it? I just went with what felt right when working through it but I'm open to a different interpretation.

[low locust]

the way I would order it is: (without () and , cause lazy)
xxx
xxy
xyx
xyy
yxx
yxy
yyx
yyy

this corresponds to counting in binary

000
001
010
011
100
101
110
111

[thin whale]

Oh cool, I didn't know that. Will take it into a count for future writings.
Also @wooden python I thought the notation of |A| represented the cardinality of set A?

[wooden python]

yeah, of set A

which in your case is {x,y}

you meant |A × A × A| not |A|

[thin whale]

ohhhhh, thank you for that clarification, I didn't even realise haha.

Thanks everyone, have a great night! Because Australia is the only country in the world and therefore it is night, globally of course.

[low locust]

if you want to step it up a bit, what would be a logical order for A^3 for A={x,y,z} ?

at some point you will definitely miss an element if you just do it randomly

[thin whale]

You are probably right haha. Thanks mate. I will soak the binary count to memory.

.close

[odd edge]

Channel closed

Closed by @thin whale

Use .reopen if this was a mistake.

[odd edge]

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.

[wide oasis]

Let $f(x) = x^2 + 6x + c$ for all real number s$x$, where $c$ is some real number. For what values of $c$ does $f(f(x))$ have exactly $3$ distinct real roots?

[clever fjord]

Prathamesh

[wide oasis]

f(f(x)) will be a polynomial of degree 4 and thus it should have maximum 4 roots. But how do I find out a value of c for which their should be 3 roots?

[unkempt lichen]

hey OP, your question looks really similar to the one going on in #help-43

maybe you can read there for a start

[wide oasis]

okayyy

[low mica]

challenge challenge problem: prove 3/n=1/a+1/b+1/c for some natural numbers a,b,c🤔

[wide oasis]

.close

[odd edge]

Channel closed

Closed by @wide oasis

Use .reopen if this was a mistake.

[odd edge]

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.

[wide oasis]

in this question, I proved that the differentiation/ slope of this function is 0 using leibniz formula. But now how do i write down the equation of the line? I mean y=mx+c where m=0, but how do I calculate c?

[wide oasis]

<@&286206848099549185>

[modern thicket]

because it is a straight line parallel y = c

solving the integral is a way to proceedd

you have to merge the to integral together

[wide oasis]

how do i do that?

[modern thicket]

we know that this is a constant therefore we can choose an arbitrary x for instance $x = \frac{\pi}{4}$

[clever fjord]

Amiso_

[modern thicket]

then you get $\int_{\frac{1}{8}}^{\frac{1}{2}} \arcsin(\sqrt{t})dt + \int_{\frac{1}{8}}^{\frac{1}{2}}\arccos(\sqrt{t} ) dt$

[clever fjord]

Amiso_

[wide oasis]

but now to integrate these terms I'll have to use substitution right?

but to calculate 'c' can I put x=0? cuz it's the intercept right?

[modern thicket]

if you put $x = 0$ you will get $\sin^2(x) = 0$ and $\cos^2(x) = 1$ you won't be able to merge the two integral

[clever fjord]

Amiso_

[modern thicket]

you have the identity

$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$

[clever fjord]

Amiso_

[modern thicket]

for $x \in [-1, 1]$

[clever fjord]

Amiso_

[wide oasis]

yeaaa that's right!

got it.

[modern thicket]

nice

[wide oasis]

i see

[wide oasis]

can you please explain how did you put x=pi/4 again?

[modern thicket]

I want to merge the two integrals

to do so the limits of the integrals has to be the same

then I look for an $x$ such that $\sin^2(x) = \cos^2(x)$

[clever fjord]

Amiso_

[wide oasis]

and how do we treat x as a constant?

[modern thicket]

it's not that x is a constant it's just that it doesn't change the result

[wide oasis]

why?

[modern thicket]

well you said it yourself

by leibniz, you know that the slope is 0

there fore for each x the slop doesn't change it stays at 0

[wide oasis]

ohh. Got it. Thanks!!!

[modern thicket]

you're welcome

[odd edge]

@wide oasis Has your question been resolved?

Channel closed

Closed by @wide oasis

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.

[waxen rain]

Hello i dont know how to do the second one

this whole making y the subject is confusing me

[vivid wyvern]

What did u try

[waxen rain]

first i made 32 into base 2

so 2 power 5

same for 4

[vivid wyvern]

What does that give u

[waxen rain]

2 power 2X+4

[vivid wyvern]

Alr cool

Now u can say thay 2x+4=5 right?

[waxen rain]

uyeah

[vivid wyvern]

So now u can solve for x?

[waxen rain]

x = 0.5

l;';'

[rapid dock]

gang 1+1=3 right?

[waxen rain]

No

but where does y come in

[vivid wyvern]

So isnt that what they wanted u to solve?

[waxen rain]

do i need y here?

[vivid wyvern]

The value of x

[waxen rain]

oh

i dont think i needed y here

[vivid wyvern]

In order to avoid negative marking u could mention the value of y as well

Since it says "Hence" as well as otherwise

[waxen rain]

how would i do that

[vivid wyvern]

What did u get the value of y from 1

[waxen rain]

i got 2x+4

[vivid wyvern]

a basically

[waxen rain]

so i sub in y

[vivid wyvern]

Yes and now u k the value of x ,subbing it in u get y

[vivid wyvern]

Yup

So what do u get for y

[waxen rain]

y = 5?

