help-23

[astral glacier]

We have (0,1) ≅ (0,1]

Add 0 to both sides

What do we get

[sly aurora]

[0,1]?

[astral glacier]

And [0,1) on the left

So we get [0,1) ≅ [0,1]

And now we're done

[sly aurora]

Ahhh, okay. I get it.

Interesting

So, the sequence we chose earlier, could we also have chosen another one that approaches 0?

Oh, and, by the way, thank you. This has been very helpful so far! :)

[astral glacier]

I mean any sequence of 1/k^n works

Cuz then you just multiply by k

[sly aurora]

Alright, thanks!

How do I close this channel?

[astral glacier]

Wait are we done

Do you understand the second one

Oh also I just realized

I gave you a proof which is completely different from the notes

[sly aurora]

That is okay.

The one in the notes kind of captures the same idea.

[astral glacier]

Yeah in the notes they're just moving over entire intervals

[astral glacier]

.close

[sly aurora]

I feel like this has been the wrong lecture to be absent at... haha

[sly aurora]

Oh, alright, thanks.

[astral glacier]

Yeah early set theory is weird

Just take comfort in the fact that you will never have to do this shit again

[sly aurora]

Haha, I do not know if it brings me much comfort right now... The exam is soon and I feel rather lost hahah

[astral glacier]

Oh god

Good luck

If you have any other questions, feel free to ask

And feel free to tag me in the future

[sly aurora]

Thank you so so much!

Anyway, I am going to go do some revision work now and hopefully understand the proof of the Cantor-Bernstein-Schroeder Theorem in a bit.

[astral glacier]

Oh god CBS

Good luck

[sly aurora]

YEAH

OH GOD indeed...

[astral glacier]

I've never completely understood it myself

Gonna be fun trying to explain it to someone else lol

Maybe I'll finally understand it

[sly aurora]

Hahah, well, perhaps later if I have questions I might tag you hahah

I am evil like that

[astral glacier]

Sure sure no worries

See ya around! 🌺

[sly aurora]

See ya around :)

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[grim plover]

is p = p_0 + rho g h (rho is fluid density, p_0 is atmospheric pressure, p is pressure at depth h) valid in fluid dynamic problems?

[zealous ingot]

no

there it depends on v and a too

bernoulli eqn or navier stokes would be valid

a rule of thumb could be to apply it when pseudoforce is absent

[safe radish]

@grim plover Has your question been resolved?

[grim plover]

hmm actually i was solving an efflux velocity problem

and the solution found the height of the container by equating the pressure at the bottom (given) to p_0 + rho g h (where rho was also given)

but i was a little off thrown since due to the hole the fluid was flowing out, and we were applying bernoulli eqn to find efflux velocity

[iron estuary]

it’s valid only if the fluid is basically at rest or moving very slowly so in that case, pressure changes mainly due to height... but in dynamic cases like efflu, you should use bernoulli’s equation since velocity affects pressure too

[grim plover]

ok ty

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[iron estuary]

yeah that makes sense, even though the fluid was flowing out, using p = p₀ + ρgh at the bottom is fine because it’s just giving the static pressure at that deptj

bernoulli’s equation then uses that pressure to relate it to the velocity at the hole

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[steep magnet]

This is a question and a sample solution

[steep magnet]

I did the exact same thing in my solution when trying it, except I chose x = 0 instead of x = 1

So I got sqrt(4 - 0) in the dominator instead of sqrt(4 - 1) like they did

So I got 4E for my answer and they got 3E

Is my way of doing it still valid or not?

[prisma scarab]

|x| should be greater than 0

[steep magnet]

Ahh yeye true

Ok thanks!

❤️

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[ripe ravine]

Does this look correct?

[safe radish]

@ripe ravine Has your question been resolved?

[patent flame]

Looks correct

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[queen mango]

solve $\in \mathbb{R}$ the equation:
$| 1-x | = 2|x| - 1$

[queen mango]

so i should solve based on three cases , 1-x<0 , x<0, 1-x>0 ?

[glacial cairn]

1-x>0 and x<0 overlap

[queen mango]

i just made a sign table

and got x=-2 , x=1/3

[glacial cairn]

Correct

[queen mango]

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[prisma scarab]

Can someone explain what's going on? Why do we take the sum of sin 2kz, why is it 2k and not just k, how do we get those sums after integration? Istg you can't learn anything from these solutions, same as trying to learn math from Wikipedia

[safe radish]

@prisma scarab Has your question been resolved?

[plucky elk]

you're leaving out too much context to answer your question

you can also just make your own more understandable solutions if you don't want to read other people's work

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[astral yacht]

Abstract Algebra - how do we go from (x + 2y + 4) * z to x + 2y + 4 + 2z + 4?

[cedar widget]

your operation is $x \star y = x + 2y + 4$

[flat frigate]

ExpertEsquieESQUIE

[still charm]

$\star$

[flat frigate]

Mr. Gamer 🇵🇸

[still charm]

Cool!

[cedar widget]

then $(x \star y) \star z = (x + 2y + 4) \star z = (x+2y+4) + 2(z) +4 = x+2y+4+2z+4$

[flat frigate]

ExpertEsquieESQUIE

[lean otter]

basically you can just note x+2y+4 as Alpha or something to not get confused

[delicate bear]

stars in maths woah

[lean otter]

it's a law of composition

[delicate bear]

ohh ok ty

[astral yacht]

Ok and then how do we go from that to z + 2y + 2z + 8?

Where does the z come from?

[cedar widget]

typo prob

[astral yacht]

lol

Of course

That is supposed to be an x right?

