help-19

[flat furnace]

i dont get it wouldnt the 21x-21x be 0

[lavish jackal]

-21x-21x

signs changes

[flat furnace]

ohh okayyy

soo.. -42x = 63+21 ??

[lavish jackal]

nope

negative and negative gives positive

if it was bigger number then its gonna be negative, do u see why?

e.g for rhs, we have -21 - 63, what do u think that will be?

[flat furnace]

-84?

[lavish jackal]

yep

so how does ur equation looks like now

[flat furnace]

42x = -84 ?????

[lavish jackal]

yes!

and ur final answer?

[flat furnace]

i divide by 42 right?

[lavish jackal]

ye

[flat furnace]

x = -2

[lavish jackal]

https://tenor.com/view/clapping-leonardo-dicaprio-leo-dicaprio-well-done-applause-gif-1867988353188143738

[flat furnace]

thank you🥹🥹

[lavish jackal]

np

[odd edge]

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[fallow geode]

for 2a I found that 32767 is not a prime (using a proof), so 2^15 - 1 is also not a prime. and since 32767 is not a prime, for 2b 2^32767 -1 is also not a prime (using a proof). Since 2^32767 -1 is not a prime, I found a factor of 32767 and got 7 and 4681 using the square root method. 7 and 4681 could be the "x" if it was asking for 32767, but here its asking for a factor of 2^32767 -1. How do I know it would be 2^7 -1? And how do i do 3 lol

[odd edge]

@fallow geode Has your question been resolved?

[late dust]

For (3), for any integer k, one of k, k+2, k+4 is divisible by 3

Up to you to prove it and use that to answer

[fallow geode]

thank you

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[magic quail]

Sorry if the work is messy but can anyone confirm that this is on the right track so far.. im at 4/49 integral cos^2theta dtheta

[graceful viper]

what was your sub

[magic quail]

I havent done that yet

[graceful viper]

your trig sub

[magic quail]

Like x=4/7 sintheta?

Pause

I wrote tan

I think thats supposed to be cosine

what the helly

[graceful viper]

your sub is correct

cos^2 is not the right term

but 4/49 is

yeah here

no idea what is happening tbh

[polar echo]

my bad, didnt see it said using trigo sub

[graceful viper]

it should of the form (sin/cos)*cos

cuz ur dx = 4/7 cos(phi) d(phi)

[polar echo]

i'm not sure how this works, can i suggest a method or that's rudely interfering 😭

[graceful viper]

first lets let the OP respond

@magic quail

[polar echo]

okayy

[magic quail]

i reworked it lemme show u

Ik it still needs to be in terms of x but

Wait

No

I forgot to do more

[polar echo]

thats the old pic lmao

[magic quail]

Wait ur right

Oops

LOL

i think i forgot a step

I didnt do u = sin

[graceful viper]

you are forgetting the integration constant but other than that, just reverse sub so u get ur answer in terms of x

[polar echo]

yeah

[magic quail]

4/49 tan theta?

[polar echo]

no no, you are at the right step just convert the cosphi in terms of x

by squaring your initial sub of x=4/7sin(phi) and then writing sin^2(phi) as 1-cos^2(phi), you'll get x in terms of cos there

[magic quail]

Oh so i wasnt supposed to set sin = u cuz ik the integral sinxdx = -cosx+c So i did that originally but i thought i messed up so i set it equal to u to get rid of the theta

[polar echo]

yep, you were just supposed to integrate sin(phi), get it in cos then reverse sub

[magic quail]

Right?

[polar echo]

🔔 🔔 🔔

yeah thats correct, add the integration constant (c) too

[magic quail]

Thankss

[polar echo]

yay

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[peak heath]

can somebody give me a hint? idk where to start really

[peak heath]

Let n be a composite squarefree integer such that (p − 1) | (n − 1) for any prime p | n. Prove that n is
a Carmichael number.

[west light]

I'd start by writing the condition of being a Carmichael number

[peak heath]

yea i have that

i'm thinking about using CRT

[west light]

Yes, good start.

[peak heath]

cuz then i can get a set of numbers, then maybe prove that all the numbers in the set have the property of a carmicahel number

[west light]

Yes, given p^k | n (and not p^(k+1) | n) you can write the Carmichael number condition as a bunch of congruences mod p^k using CRT.

(For each p | n)

So if you can prove that it works for each p then CRT tells you that it works for n.

Oh! Haha. There's a nicer way.

Oh I guess that goes through CRT anyway if you haven't proved one of Carmichael's theorems.

You can consider p | b and gcd(b,p) = 1.

[peak heath]

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[pulsar bluff]

can anyone check whether my question is correct. its about converting reccurin decimal into fractions.

[nimble blaze]

yes

[pulsar bluff]

k thanks

[nimble blaze]

dunno why your dots changed from solid to open circles though

[pulsar bluff]

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[noble ether]

https://cdn.discordapp.com/attachments/1110223603355623454/1398017640680919050/1039134259539103744.png?ex=68b0a7e3&is=68af5663&hm=cce12fb1016bd49bab3b2d32618e5fdb776e966332ec1c767635a13b237abac7&

[noble ether]

How to solve this

No induction tho

let it be

.close

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[barren gust]

Hey so I was going over a problem I missed and was wondering what I did wrong

[barren gust]

Can anyone tell me?

