help-49

[cosmic zenith]

apparently i need to make sinx = 0?

[inner wolf]

Yes, or cosx=1/3

[cosmic zenith]

And how would that work?

[inner wolf]

Sin is 0 for nπ, so you have -180, 0 and 180 as answers

[cosmic zenith]

yeah that makes sense'

[inner wolf]

For cos you need to use calculator as we don't know the exact value for cos inverse 1/3

[cosmic zenith]

do you look at it as if both sinx and cos x are x?

x(3x-1) as if its like this?

[inner wolf]

Remember to equate this to 2x+30

[cosmic zenith]

yeah

[inner wolf]

Yes

[cosmic zenith]

ahh okay

thanks man

[inner wolf]

It's just factors

[cosmic zenith]

have a nice day

im boutta have my exam

next week

[inner wolf]

If multiplication of 2 quantities is given to be zero, one of them or both of them could be zero

Glgl

[cosmic zenith]

thank sman

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[eternal pine]

what methods do you guys use to factorize stuff like this

[eternal pine]

or this

i only know the methods for binomes

[solid hornet]

Find factors of 3

And sub those values in for x

If it adds up to 0

It is a factor

It’s like

Integral zero theorem I think It was called

P(x)=x^4 + 4x +3

Factors of 3 is +-1, +-3

So trial and error those values for x

And then you can do synthetic division

Lmk when u get there if u need help

[eternal pine]

[eternal pine]

i get everything about those but the problem is the videos only show x^2

[pallid juniper]

dont go for rational roots or smth

these expressions are easy to evalutate you just need to make perfect square

[solid hornet]

Huh

[pallid juniper]

wait

[solid hornet]

Wdym

[eternal pine]

Examples shown in youtube videos are usually x^2

i can't look up the technique for higher grade polynomes anywhere

[solid hornet]

Do you know integral zero theorem

[eternal pine]

https://www.youtube.com/watch?v=KziDD1hVEMI oh sec that's a hint

[solid hornet]

Yea try watching that

If you don’t know synthetic division search for that too

[midnight plank]

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

[solid hornet]

Oh I c

I never learned like that

[eternal pine]

Where did you learn it like that

[pallid juniper]

oh i have come across such problems many times during my oly prep
so i developed this technique myself but im sure more ppl know about this .

i also got one hammer method to factor polynomials of form
x^4 -16x-12

also there is one hammer method to factorize or solve biquadratic equation with cubic terms as well
(monic)

[eternal pine]

in the first one both +1 and -1 equal 0

so how do i know which of the 2 is the correct one

or both are directly factors

[pallid juniper]

both are the factors

+1 and -1

[eternal pine]

AH ok makes sense then

[midnight plank]

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[amber abyss]

Can someone help with this? Idk how to answer

[boreal belfry]

Think about the relationship that acceleration is the change of velocity per unit time

And construct a differential equation

[amber abyss]

Velocity is the integral of acceleration

[amber abyss]

Wait wdym?

[boreal belfry]

Yep

[sage helm]

just intergate

[boreal belfry]

So workout the integral with initial value 5

[amber abyss]

But it says initial speed

s(t)

[boreal belfry]

At t= 0

It is 5

[amber abyss]

v(t)=t^2

[boreal belfry]

Yep. But there needs to be a +c

For the constant initial speed

[amber abyss]

Ohh

[boreal belfry]

Yep and you know the value for t = 0

[amber abyss]

v(t)=C

[boreal belfry]

So what is c

It's the initial speed

[amber abyss]

Integrate it again?

@boreal belfry

[boreal belfry]

No c is a constant that is lost on the differentiation. So you are setting c to 5

[amber abyss]

Ohh

Do i integrate 5=C?

[boreal belfry]

No

You have the equation v(t) = t2+ 5

[amber abyss]

Integrate t^2+5 from 0 to 5?

[boreal belfry]

You can just plug in t for the end speed

[amber abyss]

Oh ok

Thank you

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[simple falcon]

Please help me to understand this

[simple falcon]

Tried many times but cant reach the conclusion

[cursive dawn]

AB = A, and BA = B. So you can multiply both sides by B or A to get ABA = A^2 and BAB = B^2.

However, AB still equals A, and BA still equals B, so you can probably use that

[midnight plank]

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[round badger]

hi, could anyone help me understand the 4th question, it dosent make sense as we just answered it in the last question, the vertex is -3,25 and bc the leading coeficient is a - the vertex is the maximum

[round badger]

is that all we need to say ?

[latent elk]

Have u learnt calculus?

