help-10

[cursive tundra]

As in here

Bcuz for something to be an average dont u gotta like divide?

Or is it ok bcuz the coefficients beta and 1- beta serve in some way like a division?

In addition to this question, can i ask where/how i can learn more operational terminologies ?

[timid silo]

yes, this is the correct intution

ig you are confused with "weighted average"?

https://en.wikipedia.org/wiki/Weighted_arithmetic_mean

[cursive tundra]

Oh yeah, no no . I understand weighted and weighted average

[cursive tundra]

Just curious about this one “averaged with”

But thanks!

[cursive tundra]

Would u happen to know regarding this as well?

Just so im better prepared for future encounters

[timid silo]

nope, I just use context normally :/

[cursive tundra]

Kk thx!

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[vast sun]

hi, so i have this problem that i'd like to know if theres an answer. basically you have 1 is less than or equal to x, i want to find d(x;c) compared to r, where c is the center, and r is the radius. this is all in the number line. if u can give it as an absolute value as well

[vast sun]

basically a distance

the distance between x and c compared to r

if it was arabic, id explain a bit better

[obtuse pebble]

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[vast sun]

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[hybrid socket]

This is probably easier than it looks but I just cant solve this. (calculus but just gotta set up the volume for a further integration) I cant seem to understand what the volume equation might be.

[hybrid socket]

I think the circumference on top is 2pi * r and the circumference on bottom is 2pi R

is it V= 2pih(R-r)

<@&286206848099549185>

Nevemind its alot harder than it looks I got it
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[agile robin]

i forgot the name of a method

[agile robin]

for ex say 2/sqrt 5 - sqrt 3

multily both num and dneom by sqrt 5 + sqrt 3

so like you multiply but only sign changes

is there a name for that

[civic zealot]

conjugate

[agile robin]

ty

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[timid silo]

or rationalisation for the actual process https://en.wikipedia.org/wiki/Rationalisation_(mathematics)

[obtuse pebble]

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[agile robin]

cna someone pls explain why y has to be y>1

[daring rock]

Because y = csc t

[agile robin]

oh it has to due with sin pi/2 = 1 and csc pi/2 = 1/1 = 1

[daring rock]

with t between 0 and pi/2

[timid silo]

It's because you have 0 <x < 1

Which means

1/0 < y which isn't true

And then you examine to see if y is y < 1 or y > 1, with the latter being true

[agile robin]

ok i think i understand it now

ty

[agile robin]

id have to draw it too see right

[timid silo]

You can just plug in some value

And u would see

[agile robin]

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[mellow yoke]

is there a way to solve this without l'hopitals rules?

[civic zealot]

probably not without some very clever substitutions

[mellow yoke]

oh cool

too clever for me obviously

i'm just wondering if anybody else can think of any, i guess. i wouldnt know where to begin.

[obtuse pebble]

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[blissful nimbus]

I "only" need to find the limit of the sequence. i have tried the sandwich rule but i cant find 2 similar sequences that have the same limit.
My assumption is that it converges. and limit is 1.5 but I have no clue how to show that. Maybe someone has a hint 🙂

[timid silo]

@blissful nimbus

Distribute that square root in

And take the conjugate

That's your hint

[blissful nimbus]

ok i have tried what you said and now I have this

is that correct?

if yes. then I am stuck again. 😄

[tardy epoch]

missing a square root on the last n

try taking the limit now

[blissful nimbus]

yes ty!

i could factorr the sqrt n

oh no wait I fucked up in the numerator

the numerator is: n²+3n+n

is the first step correct?

ahh forget the sqrt n

its just n

[timid silo]

Make your life easier and just multiply that sqrt(n) inside the sqrt(n+3)

$\sqrt{n} \cdot \sqrt{n+3} = \sqrt{n(n+3)}$ is what I mean

Oh okay bot is in another coma

[timid silo]

Still missing something

[blissful nimbus]

is this now correct?

[obtuse pebble]

@blissful nimbus Has your question been resolved?

[blissful nimbus]

ok great. but I still cant see the limit ^^

[raven spire]

it's getting worse though

[blissful nimbus]

OK so maybe someone else has a better solution to this problem? because this fraction doesnt look better to me from what I have started

or do I miss something

[raven spire]

it should've been $\frac{3n}{\sqrt{n^2+3n} + n}$

just multiply a 'n' on the numerator and you'll be fine

[blissful nimbus]

a fuck i forgot to square the n... 3rd binomial rule 😄

yeah now i got you

ok but the same problem as before.. How can I see the limit now?

<@&286206848099549185>

[dire arch]

my best guess is to divide every term by n

I don't know if it's a sufficient clue

[tardy epoch]

,w plot sqrt(x)(sqrt(x+3) - sqrt(x)) for 10<x<100

pain

[blissful nimbus]

yeah but the thing is the sqrt. I dont think that dividing by n is any good

[tardy epoch]

start back here. the hint is to multiply the top and bottom by the radical conjugate:
(sqrt(n+3) + sqrt(n)) / (sqrt(n+3) + sqrt(n))

assume all your math done up until this point was wrong

leave the initial sqrt(n) on the outside

for the numerator: just calculate (sqrt(n+3) - sqrt(n)) * (sqrt(n+3) + sqrt(n))

the denominator is unchanged

[dire arch]

what I meant is that there is a procedure to solve inf/inf indetermination type limits by taking just the n of greatest degree (considering sqrts)

so it would be 3n/(sqrt(n^2)+n) which is = to 3n/2n so the answer is 3/2

i dont know if I'm explaining myself

in this case, the greatest degree would be 1

[blissful nimbus]

OK thx for answering me guys:

I am now trying to follow @tardy epoch . Now I have sqr(n)*(3/(sqrt(n+3)+sqrt(n))

and @dire arch yeah I got what you said. but I am not sure if that works if you have sqrt or can I now devide everything with sqrt n?