[cedar widget]

I think so

[astral yacht]

😂

Anyone else?

[cedar widget]

wdym anyone else

[plucky elk]

your opinion is not valued.

[astral yacht]

I think so after you mentioning typo warrants skepticism

I don’t think it’s a typo

Notice z is found on both lines

I had high hopes for you, Expert

[cedar widget]

its a typo in both lines

[astral yacht]

Do you think so or know so?

[dull sequoia]

One wonders hard why you don’t get help

[safe radish]

@astral yacht Has your question been resolved?

[astral yacht]

You are probably right, no sense in talking that way 😉

Why does x + e + k => e = -k?

[noble mango]

Hi!
x * e = x + e + k
By property of the Identity
x = x + e + k
Now, you could use the cancellation law (if you have proved it), then consider that ||every element has a unique additive inverse.||

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[quasi timber]

Spambots?

[safe radish]

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[carmine solar]

hey i need help with cal 2, trying to comprehend finding the limits of a sequence. pweese help me

[frigid arch]

I might be able to help. Learning the same stuff at the moment. What kind of sequence?

[carmine solar]

should i type it out or should i screenshot it and drop it here

[frigid arch]

Screenshot is better (I just learned that too. 😅 )

[carmine solar]

i understand the concept of the quotent rule which isnt in here but im not exactly sure what to do here

[frigid arch]

Hmmm... my immediate impulse is to expand the squared term then divide the top and bottom by the highest-degree term in the denominator.

Does that make sense? It's not any different than taking a limit at infinity.

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[grizzled sun]

I need help with Lagrange multipliers with multi variable calculus, here is my work for the question I’m working on and I will attach my question in a second

[grizzled sun]

here is the question

i am not sure what i am doing wrong with my work

!status

[safe radish]

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

[grizzled sun]

2

please ping me if you can help me :) i am tabbed out of discord in the mean time

<@&286206848099549185>

[silent eagle]

Yes?

I see that you are asking for help

may I assist.

[grizzled sun]

yes please!

[silent eagle]

Ok

What is the aspect that you need assistance in?

[grizzled sun]

i am mostly just confused on isolating the x and y functions but i don't know if i'm on the right track

do you need more of an explanation of where im stuck?

[silent eagle]

one secont

[grizzled sun]

here is an example i did in class

[silent eagle]

yes

slightly vague

[silent eagle]

do you need the functions to be separated?

the conversation will end in 120 seconds whom inactivity

[grizzled sun]

sorry i got side tracked

let me see

[silent eagle]

ok

[grizzled sun]

so i took the partial derivatives of Cx and Cy, set them both equal to lambda, set Cx=Cy and started solving for x and y but i think i would get that y=y if i continued to solve for y. What parts am i missing with my systems of equations?

[silent eagle]

did you miss a relationship between x and y?

[grizzled sun]

lagrange multipliers only have partial derivatives of x and y, and the constraint function.

so three equations

[silent eagle]

alr so

wait, now I am lost

[grizzled sun]

do you know about multi variable calculus?

[silent eagle]

no, I have not reached that point of extent.

i was going to ask earlier if this was calculus, but you were afk

Unfortunately, I cannot help you with something I can’t even help myself to.

[grizzled sun]

it's okay, thank you anyways

[silent eagle]

Much welcomes, I’d like for you to enjoy your day.

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@grizzled sun Has your question been resolved?

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@grizzled sun Has your question been resolved?

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[winter folio]

Its in german can somebody help me with the gauss algorhytm like how do i create one out of it

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[obsidian walrus]

I don't get it

[wind basin]

Hm

[obsidian walrus]

ikr

[wind basin]

aight hang on after the 600, what’s that number? It’s like 1010

10.90??

[obsidian walrus]

not my notes, but 18 percent

is that what you mean

oh

10%

[wind basin]

Ah ok thx

Now I can read it

[obsidian walrus]

lol

[wind basin]

Tf is ts 😭🥀

[obsidian walrus]

ikr

[wind basin]

Ok hang on what is “omen”

While 60 were engaged in ome sport? Oh it’s one.

[obsidian walrus]

LMAO

[wind basin]

Handwriting harder than the question at this point 😔

[obsidian walrus]

i can't seem to solve it i've spent like 15 min

i did the next few qs

[wind basin]

He writes his “the” as “me” and it’s pissing me off.

Lemme get pen and paper I’ll give this a try

[obsidian walrus]

lol lmk if you need me to rewrite the q

like the questions after this one are really simple so idk what's wrong w this

[wind basin]

tf is the venn diagram doing is it in percentages

[obsidian walrus]

💀

I think so

[wind basin]

fucking why 😭 nobody does that

[obsidian walrus]

ong lmk if the question is messed up

[wind basin]

Ahh

The question is solvable with just two bits of information

you can show one is definitely wrong, and I assume that since at least on of the venn diagrams is right, it means the other is right.

"School has 600 students"
"66 play hockey exclusively."

66/600 = 0.11 = 11%
The left diagram states that hockey exlusively is 12% - this is false with the actual data.

The right diagram says 11%. True.

🤷

but fucking hang on

naive by me

[obsidian walrus]

OH

I THOUGHT IT WAS 60 PLAY HOCKEY

and that's when I thought I lost it

omfg

[wind basin]

But wait.

"11% of people play hockey and cricket."

[obsidian walrus]

let me try again

[wind basin]

the right venn diagram does not accurately show this?