[vivid wyvern]

Question is whether they want u
To place 2^y=2^5 and show it or not
Either way u can do both

[vivid wyvern]

Yes

[waxen rain]

thx

how old r u

[vivid wyvern]

Why🥲

[waxen rain]

wondering what level of maths u do

[vivid wyvern]

1 year more(8 months more or less) and i m out of high school or college whatver u call it there

[waxen rain]

what age is that

aha

[vivid wyvern]

Somewhere between 17 to 20

[waxen rain]

damn im 17

[vivid wyvern]

Wbu

[vivid wyvern]

Oh so m i

Lol

[waxen rain]

ajaja

hhaha

i made ther rbacket = y

[vivid wyvern]

Alr

[waxen rain]

9ySQR

but theres no real roots

[vivid wyvern]

How come

Use 27 and 3

[waxen rain]

nvm

i did 9y

it was just y

[vivid wyvern]

Yea

[waxen rain]

gonna take a lil breeak now

did an hour might go eat something

in abit bro

[vivid wyvern]

Alr bye

[odd edge]

@waxen rain Has your question been resolved?

Channel closed

Closed by @waxen rain

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.

[winter breach]

Compute the flux of the vector field $\mathbf{F}(x,y,z)=\big(x^2+y,;y^2+z,;z^2+x\big)$ outward through the boundary surface of the region defined by

$$
x^2+y^2 \le z^2 \le 2 - x^2 - y^2,\qquad z\ge 0.
$$

[clever fjord]

dghf

[winter breach]

I get div F = 2x+2y+2z, and that the region is 0 ≤ theta ≤ 2π, and that phi is 0 ≤ phi ≤ π/4 U 3π/4 ≥ phi ≥ 0, and √2 ≥ R ≥ 0

$$\int_{}^{} \int_{}^{} \int_{K}^{} 2x+2y+2z\ dV=\left{ x=R\sin \varphi \sin \cos \theta ,y=R\sin \varphi \sin \theta ,\ z=R\cos \varphi \right}$$ $$=\int_{0}^{2\uppi} \int_{0}^{\frac{\uppi}{4}} \int_{0}^{\sqrt{2}} 2R^{3}\sin \varphi \left( \sin \varphi \cos \theta +\sin \varphi \sin \theta +\cos \varphi \right) \ dR\ d\varphi \ d\theta$$

[clever fjord]

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

[winter breach]

But how do I continue from here?

[sleek moth]

have you tried just doing it?

[winter breach]

solving it?

i'm a bit unsure about the bounds for phi tbh

[sleek moth]

oh alright

[winter breach]

since there are two different intervals for phi

if you do the math

[winter breach]

do i compute the triple integral twice for both instances of phi intervals

[sleek moth]

the lower bound smells like a cone

[winter breach]

if you take the sqrt of this $x^2+y^2 \leq z^2 \leq 2-x^2-y^2$

[clever fjord]

dghf

[winter breach]

that's what you end up with

[sleek moth]

idk what the upper part smells like

[bright oasis]

3d dist <= sqrt2

[sleek moth]

it smells like a cone going in the opposite direction

[sleek moth]

nah nvm

[winter breach]

z^2 ≤ x^2 + y^2 is just everything outside of the cone

and when we have a minus infront of x^2 it becomes that green thingy

[bright oasis]

you forgot a minus sign there no?

[sleek moth]

z^2 = 2 - r^2

[winter breach]

minus sign where? z^2 ≤ 2-x^2+y^2?

[bright oasis]

it's literally just a full sphere

[sleek moth]

So like after you're more than sqrt(2) away from the origin it's gg

[winter breach]

ah yes

[bright oasis]

of radius sqrt2

3d euclidian sphere

[winter breach]

yes you're right

[sleek moth]

well it's probably not a full sphere since z >= 0

[winter breach]

that's the region in the problem

[winter breach]

but i'm stuck here

[sleek moth]

oh yeah the upper bound is a sphere with radius sqrt(2) that's nice

[winter breach]

my prof's solution shows the solution without the interval 3π/4 ≥ phi ≥ 0

idk why

[sleek moth]

a pretty funny solution would be to find a normal vector to the cone

and then you can use that to figure out the slope

a better solution is to notice that if y=0, then we have z^2 = x^2

and if we only consider that in the positive x direction then we find that z=x

which gives a slope of 45 degrees

for the cone

it looks like your professor has phi going from 0 to pi/4

just as you'd expect

[sleek moth]

so maybe you made a typo here?

[sleek moth]

I have no idea where you get your second phi interval from because that second interval corresponds to values where z < 0, so you have like a double cone

where you're considering a case where the cone would also extend downwards

but in the problem statement they explicitly state that z >= 0

so that outrules your second phi interval

[winter breach]

ah yes youre right

[sleek moth]

It took me some time to figure out the region but I guess it wasn't that bad in the end

[winter breach]

I am not familiar with this approach, I just apply solutions that are given to previous problems and apply them mechanically to similar problems

this is how we prepare for exams here in Sweden

just go through previous exam problems and spam the solns

[sleek moth]

det är många som gör samma sak i finland också haha

[winter breach]

based

[sleek moth]

anyway a cool trick is that if you have a shape given by the equation F(x, y, z) = 0 then the gradient of F evaluated at any point on the shape will give you a normal vector