[dire arch]

wait I'll show it step by step

it works btw

[tardy epoch]

yes

[blissful nimbus]

now I can factor sqrt n?

[dire arch]

does anyone know how can I write equations on the computer? just to show you the procedure

[blissful nimbus]

I have no idea how the discord mathbot works. but the website has nice tool

https://www.mathway.com/de/Algebra?asciimath=sqrtn*sqrt(n%2B3)

which I have used for my screenshots

[dire arch]

ty

[tardy epoch]

right. now divide the top and bottom by sqrt(n)

[dire arch]

from there the limit should be easy, if you have got any problems ask me

[blissful nimbus]

@tardy epoch is ipixtxd solution then correct? or can I devide by n and sqrt(n)

@dire arch this step is not clear to me on the bottom

ah because n = sqrt n²

but is my top 3n or 3sqrt(n)?

@dire arch dw NOW i got it!!! THX so much!! ❤️

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[quick kayak]

Is there anyone here with a good understanding of generating function counting questions? I ask this most vague of questions because when I've posted the question I'm actually trying to solve, no one seems interested in helping

[latent notch]

I’ll help you out

[quick kayak]

❤️

Bless

I think I understand the basic theory, but as soon as the questions get a bit more complex, I get lost

How many ways are there to distribute 20 cents to n children and 1 parent if the parent receives either a nickel or a dime and each child can receive any amount?

[latent notch]

alrihht

let me see

[quick kayak]

I've come this far

Is this at least the right initial representation?

[latent notch]

just a sec mate

yep, that seems alright

[quick kayak]

Now, here's the solution in the back of the book

[latent notch]

ok

[quick kayak]

I notice two things about this. First, by initial state is in the form of term x term, but the solution is in the form term + term

[latent notch]

true

it’s oversimplified

[quick kayak]

The second thing I notice is that these look like coefficients extracted from these sums-of-series

So I reverse engineered his solution into this (following), though I don't see how it helps me

[latent notch]

huh

wait

let me just go through it

[quick kayak]

Sorry for dumping all that. I just wanted to share my own observations, but I'm all ears now

[latent notch]

well mate, quite honestly, it looks like youre doing a great job, it seems like you understand most of the things

however

the doubts you have come face to face with are common, you haven’t made any mistakes

basically the teacher has oversimplified the expression

[quick kayak]

He does that >_>

[latent notch]

to the point where it’s very difficult to revert it

[quick kayak]

Ad nauseam

Putting aside his solution for a moment, unless you see a clear path towards it

[quick kayak]

what would be the next step from this point?

[latent notch]

I’m going to try to write it lol

[quick kayak]

Hehe

Righto

[latent notch]

[x^15] 1-x^5 * 1/(1-x)^n

sorry if you can’t understand it

lol

it’s difficult to write on a normal keyboard

[quick kayak]

That's your solution?

[latent notch]

No

not het

yet*

let me continue

[quick kayak]

🙂

[latent notch]

[x^15] 1-x^5 / x^n - xn

[quick kayak]

Okay, hold up

[quick kayak]

(1-x^10)/(1-x^5) = (1-x^5)?

I don't remember learning this

[latent notch]

Maybe I’m overcomplicating things lol

[quick kayak]

I'm sure your solution makes perfect sense! I feel like that's something I should already know

[latent notch]

well, you can solve it with your teacher in class, and then ping me or dm me and we can check it out

[quick kayak]

🙂

Thanks!

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[wind basin]

how to write the equation of a tangent line in polar form

[obtuse pebble]

@wind basin Has your question been resolved?

[wind basin]

ni

[wind basin]

no

[obtuse pebble]

@wind basin Has your question been resolved?

[bold bane]

You substitute

x = r * cos(theta)

and

y = r * sin(theta).

eg.

[wind basin]

i got the equation like y-y1=slope(x-x1)

[bold bane]

Once you have found the equation for the tangent line, do the substitution as shown above.

[wind basin]

i used the polar curve at theta=pie

i got y-0=-sqrt(3)(x+1/2)

[bold bane]

And now just do the substitution.

[wind basin]

so for y->0=(the r equation i used)*sin(theta)

[bold bane]

$y = - \sqrt{3}(x + \frac{1}{2})\
r\cdot \sin{\theta} = -\sqrt{3}(r\cdot \cos{\theta} + \frac{1}{2})$

And then solve for r.

[warm shale]

Kookiemon

[wind basin]

i used the polar curve given to me to find the equation of the tan line at theta=pie, i found y, and x, and then i found the slope which i put together and showed u, now how do i do the substitution to write it in polar form

can u show me with example im so confused

[bold bane]

Write the original problem you are working on.

[wind basin]

find equation of line tangent to polar curve r=cos(theta/3) at theta=pi. write in polar form

@bold bane

[bold bane]

I hate to bail on you but I need to start cooking dinner. Hit me up in about an hour if nobody has helped you.

[wind basin]

ok

[obtuse pebble]

@wind basin Has your question been resolved?

[wind basin]

@bold bane u free mate

[bold bane]

yup

[wind basin]

can u check if my answer till the tan line equation was good and how do i put that in polar form

[bold bane]

Offhand, it doesn't look correct.

[wind basin]

how to do it then

[bold bane]

How did you calculate the slope?

[wind basin]

i found r=1/2

from cos(theta/3) at theta=pi

then i found x, then i found y, then i found the slope

do u understand?

[bold bane]

Do you know how to convert to x and y?