I could be missing something but yeah that says 66 should help you

like in the mathcord? or in this help discussion

yeah this question got me so cooked

ts basic venn diagrams too

[obsidian walrus]

yeah it's prpobably c

it's simple venn diagrams the english is stupid

[obsidian walrus]

yeah i think it's c

they forgot to write exclusively several times too

I was confused cus I thought it was 60% hockety ty

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[native flume]

hi, just would like to clarify is this is the right way of thinking abt this. if we consider that a subset does not contain 1, then we have need to consider the rest of the set, {2, 3, ... n}, which has size n-1, and we need to choose r elements, therefore, we get (n-1 choose r). if we consider a subset that contains 1, then we have can again consider {2, 3, ...n}, and we only need r-1 elements, so we get (n-1 choose r-1), since either a set containing 1 or not containing 1 are the only possibilities, and then this would be equal to (n choose r)?

[slow zodiac]

Sounds about right. If you want you can also prove the identity algebraically

[native flume]

thanks

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[worldly knot]

can someone do this please so i can check if im correct

[plucky elk]

not how this works

see

[worldly knot]

well im asking because im really not sure im right

do you mean i have to post my answer?

[plucky elk]

yes

it's much more efficient to show your work so someone can see if it's correct or find a mistake

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<@&286206848099549185>

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[sinful citrus]

idk why it's so difficult for me to understand how the simplification works

what rule said 3ln|x|=ln|x^3|?

[versed wave]

,tex .log rules

[sinful citrus]

yeah makes sense

so raising everything with base e

how does e^ln(x^3) + e^c turn into e^C|x|^3 ?

I don't get why it's suddenly multiplication

[versed wave]

$e^{\ln\abs{y}} = e^{\ln\abs{x^3} + C} \implies \abs{y} = e^{\ln\abs{x^3}}\cdot e^C \implies \abs{y} = \abs{x^3} \cdot e^C$

[sinful citrus]

facepalm

basic log rules that I somehow forgot made me spend forever on this problem

thanks for the help, I get it

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[lean nexus]

I'm stuck on this problem. To be honest I only know a matrix is symmetric if it equals it's transpose

[lean nexus]

and that u^T v =0 means u and v are orthogonal

[woven hound]

what have you tried

[lean nexus]

well I tried since A is symmetric A=A^T so A^Tu=lambda u and A^Tv = Bv , I just substituted it in

[woven hound]

Yeah that's not very helpful

because we want a u^T to appear somewhere

what happens if you take the transpose of both sides of $A\vec{u} = \lambda \vec{u}$?

[flat frigate]

Ari

[lean nexus]

I tried this by the way but getting nowhere

[lean nexus]

I think it's $$ (Au)^T= (\lambda u)^T means u^TA^T=\lambda u^T $$

[flat frigate]

Unknown

[woven hound]

Yeah like I said just swapping A with A^T doesn't do much since you never get u^T to appear

Consider this

If we take transpose of both sides we have

u^T A^T = lambda * u^T

[lean nexus]

I use transpose property (AB)^T=B^TA^T

[woven hound]

Yes I also used that

now we substitute A^T = A

$u^T A = \lambda u^T$

[flat frigate]

Ari

[woven hound]

Now, we want there to be u^T v

so multiply both sides by v

and see if you can finish from there

[lean nexus]

maybe try $$ u^Tv = A^-1 \lambda v u^T$$ and maybe set it equal to zero

[flat frigate]

Unknown

[woven hound]

why are you taking inverses

$u^T Av = \lambda u^T v$

[flat frigate]

Ari

[woven hound]

What's another way to write Av

[pine umbra]

May I have help with a question

[woven hound]

based on what you're given

[woven hound]

!occupied

[safe radish]

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).

[woven hound]

go to #help-10

[lean nexus]

well by definition A^-1A=I and I doesn't affect anything

[woven hound]

go from here

[lean nexus]

beta v

[woven hound]

Sure, but then you're introducing things that aren't really going to help you (e.g. A^-1) so this isn't the best approach

[woven hound]

exactly

so now we have

$u^T \beta v = \lambda u^T v$

[flat frigate]

Ari

[woven hound]

do you see where I'm going with this

[lean nexus]

maybe, I guess you can move right hand side to left and factor u^Tv

[woven hound]

that works yes

[lean nexus]

okay I think I solved it using your method but how did you know to transpose on both sides

[woven hound]

great!

[woven hound]

Remember that our goal is to show $u^T v$ = 0

[flat frigate]

Ari

[woven hound]

But we don't have $u^T$ anywhere in our given equations

[flat frigate]

Ari

[woven hound]

so you can make one by transposing both sides of $Au = \lambda u$

[flat frigate]

Ari

[lean nexus]

but isn't this just manipulating the equation

[woven hound]

Yes

I mean how else would you prove it lol

you're only given 2 equations

[lean nexus]

so what was wrong with my attempt

[woven hound]

well you probably saw that you weren't going anywhere

main reason being that you never made u^T appear anywhere

thus making it difficult to prove something about u^T is true

[lean nexus]

on the last line what if I took transpose on both sides there

I could use fact that (C+D)^T= C^T+D^T

possibly

[woven hound]

You're welcome to try it out

[lean nexus]

yeah it didn't work out

[safe radish]

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[west goblet]

Dot product hw, can anyone explain how to get 2?

[west goblet]

I assume we use the formula scal sub v (u)= u dot v /|v|

[safe radish]

@west goblet Has your question been resolved?

[west goblet]

<@&286206848099549185>

[errant finch]

not obvious what that notation means

scal_k

[west goblet]

Scalar projection

Idk if this note helps

[errant finch]

ah

usually I see that called proj_v but now that you have the definition, what's your doubt?

actually wait no

that's different

[west goblet]

We don’t know k

[errant finch]

it's the magnitude

[west goblet]

5k -4k +2k =3 K

[errant finch]

$\hat k = (0,0,1)$

[flat frigate]

gfauxpas

[west goblet]

Oh?

