help-43

[earnest wyvern]

but read my messages

bye

[digital wraith]

P(1) = (1^2 + 1 -2) Q(1) + A + B

[kind viper]

ok wonderful

now simplify

replace P(1) with 3 also

you should notice the bracket 1^2 + 1 - 2 evaluates to something special

[digital wraith]

oh okay

[digital wraith]

0 so it cancels out Q(x)?

a+b=3?

[kind viper]

yes that is in fact exactly the idea.

[digital wraith]

oh okk

[kind viper]

now do the same, only put the other x-value you know about -- in your case, -2 -- into the equation:

P(x) = (x^2 + x - 2)Q(x) + Ax + B

[digital wraith]

what does b represent btw

[kind viper]

B is the constant term in the remainder.

[digital wraith]

oh ok

[kind viper]

it is just one of the coefficients

nothing else sacred about it

[digital wraith]

i got -2A+B = -12

[kind viper]

ok yeah sounds about right

so now you have two equations:

A + B = 3
-2A + B = 12

do you see how to find A and B from here

[digital wraith]

simultaneous?

[kind viper]

well

they are simultaneous yes

[digital wraith]

a=-3,b=6?

[kind viper]

https://tenor.com/view/how-did-this-happen-bill-wurtz-planet-earth-gif-15407593

ah

crap we forgor the minus sign

or rather more specifically

i forgor the minus sign on the twelve

my bad on that

2nd eq should read -2A + B = -12

[digital wraith]

oh ok

a=5, b = -2

[kind viper]

yup

and that also now matches your answer key

[digital wraith]

oh and then do u just sub into ax + b

[kind viper]

i mean sure yeah

you recover the thing

[digital wraith]

thats what the answer says

5x-2

[kind viper]

yes

[digital wraith]

thank you ann

you're the goat

can i ask more questions or npo

[kind viper]

i'd say close this channel and claim a new one with your new question

[compact pewter]

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[rigid thunder]

I'm not getting z²=-1

[vivid breach]

where did you get z^3 = -w^14 from

[spice saffron]

Exponent laws

$x^a \times x^b = x^{a+b}$

[boreal girder]

pix

[spice saffron]

After this, it will result to -(z^-2), and you can equate it to 1 beccause again, it says |z| = 1

[rigid thunder]

z³=-conj(w⁷)=-w¹⁴ because conj of w=w²

[vivid breach]

if conj(w) = w^2, then conj(w) w = 1, so you can find w faster that way than go through this

[rigid thunder]

The thing is , is it wrong ?

Z³=-w¹⁴ right

[vivid breach]

just checked, I think it is, conj(w) = w^3 instead

[rigid thunder]

w³=1

How conjw=1?

[vivid breach]

going off of the original problem, it suggests w^4 = 1 instead of w^3 = 1 assuming z = ±i

so it would be conj(w) = w^3 instead of conj(w) = w^2 youd be using

[rigid thunder]

Wait what

[vivid breach]

where did you originally get that conj(w) = w^2

[rigid thunder]

w=(1+√3i)/2 and w²=(1-√3i)/2
So obviously conj of w=w²

[vivid breach]

again thats not possible

assuming z = ±i, that means z is order 4, so w being order 3 would imply w = -z^3 = 1 which cannot happen

[rigid thunder]

Yeah

[vivid breach]

oh w is a constant

[rigid thunder]

w is cube root of unity

[vivid breach]

then you can use that conj(w) = w^-1 since w is a root of unity

[rigid thunder]

w^-1=w² again

[vivid breach]

would make this easier at least

yea, better than considering w^14

[rigid thunder]

Okay

Always running into the same conclusion

[vivid breach]

got some bad news, z cannot be ±i

you can just plug in w and z into the equations and see that ±i cannot work

[rigid thunder]

Yes

So I have to treat omega as a constant

The book has treated omega as the cube root of unity everywhere every single time

That's why I was getting confused

[vivid breach]

I dont understand what you are referring to, but assuming that w^3 = 1,

conj(w)^7 = -z^3
w^-7 = -z^3
w^-1 = -z^3
-w^-1 = z^3
(-w^-1)^2 = (z^3)^2
w^-2 = z^6

w^11 = z^-5

w^9 = z
1 = z

w^2 = 1^6 = 1

so omega cannot be a constant

cut off, whoops

[spice saffron]

Wait a minute

[rigid thunder]

Do other countries use different notation for cube root of unity?

[vivid breach]

I use omega, so does wikipedia since thats where I got it from

[spice saffron]

z = +-i in the end solution, but somewhere between the lines it also says |z| = 1

How is that possible?

[vivid breach]

+-i has magnitude 1

[rigid thunder]

Yes

[spice saffron]

Oh yeah

Forgot bbt that this is in argand's diagram

[rigid thunder]

What should I do ?

[vivid breach]

you can treat this as a problem where you have to solve for z and w

just this once, and hopefully they dont pull this on you again

[rigid thunder]

Okay like treating omega as a complex constant

[vivid breach]

yea, since w and z are solved for to be roots of unity, you know conj(w) = w^-1 which makes life easier

from there, try doing what they did in the book (they didnt need that conj(w) = w^-1), where they raise each equation to an integer power then multiply them together to cancel out the w

[rigid thunder]

And omega is not cube root of unity because if z=±i
Then
i³=1/w¹¹
w¹¹=-1/i=i

[vivid breach]

yep

w is forced to be order multiple of 4

[rigid thunder]

Yep so I'm gonna treat omega as constant

[rigid thunder]

Maybe it's the fourth root of unity

Here

[vivid breach]

w^-7 = -z^3
w^-7 = w^2 z^3
w^-9 = z^3
-9 = 3 mod 4

w^11 = z^-5
11 = -5 mod 4

that works

[rigid thunder]

Did if and I found out I have to go with the books method

There's no other way

How to do the 2nd problem?