[winter breach]

interesting, i will jot this down for future reference

[sleek moth]

so if you were really dense like me and couldn't figure out that the sides have a 45 degree slope then you could've computed the normal vector mechanistically and used that to deduce what the slope must be

but obviously it's easiest to just think about how the cone behaves in a given direction when y=0

[winter breach]

figure out that the sides have a 45 degree slope
how could you spot that immediately, from the interval of the region?

like it being x^2 + y^2 ≤ z^2?

though i'm not sure about z ≤ -x^2 - y^2

on desmos it gives me a sphere-like shape

[sleek moth]

Well it's symmetric in all directions so it suffices to understand the slope when y=0 and x grows in the positive direction, and when that happens it follows the line z=x which has slope 1

if you mean how you would know that it's a cone

well x^2 + y^2 is the xy distance squared

so it's like r^2 = z^2 where r is distance in the xy plane from the origin

and since z >= 0, that becomes r = z

And if your height increases linearly as you move away from the origin in any direction, that's what a cone is

[winter breach]

so it's the linear relationship that makes it a cone, i see

good to know

[sleek moth]

yeahh I visualize a cone as a circle that moves in a straight line and gets scaled linearly

if it's a circular cone

[winter breach]

btw what if there was no constraint to z?

as in, if it wasn't z ≥ 0

would i have to do the integration using two different intervals for phi

or just one full interval

[sleek moth]

yes

I would compute two separate integrals I think

although depending on your function you might be able to leverage symmetry

like if the integrand is symmetric about the xy plane then you can just compute one of the integrals and the other one is gonna be the same

[winter breach]

what would that look like? i have never done it like that

[bright oasis]

yeah that's a direct consequence of: dF=gradF.dL

[sleek moth]

If you have
z^2 <= 2 - x^2 - y^2
then you can imagine moving everything to the left hand side
x^2 + y^2 + z^2 <= sqrt(2)^2
And that's the equation of a sphere with radius sqrt(2), centered at the origin. So that's why the upper bound turned out to look like a sphere

[sleek moth]

it would look like doing this where phi goes from 0 to pi/4, and then doing the same thing again where phi goes from 3pi/4 to pi

[sleek moth]

as a response to this

[winter breach]

doing the same thing again where phi goes from 3pi/4 to pi
would that be an integral with the same interval for the other two integrals, and then adding them up?

like
<three integrals of first region> + <three integrals of the other region>

[sleek moth]

yeah the other two integrals would have the same bounds still

[sleek moth]

yeah exactly

so six integrals in total

[winter breach]

ok ok thank you so much

[sleek moth]

maybe I should've worded it better

but if an exam problem looked like that it would probably have symmetry that makes both terms equal

[sleek moth]

ingen orsak!

[winter breach]

symmetry?

do you mean like when it's 0 to 2pi for theta and you've the integral of cos theta

so you cancel out those x and y terms when doing variable substitution to spherical coordinates

leaving you with the z term

[sleek moth]

not sure if I understand that situation

but I think symmetry is probably pretty unlikely actually because trigonometric functions usually aren't symmetric in a 3D sense

they can be symmetric in the 1D setting

But I was thinking like if we had the two cones, and we could somehow carry out the integration in cartesian coordinates, and the function to be integrated was sin(x)+cos(y)+z^2 (which clearly is the symmetric about the xy plane) then the integrals would turn out to be the same

[winter breach]

e.g. this [\int_{0}^{2\pi} \cos \theta \ d\theta =\left[ \sin \theta \right]_{0}^{2\pi} =\left( \sin \left( 2\pi \right) -\sin \left( 0 \right) \right) =0]

[sleek moth]

but of course once you convert to spherical coordinates you lose the symmetry

because symmetry in cartesian coordinates doesn't imply symmetry in spherical coordinates (ahhh except it should? I don't know lol)

[clever fjord]

dghf

[winter breach]

let me show you (thanks chatGPT)

[sleek moth]

yeah that's a nice application of symmetry

[winter breach]

$$x=R \sin \varphi \cos \theta, \quad y=R \sin \varphi \sin \theta, \quad z=R \cos \varphi$$
$$\int_0^{2 \pi} \int_0^{\pi / 4} 2 \sin \varphi(\sin \varphi \cos \theta+\sin \varphi \sin \theta+\cos \varphi) d \varphi d \theta$$

[clever fjord]

dghf

[winter breach]

you can then cancel those theta terms using the symmetry property in the integration

that's what i thought you meant, but good to know

[winter breach]

here is a reference to what it looked like before #help-19 message

[winter breach]

this is after the R integration of #help-19 message

leaving you with the angle integrals

[sleek moth]

yeah not much happens after the innermost integral

because you're just integrating a polynomial with a quirky coefficient in front

[winter breach]

yep

alright tack för hjälpen 😄

[sleek moth]

inga problem!

lycka till med kursen

[winter breach]

.close

[odd edge]

Channel closed

Closed by @winter breach

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.

[terse bronze]

Does anyone know a good algorithm for checking if there are roots in a specific interval of a non-linear function?

[terse bronze]

Other than newtoon method

[sleek moth]

what kind of function is it?

[terse bronze]

sum of cosines

non linear

[crystal charm]

iirc, there's the bisection method

I forgot anything about it though so might have to look it up

[terse bronze]

it just doesnt work bec the function is pretty much like spikes even at range of 1 diffrence

[crystal charm]

what function is it?

