help-0

[stone glen]

Series

I mean search on YouTube

It's all free by the way , I'm no agent

[sage locust]

oh yea got it

thank you so much for ur advice

[stone glen]

No problem

Watch it sequence by the way

[sage locust]

yea

[stone glen]

Don't start from calculus , u will get lost

[sage locust]

na xd just learning for ioqm

they mainly ask quadratic, number theory, geometry stuff

[stone glen]

Start from the 7th Lecture , and then go upto 35 , and then go to the 1st lecture

[sage locust]

ok

also should i close this channel?

[stone glen]

Oh well just watch those lectures then

[stone glen]

Yeah sure

If ur done

[sage locust]

becz it will take me some time to get back to this problem

[sage locust]

or not a problem if i keep it open?

[stone glen]

U can have as much time as u want

Swipe left and see how many channels this server has free

[sage locust]

😂 right

[stone glen]

Swipe left , scroll down

And ur clearly not done so...

[sage locust]

yep no point in closing

there are tons of available channels

[stone glen]

@sage locust i reckon don't watch that playlist

I don't think it will help for ioqm

U wanna focus on other stuff

[sage locust]

then should i watch that quadratic vid?

[stone glen]

Yeah def , watch that quadratic vid

100%

Not the playlist tho

[sage locust]

alright

[stone glen]

Alright then, ping me if ur stuck with the problem

Goodluck!

[sage locust]

yes

ty again

[stone glen]

Np

[lone heart]

@sage locust Has your question been resolved?

[sage locust]

.close

[lone heart]

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[sage locust]

got it

[lone heart]

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[dawn quail]

Is it still possible to find the extraneous solution, y = 2 from this long and overly complicated method?

Or would I need to start with the difference of two squares in the last denominator to find the extraneous solution, y = 2?

I think I took a wrong turn with my work to try and use the Quadratic formula to solve but now I am stuck. I see if I used the difference of two squares at the beginning I can find the extraneous solution much easier. But my question is more for curiosity sake:

1. is it still possible to get y = 2 using my method?
2. do I need to start over to solve this equation?

If I continue I’m not sure if this is legal…

This is the correct answer but I’m wondering if I can get to this answer using my first method?

[dawn quail]

This one

[lone heart]

@dawn quail Has your question been resolved?

[north needle]

Sorry nvm I didn’t read

You can get it with both methods just the first one takes more time and requires you to be able to factorise a cubic

[unreal summit]

Now you can equate this to zero for the roots

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[short turret]

how do I get better in rearranging formula

[worn fox]

Do more questions

[median oar]

What kind of formula

[short turret]

Like make x the subject of the formula

[gray isle]

do you have anything specific that you're struggling with?

[median oar]

Depending on the kind of formulas you got they might not be “make it the subject”-able

[gray isle]

start by doing stuff to both sides of the equation to get rid of fraction on the left

the main idea is to do algebraic manipulation step-by-step to slowly work towards isolating your desired variable

here you have no xs in the right,
so start doing stuff so that there'll be less stuff linked/attached to the x you have on the left

[lone heart]

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[stark sparrow]

How is $\sigma(\mathcal{P}(\mathbb{N}))={M\subseteq \mathbb{R} : M\subseteq \mathbb{N}_0 , \vee , \mathbb{R}\setminus M \subseteq \mathbb{N}_0}$?

[ocean seal]

Anton.

[worn fox]

What is σ? Sigma algebra?

[hazy atlas]

if thats a probability/ statistics question then i guess sigma stands for mean deviation

probably

[lone heart]

@stark sparrow Has your question been resolved?

[stark sparrow]

generated sigma algebra of the power set of the naturals

[worn fox]

Remember a sigma algebra is closed under complements, which may help make sense of the second "or" condition

[lone heart]

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[rough acorn]

i dont get how they got to the red section

howd they calculate (3, 8) or (-3, -8) thx 🙂

perpendicular to (-16, 6) ? how did they calculate it @umbral dune

ah i see

thanks

.close

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[thorn tapir]

So. I have to prove xe^x is a continuous function using epsilon delta definition of continuity.

[thorn tapir]

So far I chose all x which follow

$\left|x - a\right| < \delta$

[ocean seal]

rikusp2002

[thorn tapir]

now as f(x) = xe^x

$\left|f(x) - f(a)\right| = \left|xe^x - ae^a\right| = \left|xe^x - x + x - a + a - ae^a\right| \leq \left|x\left(e^x - 1\right)\right| + \left|x - a\right| + a\left|e^a - 1\right|$

[ocean seal]

rikusp2002

[thorn tapir]

I have no idea how to proceed after this.

[merry depot]

why not $|xe^x - ae^a| = |xe^{x-a}-a|e^a$?

[ocean seal]

Zybikron

[thorn tapir]

Trying that

hmm so, after that we can say

$e^x \geq 1 + x$

[ocean seal]

rikusp2002

[thorn tapir]

from the definition

$e^x = \sum_{k = 0}^{\infty}\frac{x^k}{k!}$

[ocean seal]

rikusp2002

[merry depot]

that seems wildly complicated

[thorn tapir]

...i'm sorry.

[merry depot]

You know $x-a \leq |x-a|$ yes? and so $e^{x-a} \leq e^{|x-a|}$

[ocean seal]

Zybikron

[merry depot]

and you can bound |x-a| by some helpful delta value that maybe just removes the e?

[thorn tapir]

I actually didn't get that, sorry.

[lone heart]

@thorn tapir Has your question been resolved?

[lone heart]

@thorn tapir Has your question been resolved?

[thorn tapir]

I'd probably not get help anymore so

[lone heart]

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[lime echo]

I need help

[alpine sable]

don't ask to ask, just ask 😄

[lime echo]

Can someone help me

If I get a 0/50

On 9.2

What’s my grade

At least 60% right?

It’s worth 50

I just wonder if I get a 0

What would the grade be

It’s summative

So it’s worth 75%

[alpine sable]

ah it's 75%

I'm blind 😄

[worn fox]

Yeah assume all summatives are weighted the same in the 75%

[lime echo]

I should have a P right

.close

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[next patrol]

why is this bit done

[sour coral]

Large cone = 125/125
Small cone + frustum = Large cone
Small cone = Large cone - frustum

[lone heart]

@next patrol Has your question been resolved?