[wind basin]

i use y=rsintheta

and x=rcostheta

[bold bane]

And r = cos(theta/3).

[wind basin]

yes

[bold bane]

So you can find dy/dx by find dy/dθ and dx/dθ.

[wind basin]

yes i found dy/dx

its -sqrt(3)

so i found y and i found x and i found slope, then i put it into the equation i first showed u

[bold bane]

Ok, so using the Point-Slope formula, you get

$y - y_{1} = -\sqrt{3}\cdot (x - x_{1})$

[warm shale]

Kookiemon

[wind basin]

yes

[bold bane]

Now you need to convert each of the y variables and the x variables back to polar coordinates.

[wind basin]

y1 was 0 and x1 was -1/2

[bold bane]

The one thing to note is the y1 and x1 will use theta = pi.

[wind basin]

how to do that

[bold bane]

And change x and y to its polar coordinate.

y = r*sin(theta), but leave theta as is.

[wind basin]

i plug in rsintheta for y and rcostheta for x?

[bold bane]

x = r*cos(theta), but leave theta as is.

[wind basin]

ok, so dont plug in pi for theta?

[bold bane]

Not for x and y, only for x1 and y1.

[wind basin]

ok so y=1/2(sin(theta))

since r i found was 1/2

[bold bane]

Leave r and θ as is.

[wind basin]

so the final answer would be (rsintheta)-0=-3sqrt((rcostheta)+1/2)?

[bold bane]

While technically correct, you should solve for r.

[wind basin]

which would give me again 1/2 cuz i just did ti

it

so its correct

[bold bane]

If you solve for r, you should get

$r = \frac{-\sqrt{3}}{2\cdot (\sin{\theta} + \sqrt{3}\cdot \cos{\theta})}$

[warm shale]

Kookiemon

[bold bane]

Which is the tangent line above.

[wind basin]

ok thats what i got thank u

[bold bane]

yw

[bold bane]

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[timid silo]

for 2) u gotta use the limit definition of derivative i suppose

[grizzled shore]

yeah i think it wants you to do it from first principles

[timid silo]

right

[obtuse pebble]

@upper jetty Has your question been resolved?

[upper jetty]

I have no idea what that is or how to do it

[timid silo]

okay ngl i forgot how to do those tables

but for 2 you have to use the definition of a derivative using limits

wait do u know whats a derivative? @upper jetty

[upper jetty]

The thing that tells u what the slope of a point in a curve is, right?

[timid silo]

yeye

so

in order to do that

we came up with a formula using limits

lim h->0 of (f(x+h) - f(x))/h

ah shit i gtg

sorry

[obtuse pebble]

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[undone yoke]

I need help with this, this problem is an polynomial synthetic division

[obtuse pebble]

@undone yoke Has your question been resolved?

[obtuse pebble]

@undone yoke Has your question been resolved?

[tardy epoch]

What's the polynomial being divided by?

[obtuse pebble]

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[cunning fox]

hello again, I'm stuck on how to proceed with this one. Obviously I need to find the distance between X axis and the circular arc, but I forgot how.

[hybrid gull]

Can you get the equation of the arc?

[cunning fox]

Well, arc length is like theta times r, but I can't find theta easily

I forgot the other equation

[cunning fox]

I can't figure it out by googling it either lol

[hybrid gull]

Can you get the equation of the parabola? Given the y coordinate of its vertex and its x-intercepts?

[hybrid gull]

That’s true - but in the x-y plane, do you happen to know? This can also be Googled just in case it doesn’t immediately come to you

[cunning fox]

not off the top of my head 😦 Couldn't figure out the right google search terms either

[hybrid gull]

Equation of a circle

No need to get too fancy with it

[cunning fox]

right, so r is obviously 15, HK I don't know just yet, and x and y... I can't remember, unfortunately. I'm very sorry, it's been years since I've taken geometry

[hybrid gull]

No problem

Actually, I think you can get the distance from the x axis to the top of the circle by getting the distance from the center of the circle to the x-axis

And then doing 15 - that

[cunning fox]

OH! triangles haha

alright, one moment

[cunning fox]

alright, I'm here now

I assume I just screwed something up with the format

yep

i figured that out

so equation of the parabola, I think I could do that

[hybrid gull]

Your goal is to get the center of the circle

That’s pretty much it

You know that x-coordinate is 0

But the y-coordinate is something you have to find

Think: y-coordinate of center - height of top of the circle above the x-axis = -15

[cunning fox]

okay, yeah I figured it out

equation of the circle would then be

hm

well, I figured it out haha

it's just that but 13.421 is positive in the equation

I think I can figure it out from there

ty!

[hybrid gull]

Np! 🙂

Take care 👍

[cunning fox]

u2!

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[proper tinsel]

I'm trying to prove the second inequality here, but I just don't know how... Do I have to use C-S?

[fathom plover]

what are you trying to prove exactly?

the C-S inequality?

[proper tinsel]

(trace(AB))^2 <= trace(A^2)trace(B^2)

[arctic ravine]

how is that @proper tinsel

[proper tinsel]

4th equality is not necessarily true I think

Or is it

Wait lemme think

[arctic ravine]

i can write it out

the fourth equal sign

here it comes

is that clearer

[unreal musk]

Hmm not really

[arctic ravine]

oh shit

i see

[proper tinsel]

No the forth equality is not true

[arctic ravine]

im trippin

[proper tinsel]

Noo but it's close

bcs

By Cauchy Schwartz

[arctic ravine]

right

[proper tinsel]

It's less than sum of (aii^2) times sum of (bii)^2

right

yep

[proper tinsel]

I'm starting to equation ur second line

Whoops

Ur second line doesn't seem right

[unreal musk]

Seconded, unless I'm being slow

$(i,j)$th entry of $AB$ is $\sum_{k} a_{ik}b_{kj}$ right

[warm shale]

chartbit

[proper tinsel]

Um inner products right so

yeah

Yeah

[arctic ravine]

no its my fault

im redoing the problem

sorry

[proper tinsel]

Hm

how can we show that

trace is an inner product

Because then it follows directly?

