help-43

[upbeat geyser]

The stuff here

I'll prob come back in 10 min maybe

Idk

[kind viper]

yeah it's cooking

if you have a pencil and a pen, or an erasable whiteboard, you can kinda replace letters with their digit values "in-place"

[upbeat geyser]

Uh

It's a fixed paper

I'm not actively teaching

I'm giving an explanation for last sessions questions

[kind viper]

ok

then you'll need like 7 lines showing intermediate steps

[upbeat geyser]

Does this look like scrambled eggs

It's messy

[bleak dock]

cryptarithms are mental torture for students

[spice path]

true

[knotty orchid]

yes

[upbeat geyser]

Yes

[upbeat geyser]

So is it uh

Somewhat readable

I'm at 9th grade and I wanna banish these questions

I ain't dealing with trial and error

[knotty orchid]

ur explanation should be as follows

a is either 8 or 9

[upbeat geyser]

Yeah

I put that on top

[knotty orchid]

d is either 1 or 2

[upbeat geyser]

This

[knotty orchid]

so the only one that fits is c or the option “none of the above”

so js test c

and if it checks out

then

[upbeat geyser]

Yeah so is this readble or not

Like can they understand my writing

Because I have 10 arrows

Going left and northeast

[knotty orchid]

its unnescessary

imo

[upbeat geyser]

Awh ok

[spice path]

still readable tho

[upbeat geyser]

Yay

Atleast they won't cause brain aneurysms now

Imma close the channel

.close

[compact pewter]

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[compact pewter]

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[mortal nimbus]

Big O 😨

[lime juniper]

what does the notation mean in full, do you know?

[compact pewter]

@deft tangle Has your question been resolved?

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[split mantle]

i dont know where to start

[misty quail]

find the number of tickets the girl collects

[split mantle]

how

[spice path]

just take the floor of 99999/78

[split mantle]

ohh ok sounds ez

[misty quail]

i think you have to add 1 because it starts from 0

[split mantle]

but the girl have less tickets

[spice path]

no she doesn't

[split mantle]

the girl has 1282 while the boy has 1298

[fast gull]

Is it bc some tickets are divisible by both but the boy doesn’t take them?

[spice path]

yes

[split mantle]

soo what's the ans

[fast gull]

So the girl takes all the tickets divisible by both 77 and 78 and so she has more in the end

[split mantle]

i thought the girl only takes tickets divisible by 78 as said in the question

[fast gull]

She does but some of those are also divisible by 77

So the boy doesn’t take all tickets he technically could and the girl gets those

[split mantle]

ohhh

[lavish gull]

hint: common multiples

[split mantle]

i undestand now

thx

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[deft tangle]

94

[junior crater]

!show

[compact pewter]

Show your work, and if possible, explain where you are stuck.

[junior crater]

What are O, H and K?

[compact pewter]

@deft tangle Has your question been resolved?

[deft tangle]

O is order, and G is group H is subgroup

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@deft tangle Has your question been resolved?

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[slim garnet]

I know that R and d/dz are 0

or

are they

the function

[slim garnet]

but yea

or is it 0

it seems like f might be a scalar function

and this is a trick question

since R and d/dz are 0 the i and j are 0

[subtle helm]

curl for 2d is kth component of curl?

so

[slim garnet]

$\frac{ \partial Q}{ \partial x} = y - y \sin(x)$

[kind crane]

no space next to $

[boreal girder]

smeagol

[slim garnet]

$\frac{ \partial P}{ \partial y} = x$

[boreal girder]

smeagol

[slim garnet]

so it would just be

(y - y sin(x) - x) k

[subtle helm]

no need for k, i think

js

y - ysinx - x

[slim garnet]

ooh I had the format wrong

It didn't like the (0,0,...)

[subtle helm]

its a 2d thing

[slim garnet]

it just wanted the z component alone

[subtle helm]

not 3d

[subtle helm]

seems right

[slim garnet]

thank you both

[subtle helm]

in 2d, u green not stoke

[slim garnet]

it took this way but not my original heh

[slim garnet]

hmm?

[subtle helm]

😔

[slim garnet]

.close

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[candid matrix]

how do i find the Nth term of 3, 6, 9, 14, ...

[subtle helm]

any pattern?

[lime juniper]

if that's all the knowledge you have, you can't find the nth term uniquely

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[proven wren]

do you mean 12?

[candid matrix]

the question only says 3, 6, 9, 14, ...

[proven wren]

send a screenshot

[subtle helm]

what the helly

[proven wren]

the whole page?