,rccw

[nimble blaze]

you didn't include 0 and 2pi in your VAs

[barren gust]

It doesn’t ask for that though

It’s (0,2pi) so I thought I didn’t include them

[nimble blaze]

they're still asymptotes to consider as f(x) approaches inf as x→ those values

[barren gust]

Oooooo

That’s fucked

Welp

Thanks

[nimble blaze]

just like how pi isn't in the domain of the actual function but x=pi is a VA

[barren gust]

Ya

I should have just put 0+ to infinity and 2pi- to infinity then

[nimble blaze]

yes

[barren gust]

Alr

Thanks

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[twin bramble]

hi can someone please clarify the answer to this question

[noble ether]

It’s asking to determine the values of k,m,n

[twin bramble]

my teacher wrote the answer was

m is +4 and n is -2

wait i mean m is -4

n is +2

mb

[wooden python]

yes, so what do you want/need clarified

[twin bramble]

this is the working out he gave

i was a bit confused on the x=0 and y=4 part

because the parabola only interesects at -4

[wooden python]

y = a(x+4)(x-2)

does this make sense to you?

also it looks like your teacher was a bit sloppy and made some sign errors all over the place

[twin bramble]

yea

[wooden python]

he should have written that y=**-**4

[twin bramble]

it should be -4=a (0+4)(0-2)

right?

[wooden python]

don't forget the a

[twin bramble]

so it gives trhe same answer write

because -4=-8a

then it gives 1/2=a

[wooden python]

right ➡️ ✅ not write 📝

but yes a=1/2 is correct

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@twin bramble Has your question been resolved?

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[odd edge]

Please don't occupy multiple help channels.

[tawdry glacier]

hi

[fierce raven]

this is similar to last one

[fierce raven]

do this using quad this time

[tawdry glacier]

i swear there is no factor

you see you seee

i didnt get what u where yapping abt

[fierce raven]

just do using quad

let lenth be l

and width be w

[tawdry glacier]

this one?

[fierce raven]

since lenght is 6 units more than witdh, l = w +6

[unkempt lichen]

8*

[fierce raven]

now we apply formula of area

[fierce raven]

yeah my bad

[tawdry glacier]

i dont think the teacher will allow that

we r doing quadric

[fierce raven]

A = l * w

[tawdry glacier]

factoring and expanding etc

[unkempt lichen]

and what's wrong with that

[fierce raven]

so, $600 = (w+8)*w$

[unkempt lichen]

just remove the asterisk

also, it's 8, not 6

[fierce raven]

let it be 6

[vernal yacht]

what?

[clever fjord]

Rex Lapis | 天妖神 🐉

[tawdry glacier]

WAIITTT what my teacher told us was x^2 +8x-600=0

and then she doesnt smth

[fierce raven]

$w^2+8w-600=0$

[clever fjord]

Rex Lapis | 天妖神 🐉

[unkempt lichen]

yeah... we're getting there

[tawdry glacier]

ok ok

[fierce raven]

hell yeah

[fierce raven]

this is basically a polynomial with 2 degree, known as quadric equation.
since the maximum power of x is 2, this eqaution have two roots. in our case, that will be length and width

get it?

[tawdry glacier]

yep

[fierce raven]

the formula for finding roots is

$x1 = (-b+sqrt(b^2-4ac))/2a and x2 = $x1 = (-b-sqrt(b^2-4ac))/2a$$

[clever fjord]

Rex Lapis | 天妖神 🐉
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fierce raven]

hmm

messy form

[tawdry glacier]

?

[noble ether]

https://cdn.discordapp.com/attachments/1110223603355623454/1410382058886271196/1039134259539103744.png?ex=68b0d026&is=68af7ea6&hm=de9deb034aa32c7f9974a2d04bf78bad1aeee54210440ef4d44631c2b73079ee&

Here

[fierce raven]

yeah

like this

[tawdry glacier]

WHAT IS THAT

i swear the teacher

[noble ether]

Two quadratic formulas

[tawdry glacier]

didnt teach us that

[noble ether]

Use whatever u like

[fierce raven]

it's the direct formula to get roots. rather than doing factors

[tawdry glacier]

oh wait i think i used that before

so is that one formular

[noble ether]

Yes

Two formulas

[tawdry glacier]

so i can just substitue numbers in this one

cause its basically the same thim

[noble ether]

This is only one root

Tho

[tawdry glacier]

ye ik

[noble ether]

Other is negative

Which can be rejected

As length is not -ve

[tawdry glacier]

whattt

[noble ether]

First solve tho

[tawdry glacier]

ok ok

brb

wait

what is 4ac

is it just -600

[noble ether]

a is leading coefficient

c is constant term

[tawdry glacier]

so u time a and c

[noble ether]

Here a is 1

[tawdry glacier]

then by 4

[tacit wasp]

Of course 😅

[noble ether]

c is -600

B is 8

[tawdry glacier]

so its 2.6

if x is positive

helllloo

[noble ether]

No

It should be

This

[tawdry glacier]

itts taking its time loading

whatt

dont u put the formular in

🤨

OK imma just ask the teacher but how do u do this one

nvm nvm

bye yall

[ashen hatch]

this question can be cleanly factored

[tawdry glacier]

actuallhy u

i am all ears

teach me plss

[ashen hatch]

have you learned the A+C, AC trick yet?

[tawdry glacier]

apparently ur suupose to split the middle term

[tawdry glacier]

oh yes

the cross?