[round badger]

nope

im in precalc rn

[echo forge]

use the vertex

[round badger]

yes

i know it

its -3,25

[echo forge]

and put it into the equation

[round badger]

i found it using -b/2a

[echo forge]

to get corresponding y value

[round badger]

its 25

[echo forge]

which will be ur maximum

which they have asked

[round badger]

its that easy ?

[echo forge]

yep

[round badger]

its basically almost the same as the c)

i just needed to tell if it was max or min

and bc leading coeficient is - its the max

thanks for the clarity anyways

have a good day and ty for the help : )

[latent elk]

WAIT

Enlight me on the vertex thing

[round badger]

wdym

[latent elk]

How do u find max/min using vertex thinggy?

[round badger]

its just -b/2a to find x value and then plug it into the function to find the y value of the vertex

[latent elk]

I didnt get taught it back then

[round badger]

-b/2a is also to find the axis of symmetry of the parabola

i have another question, when i completed the square i found 2(x+5/2)^2 -9/2

does the 2 mean a vertical or horizontal stretch ?

[echo forge]

vertical

[latent elk]

BRUH, why I didnt get taught this....

[echo forge]

if f(x) = x^2 and g(x) = 2f(x) you are basically doubling y values of f(x) by 2 in g(x)

which means that your curve will be much steeper or closer to y-axis

[round badger]

theres another way to find it, when you completed the square a(x-b)+c, the b is also the axis of symmetry

bc to find it you do -b/2a

and it makes sense because b represents the horizontal transformation

[latent elk]

Well, it probably I was taught quadratic equation in 9 grade and get taught precalc at 10

[round badger]

it will tell you if the axis of symmetry moved 2 units to the left or to the right for example

[latent elk]

Precalc offers a better way to find it then

lol

[round badger]

yep

precalc is fun

[round badger]

same here

[latent elk]

My precalc was brutal

[round badger]

yes but on the complete the square form a(x-b)+c i dont know if a is af(x) or f(ax)

so if its a vertical or horizontal

[round badger]

i think precalc trig will be brutal for me

we just started it

what grade are u in now ?

[latent elk]

undergrat

[round badger]

ig ur major is math ?

[latent elk]

Engineering

[round badger]

nice

do you like it ?

[latent elk]

Nope

[round badger]

im still not sure if i want to do enginerring or finance

[round badger]

lmfao

thats problematic

[latent elk]

Mine is civil, i dislike it so much

I should have choose CS

[round badger]

so basically architetcure ?

the math and physics behind architecture

[latent elk]

Totally different thing

[latent elk]

Lowkey yes

[round badger]

it seems fun as well

[round badger]

well civil enginerring is pretty much buildiing roads, infrastructure etc right ?

[latent elk]

yep

I hate it thats all

CS or AP is way better

[round badger]

whats ap ?

[latent elk]

Applied Physic

[round badger]

ohhh

nice

physics is fun

you basically use math to solve real life problems

best of both worlds

[latent elk]

My professor once said: "There's nothing fun about physic, only pain and suffering "

😫

[round badger]

do you still have time to change majors ?

[round badger]

lol

ig the physics i do is pretty simple compared to yours

@echo forge could you help with something ?

[latent elk]

Gonna grad next month LMAO

[round badger]

oh

fuck

so you have done the entire four years even tho you dont like it

i feel sorry for u

this year is the first year i have biology and i cant have one more class of it

you had classes everyday for four years on smth you hate

thats ruff

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[frozen totem]

Suppose $s\geq3$ is an arbitary odd number, let $p_1<p_2<\dots<p_s$ be odd primes such that $p_1+p_2>p_s$. Show that such primes always exist.\[.5em]
We start by taking $\epsilon\in\mathbb{R}^+$ such that $(1+\epsilon)^{s-1}<2$. Let $p_1$ be a~prime large enough, that for any prime $q>p_1$ there exists a~prime $q'$ satisfying $q<q'<(1+\epsilon)q$. Then let $p_2,\dots,p_s$ be consecutive primes. It follows that $p_s<p_1(1+\epsilon)^{s-1}<2p_1<p_1+p_2$.\[.5em]
My question is, why does the afformentioned prime $q'$ always exist? What theorem on prime gaps do we use? Does this proof even work?

[grand pond]

🇵🇸Mína🔆

[frozen totem]

specifically, we take p1 to be a prime, such that for a range of consecutive primes p2,...,ps, the number p1(1+ε) exceeds any gap in said range

How can we say that that'll always exist for an arbitraty s?

if anyone has a different proof, I'll be happy to see it

[midnight plank]

@frozen totem Has your question been resolved?

[frozen totem]

<@&286206848099549185>

[midnight plank]

@frozen totem Has your question been resolved?