[candid matrix]

4,a

[proven wren]

[lime juniper]

the answer isnt unique, make up an anser and explain why it's the answer

[open plaza]

likely a typo

[subtle helm]

typo i think

[lime juniper]

for example, you can say that these numbers satisfy the polynomial a(n)=1/3 n(n-1)(n-2)+3n+3

obviously

that's just as good an answer as saying

the next numbers are 1,1,1,1,1,1,...

the section says "creative section" in the assignment

be creative

[proven wren]

the next number is 69 because i said so

[lime juniper]

sure

i'm not joking

or trolling

you literally can make up the sequence you want

[open plaza]

one pattern I can make up is the sum of numbers of these 2 sequences:
2, 3, 5, 7,... (Prime numbers)
1, 3, 4, 7, ... (Lucas numbers)

[candid matrix]

i am just in 9th grade bruh what is all this

the answer in the back says 5n

[lime juniper]

I was using a computer to find a pattern for that one, it was a joke , it's not obvious by looking at it

[lime juniper]

that doesn't work for the sequence they gave lol

[candid matrix]

imma just ask my teacher

[lime juniper]

we told you the answer bro

there's no rule that sequences have to have an obvious pattern

[subtle helm]

a classic case of ts

[open plaza]

another solution is to burn the book

[lime juniper]

that's what I did with the a(n) polynomial lol

[subtle helm]

fahrenheit 451 mentioned

[lime juniper]

repeating thiss because it's important

THERE'S NO RULE THAT SEQUENCES HAVE TO HAVE AN OBVIOUS PATTERN

okay, good night

[candid matrix]

and i dont understand

[subtle helm]

(b), (c) are all ap

im guessing a typo

[proven wren]

their point is that with the current sequence you ar given, there's really no obvious rule to determine the next number

[candid matrix]

ok well imaa skip it then

.close

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[open plaza]

saw the "approved by Curriculum Development Centre" at the right bottom lol

[subtle helm]

Vedanta

[subtle helm]

i dont think they approved it 🥀

[open plaza]

looked it up and it is a mining company from India

[subtle helm]

https://vedantapublication.com.np/books

[open plaza]

got a publisher to avoid lol

[compact pewter]

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[molten badger]

Given $f(x)=4x^2\sqrt{1-x^2}-\sqrt{1-x^2}+4mx^3-3mx+m^2$ find all integer $m \in [-2020,2020]$ such that maximum of f(x) has the value greater or equal 5

[boreal girder]

Alexis_Fx

[frail bane]

Have you tried/found anything yet?

[molten badger]

Nahh, I'm clueless rn

Don't know where to start

[misty quail]

$1-x^2\geq 0$

[boreal girder]

cis(x) = cos(x) + i sin(x)

[misty quail]

so $-1\geq x\geq 1$

[boreal girder]

cis(x) = cos(x) + i sin(x)

[kind viper]

wrong way around

[misty quail]

mb

[kind viper]

you meant $-1 \leq x \leq 1$

[boreal girder]

Ann

[misty quail]

yeah

[kind viper]

anyway you can springboard off of this to let $x = \cos(\theta)$ where $\theta \in [0, \pi]$ and then $\sqrt{1-x^2} = \sin(\theta)$ (no sign fuckery)

[boreal girder]

Ann

[kind viper]

so maybe rephrasing the thing in trigonometric terms will help

[misty quail]

i think it's simpiler without trig

just use some calculus

[molten badger]

I'm not gonna take the differential all of that chunk man

[misty quail]

split it up

also it's not that bad

[celest lintel]

No trig is way simpler

[misty quail]

maybe

[celest lintel]

It’s tedious asf

[kind viper]

does that mean:
A) "[No trig] is way simpler" i.e. "You're better off trig-free"
or
B) "No**,** trig is way simpler" i.e. "No, you're wrong. Trig is the way to go here"

[celest lintel]

B

[misty quail]

lol

[kind viper]

$f(\cos(\theta)) = g(\theta) = 4 \cos^2(\theta) \sin(\theta) - \sin(\theta) + 4m\cos^3(\theta) - 3m \cos(\theta) + m^2$

[boreal girder]

Ann

[kind viper]

can we do something with this

[misty quail]

idk

[kind viper]

,w simplify 4cos^3(t) - 3 cos(t)

[boreal girder]

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[kind viper]

thought so

yeah ok cos(3t)

can we do some fuckery with the 4cs^2 - s terms

[misty quail]

idk

i think trig made it worse

[kind viper]

we get $g(\theta) = m^2 - \cos(3\theta) m + (4\cos^2(\theta) - 1)\sin(\theta)$

[molten badger]

We can

[boreal girder]

Ann

[molten badger]

,w simplify 4cos^2(x)sin(x)-sin(x)

[boreal girder]

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Requested by yobi_fx.

[molten badger]

Hmm

[kind viper]

aint this sin(3θ)

[placid bridge]

Yeah it is

[molten badger]

Yeah I thought so

[kind viper]

,w simplify 4cos^2(t) sin(t) - sin(t) - sin(3t)

[placid bridge]

Almost

[boreal girder]

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Requested by ann_dec

[kind viper]

not almost, exactly

$g(\theta) = m^2 - m \cos(3\theta) + \sin(3\theta)$

[boreal girder]

Ann

[kind viper]

this is cooking now

[molten badger]

Yeahhh, It seems okay now

Alright thank for your help everyone

.close

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[foggy mortar]

Let A denote the set of elements of G of order greater than 2, then all elements in A can be paired with their inverse, so |A| is even

Then G / A has an even number of elements, and since this includes the identity we have an odd number of elements of order 2

does this proof work?

[hushed magnet]

G \ A but yes

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[cinder fjord]

Is 1. Answer correct

[kind viper]

is this t=25.2 your answer or from the answer key?

[spark wigeon]

i don't get 25.2

[kind viper]

neither do i

[shrewd light]

hello

[serene junco]

your method is probably right though, because the correct answer is pretty similar

you should check your work

[compact pewter]

@cinder fjord Has your question been resolved?