[mortal kelp]

i need help, my task is to use this very equation to get to the calculation of pi

i know this is a unit circle, but how do i get to an infinte series that equals pi

this is the upper half of a unit circle

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.reopen

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[tawdry stirrup]

Hi, can someone help me with this exercises?

[lone heart]

@tawdry stirrup Has your question been resolved?

[rugged sun]

So for exercise 1, what combinatorics rule will you use?

[lone heart]

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[thick lynx]

Can someone please check where I made a mistake?
\
Task: Find $\frac{x}{y}$ if $\frac{3}{\sqrt{y}} - \frac{1}{\sqrt{x}} = \frac{2}{\sqrt{x} + \sqrt{y}}$
\
\
$$\frac{3}{\sqrt{y}} - \frac{1}{\sqrt{x}} = \frac{2}{\sqrt{x} + \sqrt{y}}$$
$$\frac{3 * \frac{\sqrt{x}}{\sqrt{y}} + 3}{\sqrt{x} + \sqrt{y}} - \frac{\frac{\sqrt{x}}{\sqrt{y}} + 1}{\sqrt{x} + \sqrt{y}} = \frac{2}{\sqrt{x} + \sqrt{y}}$$
$$2 * \frac{\sqrt{x}}{\sqrt{y}} + 2 = 2$$

Thus x/y = 0

[ocean seal]

Kepe

[thick lynx]

The official solution is that x/y = 1/3

[mortal trellis]

what did you do on the LHS?

to get from $\frac{3}{\sqrt{y}}$ to $\frac{3\cdot\frac{\sqrt{x}}{\sqrt{y}}+3}{\sqrt{x}+\sqrt{y}}$ ?

[thick lynx]

I wanted to get $\sqrt{y}$ to $\sqrt{y} + \sqrt{x}$

[ocean seal]

Kepe

Denascite

[mortal trellis]

you can't just add something in the denominator

[thick lynx]

I know, I multiplied by ((sqrt(x)/sqrt(y)) + 1)/((sqrt(x)/sqrt(y)) + 1)

the squareroot y's cancel out so it remains sqrt(x)

1 * sqrt(y) stays sqrt(y)

so I get sqrt(x) + sqrt(y)

$\frac{3}{\sqrt{y}} * \frac{ \frac{\sqrt{x}}{\sqrt{y}} + 1}{\frac{\sqrt{x}}{\sqrt{y}} + 1 }$

[ocean seal]

Kepe

[mortal trellis]

ok but that doesn't work for the second fraction

you would have to multiply that by sqrt(y)/sqrt(x)+1

[thick lynx]

yes

oh, and I multiplied it by sqrt(x)/sqrt(y) + 1

[thick lynx]

thx!

.close

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[woeful current]

I mean vietta's theorem? right? but how

[fervent timber]

you know x1 + x2 and x1x2

[woeful current]

-3/8 and -1/3

[fervent timber]

yeah exactly

[woeful current]

and how do I aply that then?

I'm not allowed to find the roots

[fervent timber]

try to make 1 fraction maybe

for what you are calculating

[woeful current]

for what? wdym

[fervent timber]

what are you trying to calculate

[woeful current]

x1 / x2^2 + x2 / x1^2

which is (x2^3 + x1^3) / (x1^2 + x2^2)

which is also not that helpful hmmm'

[mortal trellis]

you made a mistake there

or typos

[woeful current]

where

[mortal trellis]

also you may need (x+y)^2=x^2+2xy+y^2

[woeful current]

oh right

wait what

um how do I get that

sure I can complete the square but then I'd need -2xy

which I have...

ooo

wait nope nvm

[mortal trellis]

you have the LHS and you have xy, so you can calculate x^2+y^2

also try to factor x^3+y^3 into (x+y)(something)

[woeful current]

yeah that gets (x^2 - xy + y^2)

but what do I do from there

also what about the denominator

[mortal trellis]

yeah the denominator is wrong

[woeful current]

common denominator..?

[fervent timber]

you sure about this?

[mortal trellis]

if you know x^2+y^2 and xy, you can calculate x^2-xy+y^2

[woeful current]

-8/3

have that on paper, typo

true

[mortal trellis]

well what is the common denominator

it is not x^2+y^2

[woeful current]

typo again its multiplied

[mortal trellis]

yes

and x^2y^2=(xy)^2

[woeful current]

oh yeahhhhhhh

[mortal trellis]

now you have all the puzzle pieces, only thing left is to combine everything

[woeful current]

yep

wait we completed the square for the right side of the numerator?

(x^2 -xy +y^2)?

added -xy

[mortal trellis]

I don't know where you completed a square

x^3+y^3=(x+y)(x^2-xy+y^2)

[woeful current]

yeah,

how do I get somethinfr om the right side of that

[mortal trellis]

you know x+y and xy, so using (x+y)^2=x^2+2xy+y^2 you can calculate x^2+y^2

[woeful current]

true

x^2 + y^2 = 70/9

so Its {(-8/3)(70/6 -1/3)} / 1/9)

which is incorrect damn

-272

I get it now

.close

[lone heart]

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[zenith birch]

how should i manipulate the left side for this to be equal

[lime bobcat]

I would manipulate both same time to reach an evident identity

[zenith birch]

yea that makes the most sense but I am being forced to solve only left side

[lime bobcat]

Really?

Who forces you?

[zenith birch]

my teacher

[lime bobcat]

PFF

Maybe add and substract 4tan²x and add and substeact -3...