[ashen hatch]

it just states that a factorable polynomial can be expressed in the form
(ax+b)(cx+d)
acx²=6x²
bcx+adx=7x
bd=2

[tawdry glacier]

nv

nvm

i didn t lean that

[ashen hatch]

I believe splitting the middle term is the same as what im trying to explain

[noble ether]

If a = 1

The denominator becomes 2 not 16

[ashen hatch]

@tawdry glacier would you like to try solving this?

[noble ether]

You can use the vertex method too

[tawdry glacier]

this is confusing me

yr 9 guys

this is way to techniqual

[tawdry glacier]

ohh now that the teacher drawn before

[ashen hatch]

your goal is to find ax+b and cx+d which matches 6x²+7x+2

its given that ac=6, ad+bc=7, and bd=2

[tawdry glacier]

where does d come from

[ashen hatch]

if a thing can be factored, then it can be expressed in the form (ax+b)(cx+d)

[noble ether]

Fam

Don’t stress it

[tawdry glacier]

man guess who is failing their exam 🙂

[noble ether]

Two expressions which multiply to 20x^2 and sum to 9x

Gimme

[tawdry glacier]

ur doing the wrong question

its the second one

[noble ether]

Two expressions which multiply to give 12X squared and sum to give 7X

[tawdry glacier]

OOOOHH

so its doesnt matter abt the number before

x square

isnt that 3 and 4

[noble ether]

Yes

3x and 4x

[tawdry glacier]

so its (x+3) (x+4)

[noble ether]

To be precise

No

6x^2+7x+2

It’s

6x^2+3x+4x+2

3x(2x+1) + 2(2x+1)

=

(3x+2)(2x+1)

Done

[tawdry glacier]

so what do i search into google to get practis questions

[noble ether]

Factorization question

Quadratics

[tawdry glacier]

ok thankss

so u very literally spilt the mjiddle number

.close

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[gaunt reef]

I have some questions/

[wooden python]

send your questions here

[gaunt reef]

I have a problem with topology and more specifically with the pointwise topology. When we examine a certain convergent sequence of functions (f_n)_n in the metric space of bounded continuous functions with the supremum distance, we find that the convergence of these functions is equivalent to the uniform convergence of these functions in their codomain. Uniform convergence is stronger than pointwise convergence, so why do we need the formalism of the pointwise topology to introduce it? Suppose it is because we want to expand to situations where our codomain is a topological space. Does there also exist a topology in which the convergence is the same as uniform convergence?

[wooden python]

what's a "pointwise topology"?

[gaunt reef]

I think you can also call it the product topology. Here is its subbasis with delta_x being the evaluation function

[wooden python]

... just so we are clear: what's X and what's V?

also im not sure this topology is really explicitly needed. if you really want a topology you can speak of whatever is induced by the sup metric, no?

[gaunt reef]

My textbook on topology introduced it to use to prove that if a net of functions converges over this topology, then the net converges over the codomain of the functions pointwise. And yes sorry, the funciton space we are working with is X^V so all the functions from V->X with X being a topological space

My question is summarised, is there a certain topology who has this property but for uniform convergence

[odd edge]

@gaunt reef Has your question been resolved?

[gaunt reef]

<@&286206848099549185>

[odd edge]

@gaunt reef Has your question been resolved?

[odd edge]

@gaunt reef Has your question been resolved?

[odd edge]

@gaunt reef Has your question been resolved?

[gaunt reef]

If no one can help tell me then i can close the channel

.close

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[bronze cobalt]

$S(f(k), x)=(\sum_{k=0}^{\left\lfloor x \right\rfloor}f(k))+[(\sum_{k=0}^{\left\lceil x \right\rceil}f(k))-(\sum_{k=0}^{\left\lfloor x \right\rfloor}f(k))](x-\left\lfloor x \right\rfloor)$

[clever fjord]

Lime

[bronze cobalt]

does this work as a way to linearly approximate a continuous version of a discreet sum?

f(k) is an arbitrary summand

[manic sleet]

pardon if I'm not fully understanding but doesnt:
[\sum_{k=0}^{\floor{x}}f(k)-\sum_{k=0}^{\floor{x}}f(k)=0]
So your expression simplifies to:
[\mathcal{S}\left(f(k),x\right)=\sum_{k=0}^{\floor{x}}f(k)]

[clever fjord]

PajamaMamaLlama

[bronze cobalt]

one is roof and other is floor

[manic sleet]

ahh I see

[bronze cobalt]

first is roof

[manic sleet]

missed that

[bronze cobalt]

dw

[manic sleet]

so f(k) is an arbitrary summand, like f(k)=1/k, yes?

[bronze cobalt]

yes

[manic sleet]

well

ignoring division by zero lol

[bronze cobalt]

ignoring that

[manic sleet]

graphically it appears to be, how did you derive it?

[bronze cobalt]

i kinda just thought for a while and made a concept

this is the logic:

the first term is the whole sum, but without the incomplete term

[manic sleet]

I see, well it just appears to be a linear extrapolation between the start and end points

so it wouldn't necessarily give the correct value of say x=1/2, but it would get you a linearly approximation of it, sure

[bronze cobalt]

the 2nd and 3rd sums are the term which is partial

and the x minus floor of x is multiplying the partial term by how partial it is

[bronze cobalt]

it is just supposed to be a fun thing for me, not a serious model

thanks for the help man, I appreciate it

.close

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[manic sleet]

[manic sleet]

then you do have a pretty good intuition being able to derive that on the spot

[bronze cobalt]

thanks

[odd edge]

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[crude snow]

Can someone please tell me where I am going wrong?
I'm thinking I have the wrong equation but I know I went wrong somewhere because x cant equal a negative number for this problem.