[terse bronze]

cos (4πsqrt(x^(2)-a))+ cos (4πx)-2

or sin (2π sqrt(x^(2)-a))+ sin (2πx)

all have the same roots

the roots are finite

[crystal charm]

Yeah, I don't really know an algorithim for that.

Probably will just have to do some research or hopefully someone can answer

[terse bronze]

Thx

[odd edge]

@terse bronze Has your question been resolved?

[odd edge]

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.

[dawn hollow]

how exactly did it get (x^2 - 4x + 4)(x^2 - 5)

[sleek moth]

(a+b)(a-b) = a^2 - b^2

(a+b)^2 = a^2 + 2ab + b^2

[dawn hollow]

.close

[odd edge]

Channel closed

Closed by @dawn hollow

Use .reopen if this was a mistake.

[odd edge]

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.

[amber veldt]

they're two different variables, not sure what youre confused about

the page you showed us says it's defined in figure 6.1.3., which is not on the page

[remote axle]

That is the figure on the page

[amber veldt]

ooh

okay so the words explain it, my bad

delta ell is the change from the equilibrium length

[remote axle]

It vaguely says Δl is the difference between natural length and equilibrium point, but then i don’t see how ΔL=Δl-y, and I don’t know how they flip flop in the substitution

So then ΔL is the current change from the natural length, then I don’t know why they are defining y=0 to be at ΔL instead of at Δl+L

[remote axle]

In the figure, ΔL should be Δl? That fixes most things, the spring force equation makes sense now (not direction I think that’s a mistake too in the definition of their coordinate system), if the object is above L+Δl but below L, then the force of gravity downwards outweighs the force of the spring upwards

But I still don’t understand any of this math on how they are flip flopping between Δl and ΔL seemingly at random

specifically this

<@&286206848099549185>

removed clutter and scattered questions/thoughts

I think both the dL in the figure and in the last box should both be dl instead

[hollow quarry]

Yes, $\triangle l$ and $\triangle L$ are as you said.
$\triangle L = \triangle l - y$

If the mass falls below the equilibrium point ($y = 0$) by $d$ units then $y = -d$ and $\triangle L = \triangle l + d$

[clever fjord]

StrangeQuarkAL

[hollow quarry]

And yeah, that $k \triangle L$ seems like an error. $F_s = k \triangle l - k y$ is definitely what was meant to be substituted.

[clever fjord]

StrangeQuarkAL

[remote axle]

so then the figure was wrong as well putting y=0 at dL instead of dl?

[hollow quarry]

It may have been intended but yes for the case shown in the figure dL = dl

[remote axle]

ok, thank you, i was struggling hard to justify things as written

.close

[odd edge]

Channel closed

Closed by @remote axle

Use .reopen if this was a mistake.

[hollow quarry]

np

[odd edge]

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.

[remote axle]

why here do they only use the cos equation for phi and not sin equation for phi? they yield different values for phi

[remote axle]

it just says find a value within -pi<phi<pi

both options give a value in that range

it makes sense the one they chose because they said its in the second quadrant, but i dont know how they made that decision

[subtle raft]

I noticed something very interesting, they both add up to π rads exactly

[vivid wyvern]

Cos (theta) is negative in second quadrant

[remote axle]

but why are we in the second quadrant

[forest sky]

we must be in the second quadrant since cos is negative and sin is positive

[remote axle]

so then why do we pick the cos value instead of sin value for phi?

[forest sky]

you would get the incorrect value if you solved the sin equation using arcsin because arcsin only returns values in the first and fourth quadrants

you can still get the correct answer using arcsin if you know that the correct angle is in the second quadrant, and use the identity [ \sin(\phi) = \sin(\pi - \phi) ]
and by extension $\arcsin\bigl(\sqrt{2/7}\bigr) = \pi - \phi$

[clever fjord]

cloud

[remote axle]

i hate triangles and circles

thanks team

[odd edge]

@remote axle Has your question been resolved?

Channel closed

Closed by @remote axle

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.

[dawn hollow]

can someone explain how multiplicities work in relation to factors and graphs like im 5?

[dawn hollow]

sorry 😭 i cant seem to wrap it around my head when theres a graph

[umbral jolt]

so if a polynomial zeroes of even of multiplicities at a point x=a, it'll touch the x axis, not intersect it. For odd, it'll intersect the x axis.More the multiplicity, more flattened the curve will be at point x=a

try graphing $(x-1)^{40}$ and $(x-1)^{39}$ on desmos

[clever fjord]

Lichtbach

[dawn hollow]

odd = goes under?

[umbral jolt]

Yes

[dawn hollow]

so what abour when it like

bounces

[umbral jolt]

bounce? as in? You can show me a picture

[dawn hollow]

i mean like when its concave down

[ebon harness]

they mean when its tangent to a point

[dawn hollow]

also, how did they get (x-5)^2 for the factor when on the other ones it just solves for x

[umbral jolt]

Usually if a polynomial has zeros of multiplicity greater than 3 at a point x=a, there are no comcavity or convexity, we get a point of inflection.

[umbral jolt]

thweres a question?

[dawn hollow]

its the last coordinate in table 2.3 !

