help-49

[midnight plank]

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[tidal turret]

why does we need to multiply by the jacobian after changing from cartersian coordinates to spherical coordinates, in the context of triple integrals

[unborn stone]

Did you study linear algebra ?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

yes

[unborn stone]

When you're changing from cartesian coordinates to spherical coordinates

You're operation a transformation T from R^3 to R^3

from a base A to a base B

that mean that the vector u expressed in A is equal to M * u where M is the matrix representation of this transformation

well it's a concept somehow similar with triple integrals

because in the end an integral is a volume/surfaces/etc expressed in some base

when you're changing from coordinates types that is a transformation

meaning it'll imply some kind of matrix ^^

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[cerulean oyster]

if you remember u-sub

where you could either substitute in, or add to the previous equation the derivative of the function you chose as representative

then the jacobian is basically the same concept.

[tidal turret]

care to elaborate?

[cerulean oyster]

since ln(e^u) is an inverse function composition

here e^u is technically the jacobian determinant of the x -> e^u transformation

Basically is something we add to make the resulting function make sense

In this particular case we didnt bound the function to solve for the numerical value of the integral, but we could

[tidal turret]

.solved

[midnight plank]

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[uncut rapids]

.

,rotate

[uncut rapids]

if alpha is the root of the equation z^2+z+1 = 0

specify which of the following determinants is a real number

how should i approach it ?

[quiet hinge]

Which determinant are you having trouble with?

[uncut rapids]

so i just calculate the determinants for each ? or how should i approach it

as it is a new problem

[quiet hinge]

For the 2x2 ones, you can just expand them and use z^2+z+1=0

[uncut rapids]

wdym expand

[quiet hinge]

Calculate

For the 3x3 one, do you know that the determinant is unchanged if you add two rows? If yes then use that

[uncut rapids]

so i simply calculate the determinants until i get a real number?

[quiet hinge]

You don't know what you'll get- real or imaginary

What I'm saying is for 2x2 you just simply calculate the determinant (use det{{a,b},{c,d}}=ad-bc)

[quiet hinge]

For the 3x3 one you use this property

[uncut rapids]

for a i get 2-alpha^3 but alpha^3 - 1 = 0 thus alpha^3 = 1

so i get 2-1

[quiet hinge]

What did you get for b

[uncut rapids]

-1

for c

should i work with lines and columns ?

[quiet hinge]

Any way

Rows or columns doesn't matter since the matrix is symmetric

[quiet hinge]

Use this property

It will be much easier than calculating the determinant

[uncut rapids]

i ve used sarrus rule

it gave me 3-3

[quiet hinge]

Well 0 is correct though idk what sarrus rule is

[uncut rapids]

i mean you calculate it without adding the rows under

[quiet hinge]

If you simply add the first row to the third row and second row to the third row you get a row of 0s which immediately tells you the determinant is 0

[uncut rapids]

yeah that works

i have a question for 127

so a b c e R

show that if the matix = 0

then a = b = c

what do you say would be a better approach?

as i got nowhere

[midnight plank]

@uncut rapids Has your question been resolved?

[midnight plank]

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[midnight plank]

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[sudden tree]

Set b

[kindred bone]

hello

[sudden tree]

Hi

[kindred bone]

kiya hal hai

[sudden tree]

Bohot badhiya

Happiest person alive

[kindred bone]

kala chasma kidher hai

[sudden tree]

Pehen rakha h

Can someone help?

[sudden tree]

.

[next bobcat]

Tying to solve

[midnight plank]

@sudden tree Has your question been resolved?

[sudden tree]

<@&286206848099549185>

[next bobcat]

so x and y lie between 0 to 180 degrees

sinx and siny lie between 0 and 1

cosx and cosy lie between -1 and 1

[sudden tree]

Yes

[next bobcat]

in A

[sudden tree]

A I have already solved

[next bobcat]

find the maxima and minima of each term

Ok so do the same in B

[sudden tree]

It comes out to be R

I mean how?

[next bobcat]

Wait I can see where you could be stuck at

first do the numerator

The maxima and minima

[sudden tree]

Ok then?

[next bobcat]

maxima minima for x-y and it’s sine

as well as for the rest of the denominator

[sudden tree]

Num is dependent on den but

[next bobcat]

Just do maxima minima for everything to get a range

that gets you somewhere

@sudden tree you with me?

[sudden tree]

Yes

But I didn't get it

[next bobcat]

seee

x-y can range from -pi to +pi

ok?

[sudden tree]

Yes

[next bobcat]

So you know the range for sin(x-y) now right?

[sudden tree]

Except zero because of x-y=0 den becomes zero

[next bobcat]

From that get the maximum values

[next bobcat]

No?

Unless both x and y are pi/2

did you shove the cos terms under the rug?

you need to consider that as well

Or you could take some common trigonometric values

[sudden tree]

What I did was use sinA-sinB identity and cosA-cosB identity and write sin(x-y)=2sin()cos()

[next bobcat]

yeah that works

oh I forgot about that

[sudden tree]

Sin(x-y)/2 cancelled

On simplifying my expression reduced to this

[next bobcat]

mhm

[sudden tree]

Where alpha=(x+y)/2 and beta=(x-y)/2

[next bobcat]

ohk

[sudden tree]

And I am stuck further

[next bobcat]

you might need to backtrack a bit

try substituting common trig ratios if you haven’t already

that eliminates a couple of options most likely

[sudden tree]

I don't need elimination and all.I can do that anyday.I want the appropriate solution

U there?

