help-23

[harsh crow]

Like it feels wrong doing all that lmao

[obsidian oracle]

if we write $q = 1 - p$, then $G_X(x) = q^n + \frac{p}{q}\frac{1-q^nx^n}{1-qx}$

[flat frigate]

Raphaelisius Maximus MMIII

[harsh crow]

Im not sure it's supposed to be that long since it's supposed to be written on a whiteboard for an oral

[obsidian oracle]

that's a bad U'

[harsh crow]

Fuck

I forgot to remove the 1

I know i did

And i thought about it before writing it

And for the rest there may be a mistake too

n-1 and not +1 but i didnt make the mistake on my fraction

[obsidian oracle]

$G_X'(x) = \frac pq \frac{-nq^nx^{n-1}(1-qx) + q(1-q^nx^n)}{(1-qx)^2}$

[flat frigate]

Raphaelisius Maximus MMIII

[harsh crow]

Why the p/q tho??

[obsidian oracle]

did you forget the coefficient in front of the sum

[harsh crow]

Yep i did

Tiredness

[obsidian oracle]

wait a bit though

the sum itself looks a bit sloppy

[harsh crow]

Tho i'll remember it tomorrow

[obsidian oracle]

yeah we forgot to factor by the first term

[harsh crow]

We're missing a (1-p)x then??

[obsidian oracle]

or am I trippin

nah I'm not tripping

$G_X(x) = q^n + px\frac{1-q^nx^n}{1-qx}$

[flat frigate]

Raphaelisius Maximus MMIII

[obsidian oracle]

which I'll admit looks pretty bad from here

[harsh crow]

why px ?? why not (1-p)x ??

[obsidian oracle]

p/q * qx

simplifies to px

[harsh crow]

but tbh dw much about it i mainly needed the method to do it

tho the next questions will be annoying to deal with but i know i have them in my copybook somewhere

so thank you for the help

have a nice day or a good night or idk

.close

[safe radish]

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[harsh crow]

quick question, how can i prove a matrice cant be inverted ?? i have this one:
-2 0 -2
1 0 1
1 0 1

i want to say that by no operation with the gauss-jordan method we can find the identity matrice but idk if it's enough, and i'd like to have another method

[solar hazel]

the rows aren't linearly independent

[austere forge]

Or, the determinant is 0

[harsh crow]

we dont have determinants for matrices of order 3

we cant use it

[austere forge]

oh okay you are in precalculus math

yes, you would use the fact that there is a column of all 0s then

[harsh crow]

but how can i write it??

like what can tell me "no, you cant invert that matrice"

with numbers

[austere forge]

it depends how much work your teacher wants you to show

you can set it up as system of 3 equations, and show there is no solution

[harsh crow]

i tried but i just end up with x=-z

and idk what to say after that

[austere forge]

yeah, that should be enough to say the inverse does not exist

you set up Ax = 0 right

[harsh crow]

yeah

[austere forge]

yes, if there is a general solution (not just 0,0,0) for Ax = 0, then it is not invertable

[harsh crow]

oh and the inverse exists if the solutions are x=0 y=0 z=0 and it's not the case

yeah

and as there i dont have that it's not invertable yeah

[austere forge]

check with your teacher, but that should be enough

[harsh crow]

yeah it should tbh

thanks

[safe radish]

@harsh crow Has your question been resolved?

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[arctic gull]

Hello!

I was wondering if anyone would have free time check my proof for a Linear Algebra theorem. teacher told me that we get 10 points back as extra credit on the test if we can write it without notes (ill memorize it ig) so it will mean allot
thank you!

Theorem: If H is a subspace of V and dim V = dim H, then V = H

[arctic gull]

for ref. our teacher told us to only use the stuff learned so far, which includes:

1. Systems of Linear Equations 2. Matrix Arithmetic 3. Determinants 4. Linear Independence 5. Vector Spaces

[naive pivot]

it's a bit too drawn out

the part with contradiction is not really necessary

the fact that you have dimV linearly independent vectors is enough to say that span(B) = V

[arctic gull]

oh... i felt that too when I was writing that

[void path]

it seems a bit too complicated, maybe because you're not allowed to use some things
in n-dim space every n-element linearly independent set is a basis, so B_h is automatically a basis of V and basis determines the spanned space uniquely

what are facts about bases you're allowed to use?

[arctic gull]

So do I omit the "proof by contradiction" part and state this?

[naive pivot]

it might be worthwhile to consider this

[arctic gull]

We arent given a list of "facts" that i can actually use...

let me try to find

[naive pivot]

the following two facts are usually proven in a linear algebra class:
if dimV = n, then

• n linearly independent vectors form a basis for V
• a spanning set of n vectors forms a basis for V

[arctic gull]

we are given this information

related to dimension and bases

[naive pivot]

that's all?

[arctic gull]

yea... i dont have "facts" that i can use in the proof (maybe im not aware of them) . The fact you mentioned is probably not what our teacher proved

[arctic gull]

yea 😭 its a high school class

not that rigorous

[naive pivot]

well

there's like no results

[naive pivot]

i don't know how in depth a proof will be then

[void path]

i think this one is equivalent to what we said?

[arctic gull]

Theres also this

about standard bases

[arctic gull]

oh it is? i didnt realize it. sorry i am very new to linear algebra

[arctic gull]

our teacher did sya "You can prepare for this outside of class with your own research" so i can technically use stuff outside

just not really advanced stuff

[arctic gull]

from what i understand, i thought that statement means that you can create new basis with an existing basis with equal number of linearly independent vectors

But not necessarily extend to a new subspace right?