Why is it 001?

[errant finch]

$$\hat i = (1,0,0)$$
$$\hat j = (0,1,0)$$
$$\hat k = (0,0,1)$$

[flat frigate]

gfauxpas

[west goblet]

Ohhhh

That’s tricky

[errant finch]

it's a common notation

especially in physics and engineering

[west goblet]

Yeah, I totally forgot about this

That explains everything

[errant finch]

pure mathematicians are more likely to use e_1, e_2, e_3

[west goblet]

This is a engineering math course so that makes sense

Thanks so much tho!

[errant finch]

sure

also

the "hat" generally means normalized vector

good to know that

unit norm

magnitude 1

[west goblet]

I see

[errant finch]

so you might see something like

$$\frac{1}{\Vert \mathbf y \Vert} \mathbf y = \hat {\mathbf y}$$

[flat frigate]

gfauxpas

[west goblet]

Sorry but what is double absolute value?

[errant finch]

magnitude of a vector

[west goblet]

Ok

Got it

[errant finch]

[west goblet]

Again, thank you very much! Thanks for your time, have a nice night!

.close

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[sinful thicket]

can anyone help me with this problem 😄
i tried thiss

[silent nova]

Use the equations of kinematics

[sinful thicket]

what is ... that

[silent nova]

v^2=u^2+2as

s=ut+1/2at^2

v=u+at

These three

[sinful thicket]

hmm i never seen these equations before

[silent nova]

You must have seen if you have studied the Newton's laws of motions

[sinful thicket]

i think that is physic

[silent nova]

Yes

[sinful thicket]

yea im didnt take physic yet...

sorry, i just didnt know what those equations are

this is my calculus I hw problem

[silent nova]

These are supposed to be solved by calculus?

[sinful thicket]

ye i assume

[visual linden]

Velocity is integral of acceleration IG

[sinful thicket]

yes

[safe radish]

@sinful thicket Has your question been resolved?

[visual linden]

Okay so velocity is v(t) = -1.6t + c

Now at time t = 0, what would be the velocity

[sinful thicket]

wouldn't be 0

i dont see they give me any initial velocity

so i am assumming that initial velocity at t = 0 second is 0

[rare briar]

+c is the initial velocity

C being a constant

[sinful thicket]

right, but do i set the equation v(0) ?

[visual linden]

Yes, so what would be c?

[sinful thicket]

so c would be = 0

[visual linden]

Okay, now v(t) is simply -1.6t

Also velocity is just directional here, so we can simply say 1.6t in downwards direction

or tbh IDK how particular they are, let's just keep it in negative

So now we've answered the 2nd part

simply find v(29)

And for distance, integrate -1.6t again

[visual linden]

Nice dude!

[sinful thicket]

it was hard lol

i was tryna find initial velocity

[visual linden]

But pretty fun to debunk

[sinful thicket]

bc they didnt give me anything

but turned out it was 0 lol

[visual linden]

Yeah man, well at least now you know how to work with this

[rare briar]

I think you’d assume the rock starts from rest

[visual linden]

Yeah because it was "dropped", they play with words, if it was dropped, that means its initial velocity was 0

[sinful thicket]

yea, the other problem they gave me initial velocity

[rare briar]

Yeah

[visual linden]

Yes, if nothing is given and you're sure there's no other way to solve it or find it, then you do have to assume that

[rare briar]

Definitely

[sinful thicket]

thank you guys

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[digital burrow]

I need to find the area bounded by x + y = 2, y = x^2, y = 2
Can I do it like that:
x = 2 - y, x = sqrt(y), y = 2
Then we need to find (an)other y point(s)
2 - y = sqrt(y)
y = 1
And then we compute integral from 1 to 2, and the integrand itself is:
sqrt(y) - (2 - y)
Am I right???

[silent prairie]

yup precisely

[digital burrow]

ah thanks

ChatGPT gave some complicated ahh answer

[silent prairie]

wait what did you get as your solution?

[digital burrow]

I don't know I haven't solved it fully yet

Just got the integral

[silent prairie]

alr lmk

[silent prairie]

luckily a pretty simple one

[digital burrow]

(4*sqrt(2)/3) - 7/6

wait I haven't realized that the resulting area actually has two regions
Apparently I need to calculate the right one, since the task says it's bounded on the right by y = x^2

My solution is still correct?

[peak estuary]

imo its easier to compute two integrals. one for x from 0 to 1, the other from 1 to wherever y=2 and y=x^2 cross

(the first integral corresponds to a triangle so you dont even really need to calculate it)

[digital burrow]

1 is just where the red and blue function intersect? So is it how I can find the area for any triangular region?

[peak estuary]

yes 1 is where red and blue cross

triangle area = base*height/2

[safe radish]

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[autumn shuttle]

8/17 × 9/ 3 sqrt 34 should = 72/51sqrt34 right? Is there something wrong with my solution?

[vagrant ice]

,rotate

[vagrant ice]

why do that

when you could just do $\frac{72}{51 \sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt {34}}$

[flat frigate]

south

[vagrant ice]

actually, what's the original question?

[thin bridge]

excessively splitting components led to violating order of operations

[vagrant ice]

,rccw

[thin bridge]

you also seem to be conflating multiplication and addition of fractions

[autumn shuttle]

I rationalized the 2nd fraction into 3sqrt34/34 then multiplied that with 8/17.

Multiplied both fractions by the other's denominators to get a common fraction

Then just added the numerators from cross multiplication then adding both.