Right

By Cauchy

[unreal musk]

How exactly are you choosing to define the inner product, for my clarity? Guessing it's $(A,B) = tr(AB)$?

[warm shale]

chartbit

[arctic ravine]

its not that simple

[proper tinsel]

Yeah

[arctic ravine]

well

[unreal musk]

Hmm give me a second

Ahhh are we over the real or complex numbers btw?

If real then it's trivial I think 😂

Linearity follows from here

[arctic ravine]

ah i was overthinking

i have it

[unreal musk]

Symmetry from here

[arctic ravine]

i was skeptical about the last equality but its true

ah shit

no last equality doesnt hold

fuck me

[unreal musk]

Then as $tr(A^{2}) = \sum_{i} \sum_{k} a_{ik}a_{ki}$

[warm shale]

chartbit

[arctic ravine]

yeah this is not obvious

i will just say that\

[unreal musk]

Actually wait hmm

Maybe not quite "trivial" then, if at all true

My brain

[proper tinsel]

lmaoo wait what were u trying to do

[arctic ravine]

trying to force it lol

[unreal musk]

Prove that tr(AB) would form an inner product

[proper tinsel]

yea if u do show that then ur done but

well

Wait that's enough right

Oh and positive definite

😒'

[unreal musk]

Haha I see where you got this from now 😂

[proper tinsel]

Yeahh

ohhh

Whoops

[unreal musk]

I was wrong with my definition of inner product 💀

[proper tinsel]

I left out important info that's why

😅'

A and B are semidefinite

Didn't see that

[arctic ravine]

ahhh

[proper tinsel]

does semidefiniteness imply trace(A^2)>0

[unreal musk]

Ayy, even without that, if it's $tr(A^{T}B)$ then that should work to show it's an inner product, as they state $tr(A^{T}A) = \sum_{i} \sum_{j} a_{ij}^{2}$

[warm shale]

chartbit

[proper tinsel]

Oh yea

Then ur done

🙋‍♀️🙋‍♀️'

so their inner product is tr(A^T B) lol

[unreal musk]

Ahh it took me a bit longer than I'd like to admit to get that from assuming the inner product 😭

Actually wait a minute 🤔

Yea I see why it's a bit tough now

Oh damn, I think I need to retire myself out of the game

I'm actually soooo dumb

[unreal musk]

Thissss

[proper tinsel]

wait tr (A^T A^T) = tr(A^2)? 💀

[proper tinsel]

🙋‍♀️'

Wait wb it

Can we prove that tr(A^2)>=0 though

[unreal musk]

I was trying to see why the steps would hold for myself by doing them out, and was struggling to see why it would hold, but being positive semidefinite gives they're symmetric right?

So being symmetric gives $A = A^{T}$ and similar for $B$

[warm shale]

chartbit

[proper tinsel]

Yes ur right

[unreal musk]

Let's gooooo

[proper tinsel]

Yeah yeah

[proper tinsel]

Is this true for any matrices tho I'm thinking

I feel like it does?

tr(A^T A^T) = tr ( (AA)^T) = tr( (A^2)^T)

[unreal musk]

I think so, isn't it automatically true? You should have that $(A^{T})^{2} = (A^{2})^{T}$, so from that

[warm shale]

chartbit

[proper tinsel]

yeah yeah

Wait so

Does semidefiniteness not even matter

For that first inequality

Second*

[unreal musk]

I think it matters, that's why I was struggling when I tried it

I wrote $(tr(AB))^2 = tr((A^{T})^{T} B)^2 \leq tr((A^{T})^{T} A^{T}) tr(B^{T} B) = \ldots = tr(A^{T}A) tr(B^{T}B)$

From C/S, but my brain couldn't figure out how that would become $tr(A^{2}) tr(B^{2})$, but the positive semidefiniteness gives the symmetry of A and B

[warm shale]

chartbit

[proper tinsel]

Ah

Wait

proving that tr (A^TB) didn't require positive semidefiniteness tho right

[proper tinsel]

lmao isn't (A^T)^T = A

[unreal musk]

Yeah - wrote that so it was in the form of the inner product haha

[unreal musk]

Yea the proof of that worked for any real matrices

[proper tinsel]

Wait I don't see how u go from second to last step to the last step

[unreal musk]

Ah my bad, let me retypeset that, I forgot to add in a transpose

[warm shale]

chartbit

[unreal musk]

There, that's better, from that it should be clearer

Actually give me one sec

[unreal musk]

Ah, yeah, also w/ this

[proper tinsel]

I was thinking just

(A^T,B)^2 <= (A^T, A^T)(B, B)

Is that wrong lmao 💀💀💀

[unreal musk]

Taking that to be that trace inner product, yeah it's true (and basically what I wrote)

But $(A^{T}, A^{T}) = (A, A)$ by the $tr(AB) = tr(BA)$

[warm shale]

chartbit

[obtuse pebble]

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[sick granite]

for a problem like : lim x -> -3 |x| -3 / x^2 -9 how can i find the breakpoint?

[raven spire]

$\lim_{x \to -3} \frac{\abs{x} - 3}{x^2 - 9}$ ?