[kind viper]

!occupied

[compact pewter]

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

[cinder fjord]

Yes

I think it’s wrong

And this too

[compact pewter]

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[kind viper]

.reopen

[cinder fjord]

.reopen

[compact pewter]

✅

✅

[cinder fjord]

I got 1.5ms-1

[kind viper]

that wasn't a yes/no question

[cinder fjord]

It’s from the answer key

[kind viper]

ok well then the answer key is wrong

[cinder fjord]

Tqsm

What about the other one

[kind viper]

dunno, don't feel like doing it rn

[compact pewter]

@cinder fjord Has your question been resolved?

[cinder fjord]

.close

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[sudden spade]

Guys I’m confused on what the two F points in the middle are? I thought the idea of having two converging lenses is that there’s just one focal point in the middle.

So which focal point does the parallel ray from the object go through?

[balmy knot]

im guessing theyre focal lengths, atleast thats what im seeing from the text provided over the diagram

[sudden spade]

I drew it like this, is this the idea? That where the image created by Lens 1 that is the object for lens 2. And then the circle is the image?

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[candid matrix]

how do i find the length of BD?

[pallid rune]

You can use the pythagorean theorem, because <A is a right angle

[candid matrix]

oh ok ty

.close

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[candid matrix]

how do i find the measurement of its side and calculate its area?

[red basin]

How much trig do you know?

[subtle helm]

Let one side be 3x

[red basin]

/ do you recognise those numbers?

[stark marsh]

from the ration of sides

[cerulean steeple]

say the triangles sides are 3x, 4x, and 5x

[cerulean steeple]

trig?? 😭

[stark marsh]

a gun to a fist fight

[red basin]

Well if you don't know what a 3 4 5 triangle looks like you need trig to work it out

[candid matrix]

pretty sure its a scalance triangle?

[cerulean steeple]

it is not

[candid matrix]

oh ok

[cerulean steeple]

wait do you not recognize it?

[candid matrix]

it does not have an image so no

[stark marsh]

you don't even need to know what type of triangle it is ...

[candid matrix]

then how do i find its area

[stark marsh]

hint then
3^2+4^2=5^2

[candid matrix]

pythagorus theorum?

[stark marsh]

you can conclude that it's a what triangle ?

[candid matrix]

.close

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[novel narwhal]

can someone give me a hint on how to start any of these problems (just a hint)

[boreal nimbus]

(optional) suggested approaches

1. start with p/q, so that gcd(p, q) = 1 and p + q = 1000. can anything be said?

2. this one is a bit harder. you could try solving the problem with 36 replaced by a smaller number and see if you discover any insights

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[misty quail]

,tex
$
\text{Given } a,b,c,d > 0. ; \text{Prove:} \
\left( \frac{a}{a+b} \right)^2 + \left( \frac{b}{b+c} \right)^2 + \left( \frac{c}{c+d} \right)^2 + \left( \frac{d}{d+a} \right)^2 \geq 1
$

[boreal girder]

Nerdyasianguy

[misty quail]

C-S doesn't work, and I can't find any AM-GM pairings

.close

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[cerulean steeple]

you got it?

[misty quail]

Yeah

[cerulean steeple]

what did you use

[misty quail]

C-S but in a different way

[cerulean steeple]

titu's lemma of sort?

[misty quail]

Yes

T2 Lemma, it's called C-S Engel form in my country

[compact pewter]

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[cunning jetty]

Can anybody tell me how to proof similar objects concept

[exotic harbor]

do u mean congruency or similarity?

[cunning jetty]

similarity

[exotic harbor]

ohh do u have a question??

[cunning jetty]

This

Prove this idk how

[exotic harbor]

would be easier for me to explain

[cunning jetty]

Yes pls explain

[cunning jetty]

Pls help

[exotic harbor]

yes yes sure

ok so

first ur gonna have to mark angles

do u know corresponding angles concpt?

[cunning jetty]

Yes ik

SNK = SMP
SKN = SPM

[exotic harbor]

yesss

exactly

[cunning jetty]

It makes these 2 triangles similar

[exotic harbor]

yes

there is 1 more angle

can u figure that out?

[cunning jetty]

But why 2 similar triangles or any similar object have these "corresponding" side that--

[cunning jetty]

R u deadass🥀

MSP

[exotic harbor]

☠️ lol

[cunning jetty]

Have These formula tying them

[exotic harbor]

its bcz of the parallel lines

if u have 2 parallel lines the angle above it or under it they are always equal

[cunning jetty]

I don't understand how that gets me my answes
Pls elaborate thx

Yes

[exotic harbor]

its like a rule

[cunning jetty]

Ok

[exotic harbor]

so

first u have to take the smallest traingle

[cunning jetty]

triangle SNK

[exotic harbor]

and the largest one to prove the first one

yes

[cunning jetty]

OK

[cerulean steeple]

"corresponding angles on paralle lines"

[exotic harbor]

the side opposite of angle snk is c

[exotic harbor]

oh thnx

[exotic harbor]

so whats the corresponding angle to smp?

its c + d

[cunning jetty]

SNK

[exotic harbor]

yea yea mb sorry

so its c/c+d

the side opposite to skn is a

as spm is corresponding to skn that side will be a + b

[cunning jetty]

Yes

[exotic harbor]

so yes u do ratio so its a/a + b

[cunning jetty]

Correct

[exotic harbor]

last one is ez u have the common angle nsk

so its e/f

[cunning jetty]

Alright and how are these ratios the same

[exotic harbor]

becuase they share the same angle

[cunning jetty]

I don't understand

[exotic harbor]

suppose if snk is 20 smp is also 20

umm why not corresponding angles are always equal\

do u not recall this?