[vivid breach]

try multiplying both sides by |M + I| or by |M - I|

think of (x + 1)(x^2 - x + 1) and (x - 1)(x^2 + x + 1)

[rigid thunder]

Or try adding -I to both sides

M³-I=-I

(M-I)(M²+M+I)=-I

Like that

[vivid breach]

well if |A||B| = 0, then that means |A| = 0 or |B| = 0

if you have |A||B| = anything else, you cant as easily say something like that

that way you can split apart M^3 - I or M^3 + I into something convenient and have a part of it match the answers

[rigid thunder]

Wait the leading coefficient is 1/2M² 😭😭 if the 1/2 wasn't there then my manipulation would have worked

[vivid breach]

I completely missed that

that is a shame

got a little too excited

[rigid thunder]

Um 🥹

[vivid breach]

oh look you can at least cross out (b) and (c) by assuming M = 0

[rigid thunder]

Yes

We have assumed a matrix which satisfies this condition like a null matrix

And find the conclusion

But is there a solid solution

Obviously then a and d are correct and which is actually the correct answer

[compact pewter]

@rigid thunder Has your question been resolved?

[vivid breach]

@rigid thunder think I have an idea,
if | 1/2 M^2 + M + I | = 0,
then you can factor using complex numbers to get
| M + (1+i)I | | M + (1-i)I | = 0

for any nonzero constant c, notice the following about | M + cI | = 0
that means ( M + cI ) v = 0 for some nonzero vector v
so Mv = -cv, so M^3 v = -c M^2 v = -c^2 Mv = -c^3 v ≠ 0
however M^3 v = 0v = 0
so by contradiction, | M - cI | ≠ 0

from that, that means | M + (1+i)I | | M + (1-i)I | ≠ 0 and so (a) and (d) must be true

[compact pewter]

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[carmine garden]

part(b)

[carmine garden]

Let $v$ be an eigenvector of $T$. Then $T(v)= \lambda v$ so $S^{-1}T(v)= \lambda S^{-1}(v)$ so $S^{-1}TS(v) = \lambda v$. So this proves all of $T$'s eigenvectors are eigenvectors of $S^{-1}TS$.
\
Now let $u$ be an eigenvector of $S^{-1}TS$.So $S^{-1}TS(u) = \lambda u$ so $TS(u)= \lambda S^{-1}(u)$ so $T(u) = \lambda u$.
\
Thus they have the same eigenvectors.

[boreal girder]

wai

[ripe ether]

we dont have that $T(S(v))=\lambda S(v)$

[boreal girder]

qwertytrewq

[hushed magnet]

you cant multiply S^-1 T v = lambda S^-1 v with S from the right and get S^-1 T S v = lambda v

the v is in the way

[carmine garden]

right

[ripe ether]

instead you should try to make sure the $TS(some input)=T(v)$ so you can apply the eigenness.

[boreal girder]

qwertytrewq

[carmine garden]

got it

Well, here I use the fact that S in invertible

There exists a u in the domain of S, such that S(u)= v, where v is an eigenvector of T

[ripe ether]

You can write u more explicitly in terms of v since S is invertible

[carmine garden]

!occupied

[compact pewter]

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

[carmine garden]

hmm

S(u)=v

so TS(u) = T(v)

[copper sierra]

think simple. if Su=v and S invertible then u=?

[ripe ether]

write it in term of u=something with v.

[carmine garden]

S^{-1}v

[copper sierra]

now work with this u

[carmine garden]

so $TS(u)= \lambda v$
\
$S^{-1}TS(u) = \lambda S^{-1}(v)$

[ripe ether]

yeah its usually better to explicitly state what the element is when you know it.

[boreal girder]

wai

[carmine garden]

which is \lambda u

[copper sierra]

so what can we say about u

[carmine garden]

u is an eigenvector S^{-1}TS

[copper sierra]

with eigenval lambda

[carmine garden]

yes

[copper sierra]

problem done

[carmine garden]

huh?

right

okay yea

thanks!

[ripe ether]

not quite still need to show it is a bijection

[carmine garden]

why

[ripe ether]

you have corresponded each eigen vector to one of S^{-1}TS

but you dont know if this correspondence is a bijection

ie, why do you get all the eigenvectors this way?

[carmine garden]

As we don't know the nature of T I suppose

[ripe ether]

the argument is not too difficult

think of how to transform S^{-1}TS into T

[carmine garden]

S^{-1}TS = SS^{-1}TSS^{-1}

[copper sierra]

meh if it only asks about relationship then im fine with only establishing a one way

[ripe ether]

ig the question is quite vague lol

[copper sierra]

plus the problem basically asks about similar maps so id say smth like "the other way is proven by swapping the roles of the maps"

meh maybe thats not allowed but im too lazy

[ripe ether]

fair fair the proof is quite simple

[carmine garden]

So that's it I suppose

[ripe ether]

yeah, i remember you are tryna practice writing proof if you want i can check once you've written down the proof

[carmine garden]

I'll do that from the next problem, If it's fine. Kind of tired rn, so will break for some time

Thanks!

[compact pewter]

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[teal badge]

For i) i got 10!/(2! x 3! )= 302400
is this correct?

[kind viper]

did you mean 10!/(2!*3!)

[teal badge]

this is part ii)

is this correct?

and this is my answer to part iii)

[teal badge]

= 50400

i just wanted to know if my answers were correct

<@&286206848099549185>

.close

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[slender ferry]

this is the graph

how did we get that equation

how

[winged lion]

Maybe the best way is to write down the four equations that you get, if you resolved the absolute value.

With the respective intervals.

[slender ferry]

yes got it

.locs

.close

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[granite panther]

I know that when integrating between intersecting lines to find the area, the limits are the x values of the points of intersection. When finding the intersection area, you can combine the integrals of the same limits into one integral. However, why does the same idea not apply to volumes of revolution? Attached below is a question where I tried to do this and I got the wrong answer.