[crude snow]

I can write my math more clearly if need be, my apologies for it being hard to read

[odd edge]

@crude snow Has your question been resolved?

[normal valve]

Double check your subtraction on the right hand side of the equation.

[odd edge]

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[surreal yacht]

Hello

[surreal yacht]

Im looking for help with this

I know the symbol is a sigma notation but dont understand whats its used for or how to use it

[glass vault]

so you dont know how sigma works?

[surreal yacht]

No

Ive watched videos

But I dont understand it still

[glass vault]

it means you add up all the expressions from where j=14 to where j=44

so in this case its 14+15+16+17...+44

because its just j

you could have an expression of j^2 where it would be 14^2+15^2+16^2...+44^2

thats basically how it works

[surreal yacht]

hmmm

Ok I think I understand a bit better

so we are just adding intergers

[glass vault]

yes

[surreal yacht]

starting at 14

and ending at 44

[glass vault]

yes

[surreal yacht]

So you need to use the formula for the sum of consecutive intergers?

[glass vault]

yeah

[surreal yacht]

and that would be

n(n+1)/2?

[glass vault]

yes but be careful bc it starts at 14

not 1

[surreal yacht]

ok ok

so n(n+14)/2

[glass vault]

thats not how it works

[surreal yacht]

oph

you meant

on the sigma

44
Sigma
14

[glass vault]

yeah you can use the same formula but because the formula adds every number up to 44 you can just minus the formula adding everything up to 13

so 44(44+1)/2 - 13(13+1)/2

[surreal yacht]

ok ok

so it would be something like

44
sigma
j=14

=

44
sigma
j=1

j-

13
sigma
j=1

j

WAIT I UNDERSTAND IT

44 sigma j=1 j= 44x45/2

=990

then you do the same for the 13 sigma

get 91

899

[surreal yacht]

Thank you man

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[open mesa]

how?

[nocturne belfry]

what have you tried

[open mesa]

multiplying it by the conjugate

[nocturne belfry]

The conjugate?

[sharp oak]

What limits are you allowed to use? You'll need one that has trig in it

[open mesa]

i multiplied both the numerator and the denominator by cosx+1

[sharp oak]

This is usually taken as a "common limit" and used to prove other limits

[nocturne belfry]

I guess I mean, why

[open mesa]

sinx/x that's equal to 1

[open mesa]

i don't really know bro i'm just experimenting with it

now i'm stuck

[lone elbow]

||https://proofwiki.org/wiki/Limit_of_(Cosine_(X)_-_1)_over_X_at_Zero/Proof_3||

[open mesa]

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[fresh holly]

im looking for someone who has done vce methods unit 1 and 2

[fresh holly]

i need to learn 14e which is values of circular functions whcih is on the textbook i can send photod

[candid mesa]

deepwoken

[fresh holly]

yes what about it

[odd edge]

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[shrewd trellis]

why do we multiply by the conjugate here

I mean back in surds, we used this method to get rid of surd denominators, but why here

[shrewd trellis]

(I already got the answer to 16, but just curious)

[unkempt lichen]

i'm guessing to leverage the difference of two squares expansion

[shrewd trellis]

if that hint wasnt there, I would be clueless on how to answer this question

[shrewd trellis]

I see

but how does that make it to more simplified?

[north summit]

I guess it is to leverage the identity that $1 - \sin^{2} x = \cos^{2} x$ to further simplify the expression

[clever fjord]

James

[unkempt lichen]

and on the left side of the equation above you see a familiar pattern

[shrewd trellis]

ah, makes sense

[shrewd trellis]

wdym

[unkempt lichen]

^

[north summit]

I think the idea is to have a single term in the denominator so that you can write out the expression as the sum of two fractions

maybe that's a possible instinct that could have led you to make this specific manipulation mentioned in the hint

[shrewd trellis]

u mean multiplication of two frctions

[unkempt lichen]

no

p sure James meant the sum, cuz you need a sum to get to the RHS in the first place

[shrewd trellis]

oh yea

I confused myself w another q

got it

[north summit]

In more a unambiguous wording, $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

[wide oasis]

I guess they just want to let you know that you can write this particular thing in this way too by multiplying the conjugate. Cuz yk simplifying cos theta/ 1-sintheta can help at some places. For example integration.

[clever fjord]

James

[shrewd trellis]

yep

I understand

[north summit]

👍

[shrewd trellis]

thank u guys

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[sharp tide]

quick question bc im not really sure how to google this, how would i find the green angle for point a and any given point

[unkempt lichen]

numerically or by measurement?

if by measurement, use a protractor

[sharp tide]

i need a formula, its for code

[unkempt lichen]

then what information do you have?

i presume you have the coordinates of both points?

[sharp tide]

yes

[unkempt lichen]

ok, and you have two points. unfortunately, that will only give us one line
is your second line fixed?

[sharp tide]

the red line is the y axis so any point on that line i have as well

[unkempt lichen]

so the red line is fixed

[sharp tide]

yes

[unkempt lichen]

let me think for a bit

ok so, i think this should work (but keep in mind that there may be a library function that does this already)

find the difference in x/y-coordinates between the two points, and store them separately
then, the angle θ = arctan(y/x)

[sharp tide]

wait so what do y/x represent in this equation

the x/y difference?