Which part did i mess up?

Im going to recheck

[thin bridge]

what's "cross multiplication" doing here

you also seem to be conflating multiplication and addition of fractions

[autumn shuttle]

Followed some old notes

[autumn shuttle]

Yea that, alrighty

[thin bridge]

$$\br{ \frac ab} \br{ \frac cd}$$
is simply
$$\frac{ac}{bd}$$

[flat frigate]

ραμOmeganato5

[autumn shuttle]

Forgot abt that honestly, thanks

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[noble jetty]

is anyone here famliar with genetic algoritem code

[noble jetty]

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[hoary seal]

what did I do wrong here?

[hoary seal]

question a

also sorry for the small writing

I think it got messed up

Ill redo it

hhere you go

[inner heath]

who watches lobotomy kaisen?

[vagrant ice]

don't chit-chat here pls

this is a help channel

#discussion if you want to chat

[inner heath]

oh ok

[safe radish]

@hoary seal Has your question been resolved?

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

<@&286206848099549185>

[hoary seal]

I mean this ^

[craggy quest]

everythings correct @hoary seal

[hoary seal]

how?

I got the wrong answer

the answer key said "n=15"

[ornate wyvern]

there are issues with the term in sqaure root

it wont b such a low value

[hoary seal]

its ok ill come back to it

..close

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[lean palm]

I am by no means a math person, I have just taken an interest in Python and wanted a rabbit hole to follow for a bit, so excuse me if this is obvious to you experienced folks.

First, I calculated all of the primes up to a million.

Then, I calculated all of the primes with a prime digital sum.

Then, I calculated all of the primes with a prime digital sum that have a digital root of 2.

I took those and assigned them to a point cloud for a 100x100x100 cube.

When viewing the cube from a certain angle all of the points seem to only appear within certain bands. I counted 31 of them, which is a prime number.

I did the same process with dr =5 which yielded 29 bands, which is also prime.

Then again with dr=7 which yielded 17 bands, which is also prime.

Anyone know why this is?

[peak estuary]

I would expect this to be only coincidence

if your cube was bigger then the number of bands would also change and it seems unlikely that it would be a prime again every time

and also, dont know if you just havent hit the perfect angle for your screenshot, but saying that for dr=5 the upper right region consists of bands is also a bit of stretch

[lean palm]

I have to pan the camera to see all of the bands individually, just a constraint of a static camera. Pic was mainly just to show what I meant by bands.

But essentially, I gather it's just a coincidence brought about by the 1 million number constraint.

[peak estuary]

that the number of bands is prime, yes

that there are bands, probably no

I dont know how precisely you filled the numbers into the cube but I could easily imagine that as a consequence of that some of those slices just for example only contain even numbers

[lean palm]

If I scaled it to 1000x1000x1000 and it still produced prime # of bands would it not be coincidence anymore?

[peak estuary]

well powers of 10 are special given that the digital root is essentially working mod 9 so maybe for powers of 10 I could imagine some kind of result along those lines

but ehh I still wouldnt expect it to be primes

fundamentally almost all small odd numbers are primes so there are a lot of coincidences

its the strong law of small numbers

[lean palm]

Cool, thanks for the explanation.

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[cyan coyote]

,rccw
I don't know where to start
13th

[cyan coyote]

Nvm

Got the answer

.close

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[grand tendon]

Hi

[grand tendon]

Anyone studying trig and can help me

[junior raven]

just ask

[grand tendon]

Where are the function sinx cosx present in an triangle?

Can you help in visualisation

[junior raven]

well the most basic definition of the trig ratios in a right triangle

is sin x = opp/hyp
cos x = adj/hyp
tan x = opp/adj

[grand tendon]

Yes but are they angles or just ratios of the sides

[junior raven]

x is the angle

sin cos tan are ratios of the lengths of the legs of the triangle

[grand tendon]

so sinx means?

[junior raven]

the length opposite the angle divided by the length of the hypotenuse

[grand tendon]

opposite/ hyp × angle?

[junior raven]

no

just the ratio

[grand tendon]

Oh thank you so much

Can also tell me how trig is used in making of bridges and real life etc

@junior raven

[junior raven]

uh

im not an engineer

research on wikipedia or something

[grand tendon]

Ok thanks ig

Which grade you're in in?

@junior raven

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[versed lotus]

is 0 a part of natural number in discrete mathematics?

[cedar widget]

It is a matter of convention

usually the answer is yes

[versed lotus]

could you please elaborate on this

[cedar widget]

some people/courses assume that 0 is a natural number

and some other don't

[versed lotus]

in discrete mathematics, it is yes, right?

[obsidian oracle]

It depends on the convention

As was said

[cedar widget]

probably, its a good idea to check with your professor/teacher

[errant finch]

logicians and set theorists tend to consider it natural

number theorists tend to not consider it natural

other types of math, i havent seen a strong tendency one way or the other

[versed lotus]

thanks @cedar widget @errant finch @obsidian oracle

for explaining so well

[cedar widget]

not sure about that

[errant finch]

not sure how to prove one wya or the other without doing a survey of enough texts to show you it's more than 50%, so, shrug emoji

maybe im wrong

[cedar widget]

shrug indeed

[versed lotus]

I'll look for conventions and ask my professor

[lethal mason]

usually from what ive seen 0 isnt

[cedar widget]

if you want to use something agreed upon, you can use $\bZ_+$ or $\bZ_{\geq 0}$ or $\bN_+$

[flat frigate]

ExpertEsquieESQUIE

[versed lotus]

thanks

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[charred cedar]

hi

[charred cedar]

any tips to memorize this ?