I'm not sure

[median oar]

Usually you can’t touch RHS in proofs

At least not in high school

[lime bobcat]

One can by inyective maps

[zenith birch]

a what

[lime bobcat]

$4\tan^4 x+\tan^2 x - 3\$
$4\tan^4x-3\tan^2x+3\tan^2x+\tan^2x-3\$
$\tan^2x(4\tan^2-3)+4\tan^2x-3\$
$(\tan^2x+1)(4\tan^2x-3)\$
$\sec^2(4\tan^2x-3)$

[ocean seal]

jnkena

[lime bobcat]

@zenith birch, @median oar

[zenith birch]

oooooohhh

first step just expand and then it all falls into place

not really expand

[lime bobcat]

I added and substracted 3tan²x

[zenith birch]

is that why its called injective

[lime bobcat]

NO NO

[zenith birch]

lol

whys it called injective map

[lime bobcat]

Forget that haha, i was talking about acting both sides of equality

[zenith birch]

oh

[lime bobcat]

A function f(x) is inyective if for all x,y f(x)=f(y) → x=y

[zenith birch]

then how should i know if i need to do something like you just did

[lime bobcat]

So one can apply injective functions to both sides of equalities

[lime bobcat]

Mmm I wanted my expression to look like the right hand side

Since I had 4tan⁴x and I wanted 4tan²-3 and I saw there's like a common factor sec², I look for something similar to add and substract and then take common factor so i can get 4tan²-3, but tan⁴x has power 4 so a power ² would work and the -3 multiplying to have -3

[median oar]

Not the point, high school usually doesn’t let you even if it’s fine mathematically to do so

[lime bobcat]

Okay, but that's stupid. Maybe if the question is formulated differently, I could understand for example "apply equivalent transformations to convert this expression into this other expression" but not "prove A=B" (and prof. not allowing inyective transf. both sides)

[lone heart]

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[balmy bison]

How would I find the LCD properly

[median oar]

Do you see that the first one can be factorised

[zenith birch]

think of it like this

the numerator doesnt matter

[summer kite]

So basically finding factors?

[median oar]

It’s like 1/(2*3) + 1/(3*3)

[summer kite]

Wot is least common

[median oar]

And what’s the denominator for adding them

[median oar]

The least u have to multiply

[summer kite]

Got it

[balmy bison]

So I broke it down like this but the answer threw me off

[median oar]

The answer should be q²-36?

[balmy bison]

The answer comes out as 36q^2(q+6)^2

[median oar]

Oh it’s everything

[balmy bison]

so getting to that point had me confused and wanted to know how to do it properly

[lone heart]

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[lone heart]

@balmy bison Has your question been resolved?

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[whole wadi]

$a_{5}=55;\ a_{1}=3;\ a_{n+2}+a_{n+1}=a_{n}.\ a_{6}=?$

[ocean seal]

Theophania

[whole wadi]

Can anyone help me with this?

[merry depot]

are you sure it isn't $a_{n-2}+a_{n-1}=a_{n}$?

[ocean seal]

Zybikron

[lone heart]

@whole wadi Has your question been resolved?

[whole wadi]

@merry depot Yes

[lone heart]

@whole wadi Has your question been resolved?

[lone heart]

@whole wadi Has your question been resolved?

[lone heart]

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[astral star]

Can someone show me how to do these

[cedar field]

what exactly do you want to know?

[median oar]

,rotate

[astral star]

I just got this back and I took the test in like April so it’s been a long time

I didn’t really remember much

<@&286206848099549185>

[astral star]

I want to know what to do so I can solve it

[wary stream]

I mean you follow your work. Part a, you have a 1.2 oz box and you want to make it 4 times bigger

Part b, you want a 1.2 oz box to hold 18.75 oz, so you need to determine "what" times 1.2 = 18.75

[cedar field]

you already have the answers.

[astral star]

Then y I get 1/3

[cedar field]

oh it was from a test

[wary stream]

Because probably not enough explanation

On how to get those answers

[astral star]

I mean how else will I show it

[wary stream]

Explain your work

[astral star]

I did

I’ll just talk to my teacher tomorrow

Thanks for helping

.close

[lone heart]

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[wary stream]

Not really, you should proof that you got the answer but you didn't explain why you did that process

[lone heart]

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[loud rover]

Hey I got confused by this substitution and got nowhere with it, I know you can solve it easily by defining u = 2x and get (arcsin2x)/2 but is there a way to use the given substitution?

[raven rover]

Well, find dx, and then plug into the integral x and dx

@loud rover

[loud rover]

im getting nowhere

[near hollow]

$\int \frac {1}{\sqrt{1-4(sin^2u/4) }}dx = \int \frac {1}{\sqrt{1-(sin^2u) }}dx$

[raven rover]

It's just the derivative of x

if x = (your substitution), then dx = ?

[loud rover]

ik but when i substitute it in it looks weird

[ocean seal]

Max..

[loud rover]

oh shit

[near hollow]

Keep in mind the variable is now u and we're integrating with respect to x

So we need to do as Shen said and differentiate to get rid of dx and replace it with du

[near hollow]

Do you understand how it got to this point?

[loud rover]

ye

[near hollow]

What does the denominator simplify to?

[loud rover]

¯_(ツ)_/¯

[near hollow]

Any trig identities you recognise?

[raven rover]

All about trig

[near hollow]

Facts

[loud rover]

yea ik derivative of arcsin

[near hollow]

No no I mean this

The denominator

[loud rover]

OH

im dumb

cosx

[near hollow]

Good knowledge

[near hollow]

Cosu but yeah

So $\int \frac{1}{cosu}dx$

[loud rover]

ye mb

[ocean seal]

Max..

[near hollow]

All that is left is getting rid of dx

Just differentiate the substitution. Getting it in terms of dx (using chain rule) and sub out dx

Should be a very very very simple integral from there

[loud rover]

I have no idea wtf im doing

this shouldnt be taking this long

im just gonna move on

[raven rover]

You just needed to find dx

Knowing that x = sin(u)/2

If you know the derivative of sin(u)/2, then you're good

[woven wraith]

Hm thats missing in here

[loud rover]

what is

[raven rover]

You need dx in terms of du

$x = \frac{\sin{u}}{2}$

$dx = ?$

[ocean seal]

Shen

[raven rover]

@loud rover ^^

[loud rover]

( u'cosu du ) / 2

is it not

[raven rover]

No u'

Just the other stuff, yup

[loud rover]

o

🗿

fuck i have a calc exam in 4.5 hours and i cant differentiate sinu /2

ty

[raven rover]

Np np

Now you plug that expression into the integral and proceed

What do you get?

[loud rover]

i did everything right except the dx/du lol

.close

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[supple geode]

i need some help understanding why (2) to (3) works ?