[unkempt lichen]

the difference in y and x coordinates respectively between the two points

[sharp tide]

ah ok thank you ill try this

[west light]

In code you will likely want a function usually called atan2

You'd need to look up what order the arguments should be because they're different in different languages 🙂

[mental lotus]

you might also wanna use atan2(x/y) instead of atan2(y/x)

[unkempt lichen]

if you're allowed to use library functions, please do just use atan2 as mentioned

[mental lotus]

since the angle is wrt the y axis

[unkempt lichen]

actually i'll leave this question to you guys

[sharp tide]

i do have atan2 as a function

[mental lotus]

yea, thats the one

but the angle it returns would be wrt the x-axis

since you want it wrt the y axis, you would need to switch the parameter order

[sharp tide]

if i input it into that function with the parameters switched does that work?

[mental lotus]

you can check it with simple cases like 30deg, 60deg etc which you can verify manually with basic highschool math knowledge

[west light]

You'll also want to make sure you use B-A, and not A-B.

And check which direction your y axis goes, whether positive is up or down (pretty common to be down in computer graphics)

[odd edge]

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[safe canyon]

Hi guys

[safe canyon]

M * adj(M) = det(M) * Inverse(M)

Why is this function true when it comes to matrices

and why is it that the adjoint is equals to the cofactor transpose, and why are there alternating patterns in the first place

[south plume]

that doesn't seem to be true

isn't it M * adj(M) = det(M) * I ?

I instead of Inverse(M)

[safe canyon]

right the inverse right?

[sonic flint]

yeah

its identity

[south plume]

I is identity

[safe canyon]

AAAH forgot like the diagoinal 1's right?

[south plume]

yup

[sonic flint]

lol gave me a little heart attack when I saw your equation

[safe canyon]

Sorry just mistypo been racking my brain about that identity since the past few hours

[south plume]

it's true for M=I !

[safe canyon]

Like why does multiplying the original matrix with the adjoint is equal to the scaling factor

[sonic flint]

oooh I didnt realize that cul

[safe canyon]

And why is the adjoint of the matrix specifically needs the cofactor, like I can't visualize the cofactor really

Been trying this wrapping my head around since the past few hours with gpt but gpt can't really give me a clear picture and just give me like a circular reasoning where it just refer to the proof it stated earlier again as a proof for itself which makes my head hurt

At first I do get where the adjoint is coming from since I used a visualizer where when I aplied the adjoint to the matrix, it overlaps again to the axis and then what is left is just a diagonal matrix with a scaling applied

But then I got lost on how using the cofactor would achieve such result so I deep dive more into it and went to wedge products and it shows there

The cross product of u and v is equal to the wedge product of u and v and the cofactor of a 2 by 2 matrix

Where in some sense it's telling me about the projection but how is it that the cofactor and its transpose induce a projection, or if I am right in my visualization in the first place

wut?

[sand horizon]

<@&268886789983436800>

[copper quarry]

yakuuuuuuu

[odd edge]

@safe canyon Has your question been resolved?

[odd edge]

@safe canyon Has your question been resolved?

[odd edge]

@safe canyon Has your question been resolved?

[amber veldt]

do you mean adjugate or adjoint

oh you mean adjoint, hm, let's see

so you're asking two different questions, right?

one about adjoint, one about adjugate?

Some sources refer to the adjugate matrix of A
as the adjoint matrix of A
.

The use of adjugate may be less common than that of adjoint.

However, as adjoint matrix is also used for the Hermitian conjugate, to avoid ambiguity it is recommended that it not be used.
Oh.

that's the definition I knew, okay, re-reading the question now that I know you're using them as synonyms and the version I knew wasn't universal

so you have three questions

[low locust]

the multiplication M*adj(M) is really just laplace expansion in action

[amber veldt]

Q1)

$$M \operatorname{adj}(M) = \det(M)M^{-1}$$

[clever fjord]

gfauxpas

[low locust]

I instead of M^-1

[amber veldt]

that's your first quesiton, assuming it's true, which i dont think it is

[amber veldt]

ah

$$M \operatorname{adj}(M) = \det(M)I$$

[clever fjord]

gfauxpas

[amber veldt]

this is what youre asking?

yes this is true but it's not the form of the equation I usually see

usually it's written as

$$M^{-1}=(\det(M))}^{-1}\operatorname{adj}(M)$$

but same thing

[low locust]

not same thing

[clever fjord]

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

[amber veldt]

no?

[low locust]

this is always true. even for det M=0

[amber veldt]

if det(M)=0 then you can't write 0^{-1}

[low locust]

yes

which is why yours is not the same

[amber veldt]

ooh I see!

fair!

[low locust]

(or generally in a ring det M doesnt have to be a unit)

[amber veldt]

nevertheless, my version, even if it's weaker, I think is more intuitive

[low locust]

imo no

[amber veldt]

okay well it's a matter of opinion anyway lol

how do we answer the student

[low locust]

the multiplication on the left is the important part

its the laplace expansion

just work through what each entry of the product equals

[amber veldt]

yeah I guess it depends on how you define determinant then @safe canyon

if you define it by laplace expansion then that's why

then again if you define the determinant not by laplace expansion the question "why is it equal to the laplace expansion" is a question you have to answer anyway

sorry im not being very helpful!

[low locust]

this really is just a question where you need to take your time

and slowly work through every entry of the product

[odd edge]

@safe canyon Has your question been resolved?