[final monolith]

apparently it comes from the N/log(N) asymptote

[frozen totem]

thank you so much! Could you throw me an article to cite?

I won't prove it in the thing, I'll just quote it

[final monolith]

it's from here https://en.wikipedia.org/wiki/Bertrand's_postulate

[frozen totem]

tysm!

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[last nebula]

how do I find the diamond and square

[novel herald]

x^2 + dx + 12 = (x+3)(x+s)

distribute the right side and compare the coefficients

[junior flower]

m8of48...

[novel herald]

slayla

[radiant trout]

or just plug -3 in

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[minor turtle]

I don’t really understand this problem can someone help

[minor turtle]

I assume length is getting the magnitude

But direction I don’t really know

Is it just finding the unit vector

[novel herald]

they might want the angles

[midnight plank]

@minor turtle Has your question been resolved?

[minor turtle]

Like for example

This is a similar question with a cross product

Where it asks for length and direction

Would the direction just be the unit vector of the cross product/vector

?

Hell

Hello

<@&286206848099549185>

[hasty mantle]

Direction is the angle of the vector from standard position

Arctan(y/x) gives the angle of direction

[minor turtle]

Why is the direction here different then

Not sure about this

Direction vs angle direction

[hasty mantle]

So length is just sqrt of the sum of the terms squared

If you divide the sum of your vectors by their length you get direction

cross product*

So if the cross product evaluates to 0 there is no direction because if you divide by length it’s still 0

[midnight plank]

@minor turtle Has your question been resolved?

[minor turtle]

Alright

So the direction is basically the cross product divided by the magnitude

?

[midnight plank]

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[midnight plank]

@raw owl Has your question been resolved?

[raw owl]

<@&286206848099549185>

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@raw owl Has your question been resolved?

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<@&286206848099549185>

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[torn geyser]

does y^2+2y-1 simplify to (y-1)^2

[quaint lily]

no

[torn geyser]

how come?

i factored it and got (y-1)(y-1)

[quaint lily]

you must've done something wrong

[quaint lily]

expand that out again

and you'll see

[torn geyser]

ah

i see

well im stumped on what to do then

I moved the 4x and 21 to the other side

and the only way to complete the square would be subtracting one no?

[quaint lily]

(y+1)^2 = y^2+2y+1

[torn geyser]

ah

i see

thank you for the assistance!

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[last slate]

,, f(x)=\sqrt{2{x}^2-3{x}+1}+\sin^{-1}x

[grand pond]

yajat

[last slate]

we have to find its domain

[novel lion]

stuff inside sqrt() cant be <0 right?

[last slate]

yea

wait let me just show u what i've done

i've come to this result- ${x}\in \left(-\infty,\frac{1}{2}\right]\cup [1,\infty)$

[grand pond]

yajat

[last slate]

and also $x\in [-1,1]$

[grand pond]

yajat

[last slate]

since domain of arcsin is -1 to 1

{} is fractional part btw

[midnight plank]

@last slate Has your question been resolved?

[last slate]

astar 💔

[obsidian merlin]

{x} means fractional part right

[obsidian merlin]

ok so to satisfy arcsin x

sin-1x x has to be between -1 to 1

factorise that polynomial

3+-1/4

1/2 to 1

so in range 1/2 to 1, the polynomial is -ve so regect

answer should be [-1,1/2)

@last slate

[obsidian merlin]

yeah take the interesctio

intersection

you got it

[midnight plank]

@last slate Has your question been resolved?

[last slate]

sorry wait a min

[last slate]

but this is not the answer

[obsidian merlin]

wtf

ohhhh wait

no

[last slate]

it says -1 to -1/2 union 0 to 1/2

[] []

idk why cant we take -1/2 to 0

[obsidian merlin]

does

fractional part

be always +ve

[last slate]

fractional part is always positive

[obsidian merlin]

thaaats why

wait then

if we remove values from 1/2 to 1

we do it in -ve side too

-1 to -1/2

shouldnt we omit -1 to -1/2

[last slate]

but why?

u seems to be confused

[midnight plank]

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[last slate]

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[quaint lily]

evaluate sin of the whole thing

then you'll get a number, then draw a triangle

[gritty holly]

how would you do that

[quaint lily]

let's call that A so i dont have to write the whole thing

what do you get for sin A

[gritty holly]

the second one right

what

sorry but i dont follow

[quaint lily]

yea

[gritty holly]

ok

[gritty holly]

i know you use cos(x+y) rule but i dont know why or how you determine it

[quaint lily]

this whole thing ill call A

[gritty holly]

ok

[quaint lily]

evaluate sin A

using the sin(x+y) rule

[gritty holly]

why do you do that

[quaint lily]

you'll understand why in the next step

[gritty holly]

ok ill take your word

3/5 * 12/13 + 4/5 * 5/13

[quaint lily]

(actually you can also use cos(x+y) but I was trippin and thought it was like tan sum formula)

so then sin A would be 38/65 right?