[timid ridge]

ime memorisation in math is TERRIBLE
you should always try to UNDERSTAND things and not memorise them
that way, you can just derive whatever you need

[charred cedar]

I agree you

but I have ten hr before my emags exam

[timid ridge]

gg you are toast

[charred cedar]

[timid ridge]

i dont really know any tricks to memorise that

burn that image in your head

[charred cedar]

alr

thanks though

[timid ridge]

or maybe if you have the time in exam you can derive it from
x = p sin phi cos theta
y = p sin phi sin theta
z = p cos phi

which is a lot easier to remember

[charred cedar]

that's a good idea

I'll do that

[north fulcrum]

toast do you need a jacobian

[charred cedar]

I dont think I'll need that

[north fulcrum]

i have this for transforming coords

im having a similar test in 7 hours too

[charred cedar]

damn

are we cooked together ?

hopefully you are not cooked as me

[north fulcrum]

(yes idk how to construct the jacobian of turning xyz into some arbitrary uvw)

[charred cedar]

dw we'll survive ig

I will look into it

[charred cedar]

thank you

alr thank you everyone

imma close this

.close

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[keen moon]

"Prove that at a party where there are at least two people, there are two people who know the same number of other people there."

[keen moon]

i know i will use pigeonhole but thats about the extent ive gotten

[timid ridge]

what does "know" mean here

oh ok i got it

yeah same 💀

[keen moon]

lol

[weak spoke]

[brave wolf]

Look at how many people are known by each person

say that there are 5 people A, B, C, D, E. How many ppl could each of them know?

[keen moon]

max 4, 3, 2, 1, 0

ohh i get where its going

[brave wolf]

nice, ill give u some time to try and fiure out the rest

if u need another hint, just ask

[keen moon]

alrightt

would reasoning it out as something like
"if there is n amount of people one person could know a max of n-1 people therefore by pigeonhole... n/n-1 = 1 at the least" be enough

[brave wolf]

I see you wrote 5 numbers here, which is n

[brave wolf]

why is it n-1 here?

[keen moon]

I thought of n-1 as the number of people you can know

[brave wolf]

what if you know 0 people?

0, 1, 2, ..., n-1
this is n pigeonholes

not n-1

[keen moon]

so it is n and n..?

is it straight up n/n

[brave wolf]

yes, which isnt enough to conclude anything with pigeonhole

imagine a party with 5 people A, B, C, D, E:
what if A knows 4 people, B knows 3, C knows 2, D knows 1, E knows 0

direct application of pigeonhole doesnt lead anywhere as you can see

5 pigeons (the people)
5 holes (the n.o. people they could know, that is either 0, 1, 2, 3 or 4)

[keen moon]

but also wouldnt people know eachother? otherwise the question is wrong right

[brave wolf]

wdym by "wouldnt people know each other"?

[keen moon]

like if A knows B we assume B also knows A

[brave wolf]

yes, sure

and what's ur point?

[keen moon]

this wouldnt happen

[brave wolf]

exactly

and that's what's missing in your reasoning

you need to analyze this special case

[keen moon]

hmmm

[brave wolf]

there are 5 pigeons and 5 holes. But the case where every hole is taken cant happen

there could be a party where
A knows 3
B knows 2
C knows 2
D knows 1
E knows 0

and there could be a party where
A knows 4
B knows 3
C knows 2
D knows 2
E knows 1

but there cant be a party where
A knows 4
B knows 3
C knows 2
D knows 1
E knows 0

[keen moon]

it all makes sense on a smaller scale

[brave wolf]

so the set of holes actually changes, but there are at most 4 of them. There cant be 5 holes

[keen moon]

but how do I reason that out as a general case...

[brave wolf]

alright, so now onto the general case.

What I'd do is say that there are n people and introduce some kind of variable k_j, where k_i is how many ppl does the ith person know. Then we know that each k_i must go in some of these n holes: {0, 1, 2, ..., n-1}

and note that this isnt yet enough to apply the PHP

there re n pigeons (ppl) and n holes

now either the pigeons go into one hole twice (and the proof is done), or in the other case, the pigeons will go into 1 hole each

so every hole will be taken

i.e. youll have a person knowing nobody and a person knowing everybody

as you said, that's impossible

so you can cross out this case and you're done

[keen moon]

ok let me digest it lol

oohhh okay i get it

thanks a lot it all makes sense 😭

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[daring scaffold]

How would I do this?

[daring scaffold]

this is a practice test im reviewing in order to prepare for my actual test

[safe radish]

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

[daring scaffold]

1.\

1. I don't know where to begin.

[manic radish]

draw a venn diagram

do you know what that is?

[daring scaffold]

doing that right now one min

[manic radish]

cool

[daring scaffold]

ok

[manic radish]

not quite

wait, 1 second.

the question is not worded the best

[daring scaffold]

agreed

[manic radish]

ok, I was right

so, when it says "77 like snickers", you've put down "77 like ONLY snickers"

[daring scaffold]

ohh

How would i go about it then?

[manic radish]

the sum of this shaded region should equal 77, because these are all the cases where someone would like snickers

[daring scaffold]

okay

[manic radish]

I'll give you a hint: there's only one way someone would like all three kinds of chocolate

so you've got that inner 44 right

[daring scaffold]

yep id use it subtract?

[manic radish]

yep 🙂

[daring scaffold]

from the 77 and that equals only people that like snickers?\

[manic radish]

remember, there are 4 regions there that must add up to 77. so maybe we can do something smaller next?

when it says "46 like snickers or twix", which regions is that referring to?