[hard shard]

!noadvert

[midnight plank]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[slow thorn]

im getting a plus in the numerator..

but yeah stuck at the same place

is this advanced 😨

[sudden tree]

Yes

[slow thorn]

shiiiii

i am so cooked

i think this year?

[sudden tree]

Jee Advanced

[slow thorn]

i know i know

[sudden tree]

Not a pyq

[slow thorn]

i feel like i saw this in 2025

[sudden tree]

I tried it further assuming cosbeta=k and identifying the range of the expression in terms of k and then apply k in [-1,1].but I couldn't do so

[west ravine]

holy cow

[midnight plank]

@sudden tree Has your question been resolved?

[sudden tree]

<@&286206848099549185> pls help

[spring wave]

does B include 0?

[slow thorn]

why do you ask

yes

[spring wave]

i was giving a hint

[spring wave]

||no||

[slow thorn]

x = 60 deg
y = 120 deg?

[spring wave]

ho, i was considering 0, pi/2

then it would be easiest if you just take case y = 0

[slow thorn]

...

[spring wave]

yes, ||B is also R||

||union of x = 0, y = 0, and (45,135) will give R||

[midnight plank]

@sudden tree Has your question been resolved?

[midnight plank]

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[lost abyss]

How do I numerically represent the value of b? This is news to me

[brazen oriole]

value of b is given, that is it ig

[midnight plank]

@lost abyss Has your question been resolved?

[lost abyss]

It has 0.25 m^-3/2

It's not our standard representation

Like 0.25ms^-1

That's the part I don't get

[hybrid widget]

because $v$ has units $m/s$ that's why

[grand pond]

Annie Maqionde

[hybrid widget]

and $x^{5/2}$ has units $m^{5/2}$

[grand pond]

Annie Maqionde

[hybrid widget]

so from that you can get the units of $b$

[grand pond]

Annie Maqionde

[lost abyss]

What will be b^2

[hybrid widget]

?

the units or the magnitude?

[lost abyss]

Magnitude

[hybrid widget]

$b^2 = (0.25)^2$ from the question?

[grand pond]

Annie Maqionde

[lost abyss]

Yes

[hybrid widget]

units is $m^{-3]/s^{-2}$

[grand pond]

Annie Maqionde
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hybrid widget]

so what is the issue?

like if you're finding work treat $b$ with only magnitude if you're plotting it or doing calculations

[grand pond]

Annie Maqionde

[lost abyss]

So the m^-3/2 means nothing?

I do want to find work

[midnight plank]

@lost abyss Has your question been resolved?

[midnight plank]

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[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by americancitizen911

[pearl mist]

now here you need integration by parts

but ILATE rule is invalid here

i wouldnt be so excited and optimistic about it

nice

yeah i hate by parts worst integration method

bro there was no need

[umbral timber]

stop its the best

what type do you want? partial fractions, by parts, trig sub?

[pearl mist]

$\int_{ }^{ }\csc^{5}xdx$ ts took so much time my gawd

[grand pond]

[XRS] Xerxes

[umbral timber]

oky. let me see

[pearl mist]

yo see in this case you could split sec^3 x into secx * sec^2x and substitute such that you know the derivative of sec x and integral of sec^2x so no need to integrate secx

so its one of the final steps?

o wait it is ok then integral secx = ln(secx+tanx)+c

[umbral timber]

$\int e^{\tan^{-1} x} \cdot \frac{1+x+x^2}{1+x^2}dx$

[grand pond]

T&C

[pearl mist]

from where did you pull this bru

[umbral timber]

help idk what that is but this question has a trick to it

if you ever get the integral of e^x [f(x) + f'(x)] wrt x, you dont need to go through the long process of integrating by parts. the answer is just
e^x f(x) + c

[pearl mist]

oh yeah i remember this

[umbral timber]

cooked

[pearl mist]

also integral [f(x)+xf'(x)] = xf(x)+c

[umbral timber]

then u probably dont need to worry about problems like these. i assumed that you were given instruction on this when you learnt about by parts

[pearl mist]

substitute

[umbral timber]

u sub!

sometimes we need to use both to cut through calculation

[pearl mist]

use double angle identity

[umbral timber]

use the fact that cos2A = 2cos²A -1

[pearl mist]

generally avoid by parts because sometimes you get multiplied functions again

[pearl mist]

like in here

i didnt

im saying that avoid by parts as much as you can but in that q there was no other option

[umbral timber]

yeah i think so since cos times sec is just 1

[pearl mist]

ye

sub tanx = u^2 and proceed

remove all the inconveniences like square roots and stuff simplify as much as you can

[umbral timber]

you'd have keep to dealing with the square root then.
u² = tan x gets rid of the root

[pearl mist]

you would get u^(1/2) over some denominator atp you cant really do anything bc of the square root you'd have to substitute u as some v again

[umbral timber]

always try substitutions that gets rid of square roots, cube roots, etc (generally speaking, exceptions might always arise)

[grand pond]

Mystic

[pearl mist]

whats that supposed to be?

let u = t^2 and proceed

[pearl mist]

.

yes

whats this we have to find y?

theres a formula for 60 js learn it like i did

or trig sub

[umbral timber]

oh yeah just recalled

[pearl mist]

it looks ass but very useful

[forest iron]

Trig sub much better

[pearl mist]

who me?