[safe canyon]

thank youv ery much for the insight maybe for now I just understand first laplace expansion and so on cause I can't seem to grasp formula if theyre spoon fed to me thank you very much

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[fervent summit]

I am designing a problem. The requirements are that it should be based on the concept of The Pancake Problem. It needs to have a definitive, reproducible solution. It should not be NP Hard.
My first thought was "what's the minimum number of flips to sort this given unsorted stack of N pancakes" but I imagine there's a number of different ways to approach it. And better algorithms would give fewer flips and finding the "best" verges on, if not crosses into, NP Hard.
So how could I structure a problem, not necessarily dealing with "minimum" or "best" but still with a specific solution?

[low locust]

a variant of such problems is always "is it possible to do it with <=k moves"

but showing that its not possible is also hard

[fervent summit]

I think by "definitive answer" I should clarify that I want the solution to be a number. Not "yes/no" or "here's the sorted data"

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@fervent summit Has your question been resolved?

[dark raft]

Can this be reduced to some general form

[lean yew]

!occupied

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[dark raft]

Sry

[lean yew]

nps

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[mellow depot]

what are n and m equal to? I'd say n = m = 4, but I want to make sure, because someone else said its m = 3, n = 4

[forest sky]

you are correct

[mellow depot]

thx

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[signal yacht]

oooh greek

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[lament forge]

which part are you stuck on

[lethal musk]

where is a tab at the bottom of the screen saying its an example ill attach and show, ive been able to follow around until this part

idk how they get the 1/196

[lament forge]

do you know how to normally complete the square?

[lethal musk]

no

idk what it means to complete the square

[signal yacht]

you write $ax^2+bx+c$ as $(px+q)^2-r$.

heres an easy example

$x^2+2x+2$ is just $(x+1)^2+1$

[clever fjord]

Percy

[signal yacht]

you introduce a square term, thats called completing the square

[clever fjord]

Percy

[signal yacht]

whoops

[lament forge]

was going to ask about that

[signal yacht]

forgive me for i am sleepy

[lethal musk]

its oki im so braindead from math

im so lost

so with the problem i sent first i should have (x-1/5)^2=4/5 ???

[lament forge]

i'm not going to lie i do it a different way to what you're being shown there and i don't quite understand what it's asking you to do

[lethal musk]

how do you do it

tbh idc how i do it as long as i get the answer the system wants 😭

[lament forge]

my way of doing it doesn't help with this question as it's kinda forcing you to do it one way

like for the example you provided i'd do it like

[lethal musk]

do it like?

[lament forge]

tryna type out a way but i got no idea how to use the bot so hold up rq

[lethal musk]

oh no problem

also ill be back in about 15-30 mins, if you want you can dm me about my problem

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[wanton crown]

i have this math problem and im not sure how to solve it

-f + 2 + 4f = 8 - 3f

i combined like terms by doing (-f+4f) + 2 -> 3f + 2 = 8 - 3f

then i subtracted 3f on both sides

3f - 3f = 0

2 = 8

so i did 8 / 2 and i got 4, But this isnt right can someone explain why?

[wanton crown]

<@&286206848099549185>

[manic sleet]

!15m

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[wanton crown]

ok

[manic sleet]

also careful when subtracting -3f on both sides

-- turns into +

[sand horizon]

3f + 2 = 8-3f
-3f -3f
2 = 8 - 6f

[stable sequoia]

You add thd 3f's not subtract

[wanton crown]

why?

[stable sequoia]

Fym why 3f+2=8-3f so 3f+3f+2=8-3f+3f

6f+2=8

Erm its hard initially but focus on the

Signs

Of the terms

[wanton crown]

like -3f or something?

brro

i dont understand

why are we adding and not subtracting

this does not make sense

[sand horizon]

You can do either as long as you do the same both sides

[wanton crown]

oh shit

thx

[outer quail]

from here 3f + 2 = 8 - 3f to combine life terms futher. the -3f on the right side of the equation needs to added to the left side of the equation. add 3f on both sides so -3f goes away on the right.

[wanton crown]

i have gotten it

[odd edge]

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[graceful parrot]

for number 39 would i multiply by the conjugate

[graceful parrot]

.close

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[graceful parrot]

,calc ln(x^2)

[clever fjord]

The following error occured while calculating:
Error: Undefined function ln

[graceful parrot]

,w derivative of ln(x^2)

[clever fjord]

Results provided by WolframAlpha

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

Requested by dark945850

[graceful parrot]

actually idk what im doing

,reopen

.reopen

[odd edge]

✅

[graceful parrot]

Can someone help me with 39

it looks confusing to do, what would be my first step

i dont think i do quotient rule

or do i

do i do it

[ember oak]

You can use log rules

That will make your life better

Or, you could rationalize the denominator

even better

[graceful parrot]

yea than

.close

tnka

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[sudden vale]

can anyone help me with this i can't understand anything

[green elm]

probably because it's blurry

[short terrace]

Damn potatoes come with cameras these days huh

Trying to read that gave me a headache

[sudden vale]

sorry i was previosly using my computer

[short terrace]

Okay what do you not understand

[short terrace]

No like what about it

It's just an identity

You can multiply and check

[sudden vale]

where did a^rb^s come froom

the terms

are they the for example a^n-3b^3

they represent those terms

[short terrace]

Essentially it's trying to factorise

Do you understand how the a² - b² one works?

[sudden vale]

yeah

[short terrace]

Go on then

[sudden vale]

factorising the a^2-b^2 you get (a-b)(a+b)

[short terrace]

Girl

The question is how

Okay so

Essentially you write

[clever fjord]

Xavier 🌺

$a^2 - b^2 = a^2 + ab - ab - b^2$

[short terrace]

And then you can factorise the right side like you normally would

Do you understand this?