[void path]

so you'd have

1. H is of dimension n so it has n-element basis B_h
2. H is a subset of V so each element of B_h is an element of V
3. hence, B_h is a n-element linearly independent set of vectors in V
4. (this is kinda obvious but to match the sentence in your pictures) since every element of B_h is an element of V, it can be written as a linear combination of some basis in V
5. so B_h is a basis of V (this follows from the sentence i circled and last two points)
6. since B_h is a basis of H, span(B_h) = H, same with V, span(B_h) = V
7. ergo V = H

[arctic gull]

ohhhh, i think its starting to make a bit more sense

[void path]

but you're not proving that it's a basis for the subspace H
you're taking vectors from H - which are also vectors of V, because subspace - and proving that they are basis for V

[naive pivot]

i reccomend checking out Friedberg's Insel Spence's coverage of this topic

it's definitely accessible for your level

but they give a lot of care

[arctic gull]

ohh ic. thanks. I will rewrite the proof and make it a bit shorter

thank you for the help

[arctic gull]

ahh is that a lecture?

[arctic gull]

oh that would be helpful thanks

[naive pivot]

it's a textbook

[arctic gull]

oh

[naive pivot]

just titled "Linear Algebra"

pdfs should be easy to find

[arctic gull]

yea i found it just now

looks really detailed and informational

[void path]

in general if you plan on entering any STEM degree then linear algebra is a must
so picking up a textbook sooner or later is inevitable

[naive pivot]

fwiw, i don't think what you did originally is technically wrong, but it is morally wrong

[arctic gull]

yeaa truu. im planning on going into ML so its even more important lol

[arctic gull]

😭 morally?

im not comitting a sin? right?!

[void path]

yea ML and pretty much all numerical calculations are based on linalg

[naive pivot]

yes, in the sense that the proof should follow more immediately from more general facts at your disposal

like the two that i mentioned earlier

[arctic gull]

ye

[void path]

you're going to the linalg hell sorry

[arctic gull]

ahh ic. yea, i was def taking a long route

[arctic gull]

bro my high schooler teacher frying me bru 🙏

since its not that rigorous, it lacks allot of conceptual understanding that makes it hard for me to gain intuition

its more about "these are the steps memorize them and then do the test"

(sadly)

[void path]

but yeah, it comes from experience but it's usually the principle to use already proved facts and lemmas before trying something more elementary
sometimes you cannot escape it though and you have to modify a proof of some even known theorems, so it's not really "morally wrong", it's just inefficient to do that all the time

[arctic gull]

yea. i was looking up proofs online, and i was so confused why lammas keep appearing in the proofs 🙏 thanks for all the help tho. I will write a shorter proof now

[void path]

if you're familiar with some programming then lemmas are like functions - lets you split the proof into smaller chunks and reuse the "code"

[arctic gull]

ohh its like abstraction

[void path]

https://tenor.com/view/lama-handsome-handsome-lama-blinking-lama-blink-gif-22584915

not this lammas

[arctic gull]

LOL

MB MB

honestly tho, when i first learned about linear algebra, it was so fascinating. I have always liked math, but linear algebra is like a whole new world that i had never seen

[void path]

yeah exactly, theorems/lemmas are kind of abstractions, you use them out-of-the-box without looking inside unless you really need to

(it's useful to know how they work inside though)

[void path]

it is quite fascinating, especially in numerical part of math
you can do a LOT of fun stuff with vector spaces even in higher math

[arctic gull]

Tadaaaa

is this better?

[arctic gull]

oh, thats so similar to abstraction. im wondering, are "lemmas" universally defined? or does it change book by book or course by curse

[arctic gull]

Yea, it really opened ALLOT of cool stuff i had never known before

[void path]

no, it's just a word for helper theorems more or less

[arctic gull]

oh

[void path]

like, if you're writing a paper then you usually name only the important results "theorems" and rest as lemmas

[arctic gull]

ohh, so u can write before a proof

then use that lemma later

[void path]

seems okay at first glance

[arctic gull]

oh, wait is there smth stronger

*wrong

[safe radish]

@arctic gull Has your question been resolved?

[void path]

no i just mean that i'm a bit sleepy so i might miss something but it looks okay

but someone else can check it too

[arctic gull]

ohh, ic

[arctic gull]

alr

thanks for the help

[safe radish]

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[crystal owl]

is vf for the chp 0?

[solar hazel]

[autumn sun]

wdym, v_f is for the CHP officer here? I mean, both the speedster and the CHP would have their separate v_f's if that's what you mean

[opaque fern]

v_f is not going to be 0, no, for the chp officer

[autumn sun]

oh ok I totally misunderstood the question LMAO

sorry

[opaque fern]

The officer needs to start speeding up and must be moving significantly faster than the guy by the time he catches up to him

[safe radish]

@crystal owl Has your question been resolved?

[crystal owl]

okay so

i do this?

[opaque fern]

Ye u got it

[crystal owl]

awesome sauce

[opaque fern]

All parts correct methinks

[crystal owl]

.close

[safe radish]

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[crystal owl]

did i do this wrong? when i plug in my t into my kinematic wquations i grt 2 diffansers

[clear canyon]

wait

ill check it for ya

[opaque fern]

,rcw

[clear canyon]

(21.5 s) is WRONG @crystal owl
Correct answer:
Time to catch = 30.15 s
Speeder’s distance = ≈1296 m

[crystal owl]

how did u get thaf

[opaque fern]

You dont seem to be really including the 56 m/s top speed anywhere

He is just accelerating forever

[crystal owl]

do i need to

[crystal owl]

i used vit+at^2=delta x

[clear canyon]

@crystal owl lol you treated the officer as if he accelerates forever at 4 m/s²,
but the problem says the officer’s car has a top speed of 56 m/s.

[crystal owl]

vi is 0

[opaque fern]

If the officer accelerates at 4 m/s^2 for 21.5 seconds he would be going 86 m/s

Thats like

[crystal owl]

oh i get it

[opaque fern]

Way over the top speed

[crystal owl]

i have to us vf

[opaque fern]

Yeah

[clear canyon]

this assumes:
the officer keeps accelerating the whole time
the officer’s speed becomes greater than 56 m/s (impossible)

[opaque fern]

Just chug it in

[clear canyon]

the mistake u did is here lol

[crystal owl]

okay so t=1.2??

jeezus weezus

[opaque fern]

no

Use like

v_f = v_i + at

[crystal owl]

t=14?