[daring scaffold]

region in between snickers and twix

wait Or

the Snickers/Twix region

so 100,77 dont make sense

[manic radish]

yep

The blue region corresponds to "likes snickers and twix"
The green region corresponds to "ONLY likes snickers and twix"

because, if they "like snickers and twix", they might also like reeces cups.

[daring scaffold]

okay I see

[manic radish]

so if 46 like snickers and twix, how many like ONLY snickers and twix?

[daring scaffold]

2

[manic radish]

yep 🙂

so we put 2 in the green region

[daring scaffold]

yes

[manic radish]

keep going 🙂

[daring scaffold]

ok

how would i find the people that only like one?

[manic radish]

sorry, had to dip for a second

we know 77 people like snickers. so, the total for everyone in the "likes snickers" circle (the red bit) should be 77.

[daring scaffold]

ah

so 44+4+2 - 72?

[manic radish]

77 - (44 + 4 + 2)

other way around, but you've got the idea

[daring scaffold]

yes 27 only like snickers

nice

[manic radish]

yep

and the other three empty regions?

[daring scaffold]

19 reese and 41 twix

[manic radish]

yep

and the last region?

[daring scaffold]

wym?

the answer?

27+4+19 = 50 which is E?

[manic radish]

how many people don't like snickers or twix or cups?

[daring scaffold]

ah the outside

[manic radish]

yep

it's a large number who don't like anything XD

[daring scaffold]

add them all subtract by 500?

[manic radish]

yep

[daring scaffold]

yea i prefer peanut m&ms lol

350 outside

[manic radish]

yep

[daring scaffold]

yes

[manic radish]

now the next bit. which regions correspond to "likes reeces cups, and likes snickers, and doesn't like twix"

[daring scaffold]

27 +4+19?

[manic radish]

yep

there's your answer 😉

[daring scaffold]

50 nice

i wouldnt care about the 44?

[manic radish]

nope, because they also like twix and we only count people who like snickers and reeces cups, but not twix

[daring scaffold]

Nicee okay

thank you! this helped very much

gonna try one on my own now

[manic radish]

everyone in the red region likes twix, so we don't count them.

[daring scaffold]

yes agreed

Could we do another one?

so i can fully master this concept

or ill do it and if i get stuck ill ask you

nvm ty

.close

[safe radish]

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[safe radish]

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[daring scaffold]

I know a little of what the signs mean but i dont know how to do it properly.

[daring scaffold]

.close

[safe radish]

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[stiff nova]

guys i need your help im doing quadratic equations with absolute brackets and i just cant figure out what to do after i make the line with intervals

[fathom jewel]

can you show the problem you are working on, as well your progress

[stiff nova]

just a second

[fathom jewel]

,rccw

[stiff nova]

i got stuck here but lets say i only know how to do the left half

[fathom jewel]

i suppose the right side implies you are looking for the roots

[stiff nova]

yeah

for x1 and x2

[fathom jewel]

ok so you computed them, so what you can do is see if they are in the domain (-oo,-5) inside

now for the [-5,3) interval, it implies that x+5>=0 because of x>=-5

so x-3<0 which gives x<3, so you only add a minus to the x-3 term

[stiff nova]

i just dont know what to do with the equation after the intervals bcs how does 2x*|x-3| become -2x(x+3)

[fathom jewel]

because in [-5,3) you must have x+3<0

you get that way x<3 which fits the interval

[-5,3) is the overlap between x+5>=0 and x-3<0

[stiff nova]

and do i need to find the overlap or i avoid it

[fathom jewel]

u already did

it's [-5,3)

[stiff nova]

but when i solve both quadratic equations what do i look for

[fathom jewel]

now u just need to realize that since x-3<0 in that interval, you get |x-3|=-(x-3)

[stiff nova]

ooooohhh

[fathom jewel]

you look for solutions within the domain you are currently in

[stiff nova]

so it would be 2x*(-(x-3)+(x+5)=0

[fathom jewel]

yes

[stiff nova]

for the [-5,3)

[fathom jewel]

and the other inteval you have both expressions being positive

[stiff nova]

so when i do it within its own interval i just put a minus infront?

in the first one the +(x+5) turned to -(x+5)

[fathom jewel]

you ever put a minus in front of a term, when it would be negative

yes, because you are in (-oo,-5) which is literally x<-5

equivalently x+5<0

so |x+5|=-(x+5)

[stiff nova]

but then why did i put a minus infron of 2x(x-3)

[fathom jewel]

it's an overlap again between x<3 and x<-5

and that overlap is just x<-5 but we still have x-3<0

[stiff nova]

there may be a language gap yk but let me show you what im talking about

just a sec

why did i put the minus infront of the 2x if its not in the starting equation

[fathom jewel]

well i cant really tell what your thoughts were but what i can say is that 2x|x-3|=2x(-(x-3))=-2x(x-3)

[stiff nova]

oooh because its 2x(x-3) and bcs its under its own inteval i put a minus infront

we are on the same page i believe

so the main thing is

when i do the absolute brackets under their own interval i just put a minus infront of them

and thats it

[fathom jewel]

thats now what i wanna give away

you should check if the term inside the absolute brackets is positive or negative in your current domain

in the interval of (-oo,-5) we must have x-3<0 because x-3>0 would lead to no overlap at all with x+5<0

[stiff nova]

so the equation for the interval on the right would be 2x(x+3)+(x+5)

and i solve that to find the quadratic equation

a b and c

[fathom jewel]

you mean for the 3rd case?