[gritty holly]

what

[quaint lily]

from that

[gritty holly]

sin A become 56/65 doesn't it

[quaint lily]

oh

[gritty holly]

pretty sure its 56/65

[quaint lily]

yes that

[gritty holly]

k

[quaint lily]

so then if you draw a right triangle with angle A, you get this

[gritty holly]

k

[quaint lily]

from that you can then get cos A

by pythag

[gritty holly]

yeah

[quaint lily]

so you'd get cos A = 33/65

[gritty holly]

yeah

[quaint lily]

then A = (ii) = arccos 33/65

done

[gritty holly]

why at the start did you want sinA

[quaint lily]

.

my brain was not braining

[gritty holly]

how do you decide it's cos (x+y) then or is it trial and error

[quaint lily]

you could also do the same with cos (x+y)

cos(A) would be like 4/5 x 12/13 - 3/5 x 5/13

which gives the same result

[gritty holly]

so how do you decide to do cos(x+y)

[quaint lily]

wdym?

[novel lion]

because the RHS is cos

ig

[gritty holly]

so why cos(x+y) not cos(x-y)

[novel lion]

well u can use either sin or cos, just need to convert

[quaint lily]

because they're summing

[novel lion]

because there's + sign between them

[gritty holly]

k im stupid nvm

[midnight plank]

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[twilit field]

3 squares of a chess board are selected at random, the probability of getting two squares one colour and one square another is

[twilit field]

$\frac{2c\left(32,2\right)c\left(32,1\right)}{c\left(64,3\right)}$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

would this be it

,w $\frac{2c\left(32,2\right)c\left(32,1\right)}{c\left(64,3\right)}$

[grand pond]

Results provided by WolframAlpha

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

Requested by physicsrocks

[twilit field]

my book says it's 8/21

what am I doing wrong

[sage helm]

16/21 should be right

[twilit field]

thanks

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[twilit field]

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[midnight plank]

✅

[twilit field]

seven coupons are selected at random with replacement from 15 coupons numbered 1 to 15, the probability that the largest number appearing is 9 is

so the sample space would be $(15)^7$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

now what

[novel lion]

since the max number on coupun can only be 9

the selection is made from coupons numbered 1 to 9

[twilit field]

there are 9 choices for 6 coupns

rught

[novel lion]

so total 9^7 ways

[twilit field]

and one is fixed

[novel lion]

but that also includes when number isnt 9

which is 8^7 cases

so (9^7-8^7)/15^7

[twilit field]

hmm

ok

thanks

wait

my book says

$\left(\frac{3}{5}\right)^7-\left(\frac{8}{15}\right)^7$

[grand pond]

ƒ(Why am. I here)=misery

[novel lion]

split the numerator

9^7/15^7 - 8^7/15^7

[twilit field]

I don't get this part

why isn't it just (9/15)^6

[novel lion]

where did 6 come from

its 7

you're selecting 7 coupons

[twilit field]

yes

[novel lion]

also its not just 9/15 because

[twilit field]

but one coupn has to be 9

[novel lion]

{1,2.......9} -> 9^7

this is sample space when any seven coupons from these are selected

it could be nine or seven or one

u need probability for coupon number 9

so the sample space of {1,2........8} is 8^7

[twilit field]

got it

[novel lion]

!done

[midnight plank]

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[twilit field]

.close

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[twilit field]

Two numbers are selected at random from s={1,2,3,4,5,6} without replacement , find the probability that the minimum of the two numbers is less than 4

[novel lion]

so total 6*5 possibilities

[twilit field]

yeah

the numerator is confusing me

3 choices for the first one

[novel lion]

u can count all the cases where the minimum number is not less than 4

which would be 6

4,4
4,5
4,6
5,5
5,6
6,6

so favourable cases are 30-6 = 24

so probability = 24/30

[twilit field]

hmm

I don't get why my emthod fails

[novel lion]

what are u trying to do

[twilit field]

so there are 3 choices for the first number

1,2,3

and 3 for teh second

4,5,6

so 9/30

[novel lion]

you're missing some cases

the minimum of two numbers is less than 4 when:

(A) both numbers chosen from {1,2,3}
(B) one from {1,2,3} and one from {4,5,6}

total cases for (A) would be 3!