[opaque fern]

Yuh

[crystal owl]

howd be get 30

[opaque fern]

no like

Therr are two t's you are considering here

[crystal owl]

oh right okay now i gotta get another t

[opaque fern]

This is how long until you hit max speed

[crystal owl]

okay

how is t=14 helpful if i need 2 equations

to grt a system rifht

[opaque fern]

Bcuz like

So officer goes at a constant speed after that point right

[crystal owl]

yeah

[opaque fern]

By t = 14 you want to see the distances of both By that point

Once you find those, subtract them

Thats the distance the officer needs to cut down now

Does that make sense in some way?

[crystal owl]

i understand what ur saying with that but subtrack tbem wherev

[opaque fern]

Like

Relative to the starting point

If officer moved 500 metres

Ans thr guy moved 700 metres

Whats the distance between them

[crystal owl]

200

[opaque fern]

Yeah thats basically what im saying for u to find

[opaque fern]

First find this

[crystal owl]

pkay

[opaque fern]

Then this

[opaque fern]

Then that

[crystal owl]

okay

let me think abt that

i dont get it

[opaque fern]

Hmm?

[opaque fern]

What is it thats confusing you right now

[crystal owl]

my teacher is doing it in the board right now she has 2 phases for the spoeder and the chp

[opaque fern]

Oh damn this is live

[opaque fern]

Yes youre in the second phase now lol

First phase was when he was accelerating

[crystal owl]

I u derstand why she has it

but j dint know where im going with this

[opaque fern]

Like

[opaque fern]

Lets say i just asked you to tell me how long it would take somebody moving at blah speed to catch up to another person moving at blah speed when they are 200 metres apart

Would you be able to do that

[crystal owl]

yeah

[opaque fern]

Ok

Because thats literally what ur problem is now

[crystal owl]

but hes accelerating

[opaque fern]

He stopped

[crystal owl]

or it

oh

oh yeah

[opaque fern]

Like

By the time he finishes accelerating (t = 14 s) what is the distance between them

Thags what u need to find

[crystal owl]

392

[opaque fern]

No

Like

That's the officer's distance relative to the starting point

Not the distance between him and the guy

[opaque fern]

Do what you just did here but for the guy as well, what do you get

[crystal owl]

which guy

[opaque fern]

The speeder

[crystal owl]

602

[opaque fern]

Ok

Great

How far apart are they

[crystal owl]

210

OH I GET IT

[clear canyon]

@crystal owl did ya get the answer?

[crystal owl]

NOPE

BUT IM BRAINING

[opaque fern]

Yeah

[crystal owl]

210+x

[opaque fern]

So now whats the situation

They are 210 m apart

One guy moving at 56 m/s

The other at 43 m/s

You want to find how long until officer catches up to the other guy

How do you do that

[crystal owl]

yes um

[clear canyon]

are u still stuck in the same question?

[crystal owl]

yipperdipperdoo

[clear canyon]

ahh wait

[crystal owl]

system of ewuations?

[opaque fern]

no

[crystal owl]

56t=x+210

33t=x?

[opaque fern]

wait where is + x coming from

[crystal owl]

then what

[opaque fern]

Whats x even

[crystal owl]

distance

like

[opaque fern]

Of what

[crystal owl]

speeder goes x meters

chp went x meters plus 210

is that wrong

[opaque fern]

ok but u can like

Do this a lot simpler

Like

[clear canyon]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[opaque fern]

Do you know relative speed

Or the concept of it

[crystal owl]

not sure

my tracher doesnt use real words

[clear canyon]

yeah ig she/he might speak on egyptian i get it

[opaque fern]

If you are running 10 m/s and im running 5 m/s, what's your "catch-up" speed to me

[crystal owl]

5 m/s

[opaque fern]

Yeah

[crystal owl]

yes

[opaque fern]

So like

Phrased like that

[clear canyon]

@crystal owl What do we know about both cars at the moment the chase begins? lets start with your basic knowledge

[opaque fern]

Whats the officer's "catch-up" speed

[clear canyon]

for me?

[crystal owl]

43 m/s? 😭😭

[clear canyon]

wrong

its 0

[crystal owl]

oohoho im failing this class

[opaque fern]

I mean u subtracted them earlier here

Do the same, basically

[clear canyon]

@crystal owl first understand the situation this officer guy still helples
Speeder already moving at 43 m/s
Officer starts from rest (0 m/s)
Officer accelerates at 4 m/s²
They start at the same position (speeder passes the officer)

so the officer is cooked ryt now

[opaque fern]

You're like so not following the chat

[clear canyon]

?

[clear canyon]

catch-up speed only works when both objects are already moving. At the moment the chase starts, the officer’s speed is 0 because he was parked, So first step: what is the speeder’s distance after t seconds?
Write just that down.

[crystal owl]

okay.

hold on

[opaque fern]

You're like pretty much done almost lol

[opaque fern]

This is more or less the last thing you need to find

[opaque fern]

Which goes by the same logic as this

[crystal owl]

16.15

[opaque fern]

Yeah

So, total?

[clear canyon]

https://tenor.com/view/crying-emoji-dies-gif-21956120

finnaly

[crystal owl]

oh jeez guys it was much easier than i thiught 😔😔

i guess im just incompetent

[opaque fern]

you were just overcomplicating it at the beginning

[safe radish]

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[fiery junco]

Normally we check the endpoints for the interval of convergence, but that's not a step in this problem. Is that just because plugging 1/12 & -1/12 into the substituted version of the power series results in 1^k which diverges, and they skipped explaining that? I wasn't sure since if you put 1/12 back into the original series you get a geometric series which converges since r < 1

[still charm]

Yeah dude basically they're just asking for the radius and not the entire domain

Key distinction

[fiery junco]

Thanks!

.close

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[feral linden]

Minimal cut is removing edges not vertexes

Yeah, things like that

What do you mean?