[umbral jolt]

Ok so you are asked to find the polynomial.

should I focus on comcavity or just multiplicity only

[dawn hollow]

just multiplicity is fine

[umbral jolt]

so ur asking why they wrote (x-5)^{2} when x-5 would do the job

[dawn hollow]

yes

[umbral jolt]

So, take a limit

[dawn hollow]

i havent learned limits yet

im sorry

[umbral jolt]

Oh shoot

alright I'll explain w/o limits

so see

if x is close to 5, but not greater than 5, what should the sign of f be?

f is x(x+2)(x-5) as you claimed

as x gets closer to 5, but still lesser than 5, x(x+2) is positive, and since x<5, x-5<0, so f should be negative when 0<x<5

does the graph tell you so

[dawn hollow]

wait can we pause

[umbral jolt]

sure

explicitly state where are you stuck

[dawn hollow]

would this be the thing where its like the points increasingly get closer to 5 but the rate of change becomes smaller
eg. 4.9, 4.95, 4.99, 4.999

sorry idk if that makes sense

[umbral jolt]

Yea

but not crossing 5

actually you dont have to thinbk deep

[dawn hollow]

oh

[umbral jolt]

just see whats the sign when 0<x<5

if f is x(x+2)(x-5)

its negative

is it -ve?

no

however (x-5)^2 gives you the result

[dawn hollow]

what is -ve

[dawn hollow]

ohh i understand now

[umbral jolt]

negative

@dawn hollow a better analogy of multiplocity, if the function has a root at x=a and it doesnt change its sign aroubd x=a, then it has an even multiplicity there

and if it does change the sign

then its odd

you might wanna know why not (x-5)^4 ?

[dawn hollow]

yes 😭

[umbral jolt]

So a function with more multiplciity gets flattened out heavily

but the function is not THAT flat

so the least should do

[dawn hollow]

sorry but how am i supposed to know whether it is pos or neg without a calculator

like graphing it*

[umbral jolt]

why not

Imma teach you

[dawn hollow]

thanks

[umbral jolt]

not nessesarily graphing

If an eqn is given to you

[dawn hollow]

sorry every time u answer a question i have like 5 new ones 😭

[umbral jolt]

in factored form

Like $f(x)=(x-a)^{b}(x-b)^{c}...$

[clever fjord]

Lichtbach

[umbral jolt]

and youre analysing its sign around a point x=a

[dawn hollow]

oh

[umbral jolt]

take the nearest root to a

[dawn hollow]

can you elaborate ?

[umbral jolt]

first take b such that b is lesser than a but greater than the rest of the roots which are greater than a

[dawn hollow]

oh ok

so how exactly do i “take b to a”

[umbral jolt]

so now we do sign analysis

[umbral jolt]

take b to a? no no we wont be doing stuff like that

b is a root which is closest to a

[dawn hollow]

ohh i assumed “take the nearest root to a” meant like taking b lol

[umbral jolt]

no no

so, @dawn hollow see, the points where b<x<a, whenever there is another root of f p such that p is lesser than a, x-p>0 since p<b and x-b>0

[dawn hollow]

oh okay

[umbral jolt]

I made a lil sign mistake check again

[dawn hollow]

i understand

[umbral jolt]

can you tell me what'll happen when theres a root q of f such that q>a?

[dawn hollow]

x-q>0?

[umbral jolt]

when we define c such that c is the immediate next root of f such c>a and c is lesser than any other roots of f which is greater than a

[umbral jolt]

positive?why do you think so

[dawn hollow]

because q is less than a

idk 😭

[umbral jolt]

@dawn hollow read

read closely

what sign is it

q is greater than a

[dawn hollow]

o

[umbral jolt]

so what'll x-q be?

[dawn hollow]

less than 0??

[umbral jolt]

Yes

you get me now?

[dawn hollow]

sure

[umbral jolt]

so whenever you have a point b<x<a, you look on the sign of f at x=a

see the roots less than a will not do anything to the sign as they are positive

right?

for the roots which are greater than a at f

they will give negative numbers

if there are even number of roots greater than a, we are multiplying even numbers of negative numbers

what should be the sign of f?

[dawn hollow]

wait to clarify at f is just like on the function right

[umbral jolt]

f is a function defined here @dawn hollow

so suppose youre given $f= (x-1)^{2} (x-2)^{3}(x-4)^{5}$ and youre asked to graph f round 2

[clever fjord]

Lichtbach

[dawn hollow]

what does round 2 mean

[umbral jolt]

SORRY

around 2

[dawn hollow]

oh okay

[umbral jolt]

so

as ive said

when 1<x<2

[dawn hollow]

well find the roots and graph those first right

oh sorry

[umbral jolt]

look (x-1)^{2} is positive

and (x-2)^{3} is negative

[dawn hollow]

👍

[umbral jolt]

and (x-4)^{5} will be negative

so what'll be the sign of f

positive* negative *negative

so its positive

right?

when 1<x<2, f is positive

[dawn hollow]

oh

[umbral jolt]

Hey

im deadass

are you clear with what ive said till now

be frank with me I dont bite

[dawn hollow]

yes, except for the 1<x<2 part

[umbral jolt]

i think💔

[dawn hollow]

[umbral jolt]

when 1<x<2

x-1>0

but x-2<0

right

[dawn hollow]

yea

[umbral jolt]

so we can analyse the sign of f at 1<x<2

since x-4<0

[dawn hollow]

oh ok

[umbral jolt]

once we have determined f is positive when 1<x<2

[dawn hollow]

wait are these critical points or am i mixing that term up w something else

[umbral jolt]

I'll explain critical point later

these are roots

roots may be critical points

may be not

[umbral jolt]

@dawn hollow move to an interval such that 2<x<4

[dawn hollow]

alright

[umbral jolt]

we are making x cross 2

but we arent letting it stray further than another root

[dawn hollow]

so it will go under the x axis?