[sudden vale]

yeah yeah

[short terrace]

Now the second one is essentially just this on steroids

[clever fjord]

Xavier 🌺

Essentially you throw in a pair for every single possible $a^pb^q$

[short terrace]

And then you put the positive ones on the front and the negative ones on the back

And then factorise as normal to get the final expression

[sudden vale]

oh ok

[short terrace]

Take some time to convince yourself that this works

[sudden vale]

ok

[short terrace]

(also please tag me for any questions, I'm gonna go freshen up)

[sudden vale]

sure thanks for the help apreciate it

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[dark rain]

Its an online math homework assignment, we have unlimited amount of tries. I keep getting something wrong in this one though. Its a piecewise function. Please help,

[neat echo]

Hi can someone help me with this, I don't know how to start

[digital shore]

!occupied

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[vernal yacht]

Hiiii, you’re a few seconds late, claim another channel 😔

[neat echo]

okayyy!!

[digital shore]

i think your 3rd part needs an open circle

[dark rain]

the {2 if x>2?

[digital shore]

i think that’s the problem

[dark rain]

Na its still wrong apparently, I feel like its the point -2,-3 where there is a closed and open circle on it.

[digital shore]

so it should be closed

i think

[dark rain]

i tried that it was wrong, but apparently its no circle at all

[vernal yacht]

The function is continuous at that point

[digital shore]

well that’s weird

[dark rain]

this question took me like 20 mins 😭

anyways thanks for pointin out bout the other missed point i had before

Have a good day!

.close

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[knotty glen]

im kind of confused what this is asking me

[narrow crypt]

1/(1+t^2) is the derivative of what function whose input is t

@knotty glen

[knotty glen]

yeah

arctan

[narrow crypt]

Ye

U got it

[knotty glen]

arctan

isn't the answer tho

[narrow crypt]

Arctan(t)

[knotty glen]

for number 1

yeah thats not the answe

[narrow crypt]

Did u put ‘arctan’ or ‘arctan(t)’

[knotty glen]

on webassign it used

[knotty glen]

the expression of tan^-1(t) (ignored the ^-1 im just showing you that it uses the -1 expression of tan and doesn't use the word arctan on webassign)

but its not the answer apparently

[narrow crypt]

Tan^{-1} (t)

$\tan^{-1}(t)$ vs ${\tan^-1 (t)}$

[clever fjord]

k

[knotty glen]

yeah idk how to expresss that through discord

my bad

you get me though

[narrow crypt]

Ye

[knotty glen]

it didn't say its correct

im not sure what the issue is

[narrow crypt]

But it seems to be a notation problem for the web

[knotty glen]

oh wait now it worked

?????

okay

[narrow crypt]

Bruh

Ts magical

[odd edge]

@knotty glen Has your question been resolved?

[narrow crypt]

What

Anything else my g

[odd edge]

@knotty glen Has your question been resolved?

[odd edge]

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[granite lagoon]

I copied the solution of my teacher but I can't comprehend what I have copied
It's probably wrong

[unkempt lichen]

!xy looks like original problem statement needed

[odd edge]

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.

[wooden python]

well presumably op needs to find the sum of the first n terms of that series?

[warped glacier]

.

it's already there

[granite lagoon]

The question is to find some of n terms

[outer hinge]

sum?

[granite lagoon]

Ye

Like what would be total of this series by its nth term

[granite lagoon]

Ye

[flat sandal]

Oh

I guess use telescopic method

[wooden python]

ok so i guess step 0 is to write down the general term of the series before attempting to sum it

$\sum_{k=1}^n \frac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{2 \cdot 4 \cdot 6 \cdots (2k+2)}$

ig this is what we're summing

[clever fjord]

Ann

[wooden python]

does that make sense to you @granite lagoon

[granite lagoon]

Yup

But I do not understand how to proceed after this

[flat sandal]

Derive an expression ig

[wooden python]

this is so vague as to be completely unhelpful

[flat sandal]

Hmm

[wooden python]

based on your copy, it looks like your teacher decided to introduce a factor of [(2k+2)-(2k+1)], a.k.a. 1, into the numerator

[granite lagoon]

What did we do with the denominator after separating them into 2 different terms

[wooden python]

in the first, the (2k+2)'s got cancelled

ok i wrote the sum with k as my index variable but your teacher evidently wanted to use r

[odd edge]

@granite lagoon Has your question been resolved?

[granite lagoon]

I think I understand it now

Thx

[odd edge]

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[surreal burrow]

can someone help me

[surreal burrow]

https://discord.com/channels/268882317391429632/1411241252615229542

[meager juniper]

.close

[odd edge]

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[ebon oar]

I didn't understand the denominator part

[ebon oar]

if I set k=1

Denominator should be (2k) rather than (2k+2)

Correct me if I am wrong

[faint knot]

sounds correct, looks like a typo

(2k - 1) on top and (2k) on bottom

[ebon oar]

2×4 first term

[faint knot]

andy focus

[ebon oar]

yes mtt

[faint knot]

we're just looking at the sum right now, you referred to it

wait then why do you want it be 2k instead of 2k + 2?

are you referring to the wallis product? thats not the wallis product at all nvm

[faint knot]

ok if you plug in k = 1 into this, you do get 1 / (2 * 4)

youre supposed to view the denominator as a product of even terms that begins at 2 and ends at (2k + 2)

at k = 1, see that 2k + 2 is 4

and so the denominator is 2 * 4

[faint knot]

the denominator is correctly 2k + 2, since the original sum begins with an extra even term in the denominator

[ebon oar]

When we put k=1 we get 1/4

[faint knot]

and whats 2k + 2

[ebon oar]

What we will do with extra 2?