Write down the cut you are thinking

In this style

A cut means that by removing the edges, your graph becomes two disconnected parts

So cut 3 isn’t a cut

Yeah

Because a minimum cut has to have weight 2 in your case, all edges have weight 1, you can’t find two edges forming a cut other than the two you have given

Probably more time to get used to it you will be fine

[safe radish]

@sullen elm Has your question been resolved?

[safe radish]

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[stone fjord]

how do i get help

[stone fjord]

I need help with this true/false question

[feral linden]

T^2=id

[quasi bison]

by sending any message, ideally one containing your question, in a help channel without a name on it.

anyway, you did that

either see @feral linden's comment, or share your thoughts so far with us

[feral linden]

My intention was g•f being injective -> f is injective, g•f being surjective-> g is surjective, completely unaware that this hint itself has become an answer.

So not on me. A bad coincidence

[safe radish]

@stone fjord Has your question been resolved?

[safe radish]

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[maiden nacelle]

ok so i kn ow h ow to do a

but b is confusing me

so far i have

2lnx times 2/7lny

or 2u times 2/7v

[plucky elk]

,tex .log rules

[flat frigate]

riemann

[maiden nacelle]

ummm

how am i wronh..

help now im confuserdf

[plucky elk]

split this into a sum first using first row from the table

[maiden nacelle]

2lnx + 2/7lny

[plucky elk]

good

[maiden nacelle]

and for my domain i have

i set it greater than 0

i think

?

[split kayak]

slight consideration: In this particular case you actually have the abs value of both logarithms here.

Because x^2 and y^2 as part of your original function of x and y.

[quasi bison]

but x and y are both stated to be positive

themselves

[split kayak]

Oh yeah, didnt recall to a

mb

[quasi bison]

so you don't need any of that modulus nonsense

in any case you would put them inside the log and not outside

[maiden nacelle]

so...

[plucky elk]

What is "it"

[maiden nacelle]

the equation

like well

ok...

i trulu need helpto understand...

girl

.close

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[fading mural]

Hi, I have this one problem: Find the remainder when divide 7^2025 by 13

[kind seal]

Have you studied congruences

[fading mural]

I managed to find that t^2025=7^9 mod 13 using fermat little theorem

But I don't know how to continue

[kind seal]

What's 7^2 congruent to mod 13

Give the smallest numerical answer

[fading mural]

Like actually doing the calculation?

[peak estuary]

yes

[fading mural]

gimme a min

[vestal carbon]

You make it sound like you're calculating decimal digits of pi lol

[peak estuary]

this should take two seconds

[kind seal]

7^2 does not need a computer

[fading mural]

I suck at mental math guys...

okay so it's 10

[fading mural]

I just memorize square numbers cuz I can't do them mentally

[kind seal]

What would be a smaller numerical answer. Try a - sign

[peak estuary]

you absolutely need to be able to multiply small numbers for this

[fading mural]

that would be -3

[kind seal]

Yeez

Now use that result

I am going to have you do some squaring amd multiplication btw

[fading mural]

I even did 5+3=7 once, so that might take me awhile but okay

What should I do next

[kind seal]

Put $7^2 \equiv -3 \mod 13$

[fading mural]

mod 7?

[kind seal]

sorry

[flat frigate]

Adhi

[kind seal]

mb

[fading mural]

oh it's okay

[kind seal]

what do you get now

[fading mural]

uh.. wait

[kind seal]

ping me when youre back

Hint $7^9 \equiv 7 \times (7^2)^4$

[flat frigate]

Adhi

[fading mural]

So it's really 3^4*7 mod 13

[kind seal]

now whats 3^2 and what's it modulo 13

[fading mural]

I was afraid I did that wrong

[fading mural]

9

oh

4

[kind seal]

-4

the - sign matters

[fading mural]

Oh yeah...

[kind seal]

now put this in the equation

[fading mural]

so 4^2*7?

[kind seal]

yes

now 4^2 is ?

[fading mural]

16

[kind seal]

so 3

and you can do 3*7 mentally ig

[fading mural]

I now my square numbers thankfully

[fading mural]

21

[kind seal]

note that you could have done it directly like this without invoking fermat's little theorem but thats fine ig

[kind seal]

so what is the remainder then

[fading mural]

8

[kind seal]

yeee

[fading mural]

Okay I don't know how it took me that long to reply

I thought 21>13*2

[kind seal]

💀

it is helpful to do a little mental math sometimes.

but cool, you got your answer, any follow up questions?

[fading mural]

I don't know, I always use calculator. I rely on it too much tho

[kind seal]

nothing wrong with a calculator but it is too much friction to do say 4 * 13 or 2 * 17

[fading mural]

I'll probably, I'm new to number theory and I'm doing problem set atm

[fading mural]

it take me around a min to do each

[kind seal]

!done

[safe radish]

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

[fading mural]

okay

.close

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[safe radish]

Please don't occupy multiple help channels.

[autumn sun]

don't you already have one channel though

[opaque fern]

.close

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[verbal kite]

If sin A = m sinB and tanA = n tanB, prove that (m^2 - 1)/(n^2 - 1) = cos^2 (A)

[opaque fern]

What have you tried

This is going to be a whole lot of annoying algebra

[winged flare]

light work after dividing one equation by the other to get\\
$\cos(A) = \frac{m}{n}\cos(B)$

[winged flare]

and squaring of course

[ornate wyvern]

seems logical

[opaque fern]

Oh smart

[safe radish]

@verbal kite Has your question been resolved?

[calm peak]

@verbal kite

[winged flare]

didnt need to really give him the solution like that

[visual linden]

What the hell

[calm peak]

i have solved this once

so i know

[calm peak]

should i delete?

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[fleet zealot]

Cam someone help me solve this?

[lost jewel]

A NCERT give me bad vibe, anyway isn't there's already a solution

[quasi bison]

yeah this is a worked example and the solution is written right there

[fleet zealot]

I don't get it

Cuz

Like

[autumn sun]

then it's probably better you say which step you don't understand, and why.