[cunning jetty]

I do not understand what the word recall is

[cunning jetty]

Yes

[exotic harbor]

oh

[exotic harbor]

like let me tell u

[cunning jetty]

Does it mean Remember?

[exotic harbor]

if snk = 20 then smp is also = 20

[cunning jetty]

But like reminded by your own brain or sum

[exotic harbor]

yessss

[exotic harbor]

yea yea

[cunning jetty]

Obviously

[exotic harbor]

so we can write snk = smp

[cunning jetty]

Tru

[exotic harbor]

or even snk/smp as it is 1

[cunning jetty]

Ok that's correct

[exotic harbor]

there is also a theorem basic proportionality theorem do u know that?

[cunning jetty]

What's that

[exotic harbor]

it in easier words say u can equate sides in triangles if they are similar

so if u take the small traingle and the largest one

u can write

a/a+b = c/c+d = e/f

[cunning jetty]

a/b=c/d=e/f-e

Nvm

[exotic harbor]

XDDD

chill u can also write that in trhe same wayyy

[cunning jetty]

So the proof is basically the proof of basic proportionally theorem

[exotic harbor]

umm yes

tbh i am not 100% sure as i did it way back in 7th gradee

but i am 95% confident

[cunning jetty]

You studied ts in 7th grade?
Same lol

[exotic harbor]

yea u in 7th grade?

ok so u do understand the first proof right???

[cunning jetty]

Nah i learned the basic ratio thing and how to use it when im 7th grade
Now i wanna know the proof

[cunning jetty]

If i do i wouldn't ask🥀

[exotic harbor]

ohh npp

[exotic harbor]

ok what part is ur confusion?

[cunning jetty]

Basically simply
From our conversation

The proof of the basic proportionally theorem

[daring wren]

0/2

[exotic harbor]

hmm??

[cunning jetty]

0

[daring wren]

Wow

[cunning jetty]

Yes bruh that's what we talked about from the beginn

-ing

[daring wren]

Hello, I am new

[exotic harbor]

yea yea

so can u proof the first one atleast???

plz say yes or i will cry

[exotic harbor]

hii

[cunning jetty]

No😔

[exotic harbor]

😭

ok keep asking questions

[daring wren]

I want to be a mathematical physicist, but I'm at a somewhat low level, I'm studying online independently

[cunning jetty]

Imma search it oj google instead

Thx

.close

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[daring wren]

I started a month ago

Do you have any website that can help me study?

[compact pewter]

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[wise swallow]

Hey, I have a problem trying to construct a quartic that goes through certain points. Specifically, I have 5 points; a starting one, three turning points, and an end point. Technically the quartic that I've constructed goes through each point but the turning points are not in their correct position. How can I do this?

[kind viper]

wait can you show the turning point data as it was originally specified

bc i think if you fix five points AND require the derivative to be 0 at three of them then a quartic in general might not be enough!

[wise swallow]

The middle three being the coords of the turning points

[kind viper]

so those HAVE to be turning pts?

[wise swallow]

Yes correct

It's associated with that colour gradient you see at the bottom

[kind viper]

and the curve HAS to be a QUARTIC?

[wise swallow]

Well I guess not necessarily. I was thinking because it required three turning points, it would most appropriately be a quartic

[kind viper]

thing is

a quartic-or-under is uniquely determined by 5 points

so you will need a higher degree to get the turning points that you want

or maybe some kind of spline thing, idk

what's the ORIGINAL context for these values?

[craggy sluice]

3 turning points are quartic no?

he said the middle 3 are turning points

[wise swallow]

So essentially I'm trying to graph colour gradients utilising chirp functions and what not. The y-axis being the frequency in terahertz associated with the colours in the visible light spectrum and the x-axis being time in picoseconds

[kind viper]

i dont quite see why that relationship should be a polynomial one at all

[wise swallow]

School project 😭

[kind viper]

though i also dont know what a "chirp function" is so

hm

can you share the project instructions

[wise swallow]

yes give me one second

What I'm doing is for part B but part A gives context and fundamentals

And I decided to choose a polynomial as my first function

[craggy sluice]

y=ax^4+bx^3+cx^2+dx+e

(0,465)->e=465

(1,465)->a+b+c+d=0

y'=4ax^3+3bx^2+2cx+d

(0.25,0)'->a+3b+8c+16d=0

(0.5,0)'->2a+3b+4c+4d=0

(0.75,0)'->27a+27b+24c+16d=0

using line 3,5,6,7, you can solve for the values of a,b,c,d

[kind viper]

i think OP did that but the turning points were off

i think you actually need Hermite interpolation

[craggy sluice]

not sure if there is a more efficient way, but that is what I thought of

[wise swallow]

I think I just treated each one as a point and not the three as turning points in my first set of equations

[wise swallow]

.