[forest iron]

Me

[pearl mist]

nah

[forest iron]

Lol

Well you have to take in account that im chinese and i was 3rd in the country i live rn

But back to the question

[umbral timber]

$\int \sqrt{x^2 + a^2}dx = \frac{x}{2}\sqrt{x^2 + a^2}+ \frac{a^2}{2}\ln \abs{ x + \sqrt{x^2 + a^2}} + C$

[pearl mist]

yeahhh almighty ahh formula

[forest iron]

Bruh

[grand pond]

T&C

[forest iron]

Trig sub much easier

No need to know random formuli

In the biggest math competition of the country i live in

[pearl mist]

it is but you actually have to do the integration who does that bro just summon the formula out of you

[umbral timber]

yeah no imagine having to trig sub each time and then integrate by parts each and every time you encounter a question such as this

humans have evolved a good memory. we must utilize it

[pearl mist]

oh is that like a national olympiad or smth

[forest iron]

Yea something like that

[forest iron]

Well im in lack of that

It took me 2 years to memorize quadratic formula

Until then i had to prove it every time i needed it

And i know it bc i used it too much

@modern lynx are you finished or are there more questions

[pearl mist]

bro what

[forest iron]

Whats the problem i did it in 10s

[solid iris]

.close

[midnight plank]

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[forest iron]

Did he block me

[solid iris]

hes banned

[midnight plank]

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[elfin fossil]

how does one even solve this, do i start with M, x or y?

[fallow scarab]

what do you mean by sovle?

do you mean simplify M?

[elfin fossil]

it says find M with x and y being those

[fallow scarab]

write all terms as powers of x or y first. then get rid of the negative powers by multiplying by the reciprocal, then distribute

[elfin fossil]

what about x and y, any ideas on how i can simplify them

[fallow scarab]

,tex .log rules

[grand pond]

riemann

[elfin fossil]

alright, i'll just see what i can do

.close

[midnight plank]

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[tidal turret]

where is the mistake

[midnight plank]

@tidal turret Has your question been resolved?

[stoic walrus]

In the last image you said that $f(x,y)=x^2-y^2-12y-36$ when it was $f(x,y)=x^2-y^2+12y-36$

[grand pond]

KonoEmllikDa

[midnight plank]

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[finite jungle]

When drawing a graph like this, what things do I need to consider in able to draw a rough sketch? Heres what I came up with:
Whether its odd or even to see what quadrants its located in
The extrema and where its concave up and down to get a general idea of the shape
The roots to see where it crosses the x axis and y intercept to see where it crosses y
Is there anything else I need or could use in order to begin sketching?

[finite jungle]

I just need a rough sketch in my exam nothing perfect. Also graphing calculators (or any calculators) are not allowed

[last slate]

no calculators allowed?! wow

tough

[finite jungle]

I know 🥲

[last slate]

i was about to say "what if x = 2i"

bruh

[dawn dagger]

what about asymptotes

possibly also investigating monotony helps

[finite jungle]

Ahh ok good point

[finite jungle]

Thats increasing or decreasing, right?

[dawn dagger]

yes

[finite jungle]

Alright I see

Well thats enough I hope haha thats already a lot of work

Thanks!

❤️

.close

[midnight plank]

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[midnight plank]

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[blissful trench]

,, \frac{(14\angle25^{\circ})(42\angle 120^{\circ})^4}{(75 \angle 210^{\circ})(21\angle 32^{\circ})^2}

[grand pond]

Akιρɑ

[blissful trench]

its asking me to simplify and express in rectangle form

first step I have done is converting degrees to rad, am i doing this correctly?

[pearl hull]

What’s rectangle form?

[blissful trench]

bro adonis dont mind my prof ik this sucks ass 😭

[blissful trench]

no clue

[dawn dagger]

nasty

[blissful trench]

wait no

[dawn dagger]

a+bi

[pearl hull]

Ya have the original q?

[pearl hull]

Ah

[blissful trench]

ya

[dawn dagger]

cartesian form

can u use calculator

[blissful trench]

im doing this correctly?

[dawn dagger]

yes

then use exponentiation rules

and then try to get it into trigonometric form

[blissful trench]

do i apply the 4 and 2 exponents then the j's?

that doesn't make sense but ill try show you what i mean

[dawn dagger]

yes

[blissful trench]

like this?

[dawn dagger]

no

[blissful trench]

what is z

😭

[dawn dagger]

complex number

[blissful trench]

o

[dawn dagger]

you only applied the powers to the number factor

[blissful trench]

so 2.094j times 4

?

[blissful trench]

oh

well i forgot to apply the exponent to the j

[dawn dagger]

yes

[dawn dagger]

this is applicable, n is natural

z complex

[blissful trench]

ok now what

[dawn dagger]

you can add the exponents

[blissful trench]

even when they aren't the same?