[warm shale]

Arya

[sick granite]

yes

[raven spire]

please send a screenshot or a pic if you cannot use latex. it helps

[sick granite]

wait i just want to clarify but is it ok if this question is based on an answer i got wrong on a test?

[kind hawk]

(or at least use proper brackets)

[raven spire]

Yeah it's fine if it's a test that's passed already

so.. wdym by a breakpoint? were you supposed to evaluate this integral?

[sick granite]

this was the question

im not too sure how to figure out the breakpoint

and how its 0

[raven spire]

well, hmm. what is a breakpoint?

[sick granite]

where it is discontinuous ?

[raven spire]

Yeah so, whoever wrote the breakpoint is x = 0 needs a brain check ig

but you'll be fine if you just figure the places where the denominator is 0

like x = ± 3 for example are the breakpoints in the given expression. but, (|x| - 3) as x tends to -3 is - x - 3 so it cancels with the bottom to leave you with -1/(x - 3)

[sick granite]

yeah they trippin

i see i see

when i was previously doing this problem i tried to determine both sides of the limit and they both were positive

is there a step i should do before that ?

i was plugging in like -2.9 and -3.1

[raven spire]

ah positive or not doesn't matter. what matters is to get rid of the "breakpoint" at which you're evaluating the limit

so $\lim_{x \to -3} \frac{|x| - 3}{x^2 - 9} = \lim_{x \to -3} \frac{\cancel{-x - 3}}{\cancel{(x + 3)}(x - 3)} = \lim_{x \to -3} \frac{-1}{x-3}$

[warm shale]

Arya

[raven spire]

$= \frac{-1}{-3-3} = \frac{1}{6}$

[warm shale]

Arya

[sick granite]

I see. Would is still be able to plug in numbers from the 3rd step in order to determine either sides of the limit ?

[raven spire]

I don't see a need to plug in numbers but sure

[sick granite]

how would this problem change if the absolute value was in the denominator ?

on |x^2| - 9

[obtuse pebble]

@sick granite Has your question been resolved?

[raven spire]

|x²| = x², so the only change it'd bring would be that the top would just be (x - 3) instead of (|x| - 3)

[obtuse pebble]

@sick granite Has your question been resolved?

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[lusty gull]

how should i solve this?

i tried using both formulas for arithmetic and geometric sequence but i still cant connect it somehow

[novel knoll]

why would you use arithmetic when it says geometric?

[lusty gull]

cause in the question the original sequence is arithmetic

i thought that has smth to do with the question

[shell prairie]

when a,b,c is an arithmetic sequence 2b=a+c

now without even getting p,q you can prove tjat 27,p,q,125 is a gs

since a1,a1+d,a1+2d,a1+3d,...

d=log_8(8^d)

[lusty gull]

how did u get that?

[shell prairie]

use the property of log

[lusty gull]

i still dont rlly follow

[shell prairie]

log(a^b)=blog(a)

[lusty gull]

is there a formula i shouls use?

[lusty gull]

where did 8^d come from?

[shell prairie]

to match it with d

[lusty gull]

so after i get this what's the next step?

to prove it's geometric

[shell prairie]

log(a)+log(b)=log(ab)

[lusty gull]

elaborate?

where do i use it

[obtuse pebble]

@lusty gull Has your question been resolved?

[lusty gull]

<@&286206848099549185> ?

[obtuse pebble]

@lusty gull Has your question been resolved?

[obtuse pebble]

@lusty gull Has your question been resolved?

[lusty gull]

.close

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[elfin galleon]

can anyone tell me what 1+1 is/

[unique spear]

are you here to waste our time

[elfin galleon]

no sir, i am seeking help for my math class , i had not solved it , it very hard

[unique spear]

2

there you go

[elfin galleon]

thank you sir, my school is really trying hard to solve , i amn going to bring revoultiopn

[unique spear]

close this ticket when you are done

@elfin galleon

[high lily]

.close

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[elfin galleon]

.reopen

[obtuse pebble]

✅

[elfin galleon]

hahhahahhahahahha im a math genius

bye.

.close

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[cursive citrus]

Is it correct that if a 2x2-matrix A's eigenvalue has an algebraic multiplicity of 2 and its eigenvector has a geometric multiplicity of 1 then A is NOT diagonalizable?

[obtuse pebble]

@cursive citrus Has your question been resolved?

[obtuse pebble]

@cursive citrus Has your question been resolved?

[desert sinew]

This is a good reference.
https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk10/10_algebraic_and_geometric_multiplicities.html

If for every eigenvalue of A, the geometric multiplicity equals the algebraic multiplicity, then A is said to be diagonalizable.

[obtuse pebble]

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[timid silo]

heres my question

[timid silo]

.close

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[wanton bluff]

Whats the next step here?

[warm canopy]

times by 2

[wanton bluff]

so 2 log base 4 (2x-1)

[obtuse pebble]

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[inland mulch]

basically you can't

the antiderivative of exp(-t²) is called the error function

you can't write it in terms of high school functions

[scenic sundial]

bruh

[inland mulch]

https://en.wikipedia.org/wiki/Error_function

[scenic sundial]

how does this help them?

[inland mulch]

well uh throw it into a calculator then

however

ah wait c is 40

I'm afraid there won't be any nice expression for it

it used the error function

the antiderivative will be of the form a * erf (b * (t-c))

differentiate that and use the definition in wikipedia to find what a and b need to be

then call it a day

sure you just need to know the derivative of erf

which is written on wikipedia

[timid silo]

U can manipulate it to possibly find the result without a calculator, provided that you know that it values sqrt(pi) as it goes to minus and plus infinity

But I think this is above your current knowledge. Odd question

[inland mulch]

they want the integral from 0 to 30 or something silly like that

I think teachers will notice there is a problem if the whole class fails a question

and if not you can still complain

[obtuse pebble]

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[scarlet talon]

how do i solve this?