[fleet zealot]

Uh ok

The construction where CE||AD and CL perpendicular to AB

.close

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[fleet zealot]

nvm I figured it out

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[verbal kite]

If sin A = m sinB and tanA = n tanB, prove that (m^2 - 1)/(n^2 - 1) = cos^2 (A)

[verbal kite]

I wanna share my solution.

And want improvements for it.

m = sin(A)/sin(B) and n = tan(A)/tan(B)
m/n = cos(A)/cos(B)
m^2 cos^2 (B) / n^2 = cos^2(A) - (1)
Now m sin(B) = sin(A)
m^2 sin^2 (B) = sin^2 (A)
m^2 - m^2 cos^2 (B) = 1 - cos^2 (A)
m^2 cos^2 (B) = m^2 + cos^2 (A) - 1
Put this in (1).
(m^2 + cos^2 (A) - 1)/n^2 = cos^2(A)
n^2 cos^2 (A) = cos^2 (A) + m^2 - 1
cos^2 (A) = (m^2 - 1)/(n^2 -1)

[quasi bison]

here's one improvement:

get some line breaks in there so that the solution doesn't read like it was deliberately obfuscated

[verbal kite]

True.

There was an error; it has now been corrected.

[winged flare]

seems perfecto

[verbal kite]

No shorter way?

[winged flare]

i mean you can manipulate it some other way

but it would be the same

just the algebra part

or componendo dividendo idk

actually no idk

[verbal kite]

Ok.

Ok.

What is garbage?

.close

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[verbal kite]

Thanks btw.

[winged flare]

yeah you could find $\frac{m+1}{m-1}$, $\frac{n+1}{n-1}$ and multiply both from the first step but it might be a bit more tedious (maybe)

[winged flare]

np

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[trail otter]

hi i know that this question doesnt belong here but eletrical engeneeing server is useless

[opaque fern]

Go for it

[trail otter]

why $u_x=u_1\frac{\frac{R_1R_4}{R_4+R_2+R_1}}{R_3+\frac{R_1R_4}{R_4+R_2+R_1}}$ and not $u_x=u_1\frac{\frac{R_1(R_4+R_2)}{R_4+R_2+R_1}}{R_3+\frac{R_1(R_4+R_2)}{R_4+R_2+R_1}}$?

[flat frigate]

Slowaq

[opaque fern]

Thats messy

[trail otter]

its voltage divider

u_-=u_+=0V

R_1 is parralel with (R_4 in series with R_2) if im correct

[opaque fern]

Ur second formula assumes R4 and R2 are in series

[trail otter]

are they not?

[opaque fern]

The output of the op amp is an active node that sources or sinks current

You can't just assume it has 0 A entering it

[trail otter]

wait shouldn't currents going into inverting and non inverting inputs be 0?

[opaque fern]

Yeah

But thats input

Not output

[trail otter]

hm yes i see so then how should i express u_x?

[opaque fern]

Honestly I feel like using voltage dividers here might be a bit unintuitive

Instead apply KCL at u_x

No wait

Not at u_x but at u_-

[trail otter]

$i_-=0A\implies i_{R_1}=i_{R_2}$

[flat frigate]

Slowaq

[trail otter]

$i_{R_2}=\frac{u_O}{R_2}$ and $i_{R_1}=-\frac{u_x}{R_1}$

[flat frigate]

Slowaq

[opaque fern]

Yeah

So relate u_o and u_x

Using thay

[trail otter]

$u_O=-\frac{R_2}{R_1}u_x$

[flat frigate]

Slowaq

[trail otter]

but i dont know u_x

[opaque fern]

Yeah so thats the next goal

So you want to try calculating uh

u_x/i_4 where i_4 is the current thru R4

Because u want to calculate the effective resistance R_4.eff from u_x's side

You would think that would just be R_4 but like

Its connected to u_o which is again an active node

[trail otter]

wdym by this ? u_x/i_4

[opaque fern]

Oh sorry im tripping

Wait no im not

Yeah so its u_x/i_4

[trail otter]

$\frac{u_1-u_x}{R_3}=-2\frac{u_O}{R_2}-\frac{u_O-u_x}{R_4}$

[flat frigate]

Slowaq

[opaque fern]

No lile

[trail otter]

but i still dont get what will we get from this fraction

[opaque fern]

Its the effective resistance of R_4 looking from u_x

Because u r trying to use voltage division or whayever right

U r gonna have

,, u_x = u_1\4{R_{eq}}{R_{eq}+ R_3}

[opaque fern]

Where R_eq is the resistance of the entire circuit looking right of R_3

In turn, [
R_{\t{eq}} = R_1 \Vert R_{4.\t{eff.}}
]

[opaque fern]

I hope that makes some sense?

[trail otter]

yup i think yes but i dont get wdym by R_4.eff. isnt this circuit supplied by DC voltage source? why do we then need to consider effective voltage?

[opaque fern]

Its because of the op amp

[trail otter]

hm i havent came across this yet. how to we express it then?

[opaque fern]

Do you know Miller theorem

[trail otter]

im afraid not

[opaque fern]

R_4.eff is basically sayjng

"What single resistor, if connected from u_x to ground, would draw the same current as the real R_4?"

[trail otter]

so since $i_{R_4}=\frac{u_x-u_O}{R_4}=\frac{u_x}{R_{4.\mathrm{eff}}}$ thus $R_{4.\mathrm{eff}}=\frac{u_x R_4}{u_x-u_O}$?