[twilit field]

ooo

[novel lion]

and for (B) there are 2 cases

[twilit field]

9 right

[novel lion]

1st number chosen from A and second from B
1st number chosen from B and second from A

so 9+9

(A = {1,2,3} and B = {4,5,6})

[twilit field]

so 18/30

[novel lion]

18+6

[novel lion]

you forgot to add this

[twilit field]

ah

ok

thanks

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[plush prism]

Hey

[plush prism]

We have a system Ax=b

A and b however have errors in them

$\tilde{A}\tilde{x}=\tilde{b}$

[grand pond]

Martin

[plush prism]

$\tilde{A}=A+\Delta A$

[grand pond]

Martin

[plush prism]

Same for x and b

For ease of use i will write At for A tilde and dA for delta A

Using the maximum norm, we are given:
|dA|, |db| <= 1/100

Using this upper limit, we want to get a limit for the relative error in x, which is |dx|/|x|

,rotate

[plush prism]

This is what i got so far

I will work on something else for now, maybe i ask again later

.close

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[whole jetty]

Hello I have probability-related questions I'd like to ask

[whole jetty]

Since there are 100 printers, I'm thinking that the problem represents a sum of 100 exponentially distributed random variables

which I think means it follows a gamma distribution

[stiff pollen]

Yeah use Gamma distrbution

[whole jetty]

I just don't know how to incorporate the payments

I was thinking of using the shortcuts for the mean which is just 100 times the mean, however, i think that isn't the case here

Should I solve for the expected payment using the integral formula? If so, does the given imply that there will be no refund for years 3 and above?

<@&286206848099549185>

[midnight plank]

@whole jetty Has your question been resolved?

[midnight plank]

@whole jetty Has your question been resolved?

[midnight plank]

@whole jetty Has your question been resolved?

[whole jetty]

<@&286206848099549185>

Help me please

[whole jetty]

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[twilit field]

so I know I've posted this before

But I would like to try it again,( I work better in these channels)

so to start

$A^2=7A-4I$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

multiplying across by $A^{-1}$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

$A=7-4A^{-1}$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

hmm

maybe I factorise the OG equation

nah

that won't work

oh

[north ridge]

what if you did (A - 2I)^2 like completing the square

[twilit field]

I think I got something

$(A-2I)^2-3A=O$

[grand pond]

ƒ(Why am. I here)=misery

[strong lava]

that is the idea yes

[twilit field]

$(A-2I)^2=3A$

[grand pond]

ƒ(Why am. I here)=misery

[sage helm]

det that shit!

[twilit field]

yeah

let me TeX it, just a min

[strong lava]

that feels unnecessary

[twilit field]

hmm let $det(A-2I)=u$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

$u^2=det(3A)$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

$det(3A) is 3^3 det(A)$

[grand pond]

ƒ(Why am. I here)=misery

[twilit field]

which is 81

do the answeris 9?

[sage helm]

should be

[north ridge]

Yeah

[twilit field]

thanks!

.close

[midnight plank]

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[hard umbra]

the question is if this matrix can even exist

[strong lava]

not diagonal

[hard umbra]

the min poly needs to divide x^2 - 7x + 4

that doesn't leave a lot of choices for the char poly

and then the det can't possibly be 3

[strong lava]

it's of degree 2

since it's X(0)

[hard umbra]

?

[strong lava]

det A is the value of the char pol at 0

[hard umbra]

yeah

the char poly divides the min poly^3

and the min poly divides x^2 - 7x + 4

[strong lava]

oh right mb

[hard umbra]

so this matrix doesn't exist

question is nonsensical

[twilit field]

Damn

[junior flower]

no it means every answer choice is correct

[hard umbra]

[junior flower]

snow wtf is this name

change back

[hard umbra]

lex asked gemini a question and it replied with broseph

so now i'm snoseph

[junior flower]

wtf is gemini

[twilit field]

bard

[hard umbra]

the google ai

[junior flower]

i was never into astrology

[hard umbra]

it's ok

[twilit field]

Bard is google's answer to GPT

[hard umbra]

if you make an api key i'll add it to the server

it's free

[twilit field]

wdym

[hard umbra]

dw i'm talking to layla

[twilit field]

oh

sorry

[junior flower]

what’s that

[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[last slate]

different question

can't i just use the infinite G.P formula to find f(x)

[fresh sparrow]

not really

[last slate]

why ?