[wise swallow]

Let me try this then

Thanks for the help

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[teal vine]

i have a question
so lets say we have an equation
(x^2-25)/(x-5)=x+5
the answer to this would be every number except 5 right
so would you write
x≠5
or
all real numbers except 5
because when you say x is only real, that implies that x=i is not a solution when it actually is
so would x≠5 be a more accurate answer

[kind viper]

i mean, do you care about R vs. C at all here? like based on context, not the equation itself

[teal vine]

i do

whats the most accurate answer

is there any issues with me writing x≠5

[kind viper]

not really

[teal vine]

alright

thanks

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[lament token]

Can someone help me with question 1

[kind viper]

let $P = (c, e^{-c})$. you can assume $c>0$ because of symmetry, but be ready to negate the x-coordinate if needed to match an answer option.

[boreal girder]

Ann

[kind viper]

write down the equation of the tangent to $y=e^{-x}$ at $P$ in terms of $c$ and then find the axis intercepts of that, and from that, the area.

[boreal girder]

Ann

[lament token]

Is this right

[placid bridge]

Yes

[lament token]

The area with the intercepts will give me the maximum area?

[strange ermine]

No

You have to find the value of c that gives you the maximum area of such triangle

[lament token]

Okay

But how

[placid bridge]

Find the equation of the tangent line

[strange ermine]

By first finding the area and then by imposing it is the maximum

[lament token]

Yeh this dy/dx is

[strange ermine]

Nope

[placid bridge]

Then find the function that will give you its area and then find the maximum of that function

[strange ermine]

That is only the slope of the tangent line

[kind viper]

incomplete

you didnt yet find the axis intercepts as i asked you

[lament token]

Isn't it -e^-x?

[placid bridge]

No that is just the slope of the tangent line

[lament token]

Ahhh

So do i use thw intercepts to make the line eqn?

Tangent*

[placid bridge]

No find the equation then use that to find the intercepts

[lament token]

Ok

[placid bridge]

Good

[subtle helm]

Hi sarin

[lament token]

Finding interceptd now

[placid bridge]

Hello k

[strange ermine]

Great

Now compute the area of the triangle

But first, I strongly suggest you draw a sketch of the whole situation

[placid bridge]

Or rather find the function that will give you that area

[lament token]

And differentiate it?

[lament token]

Wdym

[strange ermine]

Just this

[strange ermine]

Even if you should know what this means but...

[placid bridge]

To find the maximum yes

[strange ermine]

Where is the graph of $y = e^{-|x|}$?

[boreal girder]

Alberto Z.

[lament token]

What do i dif it with respect to

[strange ermine]

With respect the only thing that's varying you have, i.e. with respect to the parameter c 😅

[lament token]

😭okay yea

I wrote it in tersm of x coz c are the point a which we took the tangent but x is in the whole eqn of the curve

[lament token]

Ye

Should i equate it to 0

Since we want the maxima?

[strange ermine]

Yes

[placid bridge]

Wait

[strange ermine]

Actually, that is just the necessary condition

[lament token]

Ohh

But i haven't take the mod part in consideration

[strange ermine]

You also have to make sure it isn't a minimum or an inflection point

[placid bridge]

Can you explain what you did

[lament token]

So if in the end i have to check of the part where x<0 how would i do that

[strange ermine]

I was wondering right now why they gave you that. I fear they wanted the triangle formed by two tangent lines, with symmetric points of tangency

But let's finish this and see if the result you get is among the options

[placid bridge]

@strange ermine Can you check what they have done

[strange ermine]

Oh I didn't even see the photo

Let me see it now

[lament token]

Okkayy...
But it says the triangle made by the coordinate axis and the tangent so i think it means what we are doing

[strange ermine]

Can you take a better and not blurred picture, please?

Also, don't change c with x.

[strange ermine]

,rccw

[placid bridge]

Im not able to figure out what they have done after getting the intercepts.

[lament token]

Area = 1/2 bh

[strange ermine]

Ok yes it's good

Just a notation correction

(if you don't write = 0 you can't cancel those things)

[placid bridge]

Ok my bad

[strange ermine]

Wait wait

[lament token]

Right i wrote 0 on the rhs

[strange ermine]

You can't cancel (1 + x)

Because that might give you a solution

And actually it will

[lament token]

Oh

Also not the e^x then

[strange ermine]

Instead, you can (and should) cancel e^(-x) since that'd never 0

[lament token]

Ah

[placid bridge]

excpet when x = infinity, but we don't think about that

[lament token]

Okayy

[placid bridge]

If you want to you can take the second derivative to find the points of maxima or just try whatever solutions you get and compare it with the options

[lament token]

Ph so what does the second derivative of any curve give me in general

Did i take y right?

[placid bridge]

The rate of change of that curve's rate of change. Im not really sure how else to phrase it. You can kind of think about it like the acceleration of the curve

[lament token]

Ahhh okayy

[placid bridge]

Yes seems okay.

[lament token]

No coz u mentioned it would give me the points of maxima

[placid bridge]

It won't give you the points of maxima, it just tells you whether the point you got is a minimum or a maximum

or a point of inflection

[lament token]

Ohh like whether after that point the function increases or doesn't

[placid bridge]

Yes

[strange ermine]

When at the end you write x = ±1, you should recall we've chosen c (being your x) to be positive. Hence the solution is only x = 1 giving a y-value of 1/e
Therefore, the correct option is answer A.