[granite panther]

For example the answer is supposed to be 108pi but...when I do what I've discussed above it gets a completely different answer?

[subtle helm]

,w pi integral from -3 to 15 (4/3x + 5 - 1/9x^2) dx

[boreal girder]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[subtle helm]

prolly u did ur evaluation wrong

[granite panther]

I plugged it into my calc with the limits and made sure to expand the pi across the two terms

Is this the right brackets?

[granite panther]

I put this directly into my calc

[trim plume]

I also find 108pi if I calculate what you wrote

[granite panther]

Just to clarify you can combine the integrals and take out the pi when the limits are the same

so every step upto the substition was right?

[subtle helm]

yup

ur formula for washer is correct

[granite panther]

Ok i reevaluated and I got it right I think I must have missed a bracket

another clarification you always have to expand the pi across both terms right?

so like the first chunk w the upper limit and the second chunk w the lower limit

[compact pewter]

@granite panther Has your question been resolved?

[compact pewter]

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[jagged ibex]

Could i have help with part iv pls

How do u know In is less than or equal to one from part ii

or is there a different approach

I also found out that In is less than Jn

unless im wrong

[dusky nymph]

looking at the formulas for I_n and J_n in parts (ii) and (iii), wouldn't that imply that 2^(2n) <= 1?

[jagged ibex]

How'd u deduce that??

[dusky nymph]

just set the formula for I_n <= the formula for J_n, and cancel the common stuff on both sides

[jagged ibex]

How do you know if I_n can equal J_n tho

[dusky nymph]

i'm saying, if your claim that I_n <= J_n were true, I could do that and conclude 2^(2n) <= 1

which is of course false

in fact the opposite inequalilty is true

[jagged ibex]

I_n is > J_n?

[native shard]

yuh

[ripe ether]

plug in n=1 you see that I_1>J_1

[jagged ibex]

and i can assume I_n > J_n for all n>1???

[ripe ether]

you'd need to prove it, it is not hard to prove.

[dusky nymph]

well $I_n = 2^{2n} J_n$, so you don't have to assume it, it clearly follows from that

[boreal girder]

Bungo

[jagged ibex]

hmmmm ok

Then how does knowing I_n>J_n help me prove the inequality

[dusky nymph]

look at the inequality in (iv), can you see how it's related to an inequality involving I_n?

[jagged ibex]

Yea its just the numerator of I_n <= denominator of In

[dusky nymph]

right

so in fact it's equivalent to I_n <= 1

can you show that?

[jagged ibex]

Thats what i was asking before

[dusky nymph]

can you show it's true for n = 0?

if so then it'll be true for all n, because of (i)

(for n=0 you could just use the integral to show it)

[jagged ibex]

Mmmm i see i see

ok lemme try rq

Is part i) always less than 1 because its multiplied by a fraction?

[compact pewter]

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[fathom oak]

Need help with question 5 and 6 please

[compact pewter]

@fathom oak Has your question been resolved?

[compact pewter]

@fathom oak Has your question been resolved?

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[ashen mist]

Hi! So I was having a problem specifically on the 3rd row. As you can see this is a quadratic equation problem, where one of the solution is to factorize the problems first into simple forms.

On the 2nd row, you can see that several of the equations are on the right side (of the equal sign) and some are on the left side (of the equal sign). While on the 3rd row, all of the equations that are on the right side of the equal sign have been moved to the left side of the equal sign.

I'm not specifically looking for the final answer of the equation, but rather, I am curious if there are some rules that you need to follow when ordering out the variables, constants, and coefficients when moving all of the equations to the other side of the equal sign, since I've observed that the tutor I watched from YouTube have ordered it out differently when moving it to the left side, but didn't give a reason on why or how did they change the orders.

[stuck flame]

there is no general rule for ordering them as long as you feel comfortable

personally i perfer order term depending on their degree, writing first the term of highest degree and ending in constants

because you wont fail when applying the quadratic formula

[spice saffron]

Can you explain more about "your tutor from YouTube ordering it differently?"

[compact pewter]

@ashen mist Has your question been resolved?

[ashen mist]

Well, in the 2nd row, it would be :
2p - 2 = p^2 + p - 2

And in the 3rd row, it would be :
-p^2 + 2p - p - 2 + 2 = 0

I do understand that in order to switch an equation to the other side of the equal sign means that you also need to switch their equation signs too, from - to + (and vice versa), and from * to / (and vice versa), and I do understand that constants is usually placed in the very right side for the equation for readability.

What I don't understand was when the tutor changed the orders of the variables and coefficients, like :
From p^2 + p to -p^2 + 2p - p

I thought it would still be -p^2 - p after the tutor switched the equation from the right side of the equal sign to the left side of the equal sign

[spice saffron]

The tutor most likely ordered the terms from the highest degree polynomial to the lowest degree polynomial, in this case from -p^2 to 2

[ashen mist]

Small note : in my way of ordering it, it would be :

2p - p^2 - p - 2 + 2 = 0

[spice saffron]

2p - 2 = p^2 + p - 2

To make one side equal to 0, we can substract both sides by (p^2 + p - 2), which results to 2p - 2 - (p^2 + p - 2) = 0, which then results to 2p - 2 - p^2 - p + 2 = 0, but here we can rearrange the equation so that like terms are next to each other

In this case, 2p and -p, and -2 and 2 are like terms

So we would usually rearrange the equations with those like terms together

Thus, resulting to -p^2 + (-p + 2p) + (2 - 2) = 0, or like the tutor wrote, -p^2 + (2p - p) + (-2 + 2) = 0

[ashen mist]

Ohh! So that it would be easier to subtract 2p and p, yeah?

[spice saffron]

Correct

[ashen mist]

I see I see

[spice saffron]

Don't forget about this

That's why the most often types of equations like the quadratic equation and a linear equation is written as ax^2 + bx + c = 0 and y = mx + b respectively

[ashen mist]

Thanks for the insights ^^ sorry if it sounded a bit too easy

[spice saffron]

No worries

[ashen mist]

Closing it now

[spice saffron]

Fyi

[ashen mist]

Mhm?