[midnight plank]

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[last slate]

How to do this

[last slate]

I know that $A = PDP^{-1}$

[grand pond]

Book Reader

[last slate]

and that D is a diagonal matrix with the eigen values

$D = \M[ 7 0; 0 -7]$

and P

would be the corresponding eigenvectors

so

[grand pond]

Book Reader

[last slate]

$P = \M[ ? 0 ; ? 1]$

[grand pond]

Book Reader

[last slate]

II thought it would be

$P = \M[ 1 0 ; 14/4 1]$

[grand pond]

Book Reader

[last slate]

but its not apparently

$\M[7 0 ; 4 -7] \M[ 1; 14/4] = \M[ 7 ; 4-7(14/4)]$

[grand pond]

Book Reader

[last slate]

bruh

[robust isle]

How did you get 14/4

[last slate]

$\M[0 0 ; 4 -14] \equiv \M[0 0 ; 1 -14/4]$

[grand pond]

Book Reader

[robust isle]

Ok

So that means an eigenvector (x,y) satisfies the equation x - 14/4 y = 0

Never forget what your matrix represents

A system of linear equations

[last slate]

aye

[robust isle]

So if you fix x=1 as you did

You get y=4/14

y=2/7

[last slate]

that works

how come my way didn't work tho

it has worked all the time before 😭

[robust isle]

Well you try and just wing it when you try and find your eigenvectors

[last slate]

$x = y \M[ 1 ; 14/4]$

[robust isle]

Moral of the story : don't wing it

[last slate]

wiat

[grand pond]

Book Reader

[last slate]

idk

[last slate]

nah, i'm following the method that I learnt, not winging it

the first column has a pivot position and the second doesn't

[robust isle]

Well either your method doesn't work

Or you're not following it correctly

[last slate]

so the second is a free variable

and everything can be written with respect to the free variable

idk

the top and bottom of fraction is swapped in my answer

oh wait

I see why

Yeah, I just put the 14/4 in the wrong row

[robust isle]

Yeah seems like it

[last slate]

ty

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[last slate]

How do I determine if a matrix is diagonalizable?

[last slate]

I tried to diagonalize it but couldn't figure it out so i just said it couldn't be done and got the right answer

$A - 7 I = \M[3 -3 ; 3 -3] \equiv \M[1 -1; 0 0]$

[grand pond]

Book Reader

[robust isle]

So the only eigenvalue you got is 7 right ?

[last slate]

yeah

because characteristic matrix is $x^2 - 14 + 49 = (x-7)^2$

[grand pond]

Book Reader

[robust isle]

Ok yes

[robust isle]

Well think of how many linearly independent solutions you can get from that

[last slate]

1?
$x = y \M[ 1; 1]$

[grand pond]

Book Reader

[robust isle]

Yeah so only one solution

[last slate]

So I'll just have to check each eigenvalue and see if it has a eigenspace dimension equal to eigenvalue multiplicity?

[robust isle]

Yes

That's the criterion for diagonalizability

[last slate]

Was hoping there would just be a "it doesn't look right" method

[robust isle]

It's not too bad frankly

You'd expect A-7I to be all zeros if you want a matrix with that char poly to be diagonalizable

So it doesn't look right yea

[last slate]

okay

[robust isle]

But it's a bit of a special case here

In general you'd have to go for the long checks

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

I'm trying to find the components of AB

how would that work?

I don't understand the addition or subtraction

I just know it has to do with the points of (-8,2) and (-4,4)

[weary jetty]

i think that what the text means where it sayst to find the related position vector, it means to "move" the vector so the point A is in the center

if its that the only thing you have to do is just the x component of B minus the x component of A and the y component of B minus the y component of A

[last slate]

(A,B) - (2A, 2B)?

why subtracting though

and in what order

[weary jetty]

what you have to think is that you dont care where the vectors are

you just care about the distance between the components

that in this case whould be (-4,4) - (-8,2) = (4,2)

do you understand it?

[last slate]

sorry my wifi cut out

[last slate]

but why can't you do (-8,2)-(-4,4)?

[weary jetty]

because vectors have directions

when you calculate a distance in the vector way, you first get the final point and subtract the initial point

[last slate]

final - initial = distance travelled

okay that makes sense

thanks

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[obtuse totem]

ok I couldnt type in the other channel anymore

[robust isle]

damn channels

[obtuse totem]

ok so why is [T] w respect to basis C, the inverse of the matrix formed by vectors in basis C?

[robust isle]

and what I was saying in my answer is that the typography sucks in your pic

you have to parse the statement like this

so they're just doing a change of basis here

[obtuse totem]

ohh

lmao I thought they were two seperate equations

[robust isle]

yeah it also got me at first

[obtuse totem]

ahh thank you!

thank you so much

can I stay in this channel

just in case more questions

cause theres other available channels anyways, so I dont think it matters whether I occupy this one or not

[robust isle]

if you want

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

I thought you have to prove its inner product by showing these properties..

you can just state dot product is an inner product?

thats sufficient?