[dawn dagger]

ofc

just the base has to be the same

[blissful trench]

ooo

[dawn dagger]

anyeway

these are usually proven with the series definition of the function e^z

[blissful trench]

[dawn dagger]

what happens when u multiply j

[blissful trench]

do i just convert it to rectangle form

[dawn dagger]

with itself

[blissful trench]

j

[dawn dagger]

what is j² = ?

[blissful trench]

-1

[dawn dagger]

oh wait

nvm

[blissful trench]

lmao

[dawn dagger]

my dumbass

ok lets continue

[blissful trench]

ok

[blissful trench]

here

[dawn dagger]

yea seems good so far

now you can convert into trigonmetric form, eventually into rectangular

[blissful trench]

rad to degrees?

[dawn dagger]

no

[blissful trench]

quite not sure

[solid iris]

its nicer to keep the degrees until fully simplified

[dawn dagger]

,, r \cdot e^{\bm{j}\theta} = r \cdot (\cos(\theta)+\bm{j}\sin(\theta))

[solid iris]

it also gives u less decimal error

[blissful trench]

i have rads not degrees and i was thinking of
x = 131.9 cos 4.031
y = 131.9 jsin 4.03

[dawn dagger]

it all comes down to approximating pi

[blissful trench]

[dawn dagger]

are you allowed to approximate?

[blissful trench]

yep

[dawn dagger]

ok then you can finish what u started

silly prof

[blissful trench]

im getting weird number

im gonna convert 4.031 to degrees

[solid iris]

its also much less painful to keep deg

[dawn dagger]

why is that

i mean some ppl would forget the °

oh wait nvm

[solid iris]

$(a\angle b)(c\angle d)=ac\angle(b+d)$

[dawn dagger]

stupid floats

[solid iris]

$(a\angle b)^c=a^c\angle bc$

[grand pond]

ロケット・ジャンプ

[dawn dagger]

i am so tire

[solid iris]

$(a\angle b)/(c\angle d)=(a/c)\angle (b-d)$

[grand pond]

ロケット・ジャンプ

[solid iris]

do u both agree with these?

[blissful trench]

never seen this before

[grand pond]

ロケット・ジャンプ

[dawn dagger]

i see that for the first time but looks like exponentiation rules

[blissful trench]

what about now

[solid iris]

theyre basic rules of algebra for complex numbers in exponential form

[dawn dagger]

,calc 1442/(75121)

[grand pond]

Result:

0.064793388429752

[dawn dagger]

that ought to be your rec form

[solid iris]

so use the rules to do computation WITHOUT changing deg to rad

[blissful trench]

sorry im not sure about these rules u listed bc my prof never taught us these

[solid iris]

surely u saw them in exp notation?

$e^{ia}e^{ib}=e^{i(a+b)}$

[grand pond]

ロケット・ジャンプ

[blissful trench]

I mean its almost the same

but its theta

[dawn dagger]

what's i
-# jk

[blissful trench]

stop being funny

[solid iris]

$(1\angle a)(1\angle b)=1\angle(a+b)$

[grand pond]

ロケット・ジャンプ

[solid iris]

same thing

[blissful trench]

these are the formulas

[solid iris]

theyre exactly the same as what i wrote

[blissful trench]

fair

[dawn dagger]

is this tough

[blissful trench]

no

[solid iris]

these are the rules useful for our problem

$(a\angle b)(c\angle d)=ac\angle(b+d)$

[grand pond]

ロケット・ジャンプ

[solid iris]

$(a\angle b)/(c\angle d)=(a/c)\angle (b-d)$

[grand pond]

ロケット・ジャンプ

[solid iris]

$(a\angle b)^c=a^c\angle bc$

[grand pond]

ロケット・ジャンプ

[blissful trench]

ok ill take ab rather than thetas

[solid iris]

the important result of these rules is that we deal with modulus and phase "separately"

we multiply or divide different modulus as needed

we add or subtract different phases as needed

https://cdn.discordapp.com/attachments/1018703621841494076/1445901103320531104/362561562004946945.png?ex=693207d6&is=6930b656&hm=f698930b724617850403b39f7ae30ad1a2ac9352ec15c68de0b54842eb9d9511&

what is the modulus? use the rules and do NOT simplify

[blissful trench]

no clue whats modulus

[solid iris]

$a\angle b$ modulus is $a$ phase is $b$

[grand pond]

ロケット・ジャンプ

[blissful trench]

can i start with modulus that has exponents first (42 angle 120)^4 and then multiple them by 14 angle 25?

[solid iris]

u need to apply power rule first

[dawn dagger]

distance of a complex number to the origin

[solid iris]

$(a\angle b)^c=a^c\angle bc$

[grand pond]

ロケット・ジャンプ

[dawn dagger]

,, a\angle b = a e^{b\bf{j}}

[blissful trench]

like this?

[dawn dagger]

check with a calculator

[solid iris]

where did 168 come from?

[blissful trench]

42 and 4

i multiplied them

[solid iris]

u didnt use the power rule right

[blissful trench]

i did

[solid iris]

the modulus gets RAISED to the power not multiplied

[blissful trench]

wait u mean 42^4?