[stiff nova]

theres a third case

💀

[fathom jewel]

you have - -, - + and + +

[stiff nova]

oh yeah

the Xe[3,+oo)

[fathom jewel]

so for the ++ case you have this = 0

indeed

since x+3>=0 and x+5>=0

[stiff nova]

and i just dont put minuses infront of the brackets for that one?

[fathom jewel]

yes because both expressions are now positive in the domain [+3,oo)

if you still put a minus, you would make the positive terms negative, which would defeat the purpose of the absolute values

[stiff nova]

mhm

so lets revise

the first one has a minus infront of the first absolute bracket

the second infront of the second

and third doesnt have any

[fathom jewel]

the first one has two minuses

[stiff nova]

oh yeah

beacuse its -- and --

[fathom jewel]

yes

[stiff nova]

💡

and i try to find x that fits the interval under each one

and thats the point of the whole thing

[fathom jewel]

you are trying to find solutions of x within your current domain, yea

[stiff nova]

bro

you were really helpful ngl

i got a f on this in my math class

tomorrow i gotta do something about it

[fathom jewel]

tomorrow you are gonna ace it

[stiff nova]

i believe you man

wish you all the best in life

[fathom jewel]

believe in yourself

[stiff nova]

❤️

.close

[safe radish]

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[runic ruin]

Can anyone help me out? I don’t think if (log base 2 of x)^2 is the same as log base 2 of x^2…what am I suppose to do instead? Thanks

[cedar widget]

log2(512)=9

so (log2(x))^2 = 9

if you let y = log2(x) then y^2 = 9, which should be easier to see how to solve

[runic ruin]

Ohh I see thank you!

.close

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[hoary seal]

what did I do wrong in question c?

[astral glacier]

You don't want T_n to be 945

You want S_n to be 945

He has 945 cabbages in total

Not in the last row

[hoary seal]

also I am not sure with q b as well, I got fraction again

[hoary seal]

how did u figure that out?

[timid escarp]

b is correct.

[hoary seal]

damn I thought I would get it wrong

[timid escarp]

"The farmer plants only 945 cabbages."

notice no mentions of rows.

[hoary seal]

yea but the question is asking for rows, I am not calculating the final answer because I already subbed that value in

i am solving for n, which is the number of rows

[timid escarp]

but you're solving for n using T_n, which is the nth row to contain 945 cabbages.

[hoary seal]

oh ok tN is the term I forgot

for that row

[timid escarp]

you don't want the nth row to have 945 cabbages. you want to find the nth row such that, when totaled up with the rest, you will get 945 cabbages.

[hoary seal]

got it

thanks

.close

[safe radish]

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[indigo sierra]

Hi, Im studying prob and stats for a math exam, Im really struggling with combinatorics and understanding how to use it. Im understanding basic permutations now but once it gets a little more complicated i get lost. heres the type of program that really trips me up, I know theres a total of 11!/(3!2!) possibilities for question 2.6.38 but no idea how to do the math for only when LLL is infront of the other consonants. any insight would be appreciated

[indigo sierra]

<@&286206848099549185> ^

Im not looking for an answer as ive found the answer, i just need help wrapping my head around the why

dam no one

[left gyro]

LLL S U M G U I O N
LLL S U U M G I O N
(8! ways to rearrange the non-L letters) / (2 ways the Us can be swapped)

,calc 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 2

[flat frigate]

Result:

20160

[left gyro]

should just be 20,160 wys

oh wait it says precedes all the other consonants nvm

[indigo sierra]

I know the answer, just dont know why, And yes the L's must come first

I struggle with the letter based arrangement permutations

[left gyro]

L L L S U M G U I O N
U U I O L L L S M G N
think about placing the letters one at a time
L L L first,
then 4! ways to place S M G N after it
L L L S M G N
then after that, U, U, I, and O can be placed in anywhere in between
theres 8 places to place an O:
L L L O S M G N
then 9 places to place an I:
L L I L O S M G N
then after that, 10 * 11 / 2 ways to place 2 Us
U L U I L O S M G N
so in all its 4! * 8 * 9 * 10 * 11 / 2

,calc 4 * 3 * 2 * 1 * 8 * 9 * 10 * 11 / 2

[flat frigate]

Result:

95040

[left gyro]

does this line up

[indigo sierra]

what trips me up right

is that just shows LLL first, like how does that math account for something like UUIOLLLMGNS or LLLMGNSUUIO

[left gyro]

you gotta confirm whether this matches the answer in the book or not before we can continue

[indigo sierra]

like i dont see how placing the L's first, then arranging the smgn after accounts for all the possibilities

it does

[left gyro]

the trick I used is to treat the letters as being placed one at a time

you can imagine that if we typed 3 Ls, then typed the consonants S, M, G, N after it, that would always guarantee the Ls are before the consonants

[indigo sierra]

and whyd ya do 10*11/2,

[left gyro]

try thinking about it slowly

if I explain it then youll never be able to come up with it yourself

[indigo sierra]

wait is that vowel part (11C2)?

[left gyro]

yes, thats one way to view it

[indigo sierra]

well for the U's right

[left gyro]

yes

and also because theres no condition on the Us

just as theres no conditions for the I or the O

so all 11C2 combinations are possible

[indigo sierra]

so they can be placed anywhere

[left gyro]

yep

[indigo sierra]

and the slots increase as letters get placed cause now you can place a letter on either side

[left gyro]

yes

for N letters, theres N + 1 places to place the next one

[indigo sierra]

so i understand the LLL and the 4! part, the letter placements at the end was what was tripping me up ig,

[left gyro]

more specifically its that that would make the problem easier

[indigo sierra]

I could do it manually by seeing i have a letter to place and how many slots ive got

like how you did it