[worldly otter]

Do you have a definition for sec(x) in terms of sin, cos or something?

[scarlet talon]

i just learnt about this today, so all i know is that sec is the reciprocal of cos

[worldly otter]

Perfect. Rewrite sec(x) = 1/cos(x), and try rearrange it so you get cos(x) = something

[scarlet talon]

okay let me give it a shot

my maths skills is quite weak, so i'm not sure where to go after 1/cos(x) = 2

actually do you just multiply cos(x) to both sides? i got cos(x) = 1/2.

[worldly otter]

Exactly

So you're looking for an angle x that will give cos(x) = 1/2

[scarlet talon]

inverse function?

[worldly otter]

Yep

[scarlet talon]

ahh got it. I got 1 more question tho

[worldly otter]

Sure, send it in

[scarlet talon]

my teacher explained it to me using quadrants, but I have no clue how it worked

how would you show this using a quadrant?

[worldly otter]

I'm not sure what you mean by that exactly. Usually quadrants refers to the four sections of the unit circle, being the top right, top left, bottom left, and bottom right. I'm not sure if you've done this, but if we talk about points on the circle, then cosine represents the x value, and sine represents the y value

Something like this?

[scarlet talon]

yep this is it

[nocturne sun]

oh I'm late I guess

[worldly otter]

So because cos is the x part and sin is the y part, cos and sin are positive in the top right, cos is negative and sin is positive in the top left, etc.

That can help to give a quick idea of some angles, but I'm not sure how it would give exact solutions unless it was on some specific points on the circle

[scarlet talon]

all i remember was that my teacher gave the answer in the form of a radian

pi/3 i think, and it shows up as a point on the quadrant

[worldly otter]

pi/3 is equivalent to 60 degrees

[scarlet talon]

Yep

[worldly otter]

inverse cosine is the most obvious solution to me anyway

[scarlet talon]

I have a screenshot that's basically what my teacher wrote but for sin(x)

[worldly otter]

Ah I see. So there's two solutions to each equation.

So if you're looking for cos x = 1/2, then cos x is positive, so thats the top right and bottom right quadrants. So you would get pi/3, but also 2pi - pi/3

Again, thinking about it as cosine representing the x value, and sine representing the y value, the vertical line line is going to hit two points

[scarlet talon]

right

so for cos(x), it would be positive in A and C, so after getting pi/3 for the A quadrant, I just need to do 2pi - pi/3 to get the second positive value. hopefully it's correct.

[worldly otter]

Yeah exactly

[scarlet talon]

thank you very much

[scarlet talon]

sorry i got 1 more question.

[worldly otter]

No problem, go ahead

[scarlet talon]

i'm on a different question now and i'm on cos(2x) = root3 / -2. can i just inverse that and divide the value by 2 to get x?

[worldly otter]

Yeah pretty much. But remember that you have to do the quadrant, 2 answers part before dividing across by 2

[scarlet talon]

Oh. Why so?

[worldly otter]

Let's just let y=2x for now, and solve cos(y) = - root3 / 2. Then y = 4 pi/3 and 5 pi/3. Then putting back 2x = 4 pi/3, 2x = 5 pi/3, we can divide.

5 pi / 3 = 2pi - pi / 3. If we first divided the pi / 3 by 2, and then took the answers, we'd get something different (and wrong)

I can explain that in more detail if any of that is confusing

[scarlet talon]

How did you get y = 4pi/3?

Here is what my teacher wrote btw

[worldly otter]

Oh sorry. You're right, I put sin into my calculator instead of cos.

[scarlet talon]

No problem

[worldly otter]

Yeah that's correct

[scarlet talon]

but I'm not really sure how it works (the screenshot is from my teacher)

Like, how did he get 5pi/6 for his first value of 2x

[worldly otter]

Oh right. So cos^-1 ( sqrt(3)/2) gives pi/6. But we have a negative answer, - sqrt(3)/2, so it must be in one of the left quadrants. So we take pi + pi/6 and pi - pi/6, since pi is equivalent to 180 degrees, the line between those two quadrants

that gives 5pi/6, 7pi/6

Now, usually you'd be done. But you know how the we can keep rotating past 2 pi and hit the same points? Well because we divide by 2 at the end, we actually need the next rotation, because when we scale everything by 1/2, those angles will actually be in the first 0 to 2pi

[scarlet talon]

Ah ok

so 2 full rotations?

[worldly otter]

Yep, that's where the 17pi/6 and 19pi/6 come from

Because those are 5pi/6 + 2pi and 7pi/6 + 2pi

[scarlet talon]

Rightttt

For cos(x), the values would be positive in the top right quadrant and the bottom right quadrant right? You mentioned that it's because it's negative, so does the positive quadrants now become the top and bottom left?

[worldly otter]

Well we have that cosx is positive on the top right and bottom right quadrants, so its negative everywhere else

[scarlet talon]

Yes

[scarlet talon]

Also, why is cos2x = -root3 / 2? Shouldn't it just be root3/2?

[obtuse pebble]

@scarlet talon Has your question been resolved?

[obtuse pebble]

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[hybrid tartan]

Hi

[hybrid tartan]

For (ii), i got an answer of 0.00035, and i calculated it with .6^3 * .4^7, is it correct?

or would the probability be simply .6^3

which = .216

[noble thicket]

if im correct this is binomial distribution?

[obtuse pebble]

@hybrid tartan Has your question been resolved?