[solid iris]

yes

[blissful trench]

ahhh gotcha

[solid iris]

many other algebra mistakes too

take it slow

also its good to not write down big numbers until the end so keep 42^4 as is

[blissful trench]

I did

how would you still get rid of power

do I just normally multiple them

14 and 42^2 then 25 and 120

its giving me big ass number 😭

[solid iris]

u didnt use the rule right

$(a\angle b)(c\angle d)=ac\angle(b+d)$

[grand pond]

ロケット・ジャンプ

[blissful trench]

oh wait

wrong one

im not sure what do u have to do with the powers

im doing this right or still no

[solid iris]

i wouldve kept all powers until the very end

at the end i have a single thing i can put in calculator

[blissful trench]

how

[solid iris]

,calc 1442^4/(7521^2)

[grand pond]

Result:

1317.12

[blissful trench]

i did this

[solid iris]

14*42^4/(75*21^2)

the way i apply the rules i wouldve combined the modulus together first

then combine the phases

,calc 25+1204-210-322

[grand pond]

Result:

231

[solid iris]

25+120*4-210-32*2

[blissful trench]

i have done this

no nvm

i did it differently

[solid iris]

urs is finally right

[blissful trench]

i see what u mean now

[solid iris]

u just did it one step at a time

the way i think about the rules is

i deal with modulus and phase separately

https://i.gyazo.com/7d073162fc82fefe4b9f209e76b4ed05.png

[blissful trench]

ok I think i get it now

thanks a lot 🙂

.close

[midnight plank]

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[solid iris]

np. u just needed an example to see what i meant

[midnight plank]

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[meager ore]

In $(I, \tau_u)$ the equivalence relation $R$ is defined by: $xRy$ if and only if $x = y$ or ${x,y} \subset {0, 1/2}$. Let $p$ be the canonical projection. Prove that the resulting quotient space is compact and $T_2$.

[grand pond]

Halex

[midnight plank]

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@meager ore Has your question been resolved?

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[true lark]

.close

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[green hinge]

I need help with 5

[tall lintel]

Wait

Wait no I stand by my precious statements

[tall lintel]

We dont know what the base of the ladder is,

So lets suppose the base is x

How would you find an expression for the height of the ladder?

Hint: Use similar triangles. Can you see two triangles in this image?

[midnight plank]

@green hinge Has your question been resolved?

[green hinge]

Hmm like this?

[latent wadi]

,rccw

[latent wadi]

Not quite

X/12 = (x+4) /n

[green hinge]

Ohhh I see we're using the big triangle

[latent wadi]

Yeah, it's called "similar triangles", so we only compare triangles to other triangles

[green hinge]

Ok what do I do now

[latent wadi]

Solve for n in terms of x

Then you have the right triangle with sides x+4, n, and the length of the ladder as the hypotenuse

[green hinge]

Okay and then how would I solve for the hypotenuse?

.close

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[midnight plank]

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[wispy mortar]

yo

[wraith dirge]

when n = 3^k, what is f(n/3)?

or i should say, what is n/3

[wispy mortar]

f(3^k) - 4/2?

[wispy mortar]

3^k-1

[wraith dirge]

no, i meant what is n/3, sorry about that

[wraith dirge]

yeah

so you have

f(3^k) = 2f(3^(k-1)) + 4

right?

[wispy mortar]

yess

[wraith dirge]

both terms in the relation are in terms of 3^..., so what can you do to simplify?

[wispy mortar]

make a new function?

[wraith dirge]

yes, substitute

[wispy mortar]

[wraith dirge]

what would you substitute?

[wispy mortar]

3^k?

[wraith dirge]

yeah, so your new function g(k) = f(???)

[wispy mortar]

g(k) = f(3^k)?

[wraith dirge]

yep, go back to this and see how it simplifies

[wispy mortar]

g(k) = 2g(k-1) + 4?

[wraith dirge]

yeah, now its much easier right

do you know how to solve these kinds of recurrance relations?

[wispy mortar]

yepp

[wraith dirge]

then its simple, solve for g, then you can solve for f

first you must get an initial value for g

[wispy mortar]

lemmetry

[flat spire]

laugh

[wraith dirge]

y u laughin

[flat spire]

im laufghin at vie

rn

[wispy mortar]

oh yea got the answer

damn u here again

[flat spire]

srly vie

try some adv jee

[wispy mortar]

lmao its my first maths course subjects

and i missed a lot of classes

so little cracks in methods

[wraith dirge]

you get $f(n) = 5n^{\log_3{2}} - 4$?

[wispy mortar]

@wraith dirge thank u so muchh

[grand pond]

c2b7

[wispy mortar]

YESSSS

.close

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[last slate]

hi i need help

[last slate]

need help understanding the bars and stars method in combinatorics

i wanna know the intuition behind the choose formula process

[runic hamlet]

the wikipedia article is decent

[last slate]

i read it

didnt nelp

help

[fallow scarab]

It's more useful for you to explain what part of the wiki article on stars and bars confuses you

[midnight plank]

@last slate Has your question been resolved?

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[twilit field]

Here i don't think there's a way out of using the first isomorphism theorm

So we have G/Ker(phi) ~ phi(G)

There are thus 5 distinct translates

Like does the kernel has size 6 here

so <6>

[spiral rock]

The nice thing is that a cyclic group of order n has a unique subgroup of every order dividing n

[twilit field]

oh right

so <5> is the kernel

[spiral rock]

Yes

Just one more thing

What if phi is the trivial homomorphism

Its a case you didn't consider

[twilit field]

well, then it has its kernel as Z_30

[spiral rock]

Yes

Just don't assume phi is surjective

Thats how you lose points on tests/hw

[twilit field]

Good advice ( endsem tomorrow 🙏 )

[spiral rock]

Actaully, maybe if its says a homomorphism "onto" Z5 then its surjective?