[flat frigate]

Slowaq

[opaque fern]

Hol up im kinda confused what you did there

I mean yes your expression for i_4 is right

[opaque fern]

But i expected tou to use this

[trail otter]

$i_{R_4}=\frac{u_x-u_O}{R_4}=\frac{u_x-\left(-\frac{R_2}{R_1}u_x\right)}{R_4}=\frac{u_x(R_1+R_2)}{R_1 R_4}$

[flat frigate]

Slowaq

[opaque fern]

Yeah awesome

So what is u_x/i_4

[trail otter]

$\frac{u_x}{i_4}=R_{4.\mathrm{eff}}=\frac{R_1R_4}{R_1+R_2}$

[flat frigate]

Slowaq

[opaque fern]

Yeah awesome

[trail otter]

cool so now i have everything i need awesome

[opaque fern]

Thats basically the resistor you can connect from u_x to ground that draws the same current as R_4

[opaque fern]

Yep

The rest is algebra I guess

[trail otter]

crazy thank you very much you saved me from a lot of headache

[opaque fern]

Its always good to do this bte

Like

Find the equivalent resistances of everything active like an op amp

[trail otter]

yup i can see

anyway im gonna head out but i might spawn here with another circut problem hehe

bye

.close

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[opaque fern]

Cya

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[lament salmon]

So, thanks to the common knowledge I was able to make 0,5 out of 1/2 ans 0,33 out of 1/3. But is there a more reliable way of solving these? Like if I were to do a (1/183)^-3

[opaque fern]

Well 0.33 = 1/3 isn't really true but I assume you missed a vinculum there

[drowsy moss]

(1/x)^-b = x^b

[opaque fern]

Anyways

[lament salmon]

Also, is my solution technically inaccurate? Because 1/0,33 isn't really exactly 3, it's a bit more than 3

[opaque fern]

This

[autumn sun]

I highly suggest not getting decimals involved here

[lament salmon]

Lemme see

[lament salmon]

And what if it's for example (17/x)^-b

[opaque fern]

But yeah generally you'd want to keep everything in terms of fractions at this stage of mathematics

[drowsy moss]

(y/x)^-b = (x/y)^b

[lament salmon]

So

1/(x/y) = y/x?

[opaque fern]

Yes

[lament salmon]

Thank you guys

Also, an idiot question here
I forgor how to turn 1/2 into 0,5 and vice versa 💀

[autumn sun]

long divide

[opaque fern]

1/2 -> 0.5 by division
0.5 -> 1/2 by multiplying by 10 to get 5/10 which in turn simplifies to 1/2

This is how i usually do these conversions in general

[lament salmon]

0,2837 × 1000 = 2837/10 000?

How do I divide 1 by 2 though?

[opaque fern]

2837/1000

[lament salmon]

Right
I can add a 0 and then divide by 10

[autumn sun]

https://youtu.be/do_IbHId2Os consider watching this

[lament salmon]

Are you sure? That's more than 1

[lament salmon]

Thanks

[autumn sun]

nps

[opaque fern]

Yeah but im a bit confused by what you are trying to do here. You multiplied by 1000 but proceeded to divide by 10000?

Oh wait

Dont mind me i horribly misread

Yeah thats fine, your second expression

[lament salmon]

Thanks everyone

.close

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[unkempt delta]

Prove that
$$ \sum_{n=1}^{\infty} \frac{ (n-1)! }{ n(x+1)…(x+n) } = \sum_{k=1}^{\infty} \frac{1}{(x+k)^2}$$

[flat frigate]

IntoTheUnknowm

[unkempt delta]

II don’t how how to begin this one

[cold ridge]

i'd use gamma function on the left expression

or possibly beta

because essentially (x+1)(x+2)...(x+n) IS Г(x+n+1)/Г(x+1)

and if x is a positive integer then the whole thing becomes (x+n)! / x!

[unkempt delta]

But how would I proceed from there?

[cold ridge]

Well (n-1)! is just Г(n)

So the left becomes 1/n * $\frac{\Gamma(n)}{\frac{\Gamma(x+n+1)}{\Gamma(x+1)}}$

[keen tulip]

[cold ridge]

ill write Г as G for the meanwhile

idk if theres a latex for gamma function

$\gamma$

[flat frigate]

alex <3

[keen tulip]

,,\Gamma

[flat frigate]

クーリー

[cold ridge]

ah

[unkempt delta]

Tbh I don’t know too much about gamma functions…

[flat frigate]

alex <3

[cold ridge]

I don't see another easier way for solving it tho tbh

[keen tulip]

,tex >>textgradientglow<< $\frac{ Gamma(n) Gamma(x+1) }{ n Gamma(x+n+1) }$

[flat frigate]

クーリー

[cold ridge]

yes

ty

[keen tulip]

It works out nicely with the identity

but if you cannot use the gamma function...

[unkempt delta]

We haven’t learned much about them

[cold ridge]

and the right expression to me seems to be the series representation of trigamma function that's why i thought going with the gamma approach would be good

[cold ridge]

maybe try rewriting LHS as a difference of two terms

to get it to telescope

[keen tulip]

hmmm

What immediately stood out to me is that:
[
sum_{n=1}^{oo} frac{ (n-1)! }{ (x+1)...(x+n) } = frac 1x
]

[flat frigate]

クーリー

[unkempt delta]

How? Sorry

[keen tulip]

It's a telescoping sum

textgradientglow<<[
sum_{n=1}^{oo} frac{ (n-1)! }{ (x+1)...(x+n) } = frac 1x <=> sum_{n=1}^{oo} frac{ n!x }{ n(x+1)...(x+n) }= 1
]

[flat frigate]

クーリー

[unkempt delta]

I’ll try that

But how to apply this to the question?

[lofty nebula]

I'm sorry for interrupting, but I just have to say that the latex rendering with textgradientglow looks incredible

[keen tulip]

So your left hand side is < 1/x, and we can write it as a telescoping sum. On the other hand, if we call it f(x), then f(x-1) - f(x) = 1/x^2. Should be obvious from there.

[keen tulip]

Thanks!

[unkempt delta]

I’ll give it a try thank you

[unkempt delta]

Thank you to you too!