[spice saffron]

The term teachers usually use that say "pindah ruas" does not really exist

You dont change sides, you operate both sides with the same number

e.g. x + 2 = 0, you don't change 2's side to isolate x, you substract both sides by 2 to isolate x

[ashen mist]

Yeah I have understand that recently, usually they said so that it's for us to make it easier to understand, but it really just made me even more confused when a more difficult equations came in

[spice saffron]

Exactly

If you're a beginner then that term is fine, but once you get to more complicated ones it's harder to "pindah ruas"

[ashen mist]

Usually when I watch foreigners math tutorials, they usually eliminate oth sides, and it was much more understandable than "pindah ruas"

Closing now

[spice saffron]

Alr

[ashen mist]

.close

[compact pewter]

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[teal badge]

are my sols correct

[warped oar]

Oh.

Thank you.

.close

??

[teal badge]

why is this guy impersonating me

[warped oar]

Lol.

[teal badge]

mb

[teal badge]

is the third one -6

@warped oar am i an inspirational figure u aspire to be

[compact pewter]

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[late mulch]

Problem Summary
We are given: - A set of n students S = {s, ..., s} - A set of m classes C = {c, ...,
c} where each class c has: - A time slot t {0, ..., 19} - A minimum capacity min

• A maximum capacity max - Each student s has a ranked list of all 20 time slots, where
  top-5 indices (0–4) are their preferred slots - A global constraint: at least k students must be
  assigned to a class that runs in one of their top-5 time slots
  Objective:
  Assign every student to exactly one class (respecting class capacities) such that at least k
  students are assigned to one of their top-5 preferred time slots. And each class minimum is
  filled.
  Our Attempted Ford-Fulkerson Flow Network
  We are required to solve this using Ford-Fulkerson. Our initial network is constructed as
  follows:
  Node Types:
  • SRC — Super source
  • s — Student nodes
  • t — Time slot nodes (only slots where classes are scheduled)
  • c — Class nodes
  • SN
  Edge Construction:
  • SRC → s with capacity 1
  • s → t if time slot t is allowed for the student
  • t → c if class c runs at t
  • c → SNK with capacity max
  • Lower bounds enforced on c → SNK to guarantee min
  Why I’m Stuck
  Ford-Fulkerson computes feasibility, but not satisfaction unless it’s hardcoded into the
  graph structure.
  Problem:
  All attempts to “enforce satisfaction” within Ford-Fulkerson run into one of two issues: 1.
  Node-splitting approaches (e.g. duplicating students into s_sat and s_unsat) break flow
  conservation or allow students to be assigned multiple times. 2. Routing k units through
  only satisfied paths over-constrains the problem — flow might exist that satisfies all constraints but is rejected due to forced structure.

[late mulch]

Mathematical Formulation
Let: - x = 1 if student s is assigned to time slot t, otherwise 0 - P = set of top-5 time slots
for student s
Subject to: - x = 1 for all i (each student assigned once) - For each class c at time t:
min _{i | t = t} x max - _{i, j P} x k (minimum satisfaction constraint)
This third constraint is not easily modeled with Ford-Fulkerson unless using
What i Need Help With
Im looking for a Ford-Fulkerson-compatible flow network that: - Respects one-unit flow per
student - Fulfills class bounds - Guarantees at least k students are assigned to a preferred
slot
Any structural ideas to enforce this satisfaction constraint without breaking feasibility or
duplicating student flow would be greatly appreciated.

this is a representation of the algorithm i have so far for one of the inputs as you can see all the constraints other than minimum satisfaction are taken care of here

<@&286206848099549185>

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[storm lichen]

what's next

[carmine garden]

I supposed IBP again by taking out -1/(1-x)

[storm lichen]

it returns the same function

you'll get I_n = I_n

maybe this works

[carmine garden]

$1-x=e^y$
\
$\int_{0}^{-\infty} -ye^y(1-e^y)^{n-1}$

is this more workabledoes this work?

[boreal girder]

wai

[carmine garden]

I think it should?

[carmine garden]

this can be solved via a binomial expansion at worst, IBP at best I think

[storm lichen]

nah this is more complicated

i think i solved it though

@carmine garden

lemme js do a quick check

,w int from 0 to 1 x^2 log(1-x) dx

[boreal girder]

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[storm lichen]

,w -H_3 * 1/3

[boreal girder]

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[storm lichen]

okay it works

[storm lichen]

now how do u get -11/18 from -H_3/3

oh right

-1/3 * (1/1 + 1/2 + 1/3) = -1/3 * (6 + 3 + 2)/6 = -11/18

nice

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[storm wolf]

anyone know what to do? im lost

[storm wolf]

prety sure this is the concept im supposed to use

but idk

what to do

[compact pewter]

@storm wolf Has your question been resolved?

[sick shard]

You can try pinging @ Helpers

[compact pewter]

@storm wolf Has your question been resolved?

[brazen quiver]

@storm wolf you're correct that is similar to what you're supposed to use. If (Yn - m)/(c/√n) -> N(0,1) then Y_n -> N(m, c^2/n)

This is a result of the properties of the normal distribution

Please consult your textbook for an explanation 😉

[storm wolf]

this is all he had to say

i dont understand this part

how did we calculate sigma to be 1/ 4f(theta)^2

[slate sand]

Do you know the distribution of sample median of n iid normal distribution RVs?

[storm wolf]

I know Xbar is equal to 1/n times the sum of all xi

[slate sand]

do you?

is that true?

[storm wolf]

Is it not

E(Xbar) = mu

[slate sand]

what's Xbar

[storm wolf]

The X with a line on top

[slate sand]

what does it represent

[storm wolf]

Uh like all X combined

If you have n X’s

Then Xbar = X1 + X2……. Xn ?