Gotta verify that

[twilit field]

yea, that's what I thought at first

if it's somehwere between onto and trivial we'll have an issue

[spiral rock]

It can't be

[twilit field]

hmm

yea

okay, makes sense

[spiral rock]

Maybe ask your classmates/prof about this terminology?

Just to make sure

Its not very urgent

[twilit field]

Here is \overline{G} non-finite

[twilit field]

I will

[spiral rock]

H?

It may be infinite

[twilit field]

hmm

[spiral rock]

I think it makes sense that "onto" means surjective

[twilit field]

Just checked

yes

[spiral rock]

Because if the homomorphism is trivial you can't say anything about G

[twilit field]

I think I'm close to a soln

Let $\phi(a)=g$ , where $g$ has order $8$. Then $\phi(a^8)=e$.

[grand pond]

wai

[twilit field]

okay, this won't work

[spiral rock]

This will

Continue

a has order at most 8

[twilit field]

oh right

order of an element divides order of group

so let order of a be n

[twilit field]

Am I overcomplicating

[spiral rock]

Yes

No need for lagrange

[twilit field]

suppose the order of $a$ were less than $8$, say,, $k$. Then $\phi(a^k)=g^k=e$, which would be a contradiction

[grand pond]

wai

[spiral rock]

Yes

[twilit field]

😶‍🌫️

how did I miss that

[spiral rock]

Wait

I confused myself

Ok this is a bit more tricky

[spiral rock]

What you have done here is showing a has order at least 8 (or infinity)

If a has order n>=8 but finite, then the subgroup <a> has an element of order 8

And I give to you to understand the infinite case

[twilit field]

is that necessary ?

[spiral rock]

I think so, a won't have order 8 necesseriliy

[quiet hinge]

You know that a has finite order and furthermore it is divisible by 8

[twilit field]

oh right

Fuck

thanks

.close

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[quiet hinge]

There is no infinite order case sine G is a finite group

[spiral rock]

Yes

I wanted wai to understand that himself

[quiet hinge]

Well I think thats confusing since it can be taken as there being a possibility for infinite order

Rather a better question would be can a have infinite order

[spiral rock]

👍

[midnight plank]

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[gilded current]

I just need a diagram fr pls

[gilded current]

I got various diagrams but keep getting confused b/w two

[valid ore]

how can a trapezoid be circumcised

[lyric charm]

circumSCRIBED.

it means there's a circle inscribed into it.

[gilded current]

like this i suppose>

[slow thorn]

not quite

[gilded current]

ik im bad at drawing the diagram

i just need the diagram

ITS SO hard

[lyric charm]

both are wrong

start with the circle

[gilded current]

ik all lines should be tangential

[slow thorn]

something like this

[lyric charm]

only then draw your trapezoid around it

[gilded current]

but the result isnt beautiful

[slow thorn]

ignore the side length values

[gilded current]

ty

mine kept looking like a triangle with a flat edge

.clsoe

.close

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[gilded current]

how to show that a n sided figure is cyclic

[gilded current]

<@&286206848099549185>

[paper prism]

All sets of four points should be cylic quadrilaters or I suppose that at some point, we can connect all sides to that point and get equal radii.

I doubt this is the best way though, it is highly inefficient, although the cyclic quadrilaterals can help deduce some general result maybe

[gilded current]

hmm

ty

[midnight plank]

@gilded current Has your question been resolved?

[tall crag]

there are many ways, depending on ur question

[midnight plank]

@gilded current Has your question been resolved?

[midnight plank]

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[wooden fossil]

I have no idea what even a generating system would be. Does anyone know how to approach this?

I think I understood now what the elements of U1 are ( I wrote it in the image) but really no clue how to solve this exercise

[wooden fossil]

It shouldve said "basis", sorry

[runic hamlet]

isnt this just all sequences?

!original

[midnight plank]

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.

[runic hamlet]

I can always just choose lambda_0=a_n

you at least messed up the order of the quantifiers

[wooden fossil]

Thats what I thought. But since N is fixed, at least 1, I figured thats not possible?

[runic hamlet]

why not

lambda_0=a_n, lambda_i=0 otherwise

[wooden fossil]

This is the original. There is nothing else

[runic hamlet]

yes so you fucked up the order of quantifiers

exist first, forall second

[wooden fossil]

Damn

[runic hamlet]

it really already tells you the generating system

[wooden fossil]

One second I need to process what you said

[runic hamlet]

(a_n) = sum lambda_i (n^i)

[wooden fossil]

Yes

So initially we fix n
Then we consider a_n

And for that a_n, lets say a_n = 6 for example
There have to exist N+1 scalars lambda_0 through lambda_N
Such that 6 is this sum

Correct?

But the n is fixed, or not?