[umbral jolt]

think of it as checking the train status at every station when the train is a function and roots are the stations

[umbral jolt]

thats upto you to decide

[dawn hollow]

sorry i have never been on a train, this analogy is lost on me 😭

[umbral jolt]

we have to do a sign analysis

[umbral jolt]

a bus?

[dawn hollow]

no

[umbral jolt]

bus stops?

[dawn hollow]

i dont go outside

[umbral jolt]

holy

so they exist

mom told me about them

anyway

back

now 2<x<4

so x-1>0 and x-2>0

but x-4<0

so the graph of f will be negative

so f was positive before we crossed x=2

and now its negative

so f changes sign at x=2

[dawn hollow]

so it is positive at 1<x<2 & neg at 2<x<4?

alright i think i get the concept, i can just pick intervals and do the x-(n) less than/greater than thing (for a lack of better words), right?

[umbral jolt]

Yes

can you understand the importance of multiplicity now? @dawn hollow

[dawn hollow]

yes

[umbral jolt]

🤝

[dawn hollow]

thanks!

[umbral jolt]

Im glad

np

[dawn hollow]

if you are feeling up to it, can we do critical points? 😭

[umbral jolt]

Yeah yeah, can we do that some time later tho? I'll ping you when Im available, I'll have to head out for uni

[dawn hollow]

ofc, thanks for helping once again

[umbral jolt]

npnp

[dawn hollow]

.close

[odd edge]

Channel closed

Closed by @dawn hollow

Use .reopen if this was a mistake.

[odd edge]

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.

[frank storm]

Hi! Learning long division with polynomials and wondering how to continue from here. (This is required for one of my worksheet problems, I can’t do synthetic division) I’m struggling to figure out what to do, since I can’t divide 14x by 4x^2.

[wooden python]

you're not supposed to divide 14x by 4x^2

you're subtracting

[frank storm]

Same thing/ not same thing ik

Wrong wording

1am for me

[wooden python]

why are you doing math at 1 in the morning?

anyway

[frank storm]

but

I meant subtraction

[wooden python]

these are unlike terms and so wouldn't interact

you would be left with -4x^2+14x+8

[frank storm]

I can’t subtract 4x^2 from 14x and I wasn’t taught how to deal with that

Do I then continue the long division on the latter two terms or do I leave it there

[wooden python]

a better idea would have been to write out zero terms in the dividend to begin with, to keep things a bit neater:
8x^3**+0x^2**+14x+8

and you continue! you keep going until the remainder has strictly lower degree than the divisor, which in your case means you keep going until you are left with only a constant

[frank storm]

I hadn’t learned that yet so thank you 👍

[wooden python]

learned what

[frank storm]

Zero terms

[wooden python]

it was never shown in examples?

[frank storm]

my class threw me into polynomial long division and synthetic division and binomial theorem without any warning 🥀

didn’t learn any of that last year

and no I didn’t get any examples of zero terms

[wooden python]

bweh

[frank storm]

in my material

[wooden python]

well now you know

[frank storm]

How do I know when to add zero terms

[wooden python]

generally: all terms that could be there but are missing bc of zero coefficients

ie powers of x smaller than the leading one which you don't see

in your example 8x^3+... is degree so you expect to see terms with degrees 2, 1 and 0

of these, only 2 is missing so it gets written out with a zero term

[frank storm]

so I add a zero term for every subsequent power of the polynomial degree that is missing

[frank storm]

Due dates 😔

[wooden python]

.....yes

[frank storm]

Ok ty

.close

[odd edge]

Channel closed

Closed by @frank storm

Use .reopen if this was a mistake.

[odd edge]

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.

[waxen moth]

Excuse me what out of these transformations causes angle measures to change Dilation, rotation, a vertical stretch or a reflection

[vernal yacht]

Do you have any idea?

[waxen moth]

No. I’m genuinely confused

[vernal yacht]

What’s the original question?

[waxen moth]

I don’t know how to go back to where it explains it

[unkempt lichen]

so dilation here is scaling i presume?

[waxen moth]

I think the dilation Its talking about is just enlarging the shape without changing the angle measures I think

Idk what scaling is

[unkempt lichen]

then yeah it's the same

so you stated that dilation doesn't change the angle measures

so you can rule that out

what else can you rule out?

[waxen moth]

Yeah and then the rotation just rotates it so the angle measure are the same

A reflection wouldn’t change the angles either

So I’m thinking it has to be a vertical stretch

[unkempt lichen]

so by process of elimination?

mhm

[waxen moth]

I don’t really understand vertical stretches thougu

[unkempt lichen]

let's say you have these lines

and let's say the red line is on the x-axis so it doesn't stretch

stretching is defined as multiplying the y-coordinate by some scaling factor k

so a point (x, y) becomes (x, ky)

[waxen moth]

Alright

[unkempt lichen]

so this would be an example of the new line

[waxen moth]

So then the green line would go up or down making tthe angle different?

[unkempt lichen]

the blue line is the green line stretched vertically by a factor of k

just from this you can already tell the angle will be different, but to show that it's different, consider the tangents of the angles made by the green/blue lines and the red line

tan(green) is y/x, but tan(blue) is ky/x

exceptions: there are two angles that are preserved under vertical stretching

if you want to figure it out, give it a shot

[waxen moth]

A 90 degree angle?