[slate sand]

okay okay let's start here instead

[storm wolf]

Alright

[slate sand]

Xbar is the median here

that means, in a sample of n numbers, you order them by value then pick the middle one

in general this doesn't have the same distribution as the sample mean

https://en.wikipedia.org/wiki/Median#Sampling_distribution

[slate sand]

we can look here for some more information regarding this

this says that the sampling distribution of the sample median is asymptotically mean µ and variance (4nf(m))^-1

[storm wolf]

Oh since its symmetric the median is equal to expectation

[slate sand]

yes

[storm wolf]

I understand the mu

But how is the variance

4nfm-1

[slate sand]

are you asking what that means or why it is true?

[storm wolf]

Why it is true

[slate sand]

if you scroll down there's a proof

[storm wolf]

Oh i didnt see the link

Is this only in normal distribution

[slate sand]

no

you can see this statement doesn't include anything about being normal

[storm wolf]

okay

ill just memorize it

[slate sand]

this theorem yeah

for purposes of your exam you memorise it

for purposes of actually doing statistics with it you can just google the result

"what is the sampling distribution of the sample median"

[storm wolf]

but why are they using 2 different things

[slate sand]

it is the same

[slate sand]

one is a definition to be called "asymptotically normal" the other is a theorem that states under what circumstances might we get something that is asymptotically normal

meaning, under what conditions will this thing have the properties described in the definition

[storm wolf]

ahhh

okay]

amd then we bring mu to the right side and divide by sqrtn to get distribution of Xn

right?

[slate sand]

yeah pretty much

but do note that if X ~ N(0, 1) then aX ~ N(0, a^2)

[storm wolf]

mhm

and then

we have the expectation = mu

[slate sand]

this is clear because Var(X) = 1, but Var(aX) = a^2Var(X) = a^2

[storm wolf]

ye

and the variance ^2 = 1/ 4nf(theta)^2

what am i uspposed to do with f(theta)

[slate sand]

f is the distribution's density

[storm wolf]

the pdf in the median right

[slate sand]

yeah

[storm wolf]

okay i got it tysm

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[coral iris]

can someone explain ref for the third option becuase i dont get it

[coral iris]

<@&286206848099549185>

[worn wyvern]

you'd need to solve the equation ig

and find which one corresponds to the right row

do you know the values of x,y,z or this is the entire question ?

[coral iris]

the whole question like you need to slove the maxtix for the third answer

choice

[elfin silo]

Is there any other text in the question?

[worn wyvern]

i guess the third choice is the only that have correct value for x,y,z

[coral iris]

yes thats right but you need to slove it

for valuses of x y and z

how to slove it

[worn wyvern]

you know how to solve a system of 3 equations and 3 variables ?

[coral iris]

nope noty in a maxtix form

[worn wyvern]

you know how to multiply 2 matrices right ?

[coral iris]

nope

my teacher didnt teach ref method

[worn wyvern]

interesting , wonder how would you solve it then

[coral iris]

my teacher didnt teach ref method

lol

[worn wyvern]

interesting , never heard of it

i just multiply both of the matrices , and then find the solutions

[
\begin{pmatrix}
x+y+z \
1.5x + 2y + 2.5z \
x - 2y
\end{pmatrix}

\begin{pmatrix}
15 \
27.5 \
0
\end{pmatrix}
]

[coral iris]

what do you mean

[boreal girder]

<rajel />

[worn wyvern]

[
\begin{cases}
x+y+z=15 \
1.5x +2y + 2.5z = 27.5 \
x-2y=0
\end{cases}
]

[boreal girder]

<rajel />

[worn wyvern]

by what ?

[coral iris]

by mutokiuy them together

by dong them together

[worn wyvern]

matrice A * matrice B = matrice C

ofc by respecting their dimensions

yeah since you didnt learn it nvm

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[strange pendant]

need some help with 1)

[compact pewter]

@strange pendant Has your question been resolved?

[elfin silo]

Hello again

What did you try?

[strange pendant]

dunno how to

https://tenor.com/view/luigi-stare-gif-24984189

[elfin silo]

To construct linear functions we clarified we need to assign values on a basis
And we're interested in f(S) and f(T), so we need a basis for both S and T, wouldn't you agree?

[strange pendant]

?

[elfin silo]

What is your question?

[strange pendant]

basis of S is given

[strange pendant]

we need a basis for S + T

[elfin silo]

What's one of T?

[strange pendant]

,w nullspace {{1,-1,0,1},{1,0,0,2}}

[boreal girder]

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[strange pendant]

T = <(-2,-1,0,1),(0,0,1,0)>

@elfin silo

[elfin silo]

Yes

What do you think T+S is then?

[strange pendant]

we need SnT

you follow?

[elfin silo]

Sure

[strange pendant]

@elfin silo

[strange pendant]

@elfin silo

https://tenor.com/view/luigi-stare-gif-24984189

[elfin silo]

Okay

If you add a vector outside S+T you also get a basis of R4
Then how do we define a function as required?

[strange pendant]

i am trying

can you be more explicit

[elfin silo]

To define a linear function what is enough to assign?

[strange pendant]

@elfin silo

[elfin silo]

Good

[strange pendant]

?

[elfin silo]

It's correct, what's your question?

[strange pendant]

ok

what about the other problem?

[elfin silo]

Have you tried? Because it seems you got the gist

[strange pendant]

second one is harder because of fof

[elfin silo]

The principle is the same

[strange pendant]

fof makes it harder

[elfin silo]

Start with the setup and the other requests

See what you get

[strange pendant]

is hard man

[elfin silo]

Its no so bad
See what you get first with the rest

[strange pendant]

wdym?

[elfin silo]

Set it up first
Find a basis of T
Find one of R4
Check if S is in T
Make it so that ker + S = T
Etc.

[strange pendant]

but

S is entirely contained in T

do you follow?