[hybrid tartan]

Doesnt specify, but you could use it

[noble thicket]

ye it just sounded a lot like a question where you would use binomial distribution

well

if it says less than 4 servers

that would be X<4

because it could be 3, 2, 1 or 0 servers used

but not 4 or more servers used

and X<4 is equal to X≤3

and the chance of a server being busy is 0.6

and there are 10 servers

so your binomial distribution would be X is distributed as a bionomial B 10 times with a chance of 0.6 (10, 0.6) and X≤3

hopefully this should be enough for you to figure out the answer

[obtuse pebble]

@hybrid tartan Has your question been resolved?

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[icy latch]

9b

Have I done something wrong?

I have worked it out like this aswell

<@&286206848099549185>

[obtuse pebble]

@icy latch Has your question been resolved?

[obtuse pebble]

@icy latch Has your question been resolved?

[timid silo]

1/2k - (1/(k+1)) + 1/(2*(k+2)) = 1 / 2 * k -1 / (2 * (k+1)) - 1 / (2 * (k+1)) + 1 / (2 * (k + 2))

{1 / (2 * k ) - 1 / (2 * (k + 1 ) )} - {1 / (2 * (k + 1) ) - 1 / (2 * (k + 2))}

try solving from here onwards..much easier I believe

[obtuse pebble]

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.close

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[exotic sierra]

have no clue where to start with this

its a combinatrics/permutation questions

is this just 13 choose 10 ?

10 because 3 of the heart cards have already been picked?

anyone?

<@&286206848099549185>

halllooo?

[obtuse pebble]

@exotic sierra Has your question been resolved?

[exotic sierra]

anyone?

is it 10 C 2?

[obtuse pebble]

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.close

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[timid silo]

guys
4 2 3
4 4 4
4 4 4

[obtuse pebble]

Please don't occupy multiple help channels.

[timid silo]

is this a linearly dependent matrix?

do same lines also mean linearly dependent

or does it only work for columns

[obtuse pebble]

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[distant seal]

y=sin(3x^2)

[distant seal]

It’s chain rule I’m pretty sure

[pseudo bloom]

I am having trouble on some hw. Pls help.
Given an oblique triangle wiith C=72 deg, A = 15 deg. b = 342.6, find side a to the nearest thousandth.

[nocturne minnow]

#❓how-to-get-help

[distant seal]

Um

[nocturne minnow]

For the derivative, yes

[distant seal]

Ok

Is it cos(3x^2)*6

.close

[obtuse pebble]

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[nocturne minnow]

Nope

.reopen

[obtuse pebble]

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[gentle shard]

How would I solve this?

[gentle shard]

is it 108 = x + 2y and V = x y^2?

[gilded otter]

Im thinking it would be 4y

because the length around is 4 side lengths, or the perimeter of the square

[gentle shard]

yea i think ur right

108 = x + 4y

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[rustic heath]

so im trying to prove this is a ring

and i need to show that there exists a unique additive identity denoted 0

so i said that 0=0/1, since gcd (p,1) = 1 for every prime p

so 0=0/1 is in Z_(p)

but I realize that there are multiple possible additive identities.

like 0=0/q where gcd(p,q)=1

and q need not be 1

I'm really confused because I expected this to be a ring

but it seems like the additvie identity isn't unique and that implies this structure is not a ring

would be great if someone could clear up my confusion

[median carbon]

it's not a contradiction, 0/q can be just equal to 0/1 for all possible qs, so the additive identity would still be unique if you could prove that

[rustic heath]

so i would need to prove that

0/1 = 0/q for all q such that gcd (p,q)=1?

im wondering if its right to say that 0/1 is 'the' additive identity of this ring

or if i have to say there are multiple representations

[median carbon]

I think there are multiple representations, just let q be any integer such that p doesn't divide q, and it will be the case that (q,p) = 1

and there's more than just one such integer for any p

[rustic heath]

right

i mean is it right to say that

0 is in z_(p)

because the definition of z_(p) is strictly rational numbers

and 0 isn't exactly a rational number

like it has multiple representations

[median carbon]

well isn't it so that a rational number is any number that can be represented as a quotient of two integers, m/n for example?

I mean, take 1/2

[rustic heath]

yep

[median carbon]

it's surely a rational number, but it can be represented as 1pi/2pi

[rustic heath]

yes or 2/4

i understand that two rational numbers a/b and c/d are equal

[median carbon]

and neither of those are integers

[rustic heath]

by under the equivalence relation

ad=bc

is that what you're getting at

that there are multiple representations of a ratioanl number?

like 1/2=1pi/2pi

[median carbon]

yup

[rustic heath]

i see

so i understand 0 has multipel representation

in rational number

so can i just say simply that 0 can represented as 0/1

and so 0 is in z_(p)

without having to strictly deal with all the cases where 0 can be written differently

[median carbon]

I'm not sure you know

you can just quickly show that 0 is unique for all these representations

if you already proven the existence of additive inverse this will be quick

or wait

I need to remind myself

yup

[rustic heath]

Hmm ok got it

I think

Thank you!

[median carbon]

no problem

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[scenic sable]

how was the equation simplified to find the derivative

[supple pumice]

power rule

[trail cloak]

The equation is simplified already.

[scenic sable]

wdym

[trail cloak]

And finding its derivative used this property

[scenic sable]

if i find the derivative i get 4x+1

[trail cloak]

$$\frac{d}{dx} x^n = nx^{n-1}$$

[warm shale]

VulcanOne

[trail cloak]

Our n here is 2

And the coefficient multiplied by x^2 is 3

When we find its derivative we get 3*2x = 6x

Same thing for 2x

[scenic sable]

oh wait

[trail cloak]

It will be 2•1x^0 = 2•1•1

[scenic sable]

ohh i didnt do the 2x

ty

i get it

.close

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[scenic sable]

wait i got 6x-1

thats wrong lol

.reopen

[obtuse pebble]

✅

[trail cloak]

Could you show me how you reached it?