[unkempt lichen]

and?

[waxen moth]

180 degree I’m pretty sure

[unkempt lichen]

i'll give you the credit

i was thinking 0 degrees

but that works

but you are completely correct

[waxen moth]

Wait

What’s zero degrees?

Wouldn’t that just be nothing

[unkempt lichen]

two lines stacked on top of each other

[waxen moth]

Oh

[unkempt lichen]

aka collinear

[waxen moth]

So would that make it three angles?

That are unaffected

[unkempt lichen]

well in that case yeah

[waxen moth]

Well I understand it now thank you

[unkempt lichen]

glad to help

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[waxen moth]

.close

[odd edge]

Channel closed

Closed by @waxen moth

Use .reopen if this was a mistake.

[odd edge]

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.

[tulip cradle]

x1 and x2 are two different solutions to the equation sin(2x + π) = sqrt(2)/2.
What can x1 − x2 not be equal to?
A) π/8
B) π/4
C) 3π/4
D) π

[wooden python]

!status

[odd edge]

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

[tulip cradle]

2

I have x1 and x2 but I don't know how to check the answers what cant be equal to

x1 = -3π/8 + k2π
x2 = 11π/8 + k2π

Ok so I know the answer is A

but I dont understand how x1-x2 can be π

nvm

I got it

.close

[odd edge]

Channel closed

Closed by @tulip cradle

Use .reopen if this was a mistake.

[odd edge]

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.

[glacial grove]

I have a question regarding derivatives

[glacial grove]

If you're given this type of equation
y= 3x(2x²+6x-10)

Do i have to simplify it first then derive or use the power rule?

[zenith mountain]

you expand it, so u would get 6x^3 + 18x^2 - 30x

and then use power rule on each term

[crystal charm]

I mean you can use product rule if you want

but it's better to expand it and use power rule

[zenith mountain]

yeah, that too, but if u wanna stick to power rule then this

[glacial grove]

OHHHH alright2 thank you so muchhh

Ill see you guys next time

.close

[odd edge]

Channel closed

Closed by @glacial grove

Use .reopen if this was a mistake.

[odd edge]

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.

[toxic meteor]

Hi guys

[odd edge]

Please don't occupy multiple help channels.

[toxic meteor]

Is this right

.close

[odd edge]

Channel closed

Closed by @toxic meteor

Use .reopen if this was a mistake.

[odd edge]

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.

[somber sparrow]

how do you show that the sum and product of algebraic numbers is algebraic

[somber sparrow]

this problem seems so trivial but im having some trouble

can i get a hint

[vivid wyvern]

If u can express it as a polynomial
Uk variable with non negative integer,ig then u could call it algebraic

[crystal charm]

what do you mean by algebraic numbers? (nvm my dumbass just realised it what you're asking for)

[vivid wyvern]

Js a hint but if u r given 0 or any other number u can still express it as a polynomial just by considering the power of the variable as 0

[clever fjord]

LY

[red surge]

wait fuck algebraic numbers ignore me

[modern sundial]

Maybe I am missing something, but I thought your first suggestion was a fine strategy, just that it was probably dressed up in fancier language than necessary

[red surge]

nah my suggestion doesn't work since the characterisation is of algebraic integers not algebraic numbers

[red surge]

although now that i am thinking about it the only proofs ^ that i am aware of are like using basic abstract algebra

[modern sundial]

Like, let x be some algebraic number, then it is the root of some polynomial p_n(x)=0 for some finite n with not all coefficients equal to zero. That is the same thing as saying that 1, x, x^2,.., x^n are a linearly dependent set with F = Q.

So now just take (x+y)^n and apply the Binomial Theorem and stare at it until you can see a finite dimensional space in there. Isn't that basically what you are saying?

(where x, y are algebraic)

I haven't slept, so I'm just going to stop thinking about this now

[red surge]

well it's like the same strategy but like what i was originally thinking of was for algebraic integers so it's not 100% the same

[red surge]

yeah uhh i'm not actually aware of any "elementary" olympiad-style proofs of this, the most natural ones come from like basic field theory

[odd edge]

@somber sparrow Has your question been resolved?

[odd edge]

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.

[full meteor]

I understand why D is correct but why isn’t B?

[faint knot]

this depends on how you use asymptotes, but its just more convenient to have E be correct

[sturdy cape]

Because that's not the definition of an asymptote

[faint knot]

B is not necessary for E to be correct

[full meteor]

So is it a bad question?

[faint knot]

theres a few "bad questions" on the AP test which you just have to accept as required

for this question though its possible to create an example that has an asymptote

[full meteor]

but is still defined for f(x)=2?

[faint knot]

if you had a graph that looks like this for example,

this is considered to have an asymptote of y = 4

[sturdy cape]

(Are you sure?)

[faint knot]

you often dont see graphs like this because you only ever hear asymptotes for graphs like the black one

[faint knot]

yes?

[sturdy cape]

You need something to shoot off to infinity

[faint knot]

let me speak alr?

[full meteor]

So now functions can be defined at asymptotes? I thought that was the whole point

[faint knot]

no it isnt

[sturdy cape]

OH SHIT MB

I'M THINKING YOU SAID x = 4

[faint knot]

???!??!?!?!?!

[sturdy cape]

FUUUUU

[full meteor]

What is it then?