Rest is fine

[faint knot]

the extra 2?

[ebon oar]

Yeah at the start

[faint knot]

its supposed to be 1/(2 * 4), and indeed it is

yea thats for the 2 * in the 1/(2 * 4) for the first term

do you understand what I said here?

[ebon oar]

Yeah. I got it

I can use double factorial

.close

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[ashen hatch]

Find all positive real numbers x,y, such that x³+y³+(x+y)³+30xy=2000

[ashen hatch]

factored stuff around and got 2(x³+y³)+3xy(x+y+10)=2000

am I supposed to factor out even more?

[faint knot]

factor x^3 + y^3 before grouping things up

[ashen hatch]

(x+y)(x²-xy+y²)?

[faint knot]

then ideally try having things in terms of x + y

that includes doing + 2xy - 2xy on the inside of the second parentheses there

[ashen hatch]

ohh

[faint knot]

youll get this

then notice that 2000 = 2 * 10^3, and consider that to factor this down further

[ashen hatch]

makes sense

thanks

[faint knot]

np

spoiler, when fully factored it looks like this, showing that since x > 0 and y > 0, ||only x + y - 10 can be 0||

[ashen hatch]

I got the same answer as you

yippie

so, is this true for all pairs
(0,10),(1,9),...(9,1),(10,0)

.solved

[odd edge]

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[faint knot]

@ashen hatch theyre asking for positive real numbers, not positive integers, as youve translated it

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[spare crater]

Hello, I need help on pointing out my mistake, the last line I try verifying using my calculator return False

[spare crater]

,help

[clever fjord]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[spare crater]

The 6th line is correct

But the 7th line is incorrect, I can't find what went wrong though

(sorry for bad handwriting)

🙏

[carmine idol]

can you show how you are keying it into the calculator?

[spare crater]

Lemme picture my calc

[carmine idol]

that's a mistake

[spare crater]

Oh

Now that worked

Tysm

[civic wasp]

if you're done, you can type '.close' to close the channel.

[spare crater]

.close

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[viral pumice]

I understand the LCM method we learn in HS, but this is apparently a shortcut of some sort but I don't understand how it works

[tacit haven]

whats supposed to be the method here..

[viral pumice]

like, how did they find the lcm that quick

[woeful briar]

they don't show how, they kinda just say what it is

[viral pumice]

what I use is the table thing
2 12, 16
2 6, 8
2 3, 4
2 3, 2
3 3, 1
1, 1

[viral pumice]

I don't get it

[tacit haven]

take the highest number and divide the rest by the list's highest common factors, then multiply all of them together..?

[viral pumice]

see I sorta get how, they take the largest number and if the other numbers are factors thats the answer, but if not, what do they do then?

[polar echo]

i'm not sure about this but the observation i can make seeing 'these' examples is, they took the bigger number and divided it with the hcf of the other numbers and then whatever remains in denominator is multiplied with the bigger number

[woeful briar]

this is messy

just find the prime factorization of all these numbers

[viral pumice]

okay yeah got it, so for 12 and 16, why did they do 16x3?

[polar echo]

to elaborate, (12,16) , bigger number - 16, divide 16/12 with their HCF, i.e 4, you're left with 4/3 , now they took this 3 and multiplied with 16

[viral pumice]

why isn't it 16 x 6

[polar echo]

similarly works for the other examples in these case

[polar echo]

yeah thats true but op is asking for this particular method, i'm not sure about this either, just tryna work it out

[viral pumice]

OH

[polar echo]

cause 16 isn't divisible by 6 (by my logic/example i elaborated just above)

[viral pumice]

okay how about the second example then, 9, 12, 15

[woeful briar]

gcd(15,12)=gcd(15,9)=3

[polar echo]

right, what i concluded is that they take the numbers pairwise

that is 15/12 , hcf 3, you're left with 4 in denominator then 15/9, you're left with 3 in denominator, hence 15x3x4

now better

if that makes sense

i feel like what we're doing here is the same as the prime factorization method, it's just being done in a peculiar way

[viral pumice]

damn

yeah that makes sense

lemme try on the rest

[polar echo]

do let me know if this logic fails somewhere!

[viral pumice]

what do you use?

[polar echo]

more or less this, the typical method

[viral pumice]

okay so you just divide the largest number with the rest and whatever you get as the denominator you multiply

[polar echo]

is what i can observe, yeah

[viral pumice]

yeah makes sense tysm

i was so confused

tyy

[polar echo]

you're welcomee

[viral pumice]

.close

[odd edge]

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[odd edge]

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[spare crater]

Hello, how can I continue the last (3rd line)

[vivid wyvern]

What were u asked to do?

[strange aspen]

,rcw

[strange aspen]

@spare crater trying to expand ?

[spare crater]

Number 8

Write expressions as polynomials

[strange aspen]

$a^3 \pm b^3 = (a \pm b) (a^2 \mp ab +b^2)$

[clever fjord]

qimmah

[spare crater]

$why is there \pm

[strange aspen]

if it happens that it's negative the expansion would differ

[spare crater]

Ok