[elfin silo]

Yes

[strange pendant]

so what do i do? exercise is fucked

[elfin silo]

Why u think it's fucked?

Did you find a basis of T?

[strange pendant]

Nuf contained in Nu fof

[strange pendant]

yes

,w nullspace {{1,-2,1,-1}}

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[strange pendant]

https://tenor.com/view/luigi-stare-gif-24984189

[elfin silo]

Did u find what to impose to get kerf + S = T?

[strange pendant]

yes, but

dim nu fof = 3

[elfin silo]

Sure

[strange pendant]

dim nuf + 1 = 3

dim nuf = 2

do you follow?

[elfin silo]

Okay?

[elfin silo]

What did u get here?

[strange pendant]

like 2 or 3 conditions, let me show the sketch

[strange pendant]

@elfin silo

[elfin silo]

Might be convenient to use the basis of S to build a basis of T, so u can control where S goes (in particular so that its not in the kernel)

[strange pendant]

?

[elfin silo]

You wrote a basis of T that doesn't use the given basis vector of S

That's inconvenient since we want to know how f behaves on S

[strange pendant]

i told you it's hard

[elfin silo]

Well now think about it
You need a kernel to be 2-dim
Can (2,1,3,3) be in the kernel?

[strange pendant]

idk

[elfin silo]

You wrote the answer there

Nu(f)∩S = {0}

[strange pendant]

it can't

[elfin silo]

So (2,1,3,3) can't go to 0

We only have 2 more vectors in the basis that belong to T

And the kernel is all contained in T and is 2-dim

So where do they go?

[strange pendant]

@elfin silo

[elfin silo]

Yes

[strange pendant]

was this one hard or easy?

https://tenor.com/view/luigi-stare-gif-24984189

[strange pendant]

i think this was really helpful, i appreciate it luigi

prolly couldn't have done it without u

for me it's hard, but idk, I'm just starting my linear algebra journey

idk, this exercises are good, makes you think hard, no?

I appreciate it

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[quartz yoke]

help

[quartz yoke]

wait maybe its better if i use my old channel

.close

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[quartz yoke]

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[serene blade]

i dont know where i went wrong

[sick shard]

It's open brackets (-π,π) so π and -π aren't included in domain

[serene blade]

arnt the critical numbers the high and low pretty much

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[flint wing]

so for this example if i were to expand along the 2nd row or third row or third column for example i would still end up with the same determinant?

[rigid perch]

the determinant is the same no matter which row or column you pick

[flint wing]

okay thanks

.closed

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[flint wing]

.reopen

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✅

[flint wing]

i dont get how they picked these

the determinants for those

like i get for expansion along first row there was a general pattern

[dusky nymph]

mentally remove the row and column containing the coefficient

what remains is the submatrix that goes with that coefficient

[flint wing]

like removing the second row and column?

[dusky nymph]

for the 2x2 matrix associated with the 0 coefficient, yea

[rigid perch]

for the first one, it's row 1 column 2 so remove that row and that column

[dusky nymph]

poor man's markup

[flint wing]

i dont think i understand

[dusky nymph]

if you're expanding along column 2, you start with row 1 and column 2

the coefficient (circled) is -3

times -1 since it's an odd number of steps from the upper left

the submatrix that goes with that is everything that doesn't have a red line through it

[flint wing]

they mentioned the checkerboard thing of +-, is that what ur referring to here?

[dusky nymph]

yes

[rigid perch]

yes, you choose the entry of the checkerboard corresponding to the expansion point

[flint wing]

okay i get the first part of -(-3) * det|sub matrix 1| forgive my notation and then

the second part we go down the column?

so we look at the 0 coefficient?

[dusky nymph]

yep 0 is your next expansion point

this one doesn't get a - sign (due to the checkerboard pattern)

and the submatrix is what?

[flint wing]

hold on let me get microsoft paint

[dusky nymph]

sure

[flint wing]

ok im definitiely doing somethign wrong

[dusky nymph]

well your red lines are in the right place

[flint wing]

ohh wait

[dusky nymph]

so just read off the remaining entries to form your submatrix that goes with 0

[flint wing]

so my sub matrix is gonna be 5,2,2,3 respectively

[dusky nymph]

yep

[flint wing]

ohhh okay

and for the last coefficient its gonna have a -(-1)

[dusky nymph]

yep you got this

[flint wing]

ohhhhh ok that makes sense now thank u bungo

[dusky nymph]

btw

[rigid perch]

one thing we notice is that we don't actually have to calculate the determinant of the second submatrix, since it will be multiplied by 0 anyway. so in general can be useful to expand out along whichever row or column has the most 0's

[dusky nymph]

oops sniped by cloud haha

yep zeros are your friends when doing determinant expansions

[flint wing]

ohhh yeah i was about to say if there was like specific things i should be looking out for to choose where to expand

that makes sense

[dusky nymph]

generally pick whichever row or column has the most zeros

[flint wing]

thank u guys

[dusky nymph]

yw cheers!

[flint wing]

.solved

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[flint wing]

just checking these are triangular matrcies because of where the zeroes are, that i highlighted in the image?

[rigid perch]

well you didn't show all of the requisite 0's, but those are most of them

[flint wing]

what else am i missing?

[strange pendant]

yeah, all zeros below the diagonal, no? and above

[rigid perch]

for an upper triangular matrix, every entry below the main diagonal should be 0

and similarly for lower triangular matrices

[flint wing]

can it be any diagonal?

[rigid perch]

just the main diagonal is what's important

[flint wing]

which one is the main diagonal

[storm lichen]

a_ii

[rigid perch]

as flux mentioned, the entries of the main diagonal are the ones where the row and column numbers are equal

[flint wing]

so here we are looking at the 2 and 0 in the middle column and middle row?