.close

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[cold ridge]

np

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[blazing cloak]

Need lots of help with trig derivitives

[blazing cloak]

Missed a week of classes, havent done trig in a week

Im lowkey still struggling with highschool trig

[plucky elk]

,tex .unit circle

[flat frigate]

riemann

[plucky elk]

,tex .sohcahtoa

[flat frigate]

riemann

[plucky elk]

those are the two main tools. you'll need csc = 1/sin and sec = 1/cos and cot = 1/tan

[blazing cloak]

Yea i got those down a bit

My teacher said the unit circle isnt really necessary if u know the basics

Sin30 = 1/2

[austere forge]

this cheat sheet is really good for trig https://tutorial.math.lamar.edu/Extras/CheatSheets_Tables.aspx#TrigSheet

[plucky elk]

sure depends what you consider basics are

[blazing cloak]

Just this

And sin60 is root3/2

Then cos is the inverse

And pi/3 is 60 etc

[austere forge]

do you need help with any specific problems here?

[plucky elk]

no

[blazing cloak]

Yea, how do u 1e?

[blazing cloak]

I meant vice verse

Not like mathmatecally inverse

[austere forge]

what is cotangent?

[plucky elk]

you cropped out the number

[blazing cloak]

Oh sorry

Its tan^2(2pi/3)

That one at the top

Ref angle would be pi/3

Which is tan(60)

Right

[austere forge]

yeah

then what do you get?

[blazing cloak]

And then is this where u do cos/sin

Or whatever

[blazing cloak]

Idk where to go from
Here

[austere forge]

you drew the triangle for 2pi/3, correct?

[blazing cloak]

No i havent, lemme draw it

Ignore 1/2

I lowkey dont know if i ever solved this way

[austere forge]

this is incorrect, think about 2pi/3 (or 120 degrees) in a unit circle

this tool might help https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html

[blazing cloak]

This is all i woukd be able to do

[austere forge]

well, do you know what sin(2pi/3) and cos(2pi/3) are?

from the unit circle

[blazing cloak]

1/2 and root3/2?

Sin60 and cos60 right

[austere forge]

no, 2pi/3 is 120 degrees

pi/3 is 60

look again at the unit circle

[austere forge]

.

[blazing cloak]

Wait am i looking for these fractions or what

[austere forge]

remeber, cosine is the x coordinate, sine is the y coordinate

cosine is the base, sine is the height

[blazing cloak]

Root 3/2 and 1/2

[plucky elk]

read the top right

[austere forge]

yes, -1/2 to be specific

[blazing cloak]

Oh

[plucky elk]

match the colors together

[austere forge]

so what is tan(2pi/3) then?

[blazing cloak]

Root3?

Wait

Negativr

[austere forge]

yeah, so what about tan^2(2pi/3)?

[blazing cloak]

Does the square just mean u square it?

so 3

[austere forge]

yep

[blazing cloak]

Ok got it

How would i do something like 2b, 2 being the 2nd set of questions

Cos = -1/root2

[austere forge]

what is -1/sqrt(2)?

[blazing cloak]

A was easy

[austere forge]

can you simplify that?

[blazing cloak]

Ah i see

The root2 over 2 one

I didnt recognize it

[austere forge]

then its again just applying from the unit circle

[blazing cloak]

Hm what about c?

Do u write it in terms of sin/cos

Wait

[austere forge]

you could do it that way, or you could think about tan(theta) as the slope of the angle

[blazing cloak]

Well i thought the -1 just meant like

[austere forge]

what angles have a slope of -1?

[blazing cloak]

Not sure tbh

[austere forge]

look back to the unit circle

[blazing cloak]

Where would i find it

Ok

Like the 45 degree angles in terms of slope of a line?

[austere forge]

yeah, but which angles have lines with slope of -1?

[blazing cloak]

3pi/4 and 7pi/4

Cool u can do that

Is that only with tan

The next question is csc=1

[austere forge]

yeah, tangent is the slope of the angle, which is also why tangent of pi/2 and 3pi/2 are undefined

[austere forge]

what is cosecant?

[blazing cloak]

1/sin

[austere forge]

can you use that to solve?

since 1/sin = 1

[blazing cloak]

Wait

U can also do this right

When y is 1

[austere forge]

that works, but you also dont have to draw an image to solve it

[blazing cloak]

How would u do it this way?

[austere forge]

csc = 1
-> 1/sin = 1
-> sin = 1

right?

[blazing cloak]

Yea

[austere forge]

then how do you solve?

[blazing cloak]

Like mathmatically? Cuz i would have thought that sin = 1 at 90 deg

Pi/2

[austere forge]

yeah use unit circle knowledge

[blazing cloak]

Imma try e

Cot = -root3

Idk what to do actually

Actually i got the answer from guessing 5pi/6 and 11pi/6 just cuz it made the most sense

But idk how u would solve it

[austere forge]

one way is to know cot = 1/tan

then tan = -1/sqrt(3)

then you can think about it as which angles have lines with that slope

[blazing cloak]

Ah i see

Are trig basics like this super important for deriving them

[austere forge]

yeah, its very important to know the unit circle, definitions, and formulas in trigonometry

the cheat sheet i posted above has all of that which helps memorize these things

[austere forge]

.

the rest is essentially just algebra

[blazing cloak]

.close

[safe radish]

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[soft light]

do inverse cos no??

[safe radish]

@soft light Has your question been resolved?

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[viral stream]

if anyone here is good at ap stats i need you asap

[viral stream]

please help in dms if possible

[solar hazel]

why not here?

[viral stream]

idk id rather have streamline dms

to look back at easier

than in a channel w constant flow

[astral glacier]

Just save a link to the conversation

[solar hazel]

personally i always prefer to help people in help channels rather than dm so people can add things on or correct me

i think most people are the same

[hard crest]

we prefer to keep conversation here due to moderation purposes as well

you will have control of this channel for as long as you're still actively working on the problem (up to 3 days i think)

[astral glacier]

-# (I'm pretty sure it's longer, pajamamallama had a channel going for two weeks at some point)

[solar hazel]

back in my day we had months long channels

until an evil man (riemann) came along

sorry that sounds meaner than i wanted it to

[astral glacier]

Now I want this lore

[viral stream]

oh

i mean id really feel more comofrtable in dms

tbh

its fine ill ask the question

A survey is conducted ar a local high school to determine the effectiveness of school commhnication with students families who speak a language other than English at home. To collect data, the schools statistician plans to send a questionnaire home with students to ask for the parents opinion about the schools communication effectiveness. Which of the following sources of bias is apparent in the survey?