[runic hamlet]

they say that they have earlier shown these things for the dot product

[tribal temple]

(Yea in this case you’d have shown the dot product satisfies the inner product properties, and with the inner product defined here [on those polynomials], it’ll be exactly the same steps as showing the dot product [on R^3] is an inner product)

[obtuse totem]

ok

I think lin alg this semester has made me realized that

I do not want to research linear algebra lmao for the rest of my life

no wayy

one semester is enough trials and tribulations

nah nah I just wanna jump intto a lake or ocean or whatever and call it a day

cliff jumpiing

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

LMAO not me listening to clubbing music while doing math😭 😂

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

I think theres a typo

should be [2t 3t]

Please let me know if there isnt... cause I really dont trust my thinking

[robust isle]

@obtuse totem looks like a typo yes

2+3x should work

[obtuse totem]

ok

thanks!

[obtuse totem]

studying is surprisingly going well

sweet

[obtuse totem]

I think answer to a is wrong

I got 4+7i

if someone can cross check that'd be great

Im pretty sure its right...

[tribal temple]

-did they use the dot for the dot product to instead represent the complex standard inner product

If you use the “normal” (real) dot product definition I agree you get 4 + 7i

Though it seems they’re using the complex inner product (defined by bar{u}.v) which works out as they have

[obtuse totem]

ohh

what does the bar do?

ontop

[tribal temple]

Complex conjugate (of all entries)

[obtuse totem]

so it negates only complex numbers?

i bar would be -i

but reals keeps the same

alright

okie

thank you

I think I'll call it a day today

thank youuuu

gn 🙂

[tribal temple]

Goodnight hope you rest well

[obtuse totem]

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[forest karma]

no one is answering my f question

please help me

[midnight plank]

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[sage helm]

Consider a rectangle of length 4a and breadth 3a. If a circle of radius 1, completely inside the rectangle, is drawn, what is the probability that it does not touch nor intersect the diagonals?

[sage helm]

Kind of confused on how I'd even start with this one

i think it has to do with area

maybe a 1 cm clearance around the diagonals?

1 - that area/total area?

[vague reef]

Pythogaras theorem

[sage helm]

huh?

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[sullen scaffold]

Quick question for understanding a concept

Cramer's Rule for the resolution of a linear equation system.

Say we have a matrix A that is not inversible. We will have to use the usual resolution with Gauss-Jordan?

[crude pier]

yes, because cramer's rule requires det(A) =/= 0

[sullen scaffold]

Exactly. Perfect, just making sure to know what to do if I get det(A) = 0.

If that is the case, gauss-jordan resolution.

[crude pier]

yes

[sullen scaffold]

Perfect. Thank you.

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[midnight plank]

@thick grotto Has your question been resolved?

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[thick grotto]

.reopen

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✅

[thick grotto]

.reopen

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[topaz wolf]

whats the fastest way to find inverse matrix

[topaz wolf]

or a simple way idk

a good way

like ik that A^-1 = adj(A)/det(A)

that will take forever to find

and theres this way

[subtle blaze]

It’s a very hard task

[topaz wolf]

where AB=I

[topaz wolf]

wdym

[subtle blaze]

There’s no easy way to find the inverse of a generic matrix

[topaz wolf]

oh ok so only those 2 ways right ?

[subtle blaze]

If there were we’d use it

[topaz wolf]

ye but like those r the only 2 ways

?

or there r others

[subtle blaze]

There are others

https://en.m.wikipedia.org/wiki/Invertible_matrix

[midnight plank]

@topaz wolf Has your question been resolved?

[topaz wolf]

damn tysm

appreciate it

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[last slate]

Can someone please help me out with this?

[robust isle]

it's a set containing relations inside of it

so sure you don't see that everyday

[last slate]

Oh, C contains congruence modulo relations

[robust isle]

but it's not something completely outlandish

[last slate]

Yeah, right

[robust isle]

yes

[last slate]

The question itself is throwing me off though

[robust isle]

now the question a) itself starts getting quite abstract yes

[last slate]

"Notion of refinement defines a binary relation on C"

[robust isle]

they're defining a relation between congruence relations

[last slate]

That means that for all $m, n \in \mathbb{N}^*$, we must have $\equiv_m$ refines $\equiv_n$?

[grand pond]

Omega

[robust isle]

no

[last slate]

Are we taking this as a given?

Oops

[robust isle]

that would be like saying "a=b mod n for all integers a, b"

[last slate]

Which obviously isn't true

[robust isle]

so no, not all congruence relations refine one another

[last slate]

For some values of m and n, I'm assuming they DO though

And that's the whole point of the problem

[robust isle]

to give an example, $\equiv_4$ refines $\equiv_2$

[grand pond]

aPlatypus

[robust isle]

if a=b mod 4, you know that a=b mod 2

[last slate]

Really?