[runic hamlet]

no

we first find the lambda's

and then it holds for all n

so eg N=2

we find lambda_0, lambda_1, lambda_2

[frosty plume]

Help me please

[runic hamlet]

and then a_6 = lambda_0*6^0 + lambda_1*6^1 + lambda_2*6^2

and a_17 = lambda_0*17^0 + lambda_1*17^1+lambda_2*17^2

etc

[runic hamlet]

#❓how-to-get-help

[wooden fossil]

I hate this. Sometimes people denote the quantifiers after the statement. Thats why I thought this was the case here as well.

You know, like when someone writes

P(x) holds, for all x in U

But I think I got your point with the quantifiers now

Let me read the other things you said

[runic hamlet]

yeah but thats only done for the last quantifier

but yes it sometimes sucks

[wooden fossil]

Okk. So thanks for that already!

I will think about it for a moment again with this new insight

Ok no I still have no clue.
:/

I think I understand why the set of all real sequences would be a generating system for this subspace.
But thats quite far from a basis ig

[runic hamlet]

yes very far

consider the sequences (1,1,1,1,1,...), (1,2,3,4,5,6,...), (1,4,9,16,25,...)

[wooden fossil]

I am not even sure why the last two are in U1 tbh

For the first we'd find lambda_0 = 1 and
lambda_k = 0 for 2<= k <= N, right?

But for the second sequence I don't even see how we could choose the lambdas to construct 1 for n =1 and also 2with n =2

We know for sure that lambda_0 = 1, otherwise a_1 =1 shouldnt be possible

But then for a_2, wouldnt we have to add an even number onto that, so how can we achieve 2 as a result?

Oh damn sorry, I messed up the n

[runic hamlet]

oh wait 0 is in N for you

(1,1,1...), (0,1,2,3,...), (0,1,4,9,...) then

[wooden fossil]

Ok ig then thats the same problem I described

I hope you can understand what I am trying to explain

We know for the second sequence that lambda_0 is 0 right?

Because we have
a_0= 0 = lambda_i x n^0
= lambda_0 x 0^0

[runic hamlet]

yes

[wooden fossil]

Then
a_1 = 1 = 0 x 1 + lambda_1 x 1 +... +lambda_N x 1^N

= sum over lambda_i for 1<= i <= N

[runic hamlet]

we still have N=2

for simplicity

[wooden fossil]

Oh ok

So a_1 = 1 = lambda_1 + lambda_2

So lambda_1 = 1-lambda_2

Then a_2 = 2 = (1-lambda_2) x 2 +
lambda_2 x 4

2 - (2 x lambda_2)

So 2 x lambda_2 = 0
So lambda_2 = 0

?

Then lambda_1 = 1

Im sorry, I still dont see in which direction you wanted to nudge me

[runic hamlet]

write down for n=0,1,2,3,4,5,6 explicitly what a_n = sum lambda_i n^i means

N=2

distinct lines

and try to align the terms

[wooden fossil]

Hang on

Does it have something to do with being a linear system of equations?

[runic hamlet]

well all of LA has to do with linear systems

[wooden fossil]

Lol

Im sorry Im still slow in LA

[runic hamlet]

dw

no thinking required now

only writing

[wooden fossil]

Is this what you meant?

[runic hamlet]

yes

[wooden fossil]

So we know a_n for a fixed sequence.

Hence we can solve for the coefficients?

[runic hamlet]

how would you write this using vectors with length 6?

whoops I cant count

length 7

[runic hamlet]

no

where are all these lambdas from

[wooden fossil]

:/

Uhm

Idk

I dont understand the question rn, sorry

Oh

Ok i get it

Seven lambdas

But we have only 3

Damn

Sorry, I'll retry

I dont think I know how to do this

You want me to write the system as a vector matrix product?

[runic hamlet]

when I have equations like
x = 3a+4b
y = 7a-15b
then I can write those using vectors as
(x,y) = a(3,7)+b(4,-15)

yes?

[wooden fossil]

Yes

Ok I'll retry

[runic hamlet]

good

so this tells us that the vector a is in the span of the vectors (1,1,...), (0,1,2,3,...), (0,1,4,9,...)

yes?

[wooden fossil]

Ohhh

Those vectors you denoted

Are those exact vectors

I didnt see this until just now

Yes

[runic hamlet]

but was there anything specific about the number 6?

[wooden fossil]

No

Ok I'll try to connect the pieces in my mind now

But if we now generalize to general n, wouldnt we run into a problem?

I mean here we fixed n

So our vectors have length n+1

But our sequences are not finitely long

Only N is a fixed number

[runic hamlet]

we can just imagine that in this image you wrote ... under all the last terms from the vectors

nothing would change

[wooden fossil]

Like this you mean, right?

[runic hamlet]

yes

this is just all those equations a_n = sum lambda_i n^i written in a different form

this is the "defining equation" of the allowed sequences (a_n)

[wooden fossil]

Yes

[runic hamlet]

so now tell me a generating set of the subspace

[wooden fossil]

Ah ok, our basis vectors are infinite sequences, so they have to be like this, right?

I was kind of confused about the fact theyre infinite

[wooden fossil]

These vectors on the right?

[runic hamlet]

yes

[wooden fossil]

Now to show that this is a basis I need to show that if (a_n) is the constant 0 sequence (0,0,0,...) then all the coefficents are 0, right?

Thats just the definition of linear independence

[runic hamlet]

yes

[wooden fossil]

An immediate idea I have would be to prove by induction that lambda_i is 0.