Wording effect
Non random sampling
Undercoverage
Nonresponse
Sampling error

[prisma wren]

What have you thought of so far?

[viral stream]

under coverage or nonresponse

[prisma wren]

why? + define them

[viral stream]

they js seem the most probable

its honestly guesswork

those r the common anseers

[prisma wren]

surely the trick to getting a 5 in ap stats cannot be to coinflip every question as undercoverage or nonresponse. justify how it could be one (or either). IMO there's a difference between the target demographic & the sampled group, which could lead to some type of error

[viral stream]

yea

[solar hazel]

a little over a year ago there was a very long running and active help channel: #help-16 message
probably the most famous help channel ever, though i never worked in it so i could not tell you much about it

concurrently there was also this one:
#help-47 message
which is not as famous and wasn't as long running but it holds a special place in my heart because i loved the question so much and worked on it a lot in that channel

sadly one day they were both closed

[astral glacier]

I see

Thanks for the lore 🌺

[solar hazel]

looks like they were closed 9/17/24

so, about 2.5 months and 1 month for them

[safe radish]

@viral stream Has your question been resolved?

[safe radish]

Channel closed

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[ornate wyvern]

Smone held to a channel for a week in the not so distant past

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Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mighty mango]

f(1/2 x) = (abs(x))^3
f''(-1) =...?
-48
48
14
1

[mighty mango]

can someone solve this and tell me what they get? bc i keep getting 24.

[ornate wyvern]

By replacement property replace x by 2x in both LHS n RHS ull get f(x)=|2x|³

[mighty mango]

mhm

[plucky elk]

You should be showing your work

[magic junco]

$f(\frac12x) = |x|^3$\
Find $f''(-1)$\

-48; 48; 14; 1

[flat frigate]

Sky

[magic junco]

You only diff it once

You should diff twice

[mighty mango]

1/2 f'(1/2 x) = 3 (abs(x))^2 * (x/abs(x))
which is
1/2 f'(1/2 x) = 3 * abs(x) * x
second derivative
1/4 f''(1/2 x) = 3 * ( (x^2)/abs(x)) + abs(x) )
f''(1/2 x) = 12 ( (x^2)/abs(x))+ abs(x) )
subsituting x=-1
12 * (1/1 + 1)
12 * 2 = 24

[mighty mango]

but... i did differentiate twice

[magic junco]

Oh, you’re using different approach

[mighty mango]

yeah i also tried this

got 12 (forgot to differentiate f'(1/2 x) properly... then 24...

but yeah both methods give me 24

[magic junco]

How did you get 24 w the second approach?

Mind briefly explain?

[mighty mango]

I differentiated both sides w3ith respect to x

1/2 f'(1/2 x) is the derivaitve of f(1/2 x)

3 (abs(x))^2 * (x/abs(x)) is the derivative of (abs(x))^3 with chain rule

[magic junco]

No, I’m referring to the second approach that emoji name suggested

Which is to find f(x) first

[tardy mango]

hi

[magic junco]

hi cutie

[mighty mango]

I set u = 1/2x
f(u) = ( abs(2u) )^3

[mighty mango]

then i differentaited both sides with respect to u.

[magic junco]

It doesn’t need to be that complicated

no chain rule required at all

[mighty mango]

how else would you go on about differentiating abs(2u)^3

[magic junco]

Here’s my idea

[mighty mango]

go ahead

[plucky elk]

|ab| = |a| * |b|

[magic junco]

You first find f(x), and diff it twice

[magic junco]

We already know that f(x) = |2x|^3, and we’ll looking for the differentiation at -1

Which is equivalent to diffing -8x^3

[winged flare]

seems alright, although not very sure if youve reached the second derivative through the right means (the second derivative itself is correct however)

[mighty mango]

24 is correct..?

[magic junco]

I got -48

[mighty mango]

elaborate how you got rid of the.... wait is absolute value of x^3 replacable with that

[winged flare]

but the problem is that you found f''(-1/2)

substituting -1 on both sides

[mighty mango]

...

[magic junco]

Quick question
Do you know how to remove the abs of
$|-2x+7|$ if x>4?

[flat frigate]

Sky

[mighty mango]

square of square root

or

the opposite

square root of that squared

[winged flare]

actually no the second deriv is not right

[mighty mango]

give me a second.

[mighty mango]

or if you mean- in an equation, we can just square.

[magic junco]

No squaring, just simplify it

[mighty mango]

|x|x with product
( |x| + x/|x| * x)

[mighty mango]

+-

[magic junco]

blurple’s approach first as it’s closer to your original attempt

[magic junco]

idk what is this 😔

[mighty mango]

did i fuck up the product rule 😭

[mighty mango]

positive negative 😭

y= abs(x)
y=+-x

[magic junco]

Well talk about that later, blurple first

[winged flare]

this is correct

[mighty mango]

that simplifies into |x| + x^2/|x|

[winged flare]

i have no clue how that simplified to $\frac{x^2}{|x| + |x|}$ though

[winged flare]

f''(1/2 x) = 12 (x^2)/(abs(x)+ abs(x) )

id assume you messed up fraction addition

[mighty mango]

i messed up the

partenthessessss

n discord

on

i fixed it

[mighty mango]

I did not do this 😭

ok i did it again

using the same method

idk what went different this time

but

yeah i got 48

[winged flare]

right

[mighty mango]

with f''(1/2 x)=12( |x| + x^2/|x|)

1/2x = -1
x=-2

subbing that in gives 48.

[winged flare]

and its always positive since the function is concave upwards

[mighty mango]

lovely!

ty.

@magic junco what was your approach?

[magic junco]

First of all we need to reach the common ground where you can remove abs if the information is sufficient

[magic junco]

Can you simplify that again?

[mighty mango]

ok yeah i get what you mean now

if x=4 then its negative

if x=anything above 4