[supple geode]

i dont really know how to simplify from (2) to (3)

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[supple geode]

<@&286206848099549185>

[supple geode]

only thing I can think of is for the term on left look at when j = k, and for the term on the right look at when i = k

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[alpine sable]

I have some homework questions I need some help with

[simple current]

post them

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[tawny relic]

part d

[alpine sable]

If you are working in Real numbers (not complex*), then (2x +1) >=0.
Solve for x: x >= -1/2.

• From part c, I derive that you haven't learned about complex numbers yet. sqrt(-1)=i.

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@tawny relic Has your question been resolved?

[carmine oar]

Or for a more rigorous but complex way

k’(x)

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[fervent cloak]

why cant u simplify this

[gray ingot]

how do you want to simplify this?

[fervent cloak]

gcf of 3?

idk bc

a gcf was pulled for this one

jw why it doesn't apply to the first one i sent

[gray ingot]

you can't factor 3x - 1

well you could

3(x-1/3)

[lime bobcat]

Because $6\frac{\sqrt{3x}-1}{3x-1}$ so numerator and denominator are not equal, neither have they factors in common and denominator has no roots

[fervent cloak]

o so if there's a variable

[ocean seal]

jnkena

[fervent cloak]

it'd just leavei t

?

so in which instances

would i pull a gcf from a radical with

a variable

if ever

[median oar]

i would argue that this is easier to graph

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[buoyant sage]

Hi, I have been struggling with this problem. I have done it from scratch a day after originally attempting it without looking at my work and still got the same incorrect answer. Is there anyone who knows how to do this? I could send work but its pretty messy.

[lone heart]

@buoyant sage Has your question been resolved?

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@buoyant sage Has your question been resolved?

[balmy warren]

@buoyant sage is this some sort of differential?

[buoyant sage]

yea its differential equations

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@buoyant sage Has your question been resolved?

[vale wigeon]

this person has left

.close

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[lone heart]

Please don't occupy multiple help channels.

[balmy warren]

what is the psi function?

[worn fox]

Eulers totient function

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[south imp]

A is a square matrix of order 3 with IAI=9, then find the value of I2 adj AI

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[lone heart]

@south imp Has your question been resolved?

[lime bobcat]

$A^{-1} =\frac{1}{|A|}\left(\text{Adj}(A)\right)^T$

[ocean seal]

jnkena

[lime bobcat]

Does this help you, @south imp?

[south imp]

nah

[lime bobcat]

Apply determinant both sides and use the fact determinant is a morphism of rings

$||A|\cdot I_n\cdot A^{-1}| =|(\text{Adj}(A))^T|$

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[ocean seal]

jnkena

[lime bobcat]

$\left|\text{Adj}(A)\right|=|A|^{n-1}$

[ocean seal]

jnkena

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[spare vine]

If $\langle x \rangle$ is the group generated by x, and x has order p, then why is it true that:
$$\langle x \rangle y^p = (\langle x \rangle y)^p$$

[ocean seal]

ryep

[placid zinc]

The right doesn't parse for me. <x> is a group, so <x>y can be considered a set. But then what's (<x>y)^p?

[spare vine]

I believe it's multiplying the set <x>y with <x>y, p times.

Similar to how set multiplication is defined in quotient groups "i think"

[placid zinc]

If we can put a group structure on <x>y then that makes sense. Maybe I don't know some common way this is done. I'll let someone with a better head in

[broken hearth]

use gauthmath or mathway

[spare vine]

lol

[broken hearth]

but I recommend gauthmath because your just gonna scan the problem

[spare vine]

i didn't know they supported group theory

[broken hearth]

its just my opinion 🙂

[placid zinc]

Yeah, wow that must be a powerful app

[broken hearth]

ahm yea ig

most highschool and elementary students use that

[spare vine]

Hmm maybe i should try

[broken hearth]

when we are confused

[placid zinc]

And apparently a lot of pure math university students too!

[broken hearth]

since it gives a solution and answer

yeaa

[spare vine]

Interesting...

[broken hearth]

you guys should try

tbh sometimes math is sometimes the reason I wanna end my life

[spare vine]

Thats not good

[broken hearth]

ikr

[spare vine]

I recommend maybe taking a break

[broken hearth]

I've been

taking a break

but I can't stop thinking abt it

[spare vine]

About what?

[broken hearth]

math

[spare vine]

What kind of math?

[broken hearth]

I can't teak a break until I solve it

[spare vine]

I mean i guess thats good, but you want to find math that you enjoy doing, or is fun. And not something a teacher or someone has told you to do.

[broken hearth]

I really dislike math

it gives me headache

[spare vine]

That's probably because you've only done ones which people have made you do.

[broken hearth]

yea..

[spare vine]

It's a lot more fun when you just make up problems or do whatever you want to do with it.

[broken hearth]

I don't understand math haha

like whenever we have class and its math time it makes me want to sleep

[spare vine]

Yeah math classes are boring

[broken hearth]

that's also the reason why I don't understand math

[spare vine]

Neither did I

[broken hearth]

oh ok ok what grade are you in?

[spare vine]

:^)

[broken hearth]

huh?

[spare vine]

We've kinda gone off topic ig

[broken hearth]

I mean if its ok to ask

[spare vine]

It's been nice chatting though

[broken hearth]

yea

[spare vine]

I mean I came here to get help with a math problem

[broken hearth]

sorry sorry

[spare vine]

Good luck with your studies though

[broken hearth]

🙂

[spare vine]

Let $x$ have order $p$, $y$ have order $2$, and $xy$ have order $p$

Then $(xy)^p = e$

$\langle x\rangle = \langle x\rangle (xy)^p = (\langle x\rangle xy)^p = (\langle x\rangle y)^p = \langle x \rangle y^p = \langle x\rangle y$

A contradiction - "Groups and Symmetry" by M.A Armstrong.

The problem is how is:
$$(\langle x \rangle y)^p = \langle x \rangle y^p$$
(same with the earlier part with xy.)