But, by the simmetry of the situation, also answer B is acceptable

[lament token]

Its a multiple option question so both are the answers

[strange ermine]

Make sure as a final step to check that 1 is indeed a point of max (and not of min, for instance)

[lament token]

By double differentiation right?

[strange ermine]

That's one way, yea

[lament token]

Okay

[strange ermine]

Another way is studying the sign of the derivative before and after x = 1

If 1 is really a point of max, you need y' to change sign from positive to negative

[lament token]

Ohhh okay

[lament token]

But that will be at x+ h right

[strange ermine]

What is h? 🤔

[lament token]

h being as small quantity

[strange ermine]

Mmmh i don't know what you're talking about honestly

[lament token]

No i didn't get why the sign changes where

Here*

Like i know if its a Maxima if we go a little ahead the slope becomes negative

[strange ermine]

Because that's how a point of max is described graphically

[strange ermine]

Exactly, and a little before the slope is positive

[lament token]

But at y' aka dy/dx the maxima point the slope

[lament token]

Yes

[strange ermine]

Huh?

[lament token]

Ohh i got what u meant...
U meant that as it turns around the maxima point the slope goes from positive to negative

. close

[strange ermine]

In between those there's the point of max.

I hope that's a bit more clear why it has to be that @lament token

[strange ermine]

Exactly

[placid bridge]

You didn't type the command correctly, do it without the space

[compact pewter]

@lament token Has your question been resolved?

[compact pewter]

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[wary mulch]

What is a coterminal

[subtle helm]

Read the second sentence

[wary mulch]

I need elaboration

[wind grove]

-60 degrees basically means youre going counterclockwise for 60 degrees. on the unit circle, 0 degrees and 360 degrees are the same. so if u subtract 60 from 0 u get -60. if u subtract 60 from 360 u get 300

thats basically what it means

[lavish gull]

-60deg and 300deg are exactly the same angle, from different directions

[wind grove]

wait no mb i meant clockwise

[lavish gull]

time to break out the ms paint!

[wind grove]

youre going CLOCKWISE* for 60 degrees

[wind grove]

yeah

[compact pewter]

@wary mulch Has your question been resolved?

[lavish gull]

you there OP?

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[wary mulch]

Sorry im reading it rn

[wary mulch]

So the other angle is facing out wards and the other inwards

[lavish gull]

would rather you think in terms of clockwise and counterclockwise

[compact pewter]

@wary mulch Has your question been resolved?

[wary mulch]

So one is positive

And the other is negative

[lavish gull]

kinda, yes

the negative angle is just measured in the opposite direction

in this case, the opposite direction is clockwise

[wary mulch]

But they are not the same angles right

[subtle helm]

what can u infer from this

[lavish gull]

they aren't exactly the same angles, but they are formed from the same lines

that's the whole point of coterminal

[kind viper]

they differ by 1 full turn

coterminal in general means angles that differ by a whole number of full turns

[lavish gull]

put another way

[lavish gull]

if you have two coterminal angles and two "clock hands", and you take each hand and rotate them by one of the two angles each

[kind viper]

same thing as what i just said

[lavish gull]

they will overlap

[kind viper]

only a slightly confusing usage of "integral" as an adjective

[wary mulch]

Makes sense now

[subtle helm]

wonderful

[lavish gull]

whenever in doubt

break out the unit circle

[subtle helm]

https://tenor.com/view/one-more-left-gif-200125183678896711

[lavish gull]

https://www.geogebra.org/m/S2W46Thv
i highly recommend checking this out

since you seem to struggle with trig

you can stop the animation and play around with it yourself

[subtle helm]

thats cool

[compact pewter]

@wary mulch Has your question been resolved?

[compact pewter]

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[coral iris]

how does chnage of base work

[coral iris]

ii we have logbaseb(c)

it is

[kind viper]

you can write that as log_b(c) btw

[coral iris]

logbase(c)/logbase(b)

is that right

[kind viper]

no, now you've gone on to not even state a base despite writing out the word "base" in full 2 times

so what you've just said falls flat on its head somewhat

fix the typo or write it on paper

[coral iris]

if log_b(c)=log_c(c)/log_c(b)

in log_b(c)

it is log(c)/log(b)

right

plus the new base

[kind viper]

the full change-of-base formula is $\log_b(c) = \frac{\log_a(c)}{\log_a(b)}$

[boreal girder]

Ann

[kind viper]

you tried to state it several times but fumbled it in different ways unfortunately

here a can be whatever you want that's positive but not 1

[coral iris]

this is what i menat

so thanks then

where logc/b and base chnage be rewrriten

right?

and if base is e then it is ln

@kind viper

[kind viper]

i am having great trouble understanding you right now.

yes, if the new base is e, then the right-hand side has ln's.

i.e. log_b(c) = ln(c)/ln(b).