[rigid perch]

not entirely sure what you're referring to

[flint wing]

allg i got it

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[round tulip]

I understand what the question is asking, but just have no clue how to start with representing it as a PIE

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<@&286206848099549185>

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[marsh void]

bit of an weird question but

[marsh void]

1/[x* ln^2(x)] dx

means

this right

[proud halo]

yeah

it would appear so in this case

sometimes f^2(x) could mean composition for pedants

[kind viper]

2c25?

[proud halo]

2025

[kind viper]

what's with that zero 😭

[storm lichen]

ln(x) = k

[marsh void]

?

[worn wyvern]

ibp

[storm lichen]

substitute ln(x) = k

[flint wing]

u is better than k

[marsh void]

when solving for it?

[storm lichen]

yes

[marsh void]

cus i got the answer with just integrating by parts

[storm lichen]

why integrate by parts when you can substitute

[worn wyvern]

@subtle helm

[storm lichen]

,w int from 2025 to inf 1/(x(lnx)^2) dx

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[storm lichen]

oh yeah

log c

[marsh void]

alr tnx everyone

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[marsh void]

.reopen

[compact pewter]

✅

[marsh void]

are there any other way to form this from

with ln 2025

than this

can't be cus the multiple of 2025 doesnt contain anything except 3 and 5 right?

[marsh void]

mustve been typo or smth cus i submitted it with the ans i found and that was correct

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[subtle helm]

based

[subtle helm]

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[coral halo]

am i supposed to start reading from a blank or a

[coral halo]

for (b)

[river portal]

a

Initially a Turing machine is always looking at the left-most part of the non-blank portion on the tape

[coral halo]

well then my lecture slides has 0 consistency at all

diamond symbol for blank btw

[river portal]

Well yeah even as a convention starting at a blank makes no sense whatsoever

[coral halo]

but if i don't start with a blank this machine gets stuck at literally anything lol

since it only has transitions when reading blanks

at the start anyway

[river portal]

That's weird, what does the definition that you use say?

[coral halo]

yeah its kinda contradicting itself rn

[river portal]

Alright, I would inform the professor about that

[coral halo]

yeah

but fuck it i'll start at the blank since theres like no answer otherwise

[river portal]

Yeah as for the question itself I would do the same

[coral halo]

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[subtle dagger]

claim

[subtle dagger]

my mathbook has this as an example for polynomial longdivision and they say they do 2x because 2x(x+1) is the largest thing that fits in 2x^2+5x. But I do not understand where they get the +3 from

[subtle helm]

what is the largest thing that fits 3x+5

[subtle dagger]

3x+3

ohhhhhh

the 3 is from like the second round

[subtle helm]

yup

[subtle dagger]

ye ok then it makes sense

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[carmine garden]

You want to solve this?

[ebon lotus]

Is that greater than or equal to?

[carmine garden]

≥ I think

[ebon lotus]

Hmm

Ok

This is gonna take a lotta time to do

[carmine garden]

not really IMO

[ebon lotus]

Considering the base of the powers aren't the same

[carmine garden]

3 functions can be positive or negative

[ebon lotus]

Oh this is from IMO?

[carmine garden]

no

In my Opinion == IMO

[ebon lotus]

Ah shi

Ok

[warped oar]

lol

What to do?

[lime juniper]

if you had an inequality like 5y<0 how would you solve it

[warped oar]

I would solve it using the RULES OF MATHEMATICS.

y must be negative

[lime juniper]

howd you isolate y?

[warped oar]

Because if y were 0 or positive, we would get a contradiction

idk

[lime juniper]

you could divide both sides by 5

agree?

disagree?

[warped oar]

well

certainly

I agree

[lime juniper]

what about

ye^y > 0

[warped oar]

e^y is always positive.

So I think this depends on the sign of y

[lime juniper]

oh, so what youre saying is

[dire fiber]

Wavy curve merhod

[lime juniper]

you can divide both sides by functions that are always positive?

[warped oar]

yes

[dire fiber]

You dont need to

Multiple even powers on both sided and for odd powers, multiply their squares on both sides

the inequality will hold

And the critical points remain same

Then check multiplicity of the roots and get the domain

[lime juniper]

multiplying even powers on both sides is the same as dividing even powers on both sides

[warped oar]

critical points?

[dire fiber]

Nah i know but i assume the natiural first step is to get rid of the denominator

[lime juniper]

dividing both sides by 1/x² is the same as multiplying both sides by x²...

anyway take over for me then, you seem enthusiastic

[dire fiber]

For the odd powers in the denominator, square it and multiply on both sides

[warped oar]

[(x+3)(2x+7)^5]/[x-2)^3] >= 0

[dire fiber]

Multiply both sides by (x-2)^6

[warped oar]

interesting

(x+3)(2x+7)^5 (x-2)^3 >= 0

[dire fiber]

yes now use wavy curve method

[warped oar]

Oh.

Thank you.

Very much.

[lime juniper]

keep in mind

[warped oar]

Isn't x = 5 also a solution

[lime juniper]

division by zero in the origjnal form limits the domain

and in any intermediate steps

[dire fiber]

Btw if its a strict inequality do not get rid of the even powers

you will need to know all roots

[warped oar]

okay, thanks y'all

[lime juniper]

how did you get rid of (x-5)²?

how'd it disappear

[dire fiber]

i guess he divided it on both sides

[lime juniper]

yeah that's what im.guessing. if x=5, you divided by 0

[dire fiber]

His answer wont change in this case i guess

If he hadnt divided it he would have gotten one extra root

but the graph would just bounce of zero

at x = 5

[lime juniper]

it's good practice to.deal with the division by zero poiunts seperately anyway

trains oneself to watch out for such mistakes in the future

[warped oar]

🆗

[warped oar]

certainly

[lime juniper]

also x=2 and x=-7 are not in the solutiom

my approach would have been to multiply both sides by

(x-2)²(x+7)⁴/((2x+7)⁴(x-5)²)

and then examime x=-7/2 and x=5 separately

[dire fiber]

Isnt it better to avoid division so that there's no need to look at domain restriction ither than the ones in the original equation

[lime juniper]

that's a matter of opinion. you thought yes, i thought not if youre careful

[dire fiber]

yeah

[lime juniper]

just offering a different approach, your approach isn't wrong

[dire fiber]

Nah its alr

[compact pewter]

@warped oar Has your question been resolved?