[scenic sable]

sure

so

-2•1•1 +1

[trail cloak]

Oh I now know why

Ok so

$$\frac{d}{dx} c = 0$$

[warm shale]

VulcanOne

[trail cloak]

C can be any number

If it is not multiplied by x or any function of x, it means that it does not change with respect to x, making its derivative 0

$$\frac{d}{dx} (3x^2 - 2x + 1) = 3(2)x^{2-1} - 2(1)x^{1-1} + 0 = 6x - 2$$

[timid silo]

[h(x) = 3x^2 -2x +1]
[\frac{d}{dx}(h(x)) = \frac{d}{dx}(3x^2-2x+1)]
[\frac{d}{dx}(h(x)) = \frac{d}{dx}(3x^2) +\frac{d}{dx}(-2x) + \frac{d}{dx}(1)]
For all the details I guess

Oh ahaha didn't see u write it

[warm shale]

VulcanOne

♡LexQa♡

[trail cloak]

Haha

[scenic sable]

thats confusinggg

[trail cloak]

What is?

[scenic sable]

like it doesnt make sense

[trail cloak]

Keep these 2 rules in mind:

$$\frac{d}{dx} x^2 = 2x$$

[warm shale]

VulcanOne

[trail cloak]

$$\frac{d}{dx} 5 = 0$$

[warm shale]

VulcanOne

[trail cloak]

The rule that works with 5 works also with 1

Works also with any number

Also the rule that works with x^2 works with x raised to any number

These 2 rules are essential in taking derivatives

[scenic sable]

ohhh

so 1 = 0

and -2x = -2

[trail cloak]

Derivative of 1 is 0 yep

Yep

[scenic sable]

ok got it

thanks

[trail cloak]

[scenic sable]

.close

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[static patio]

can someone help with 2nd task? RE is recursively enumerable. I need to describe a turing machine.

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[steel harness]

what is the equation of the line through (4,-1) and whose segment intercepted between the axes in the fourth quadrant is equal to 2sqrt17?

[steel harness]

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[meager root]

I need to rewrite the top line using properties of logarithms. I have the answer on the bottom line, but I'm not sure what steps I'm supposed to be taking to get to it.

[timid silo]

[
\log_{a}(b \cdot c) = \log_{a}(b) + \log_{a}(c)
]

[warm shale]

♡LexQa♡

[meager root]

hmm lemme look at this a sec

the parenthesis aren't around the 5, just the 2^x. Does that still conform to (b * c) ? rn it'

[urban patrol]

Yes

There's no parentheses that indicate its in the log function

[meager root]

c = 2^x

oh so 5 is separate from log? log_2 is the log function, and then it's being multiplied by 5?

i'm thinking of 5 as a part of log

[urban patrol]

logsub2(5*2^x)

[meager root]

yes

[urban patrol]

part of the log

[meager root]

as in (5*2^x) is part of the log as a whole?

[urban patrol]

Mhm

I was trying to say that the parentheses around 2^x weren't indicating being part of the log

[meager root]

and logsub2 * 5(2^x) = logsub2(5*2^x)

yea?

[urban patrol]

No

[meager root]

oh

[urban patrol]

Logsub2 isn't a thing alone

[meager root]

5 is a part of it?

[urban patrol]

What

Multiplication and being in the function aren't the same

just like you can't divide out x from sinx

but you can from xsin(x)

Same concept

[meager root]

im not familiar with sin usage yet

[urban patrol]

oh

well basically those numbers and variable are in the function

Its not the same as multiplication

[meager root]

let me plug the original 1st line into the formula i was given and see if i can work it out one sec

well just starting out im seeing something familiar. I've got this:

thats good so far right?

[urban patrol]

Yea

[meager root]

i've got the same base for both of the logs, so there's somethin I can do with combining em. lemme check my notes

[urban patrol]

Do you know the other formula you need

No not combining

You just seperated it it

Basically log(x^n)=nlog(x)

you can move the exponent out to the front

[meager root]

this i know yee

oooooh okay, so that would be step 2? movin the expo to the front of logsub2 * 2

[urban patrol]

Yes

Then the third step

Is just cancelling out that log

logsubn(n)=1

because you can turn it into logn/logn

[meager root]

oh right right

[urban patrol]

change of base formula

logsuba(b)=logb/loga

[meager root]

OH!

i see it now

[urban patrol]

So you're left with logsub2(5)+x

[meager root]

the second line / answer is just showing two steps. One step before the final answer, and the final answer - which is logsub2 * 5 + x

[urban patrol]

No

It does one step then two steps to get the final

[meager root]

yer right, not directly one step before

[urban patrol]

First split the log

Then take the exponent to the front

and simplify

[meager root]

i understand how i'd get logsub2 * 5 + x from the original function now 🙂

thank you!

[urban patrol]

Np

[meager root]

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[timid silo]

is it sill a function if the vertical line in the graph passes through the origin?

[gilded otter]

Yes

The only time something is not a function is if there is 2 y values for one x value

[timid silo]

Oh yeah, got it thanks so much, appreciate it

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[timid silo]

hello i got a small math equation, it should be easy but i skipped some classes due health problems so now i need jsut 1 solution to get back so this is the problem?

[sudden hearth]

Bro thats only equation what the question?

[trail musk]

There's no problem here

[timid silo]

ye i dont even know how to start it so i was wondering if some1 can make it so i got an example so i can do my homework alone

[neon eagle]

yes but whats th equestion