[coral iris]

alright nevermind

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[forest token]

can someone help with this

im somehow getting x= 2^infinity

[kind viper]

where and how 💀

[forest token]

shit i dont have my phone so i cant send

but i multiplied and divided by 2 sin x/2, 2 sin x/4 etc

and things got cancelled

[kind viper]

uh.

prod cos(x/2^k) = sin(x)/x is actually like, a thing that's true for all x lol

[forest token]

oh

[kind viper]

idk what you were trying to solve for

💀

[forest token]

😔

first i was trying to get tan somehow

so that i can differentiate and get sec^2

[kind viper]

you are working with sum 1/(2^(2k) cos^2(x/2^k))

ig think about the integral of this?

[forest token]

yeah thats what i was thinking abt

[forest token]

^

[minor snow]

hi

[forest token]

hey

[compact pewter]

@forest token Has your question been resolved?

[sage oxide]

To get this, we need to differentiate the first equation twice.

[forest token]

twice?

could u explain why tho

wont creating a tan in the series be enough

so that we can differentiate it only once

[sage oxide]

in order to get 1/2^2n sec^2 x/2^n term

[forest token]

yeah im not able to understand how we get that term

if we double differentiate

[sage oxide]

did you try that?

[forest token]

no i didnt

also i wouldnt know how to differentiate that..take log both sides?

[sage oxide]

yeah

absolutely!

[forest token]

okay let me try that

[sage oxide]

yeah sure!

[forest token]

oh

yeah i can see the tan term coming now

which on differentiation will give us sec^2

okay thanks bro

i didnt think of that for some reason

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[sage oxide]

Yw mate

[compact pewter]

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[fair thunder]

Show that 2(7^n) + 3(5^n) - 5 is divisible by 24 for all positive integers.
im using induction and im really stuck with the manipulation

[fair thunder]

did till here

[prisma tulip]

I try

[boreal dawn]

Just fyi, please do not share solutions.

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[prisma tulip]

oh ok yes

[boreal dawn]

thanks

[prisma tulip]

have you seen congruences in class ?

[fair thunder]

nope i do have it later

like just after this

[prisma tulip]

because this would be much easier I think

[fair thunder]

i mean

im supposed to do this with induction

[prisma tulip]

ah ok

I see

I will try to search but I think it's harder

[fair thunder]

i tried checking the solution after so much, and they missed out a lot, like ig some technical glitch, its only half

[prisma tulip]

oh okmaybe send it here if I can understand

i can try to explain it to you

[fair thunder]

thats it, i already did this

they didnt give the n+1 part

[prisma tulip]

wow it quick as a correction lmao

iwill try wait

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[wind spindle]

It's easy to verify the two examples are two-sided ideals and I believe then that we only need to deal with L(V) of dimension 2 or above. So I started by taking an arbitrary subspace U(V) of L(V) s.t. U(V) is not {0} or L(V)

[safe oyster]

Well i'd try to show that if I is a nonzero two sided ideal, then I contains the identity map and hence must be L(V)

Thats probably simpler

[wind spindle]

uh

alright

so

ig i'll think about that

[safe oyster]

If you want to stick with your approach, take a map M thats not in U(V) and then take a map M' ≠0 that is in U(V). Then try to construct M from M' giving you a contradiction

[wind spindle]

hold up

I already did that

what I did write and I'm not convinced is true is the following
Let L be in L(V) but not U(V)
then there exists some U in U(V) and T in L(V)
s.t. TU = L
my reasoning was that U takes the basis vectors v1,...,vn to some list of v's
and we can just make T
take that list of v's to whatever list of v's L takes the basis vectors to
I'm just not sure if this T actually will exist, although idk a glaring reason why it wouldn't

tbh I don't really get subspaces of L(V)

since its a new type of vector space I'm not used to

[safe oyster]

But what if the maps in U(V) take "too many" vectors to 0? Then U is also forced to send them to 0

[wind spindle]

yes

that's why its not correct

[wind spindle]

I think i'll stick with this but I don't have any good ideas

First of all, why does containing the identity map imply being L(V)

[safe oyster]

You try to prove that 😁

[wind spindle]

well

span(Identity map)

[safe oyster]

Its precisely 1 line

[wind spindle]

surely isn't L(V)

[safe oyster]

Ofcourse

But U(V) is a two sided ideal

[wind spindle]

oh right

[safe oyster]

You arent using that fact

[wind spindle]

so two sided ideal + containing identity map = L(V)

[safe oyster]

Yes

[wind spindle]

oh wait

lets say T is in L(V)

and I (identity map) is in U(V)

if U(V) is a two sided ideal

then TI is in U(V)

and TI is just T

[safe oyster]

Yes

[wind spindle]

okay cool

so now, assuming U(V) is not {0}

we somehow gotta show it has I

TU and UT are in U

well I haven't done any work with inverse linear maps but

we could just pick T to be the inverse of U

and maybe TU would be I?

[safe oyster]

What is T, what is U here?

[wind spindle]

T is a linear map in L(V)

U is a linear map in U(V)

[safe oyster]

Why does U have to be invertible

[wind spindle]

exactly it doesn't

[safe oyster]

It requires some more thought

[wind spindle]

yeah

[safe oyster]

Its not so simple

[wind spindle]

oh lord

since the book never went over this

is a subspace U(V) of L(V)

the set of all linear maps from U to U

where U is a subspace of V

[safe oyster]

No, that is not true

[wind spindle]

okay

[safe oyster]

Just take the subspace spanned by a single nonzero map

Its still consists of maps V to V

[wind spindle]

oh

so L(V) has some "basis linear maps"

if you will

what are the standard basis linear maps of L(V)

if v1,..., vn

[safe oyster]

Definitely, since its a vector space

[wind spindle]

are the standard basis vectors of V

would they be constructed as follows

Tk(a1v1 + ... + anvn) = vk

[safe oyster]

Well that doesnt span all maps, and that map is not linear

You are close though

[wind spindle]

wait what

[safe oyster]

Do you know the correspondence between linear maps and matrices?