[ocean seal]

ryep

ryep

[spare vine]

Ok stuff this, lets stick with p = 3

Let $ord(x) = 3, ord(y) = 2, ord(xy) = 3$

Then $\langle x \rangle = { e, x, x^2 }$
and $\langle x \rangle y = { y, xy, x^2y }$

If so then:

$$\langle x \rangle y^p = { y^p, xy^p, x^2y^p } = {y, xy, x^2y}$$

Now $$\langle x \rangle \cdot \langle x \rangle = {
ee, ex, ex^2,
xe, xx, x^2x
x^2e, x^2x, x^2 x^2
} = {e, x, x^2}$$

[ocean seal]

ryep

[spare vine]

Ok this means:
$$\langle x \rangle y \cdot \langle x \rangle y = {
yy, yxy, yx^2y,
xyy, xyxy, xyx^2y,
x^2yy, x^2yxy, x^2yx^2y
}$$

Ok this doesn't make sense

[ocean seal]

ryep

[spare vine]

$xy \in \langle x \rangle y$, then $(xy)^3 = e \in (\langle x \rangle y)^3$

However $\langle x \rangle y^3 = { y, xy, x^2y }$

[ocean seal]

ryep

ryep

[spare vine]

Ok so the author of the book seems wrong...

Let me clean this

Let $x$ have order $p$, $y$ have order $2$, and $xy$ have order $p$

Then $(xy)^p = e$

$$\langle x\rangle = \langle x\rangle (xy)^p = (\langle x\rangle xy)^p = (\langle x\rangle y)^p = \langle x \rangle y^p = \langle x\rangle y$$

A contradiction - "Groups and Symmetry" by M.A Armstrong.

I believe the proof is flawed since, if we let $p = 3$ then $(xy)^3$ = e:

$$ \langle x \rangle = {e, x, x^2}$$
$$ \langle x \rangle y^p = \langle x \rangle y = {y, xy, x^2 y }$$

Notice $xy \in \langle x \rangle y$, and $e \notin \langle x \rangle y^3$. Now:

$$xy xy \in (\langle x \rangle y)^2 \implies (xy)^3 = e \in (\langle x \rangle y)^3$$

So $e \in (\langle x \rangle y)^3$ and $e \notin \langle x \rangle y^3$. Therefore:

$$\langle x \rangle y^3 \neq (\langle x \rangle y)^3$$

[ocean seal]

ryep

[spare vine]

My best guess is that I've misunderstood what they mean by $(\langle x \rangle y)^3$.

[ocean seal]

ryep

[spare vine]

$Hx \cdot Hy = Hxy$

[ocean seal]

ryep

[spare vine]

Ok let me fix this

[clear stump]

Are you trying to write a book here

[spare vine]

lol sorry I was trying to understand what a book was saying

[clear stump]

oh

good luck

[spare vine]

I thought the author was wrong, but they weren't

[clear stump]

they smart

[spare vine]

im just dumb lol

[clear stump]

I have no idea what thise even mean

[spare vine]

moon runes

[clear stump]

yeah

[spare vine]

Let x have order p, y have order 2 and xy have order p, and $(xy)^p = e$

Let $H = \langle x \rangle$ be a normal subgroup.

Then $H = H(xy)^p$, recall since $Hx \cdot Hy = Hxy$, then $$Hxy \cdot Hxy = H(xy)^2$$

Therefore $$H = H(xy)^p = (Hxy)^p = (Hy)^p = Hy^p = Hy$$

Addressing the earlier concern, we let $p = 3$. Then indeed $(xy)^3 = e$

$$ \langle x \rangle = {e, x, x^2}$$
$$ \langle x \rangle y^p = \langle x \rangle y = {y, xy, x^2 y }$$

And now: $g \in \langle x \rangle y \cdot \langle \rangle y $, should have form:

$$g = (x^n)y (x^m) y = (x^n)y^{-1} (x^m) y$$

Since $\langle x \rangle$ is normal, $y^{-1} (x^m) y \in \langle x \rangle$. This means:

$$g = (x^n)y^{-1} (x^m) y = (x^n) (x^m)$$

Therefore $e \in (\langle x \rangle y)^3$

[ocean seal]

ryep

[spare vine]

Ok fuck it whatever, I think I understand it now

.close

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[alpine sable]

Hi, i will be doing exam in the next week and i need to find some topic. The task is based on limits and how to overcome them, can you help me in finding this argument related to math and physics? I don't want to tackle limits in math cause all of my classmates do this... THX a lot

[clear stump]

prove that 1+1≠2

[alpine kiln]

What do you mean how to overcome them

[alpine sable]

it is the trace assigned by the teacher

[alpine kiln]

?

[alpine sable]

for example in english i speak about joyce and his paralysis

however it is not fundamental

[alpine kiln]

Sorry I’m really confused

[alpine sable]

in physics i wanted to speak about relativity

[alpine kiln]

Oh

Sry idk

[alpine sable]

no problem

[alpine sable]

could be an interesting topic

[clear stump]

yes

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[north scaffold]

the function is e^-alpha into t , after plugging this in some intergral (laplace transform) which now contains (e^alpha into t ) * (e^-st) cant i interchange the e^-st and e^aplhat places
like (e^-st ) (e^alpha into t)

[worn fox]

You can yes

Multiplication is commutative

[north scaffold]

so now i can write e^t into e^a-s

e^a-s just const i take out that

can i write this as e^(a-s) and e^t

@worn fox

[placid zinc]

"into" is awkward here

(e^x)(e^y) = e^(x+y)

[native cloud]

You mean like this?

$$\int _0^{\left(a-s\right)n}\frac{e^u}{a-s}du$$

[ocean seal]

AirToastie

[placid zinc]

I think you may be trying to multiply the exponents, which is no good

[native cloud]

Take the constant out and use the common integral

[north scaffold]

np thanks

.close

.close

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[runic gate]

How to do 22.b

22a given

[winter tangle]

Substitute pi and 0 then find the difference

[runic gate]

咁我都知😂

By part a la ofc

[stone glen]

Is ur question solved?

[runic gate]

Nope

[winter tangle]

which specific part are you struggling with then?

[runic gate]

The first step

Idk what the first step need to do

[winter tangle]

erm

where there are x, input a pi

and then subtract the same for 0

[runic gate]

Emmm you need to do the integrate first

[winter tangle]

you already done that in part a

[runic gate]

what ?It just a prove

[winter tangle]

oh oops

okay

do you know any of the methods of integration?