[runic hamlet]

induction on what

on i?

[wooden fossil]

Induction on the lambdas' indexes

So first lambda_0

[runic hamlet]

imagine all the a's are 0

[wooden fossil]

Yes

[runic hamlet]

how would you find the lambdas

[wooden fossil]

Well in the finite case solving for one after the other or solving the linear system of equations

[runic hamlet]

good

we can still do that here

if all a's are 0, then also the first N+1 are 0

and those are enough to find our lambdas

and at some point you need the magic word vandermonde which you have hopefully done in your lecture

[wooden fossil]

No

I have heard of vandermonde but in a very different context

It was vandermonde identity for binomial coefficients

If I remember correctly

[runic hamlet]

never heard about the vandermonde matrix?

oh boy

[wooden fossil]

Not that I can remember

Definitely not in the lecture

[runic hamlet]

its the matrix that appears in the linear system of the first N+1 rows

[wooden fossil]

Ok

Hm

[runic hamlet]

I dont know how to do this without knowing about the vandermonde matrix

[wooden fossil]

I dont know how to do this with vandermodne matrix

Lmao

But ok I'll try

Thank you very much

[runic hamlet]

well the point is, the vandermonde matrix is invertible

done

[wooden fossil]

Actually

We didnt even have matrices yet

[runic hamlet]

then I dont know what they expect from you

you have the generating set of the space

[wooden fossil]

Yeah me neither

[runic hamlet]

but I dont know how to show that its a basis without using some knowledge about vandermonde

[wooden fossil]

Maybe maybe there is some really handy theorem in the notes that I could use and am too ignoratn to spot

[runic hamlet]

hmm

[wooden fossil]

I'll check that out as well

[runic hamlet]

well you can restate everything in terms of polynomials

what have you done about polynomials

that is actually much easier

[wooden fossil]

Idk actually but not too much

Stuff like polynom ring

Cant find much rn

[runic hamlet]

what does each of these rows say about the polynomial p(x)=lambda0+lambda1 x + lambda2 x^2 ?

[wooden fossil]

Ah man, I have to go now

I will think about it later

Sorry

Ah man

[runic hamlet]

keywords to look at later:
||roots||
||number of roots of a polynomial||

[wooden fossil]

But thanks again so much for your help

I highly appreciate it!

[wooden fossil]

Thanks!

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[wanton spade]

i took g(x)=f(x)-x
then g'(x)=f'(x)-1 and it is never 0, so g'(x) is 1-1 in the interval
dunno what to do from here

[prime hornet]

an injective function on an interval of R must be strictly increasing or strictly decreasing

[trail sphinx]

whispers: “mean value theorem“

[prime hornet]

||so how many times can it cross the x-axis?||

[wanton spade]

if I assume there's two fixed points then i use MVT to show that there's a point where f'(c)=1?

[trail sphinx]

[wanton spade]

once

[prime hornet]

at most once*

[junior flower]

no need to do that

[wanton spade]

right

[wanton spade]

which?

[junior flower]

the assume there are two fixed points thing

[wanton spade]

oh yeah

thanks slayla, dotdoc and higher

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[full grail]

did i do this correctly?

[fallow scarab]

no, you completely left out |x-2|

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@full grail Has your question been resolved?

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[buoyant fjord]

im tryna figure out how to prove differentiability at -2

[buoyant fjord]

oh shoot

i messed up my arithmetic

oh this is easy cuz the x+2 factors bruh

i think

$\frac{3(x+2)^3}{(x+2)}$

[grand pond]

Julian

[buoyant fjord]

hey so i know how to do it know but i have a problem

somehow i need $3(x_n+2)^2 < \epsilon$

[grand pond]

Julian

[buoyant fjord]

but im gonna have some weird square quadratic solution root expression of epsilon

for the bound of x_n or something

[fluid dust]

U talking to yrself or what?

What's the question?

[buoyant fjord]

well i was just typing what i was thinking but also waiting

so

is there a cleaner way to do this cuz my way is kind of scuffed

[junior flower]

are you doing this straight from the definition of differentiable?

[buoyant fjord]

yea

wait lemme see

it says i can use anything in rudin up to and including chapter 5

but theres nothing useful in there really i think for this

[junior flower]

ok this should be fine i was just wondering

[buoyant fjord]

i have an example of what i mean by messy one sec

this is unrelated

but the epsilon mesiness is there

[junior flower]

if you take a point b near -2, the secant line between (-2, 0) and (b, f(b)) will be very similar to the line y = 0

[buoyant fjord]

yea thats why i wanted to use a sequence with limit -2

[junior flower]

a sequence doesn’t seem that helpful?

like i’m sure you’d need to use a statement involving all sequences that converge to -2 (if you use sequences)

but maybe that’s what you were saying anyway

[buoyant fjord]

i mean a general sequence

yeah

i feel like the sequential limit of a function definition is easier to work with

usually than epsilon delta

[junior flower]

do you know a statement of “f is differentiable at a” in terms of sequences?

[buoyant fjord]

f is differentiable at a if for every sequence $(x_n) $ in $\mathrm{dom}f $ whose range doesnt include a , the sequence $\frac{f(x_n)-f(a)}{x_n-a}$ converges?

?

oops

[grand pond]

Julian

[junior flower]

x_n converges to a right