[wind spindle]

haven't gotten to matrices yet

Just say that T_k is the linear map that takes all basis vectors to v_k

so T_1 takes em all to v1, T_2 takes em all to v2

then with all the T_k's

why not

[safe oyster]

You need to prove that it is linear

[wind spindle]

nah it is cuz

Just let the list of w's be all vk

ik the notation is now confusing cuz there is another vk there that is not a fixed k

bro realized

[safe oyster]

This is different from the map you wrote here

#help-43 message

[wind spindle]

ik i realized that

the first one wasn't linear

[safe oyster]

Yes this map exists and is linear

But that list isnt a basis

[wind spindle]

ah

maybe this topic will be explored later and tbh its kinda off topic to the question

i searched it up

T_k

maps v_k to v_k

and all other v's to 0

that makes sense tbh

[safe oyster]

You are claiming thats a basis?

[wind spindle]

yeah

if T_1 maps v1 to v1 and all other v's to 0

then the list T_1, ... T_n would be the basis of L(V)

[safe oyster]

But for example, you cant construct that map that takes v1 to v2, v2 to v1 and keeps the rest of the basis as identity from that list of T_ks

[wind spindle]

I think you can

[safe oyster]

How

[wind spindle]

Lemme think about it real quick

[safe oyster]

I can tell you it isnt a baisis because it has the wrong cardinality

There should be dim(V)^2 many maps

[wind spindle]

oh they aren't the same dim?

ahh

AI must've been capping

[safe oyster]

The real basis looks like:
Tij which take vi to vj and keeps the rest 0

[wind spindle]

yeah i just asked deepseek

so there's one that takes v1 to v2 and the rest to 0

and one that takes v1 to v3 and the rest to 0

[safe oyster]

Yes

[wind spindle]

okay

thats great that makes sense

idk if I can use that

for the identity question

or proof

rather

[safe oyster]

You could

There are many ways to construct the identity. Here is the one i came up with:

Let U be a nonzero two sided ideal. So you know there is a nonzero map N in U

||Let v1,v2,..,vm be a basis for V. Then the image one of these under N is not 0, say v1||

||Let A be the map sending N(v1) to v1 and everything else to 0||

||For i in 1 to m, define the linear maps Ti as follows: Ti sends vi to v1 and everything else to 0||

||Similarly let Pi be the map sending v1 to vi for i in 1 to m||

||then the map P1ANT1 + P2ANT2+...+PmANTm is in the two sided ideal by the properties and it is the identity map, so we are done||

[wind spindle]

but U is arbitrary subspace

so idk what the bases are

Wait why is A1 there

u just define it but don't use it at all

[safe oyster]

Ah sorry i changed it now

I meant to include it in the last part

[wind spindle]

and the crazy construction is in U because

P1A is in L(V)

so P1AN is in U

and T1 is in L(V)

so P1ANT1 is in U

and etc. with the rest

[safe oyster]

And their sum is in U because its a subspace

[wind spindle]

yes

bro

how do you come up with this 😭

I swear everybody in this discord makes me feel dumb

Is everyone here just super good at math

[safe oyster]

The Tis basically take a vector, take the ith coordinate and move it to the first coordinate. Then N does something other than map it to 0, A "undoes" that action, then Pi "undoes" the action of Ti and moves the ith coordinate back to the ith slot

Its actually very intuitve

If you prove that the construction is actually the identity, you will see what is going on here

[wind spindle]

no I understand it

I just feel like I would've never thought of this

[wind spindle]

especially because I would never think in coordinates like this

since it's not necessarily coordinates at all

it could be other abstract objects

but yeah I see how it works and thanks for your help

[safe oyster]

Well now you know

[wind spindle]

what was ur sat math

just curious

[safe oyster]

Sorry, not sure what you are referring to?

[wind spindle]

oh i guess ur not american

well nvm

thanks anyways

.close

[compact pewter]

Channel closed

Closed by @wind spindle

Use .reopen if this was a mistake.

[safe oyster]

No problem, glad to help

[compact pewter]

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[toxic helm]

could someone help with part b please i dont even know how to get started?

[toxic helm]

.close

[compact pewter]

Channel closed

Closed by @toxic helm

Use .reopen if this was a mistake.

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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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.

[boreal nimbus]

<@&268886789983436800>

[snow finch]

they don't even try to disguise it

[brazen quiver]

@sussyzeus121 please don't abuse help channels

Oh, nevermind lol

[mortal bay]

https://tenor.com/view/ltg-mods-crush-his-skull-low-tier-god-fish-react-gif-4333739735553177172

[brazen quiver]

.close

[compact pewter]

Channel closed

Closed by @brazen quiver

Use .reopen if this was a mistake.

[brazen quiver]

🐟

[compact pewter]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[near swan]

Can someone enlighten me no sols please

[near swan]

Also