[runic gate]

Yes but the problem is what is the first step to do

To make use of part a

[winter tangle]

what are the methods of integration?

what?

[runic gate]

Substitution by part trigo ...

U mean??

[winter tangle]

what?

[runic gate]

Wdym

Anyways I found sby to help wth

Thx

. close

.close

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[junior obsidian]

I have used a python program to brute force close solutions, for example at x = 10, the amount of pours is around 27 needed for a very close 1:1 ratio

at number 27 on the left here , the next two numbers are the ratios in either cup, and the four numbers on the row below are the amount in either cup starting with y in cup A, z in cup A, y in cup B, z in cup B

<@&286206848099549185>

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@junior obsidian Has your question been resolved?

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@junior obsidian Has your question been resolved?

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[strange jay]

help

[native cloud]

What number

[strange jay]

2

[native cloud]

What sequence should we be using?

Arithmetic or Geometric

?

[strange jay]

ARITHMETIC

sorry caps on

[native cloud]

Should we be using explicit or recursive?

[strange jay]

idk

[native cloud]

First of all, what is the common difference

And what is the first term?

[strange jay]

the first term is 777

and the common diff is 5

[raven dagger]

Divide by 5
The remainder is the answer

Because subtracting 5 from that will give negative

So answer is 2

[native cloud]

Oooo easy way

Noice

I don't think the teacher wants he or she to solve it like that though

[strange jay]

okay ty

can someone help me with question 3 and 4?

[native cloud]

Just do f(n) = a + d(n - 1)

[strange jay]

can u tell me what the varibles mean?

[lone heart]

@strange jay Has your question been resolved?

[strange jay]

ye

nope not anymore

I need help with question 3

arithmetic sequence with a common diff of 3 1/3 and the first term is 4

what are the first 4 terms

[native cloud]

What's 'a' here

[lone heart]

@strange jay Has your question been resolved?

[strange jay]

idk u wrote it??

[lone heart]

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[native cloud]

smh

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[tame hound]

$$\frac{\frac{d}{dx} [sin(x)]\times cos(x)-sin(x)\times \frac{d}{dx} [cos(x)]}{cos^{2}x}$$ $$=\frac{cos^{2}x-sinx\times -sinx}{cos^{2}x}$$
$$=-sin^{2}x$$

[ocean seal]

AuHasard

[tame hound]

I'm having a bit of a brain fart, but can someone explain why we can't do this?

[tacit arch]

can't even tell what you're trying to do

[tame hound]

simplifying

d/dx tan(x) is what i have to do

[tacit arch]

oh

this is right

no idea how you got the next line

[tame hound]

$\frac{cos^{2}x-sin^{2}x}{cos^{2}x} =\frac{cosx\times cosx-sinx\times sinx}{cosx\times cosx}$

[ocean seal]

AuHasard

[tacit arch]

nope

[tame hound]

the cos cancels?

why

[tacit arch]

you lost a minus sign

[tacit arch]

the first fraction here is already wrong

[tame hound]

$\frac{cos^{2}x+sin^{2}x}{cos^{2}x} =\frac{cosx\times cosx-sinx\times -sinx}{cosx\times cosx}$

no

give me a minute lel

[void niche]

you seem to also want to do $\frac{a+b}{a} = b$, which is not true (generally)

[ocean seal]

Migillope

[tame hound]

$\begin{gathered}1.\ y=tan(x)\ \ \ 2.\ \frac{d}{dx} \left( \frac{sinx}{cosx} \right) \ 3.\frac{\frac{d}{dx} [sin(x)]\times cos(x)-sin(x)\times \frac{d}{dx} [cos(x)]}{cos^{2}x} \ 4.\ \frac{cos(x)\times cos(x)-sin(x)\times (-sin(x))}{cos^{2}x} \ 5.\ \frac{cos^{2}x+sin^{2}x}{cos^{2}x} \ 6.\ cos^{2}x\ cancels\Rightarrow we\ get\ sin^{2}x\ \ \ \end{gathered}$

[tacit arch]

like i said, you lost a minus sign

(-a) * (-a) != -a^2

[ocean seal]

AuHasard

[tame hound]

but this is wrong and i'm not sure why

[tacit arch]

then it's this

[tame hound]

ah

makes sense

this is a common error, i've been told

[native cloud]

Wait, how did you get the second line?

[tame hound]

tanx=sinx/cosx

correct me if i'm wrong

$\frac{a+b}{b} \neq \frac{a}{b} +\frac{b}{b}$

[ocean seal]

AuHasard

[void niche]

this should be equality

[tame hound]

oh ok

$\frac{a}{a+b} \neq \frac{a}{a} +\frac{a}{b}$

[ocean seal]

AuHasard

[tame hound]

so it's this right?

[void niche]

generally

[tame hound]

but like

[native cloud]

((cos(x) * cos(x)) - (sin(x) * -sin(x)))/cos^2(x)

[tame hound]

$\frac{a\times a+b\times b}{a\times a}$

[ocean seal]

AuHasard

[tame hound]

we can't eliminate the a's?

[summer kite]

Quotient rule fun

[native cloud]

It's ew

[void niche]

use this

(with equality)

[tame hound]

ok one sec

[native cloud]

What question are you solving?

[summer kite]

Isn't it sec^2 x?

[tame hound]

$\frac{cos^{2}x+sin^{2}x}{cos^{2}x} =\frac{cos^{2}x}{cos^{2}x} +\frac{sin^{2}x}{cos^{2}x} =1+tan^{2}x$

[ocean seal]

AuHasard

[tame hound]

so this is sec^2(x)?

[summer kite]

Numerator is 1 btw

Instead of breaking the stuff down

[void niche]

yes that's an identity

[native cloud]

Where did sec^2(x) come from?

[summer kite]

The identity thing

[tame hound]

https://www.derivative-calculator.net/

[summer kite]

Bc numerator is 1, 1/cos = sec

[tame hound]

you can check the steps here if you want

ok thanks

[native cloud]

Ohhhh

Right

The numerator equals 1

But why doesn't his work also work?

[native cloud]

Shouldn't this be also correct?