[faint knot]

we're going to restart this so we dont have anything confusing in the middle

[sturdy cape]

I can't read a fokken graph I've fallen so far

please do I'm so sorry 🤣

[faint knot]

for some functions, its harder to see what their behavior is

a big example is the prime counting function

it counts the number of primes up to and including x, notated as pi(x)

needless to say youre not going to find a nice expression for this function

you can cheat, or you can approximate

and a well-known approximation is x / ln(x)

so pi(x) here has an asymptote

an asymptote of x / ln(x)

[opaque hearth]

wtf

[full meteor]

i’m so lost 😭

[faint knot]

brb

[opaque hearth]

i think the answer is e

[tidal matrix]

there are a couple of oscillations earlier on where f(x) = 4, but the asymptote still exists

[opaque hearth]

f(x) may not seem to go to 2 but as it goes to infinity it's basically closer and closer to 2 to the point where it's 2

[tidal matrix]

its possible for f(x) = 2 for some finite x

[tacit wasp]

Because, for example, you could have the constant function f(x) = 2

[amber schooner]

no

[tacit wasp]

Ups sorry

[amber schooner]

people just never learn what asymptotes are formally

they think it’s just "a line that you approach but never touch"

[tacit wasp]

Or they do but then forget about it 😬

[tacit wasp]

Yeah exactly

[amber schooner]

no in my high school this is literally what the teachers would say

😭

[tacit wasp]

In mine as well

[opaque hearth]

that's what i learned

[full meteor]

That’s what the teacher said sooo

[amber schooner]

sort of like the whole continuity is being able to draw it

[amber schooner]

yea it’s just not rigorous and is not true

[tacit wasp]

But then through examples they (usually) show that this is only the most frequent and intuitive case, rather than the definition

[full meteor]

So is a horizontal asymptote just where x tends to infinity?

no necessarily that there isn’t a value?

[faint knot]

I have to take out the trash one time

[tulip palm]

There is only Y value

[faint knot]

and I come back and this is what I see?

[tacit wasp]

Of course
Actually, +∞ or -∞

[crystal charm]

I was taught before that discontinuity is just "if you lift the pen off, that's discontinuity"

[tulip palm]

lim x---> ♾️ = L (horizontal asymptote)

[tacit wasp]

Wdym?

[amber schooner]

yes it’s just the limit at +inf or -inf

[crystal charm]

anyways I'll stop interupting

[faint knot]

yea, in this case the asymptote is a horizontal line

[full meteor]

Is it the same for vert asymptotes?

[faint knot]

yep

but asymptotes dont need to be horizontal or vertical lines

[tulip palm]

Discontinuation is where limit doesn't exist

Or where limit ≠ f(x) at c

[crystal charm]

there's an epsilon delta definition of continuity, which is more rigorous

[tulip palm]

In vertical asymptotes

y tends to infinity

In horz asymptote x tends to infinity

[amber schooner]

well that would be the limit at a particular point (considering left and right sided limits of course because it need not approach the vertical line the same way from both directions)

[full meteor]

o wait i’m seeing

[summer wave]

it's equivalent I think (so both are rigorous)

[full meteor]

and um for this one

it says “f has a limit at x=2”

does that mean it could just be a one sided limit

[amber schooner]

just means the limit exists

[faint knot]

if they define both sides, then it has to be the doubled sided limit that exists

[amber schooner]

lim x -> 2 f(x) = 4

[faint knot]

if they only define one side, then it should be that only that one sided limit exists

[amber schooner]

we don’t care about what happens at x = 2 until we consider continuity

[faint knot]

so here theyve defined f(x) on both sides of x = 2, so both sides are required to converge to the same value

[faint knot]

B here is wrong since the limit converges to 4 as x approaches 2

[amber schooner]

you mean 4 and 2

yea

[full meteor]

I know why b is wrong i’m just having trouble understanding the uh

[amber schooner]

the what

[full meteor]

how would i graph that in desmos

i wanna see what it looks like

[amber schooner]

x + 2 for x ≠ 2 and 1 for x = 2

[faint knot]

you dont, the discontinuity is just a single point so it wont show up

you can at least type the function in to test out points

to do that, you use what I think are called restrictions

if youre looking for this,

[amber schooner]

note that them writing (x^2 - 4)/(x - 2) is really just there to throw you off because where it’s defined (x ≠ 2) it’s identically x + 2 for x ≠ 2

[faint knot]

you have to enter in the points manually

thats just me placing the points, desmos wont show you a discontinuity especially in the exam version

you can at least see the discontinuity by dragging the cursor on the graph until you get to the discontinuity, where this will appear

[full meteor]

oh i c

[faint knot]

this will definitely not be available on the exam

[full meteor]

it’s just removable

right

?

[amber schooner]

drill it into your head that when you’re considering limits you don’t give a single fuck about what happens at that particular point

[faint knot]

do you know how to simplify $\frac{x^2-4}{x-2}$

[clever fjord]

mtt

[full meteor]

yes

[faint knot]

so that would mean its a removable discontinuity if the discontinuity can be removed, doesnt it

[full meteor]

i’m seeing

[full meteor]

yes

coolio

thank you

i’m too tired for this

[faint knot]

np

[full meteor]

and my math class that i took the test in was 108 degrees

tyy

.close

[odd edge]

Channel closed

Closed by @full meteor

Use .reopen if this was a mistake.

[faint knot]

thats impressive

[odd edge]

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.

[final grove]

Okay, the other one is occupied