[robust isle]

the idea for refinement essentially is that a=b mod 4 gives you more info on a and b than a=b mod 2

[last slate]

(7 = 3) mod 4

(7 = 1) mod 2

Is that incorrect

[robust isle]

you still have 7=3 mod 2 tho

their difference is a multiple of 2

[last slate]

Yes

[robust isle]

we're working with equivalence relations

[last slate]

Yeah

[robust isle]

so embrace them fully, and don't think of mod as only the comp sci remainder operator here

[last slate]

Ok

Sure

[last slate]

I didn't even know we were allowed to write this

But that's besides point

[robust isle]

well you don't have to reduce everything to the smallest representative all the time

7=3 mod 2, 7=5 mod 2, 7=54867434984771617 mod 2, ...

these are all true

[last slate]

Got it

[robust isle]

so yeah do you get why =_4 refines =_2 ?

[last slate]

Yes, I do

That's fine

So the problem is asking me to show that the REFINEMENT relation is a partial order, i.e. reflexive, antisymmetric, and transitive on C?

[robust isle]

yeah

[last slate]

So we go back to the definition of Refinement right

aRb implies aSb

Actually, I feel like I'm getting ahead of myself

I'm not sure

[robust isle]

there's not a lot else you can work with anyway

[last slate]

Yeah

True

[robust isle]

if you want a graphical representation of the situation regarding =_2 and =_4 here it is

each of the equivalence classes from =_2 is neatly partitioned by some classes of =_4

[last slate]

Yes

[robust isle]

and there's no equivalence class from =_4 that spans two classes of =_2

that's refinement

now if you took =_2 and =_3

[last slate]

There's no refinement

[robust isle]

yeah

each of the =_3 classes is split between the two =_2 classes

so yeah it prolly won't help directly for the abstract question

but it's nice having some examples in mind

[last slate]

Yup

[robust isle]

you can try solving the q yourself for a bit

[last slate]

Yeah, I'm thinking

I'm confused

[last slate]

Do we need to show that the congruence modulo relation is a "refinement relation"? How does that relate to the fact that the refinement relation is a partial order?

Naturally, with all the information we have at our disposal, we are inclined to show that congruence modulo relations can be refinements of each other

But as we said, that is not NECESSARILY the case

And I'm not sure how to use that fact to answer the actual question

Here's the start of my write-up

[robust isle]

yeah so what

not all integers are congruent to each other mod n, but you can still talk about the "congruence mod n" relation

[last slate]

Yes, sure

[robust isle]

the question a) isn't really specific to congruence relations mod n anyway

[last slate]

Indeed, which is what is confusing me

[robust isle]

you should try to answer it without referring to the internals of the congruence relations

[last slate]

They're asking me a question that's pretty specific to the refinement relation, it seems like

[robust isle]

I think they just said "ok let's look at congruence relations mod n" to make the problem more digestible

[robust isle]

yeah sure

but it's not specific to C at all is what I'm saying

[last slate]

But I need to refer back to C though

The question is asking "on C"

[robust isle]

C could have completely other relations inside it

the proof would be the same

[last slate]

Okay, noted

So wrong write-up

[robust isle]

let's just get into it shall we

[last slate]

Shouldn't refer to what's inside C

Sure

[robust isle]

so yeah we want to show the refinement relation is a partial order

i.e. that it's reflexive, antisymmetric, transitive as you said

[last slate]

Indeed

That I'm okay with

[robust isle]

can you rephrase these three properties in terms of the refinement relation ?

like just write what reflexivity would look like in this situation

[last slate]

Give me a minute

[last slate]

$\forall \equiv_n \in C, \equiv_n [_{ref} \equiv_n$

[grand pond]

Omega

[last slate]

It seems a little iffy notation-wise

[robust isle]

yeah or "forall R in C, R refines R"

they're really insisting on the congruences mod n in the question for some reason

[last slate]

Yeah

For antisymmetry and transitivity, would be a similar thing

[robust isle]

yeah, just state them so that we really know what we have to prove

[last slate]

Ok, let me write this down

Should I delete reference to the congruence modulo relations in my write-up?

Or does it not matter

[robust isle]

it's gonna be easier to follow (and to write) if you don't refer to them

[last slate]

So, remove the first paragraph?

Isn't that going to be confusing

Ok, it's fine

I'll write up later

What we need to prove

[robust isle]

ok yeah

pick one and try to prove it

the whole proof is gonna rest on the fact that set inclusion defines a partial order already

[last slate]

Don't we need to bring back the congruences to prove them

[robust isle]

no, use the definition of refinement

[last slate]

So, if we use the definition of refinement