[tame hound]

yes you're right abt that

cos^2x + sin^2x = 1

so it's just 1/cos^2x = sec^2x

[native cloud]

Yeah, but your previous work process also makes sense

[tame hound]

use this identity #help-0 message

[worn fox]

1+tan^2(x) = sec^2(x)

That's why it works the other way

[tame hound]

i see

[worn fox]

Divide this by cos^2(x) to see that

[native cloud]

Oh, so 1 + tan^2(x) can be turned to sec^2(x)?

[tame hound]

i did here

ooh you mean that

[summer kite]

Yes it's also an identity

[native cloud]

Oh, why didn't I learned that b4

[worn fox]

Divide this by sin^2(x) to get another handy identity

[tame hound]

like that?

$\frac{cos^{2}x+sin^{2}x}{sin^{2}x} =\frac{cos^{2}x}{sin^{2}x} +\frac{sin^{2}x}{sin^{2}x} =cot^{2}x+1=cosec^{2}x$

[ocean seal]

AuHasard

[tame hound]

that?

[tacit arch]

👏🏻

[tame hound]

noice

.close

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[native cloud]

I'm quite confused on how you got cot^2(x) + 1 = csc^2(x)

[summer kite]

Steps are mentioned above

[native cloud]

Sorry, but still lost

[summer kite]

In which part you don't get it

Am on mobile so can't type all those steps

[native cloud]

I get that 1/sin^2(x) = csc^2(x) but

1 + cot^(x) = csc^2(x)

[summer kite]

It's basically the sin^2 + cos^2 equation

But both sides divide by sin^2

[native cloud]

so csc^2(x) + sec^2(x) = 1?

[summer kite]

Wot

No

Oh Bruh what stupid shit I just said

[native cloud]

I..iii

confused

[summer kite]

About what

[native cloud]

Oh nvm

I'm stupid

Thanks

[worn fox]

Start with the top equation, divide both sides of it by sin^2(x), you get the bottom equation

[native cloud]

👍

Yeah since both sin^2(x) cancels to 1

and cos^2(x)/sin^2(x) = cot^(x)

Gotcha

.close

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[tame hound]

A drop of water falls on a completely smooth water surface. Waves are formed in all directions that spread circularly. The radius of a circle that spreads increases at a certain time by 1 meter per second. Just then the radius is 20 meters. How fast does the area of ​​the circle increase at that time?

[tame hound]

$\frac{dr}{dt} =1,\ \frac{dA}{dt} =?,\ ...$

[ocean seal]

AuHasard

[lime bobcat]

Yes

[tame hound]

I don't know what's missing here

[lime bobcat]

You has to use the fact the radius is just 20 m

[tame hound]

Yes.

area = πr^2

[lime bobcat]

Yes

A(t)=πr(t)²

[tame hound]

so we want to do the derivative of π20(t)^2?

what's t here, i'm not sure

[lime bobcat]

A'(t) = 2πr(t)r'(t)

We know for the current time t_0 we have that r(t_0)=20 m so

A'(t_0) = 2π·20·1 = 40π

I think that's the solution.

At that time the rate of change is 40π

[tame hound]

What are the steps to solve these kinds of problems?

$$\frac{d}{dt} \left[ A\right] =\frac{d}{dt} \left[ \pi r^{2}\right]$$
$$\frac{dA}{dt} =2\pi r\frac{dr}{dt} =2\pi r(1)=2\pi r$$

[ocean seal]

AuHasard

[tame hound]

I've understood until this part

related rates is some difficult stuff

[whole wadi]

Solving a related rates problem.
(i) Sketch a diagram showing the ongoing situation and label
relevant quantities.
(ii) Express the given information and the quantity to be
found using symbols.
(iii) Write an equation expressing the relationship between the
quantities.
(iv) Use implicit differentiation to find the desired derivative.
(v) If required, evaluate the derivative at the specified
value(s).

[tame hound]

Thanks.

So, I think I have to find the the rate of change of area dA/t at radius r = 20 meter.

So, I should plug in r = 20 and calculate dA/dt = 2π(20)?

So we multiply the circumference of the circle by the rate of which the radius of the circle is increasing (1m/s), then plug in r = 20?

[tame hound]

Here, we're given A drop of water falls on a smooth water surface. So waves are formed in all direction. Here, the radius of circle is increasing with time, which means the radius is a function of time.

[tame hound]

(This is my attempt at sketching it)

Continuing: And the function of t is represent in notation r(t).

As radius change with time 't', the area of circle also change with time 't'. This means area function is A(t).

[lime bobcat]

Beautiful representation

[tame hound]

@lime bobcat dA/dt = dA/dr * dr/dt, so dA/dt = 40π * 1 = 40π m^2/s

(Where 20 is r, and then you've the dr/dt that's 1m/s, so we get m*m = m^2)

Is that right?

[lime bobcat]

yes

But I'd write it better notationally, specifying the point where the derivatives are calculated

[tame hound]

Where would you say needs to be touched upon

[lime bobcat]

Sorry, I don't understand that sentence

[tame hound]

Where do you think need to be written better notationally

And which point needs to be specified where the derivatives are calculated?

[lime bobcat]

Ah

[tame hound]

$$A=\pi r^{2}$$
$$\frac{dA}{dt} =\frac{dA}{dr} \times \frac{dr}{dt}$$
$$\frac{dr}{dt} =1\frac{m}{s}$$
$$A=\pi r^{2}\Rightarrow A^{\prime }(r)=2\pi r$$
$$r=20\Rightarrow \frac{dA}{dr} =2\pi \times 20=40\pi$$

[ocean seal]

AuHasard

[tame hound]

I don't know if this is what you had in mind

$$\frac{dr}{dt} =1\frac{m}{s} \Rightarrow \frac{dA}{dt} =40\pi (m)\times 1(\frac{m}{s} )=40\pi \frac{m^{2}}{s}$$

[ocean seal]

AuHasard

[lime bobcat]

From $A(t)=πr(t)^2 m^2$ we deduce
$$\frac{dA}{dt}(t)=2πr(t)\frac{dr}{dt(t)} \frac{m^2}{s}.$$

Now we take $t_0$ such that $r(t_0)=20 m$ and $\frac{dr}{dt}(t_0)=1 \frac{m}{s}$, we have that $$\frac{dA}{dt}(t_0)=2πr(t_0)\frac{dr}{dt}(t_0) = 2\pi\cdot 20 m \cdot 1 \frac{m}{s}= 40\pi \frac{m^2}{s}.$$

[lime bobcat]

This is waaaay better

[tame hound]

Lol what's with all the edits

[ocean seal]

jnkena

[lime bobcat]

The last I promise

HAHA

[tame hound]

😂

[lime bobcat]

I wanted it to look cool

And still have a typo

can I correct?