[warped oar]

All-right

Thanks guys

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[tropic agate]

Given two rotation matrices $A,B$ in $\mathbb R^3$, I'm trying to find the new axis of rotation of their product. Is there some efficient way of going about this. The matrices are $$A=\begin{pmatrix}1&0&0\ 0&\cos\phi&-\sin\phi\ 0&\sin\phi &\cos\phi\end{pmatrix}\text{ and } B=\begin{pmatrix}\cos\psi&-\sin\psi&0\ \sin\psi&\cos\psi&0\ 0&0 &1\end{pmatrix}.$$ I know the axis of rotation is the null space of $AB-I$, but this gets messy pretty quickly.

[boreal girder]

psie

[quartz yoke]

Where did you get that question from?

[tropic agate]

A linear algebra book.

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[worn wyvern]

9 balls , 4 red 3 green ,2 black
We draw 3 balls at random; what is the total number of possibilities for drawing 3 red balls in a successive draw without replacement

[worn wyvern]

i know it can be done with $A_3^9=\frac{9!}{(9-3)!}$

[boreal girder]

<rajel />

[worn wyvern]

but i dont understand how the value makes any sens

,calc 987

[boreal girder]

Result:

504

[shell horizon]

Do you mean to ask what is the possibility that you draw 3 reds?

[worn wyvern]

not the possibility but , how many ways we can draw 3 red balls at once

[shell horizon]

I'm guessing you draw 3 balls at once 3 times?

What's the original question word for word

[worn wyvern]

what i wrote is entirely wrong

504 is the number of all posibilities

i.e if we draw balls in a successive way without replacement , than the possibilities number decreases

9,8,7

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[near swan]

allo a question for linear algebra

[near swan]

T(v) = T(u1+u2) = T(u1) +T(u2) = 0

T(u1) = -T(u2)

from here i wanna say that this is only possible if T(u1) = T(u2) = 0

am I allowed to claim this?

[eternal marten]

nah

[near swan]

can you explain?

[eternal marten]

ye sure

t(u1) = t(u2) you should only be able to knojw that (u1-uw) is in ker(T) for exmaple lets say hmm T(x,y) is x then T(1,0) is t(1,5) = 1 not equal to 0 so if t(u1) = t(u2) means the vector map would achieve the same output and can be any vector in this sense in the reange of T but not necessarily 0

sorry if my explanation is a bit weird

[near swan]

I get it no worries

mm thank you

I think there's probably a part I need to use U1 intersect U2 = {0}

oh!

I GET IT

T1 and T2 map within their own subspace

[eternal marten]

now u do?

[near swan]

so T1(u1) is in U1 and T2(u2) is in U2

[eternal marten]

yup

nice

[near swan]

their intersection is {0}

so they must be 0

[eternal marten]

you are in college?

[near swan]

yuh

[eternal marten]

I see

[near swan]

ty byebye

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[eternal marten]

ye np

[compact pewter]

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[ruby mason]

T1 and T2 weren’t even defined wtf

[ruby mason]

oops

!clos

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[near swan]

mbmb it was in another part of the q just forgot to send it 😅

[ruby mason]

i see, cherry guy was kinda sus anyway

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[north maple]

hello, can anyone explain why the answer for part d iv is that it cannot be evaluated?

[subtle helm]

Consider end reuslt’s dimension

[north maple]

can u explain further?

[subtle helm]

It would be 2x3 right?

DC^T E

[north maple]

yes

[subtle helm]

Can u multiply 2x3 with 2x3?

[north maple]

nope

ohh i get it now

thank you very much

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[north maple]

hi, can anyone help with this qn, the second part

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[north maple]

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[chrome shadow]

kp + jq = GCD(p, q) where p, q are natural numbers and k, j are integers

[chrome shadow]

Why do k, j always exist

[subtle helm]

Wdym?

[chrome shadow]

Why can we always have such k, j such that kp + jq = GCD(p, q)

[kind viper]

i take it you haven't seen the proof?

[chrome shadow]

Proof of ?

Their existence?

[kind viper]

Bézout's lemma

[chrome shadow]

No I haven't

[kind viper]

ok

are you familiar with the Euclidean algorithm

[chrome shadow]

For gcd?

I need to refresh

I'll check it out and be back and ping you is that fine

[kind viper]

sure but i was gonna link a convo in a different help channel where i explained it to someone else

#help-30 message

[chrome shadow]

Thx

[compact pewter]

@chrome shadow Has your question been resolved?

[chrome shadow]

Ah I got it

So I suppose the idea is that

since GCD of p and q is also GCD of p and |p-q|, you can continue this until you reach the GCD itself. And so you can reverse it and express GCD as a linear combination of p and q

GCD(p, q) = GCD(lc(p,q), lc(p,q)) = GCD(GCD(p,q), GCD(p,q))
=> GCD(p,q) = lc(p, q)

[chrome shadow]

Is this reasoning accurate? @kind viper

lc being linear combination

[kind viper]

notation is kinda shoddy but yes you have roughly the right idea

[chrome shadow]

Thanks a lot;

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[subtle helm]

Test converge for
[ \int_0^\pi \frac{dt}{\sqrt{t} + \sin t} ]

[boreal girder]

k

[dusky nymph]

any thoughts?

[subtle helm]

Tried conjugating

But doesnt look promising

Even after splitting the fraction

[dusky nymph]

maybe try a simple comparison test?

[subtle helm]

${\frac{1}{\sqrt{x}}}$?

[boreal girder]

k

[dusky nymph]

yea that's what i had in mind