[tame hound]

Sure

[lime bobcat]

From $A(t)=πr(t)^2 m^2$ we deduce
$$\frac{dA}{dt}(t)=2πr(t)\frac{dr}{dt}(t) \frac{m^2}{s}.$$

Now we take $t_0$ such that $r(t_0)=20 m$ and $\frac{dr}{dt}(t_0)=1 \frac{m}{s}$, we have that $$\frac{dA}{dt}(t_0)=2πr(t_0)\frac{dr}{dt}(t_0) = 2\pi\cdot 20 m \cdot 1 \frac{m}{s}= 40\pi \frac{m^2}{s}.$$

[ocean seal]

jnkena

[lime bobcat]

I copied instead

Now it is

[tame hound]

I just don't get why we introduce t when we can do it my way (with one less variable)

Here, r is a function of t, r(t). Yes?

[lime bobcat]

Yes

The radius is a function of the time

[tame hound]

So, as time change, the radius also change

Yes

[lime bobcat]

Yess

[tame hound]

Then we have r = 20m

[lime bobcat]

yes

The current instant t_0 is such that r(t_0)=20m yess

[tame hound]

If it's r(t), why don't we just write r(0) = 20m?

[lime bobcat]

Yes, you could do that, the t_0 does not matter

But it is the current instant of time

You could avoid expliciting it

I personally like expressing all dependencies on variables, but an engineering might write something like A'=2πrr' with r=20 and r'=1

[tame hound]

Is it possible to write dr/dt (0) there, or would it be not the same thing as t_0?

[lime bobcat]

But mathematically I like to write t and t_0 because for me r' is a function, not the derivate of r at an instant

[tame hound]

When I think of r(0), I think it represents the radius at time 0, which means radius when the drop is touched to water surface.

[lime bobcat]

You can write 0. I prefer t_0 because if you write 0 you're implying negative time because the wave is increasing from an initial position and you're calculating at the current one

But all works the same

No problem

[tame hound]

Oh you mean 0 implies that the wave grows from negative to current?

Sorry if I misinterpret

[lime bobcat]

Yes, from negative time to time 0

which is current

[lime bobcat]

You yourself said that

0 makes you think on the first instant

[tame hound]

Okay

[lime bobcat]

so if you take 0 as current the first instant would be negative

[tame hound]

I think t_0 would be better then

I was under the impression that it's also fine to not use any subscript notation, maybe

[lime bobcat]

in my opinion yes, but i insist this is not too important

[lime bobcat]

Yes, maybe another letter

Because t_0 sometimes means the starting point

[tame hound]

Okay thanks, I will try other problems and see if I still got this.

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[amber adder]

is the answer (14sin(7t)) / (12t^2 + 8t)

[lone heart]

@amber adder Has your question been resolved?

[amber adder]

please help

how do I know what's my lower limit and upper limit

[lone heart]

@amber adder Has your question been resolved?

[lone heart]

@amber adder Has your question been resolved?

[abstract fractal]

Well, you know it starts and ends when r = 0. You can also tell that its the smallest positive angle and the largest negative angle that gives you that area. What values of θ give you that?

[lone heart]

@amber adder Has your question been resolved?

[lone heart]

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[magic terrace]

So I did the steps I thought I was supposed to do

1. Find the different variables involved

V = Volume
C = Surface area (Cost)
r = Radius
h = Height

2. Find which variable needs to be optimized and any constraints

C needs to be optimized, and V is a constraint (250)

3. Create equation definitions for all relevant variables using as few variables as possible in each equation

$V=h\pi{r^2}$

[vale wigeon]

space after \pi

[ocean seal]

Eyesonjune

[vale wigeon]

...or that

[magic terrace]

thx

$C=2\pi{r}h(0.02)+2\pi{r^2}(0.07)$

[ocean seal]

Eyesonjune

[magic terrace]

I can find the definitions of r and h by solving the volume equation for each variable

$r=sqrt(\frac{V}{\pi{h}})$

[vale wigeon]

$r = \sqrt{\frac{V}{\pi h}}$, but this is extraneous and will make your life difficult.

[ocean seal]

Ann

[magic terrace]

$h=\frac{V}{\pi{r^2}}$

[vale wigeon]

better to have $h = \frac{V}{\pi r^2}$ and express everything in terms of the radius, that way you're not dealing with square roots

[ocean seal]

Ann

[vale wigeon]

also capital V not lowercase v. be consistent.

[magic terrace]

that is what I did

[ocean seal]

Eyesonjune

Eyesonjune

[vale wigeon]

and what'd you do afterward?

[magic terrace]

ok

Now I take the equation I'm trying to optimize

Well I'm gonna write the step formally

[vale wigeon]

$C = 0.04 \pi rh + 0.14 \pi r^2$

[ocean seal]

Ann

[magic terrace]

this is mostly for my own purposes because I've not written them out explicitly before

4. Now take the equation that needs to be optimized and plug in one of the definitions you found, also taking care to plug in any constraint values

So I used h

$C=2\pi{r\frac{250}{\pi{r^2}}}(0.02)+2\pi{r^2}(0.07)$

lemme edit that rq

[ocean seal]

Eyesonjune

[magic terrace]

Some algebra later

$C=0.1r+0.14\pi{r^2}$

[vale wigeon]

bad

[magic terrace]

why is that

[vale wigeon]

please do not write decimals without the leading zero

it makes my eyes bleed

[magic terrace]

🙄

[vale wigeon]

also you are just ASKING for a chance to lose those decimal points accidentally

[ocean seal]

Eyesonjune

[magic terrace]

Ok 1950s math teacher

[vale wigeon]

also you screwed up