help-49

[twilit field]

let a be the smallest positive power

finally got it

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[runic hamlet]

.reopen

[midnight plank]

✅

[runic hamlet]

note that this is the exact same proof as showing that the subgroups of Z are of the form aZ

can you tell me why its the same proof?

@twilit field

.close

[midnight plank]

Channel closed

Closed by @runic hamlet

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

Because we want to show practically the same thing, everything is generated by adding or in this case multiplying the same element multiple times

[runic hamlet]

yes

even more explicitly, Z is a cyclic group

so its literally an example of this lemma

[twilit field]

Got it

Thanks!

Can I close this now

[runic hamlet]

yes

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

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[oak fable]

hey

[oak fable]

so

there were 4 black, 5 white balls in a bag. If u pick 2 balls randomly..

1. whats the probability u will get 2 white.

2. whats the probability u get 2 black

3. whats the probability u get 1 black and 1 white

[spare seal]

Absolutely crucial question: are you picking the balls with replacement or without replacement?

[oak fable]

hmm

like the questions 1 , 2 and 3?

oh their complete different questions so with replacement

[spare seal]

well, I guess the question implies you're picking them at the same time

okay, that's fine

What have you tried?

[oak fable]

i mean like if u get 2 out of the bag at the first question, itll all just reset at 2

okay uhm

i tried like 2/9.. 💔

[spare seal]

i mean ... what steps did you try to get the answer?

[oak fable]

like

i added 4 and 5

so 9

and u pick 2 balls randomly

so 2 over 9

[spare seal]

adding 4+5=9 is good

I don't think your next step makes much sense

Okay so ... what's the probability of getting 1 white ball if you pick just one ball?

[oak fable]

hm

like 1/9?

[spare seal]

it's not. it doesn't help to just divide whatever numbers I put in the question

here's a sanity check: if there were all white balls the probability of picking a white ball would be 1 (or 9/9)

[oak fable]

yea

[spare seal]

so like there are 9 possible balls you can pick, and 5 of them are white

[oak fable]

ohh

so 5/1

i mean 1/5

[spare seal]

not that either

[oak fable]

so like 0.20%?

[spare seal]

the probability should be bigger the more white balls you have, right? because the more you have the more likely you are to find one

[oak fable]

yeah

[sudden yacht]

1/5 is not 0.20% by the way

[oak fable]

ok

[spare seal]

I'm trying to figure out how to explain the intuition of how to work out these probabilities ...

Okay so let me list down all the events that could happen if you picked one ball

Let W_1 be the outcome that you pick the white ball number 1, W_2 the outcome you pick white ball number 2, ect.

Then here's the list of all possible outcomes if you pick just one ball

W_1
W_2
W_3
W_4
W_5
B_1
B_2
B_3
B_4

the total number of outcomes is 9

out of those 9 outcomes, 5 of them are picking a white ball

[oak fable]

ye

so 5/9

[spare seal]

exactly yes

it's all about the relative number our different outcomes

[oak fable]

and black is 4/9

[spare seal]

yes

[oak fable]

oh alr

soo

[novel lion]

you can also do like: total 9 balls, ways to pick 2 out of 9 balls is 9C2 so 36, you want both of them to be white, there are 5 white balls so the number of ways is 5C2 which is 10, so probability of picking 2 white balls is 10/36

[oak fable]

whats C?

[spare seal]

this is also true but I kind of assumed the choose function wasn't something ACE_Blade had been introduced to yet

and didn't want to introduce too many new things

[oak fable]

im 14

[spare seal]

the choose function (C) is notation for the number of ways "choosing" things

So 7C2 is the number of ways of choosing 2 items out of 7 options

[oak fable]

oh

like 7 times 2?

[novel lion]

ignore it

[spare seal]

nope it's got a more complicated formula

[novel lion]

you will learn it later

[spare seal]

but yes don't worry for now

back to the main Q

[oak fable]

ok

[spare seal]

so probability of getting 1 white ball is 5/9

What's the probability of getting a second white ball given that we already got one white ball? (keep in mind the number of balls in the bucket has changed since we took one out already)

[oak fable]

oh

so like

5/8

[spare seal]

almost

[oak fable]

4/8?

[spare seal]

yes

[oak fable]

why

[spare seal]

8 total balls, 4 of them white

[oak fable]

ohh

[spare seal]

because we took out one of the white ones already

[novel lion]

because u picked 1 white ball already

[oak fable]

yeah mb

[spare seal]

okay so the probability of getting two white balls is

P(two white balls) = P(get white ball fist pick)*P(get white ball second pick | got white ball on first pick)

does that formula make sense/seem familar?

[oak fable]

hm

so 5/9 times 4/8?

[spare seal]

yes exactly

[oak fable]

oh

there were 4 black, 5 white balls in a bag. If u pick 2 balls randomly..

1. whats the probability u will get 2 white.
2. whats the probability u get 2 black
3. whats the probability u get 1 black and 1 white

so the answer is

wait

5/1

5/18

[novel lion]

yes

[oak fable]

oh

[spare seal]

5/18 is correct yes

[novel lion]

second one is the same, just with black balls

in third question you have to consider two scenarios

they are?

[oak fable]

18

[novel lion]

what

[oak fable]

theres 9 balls

so 18/18

[spare seal]

(which is equivalent to what Astar777 got with the choose method. both methods do essentially the same thing in different ways; it counts the number of outcomes that are in the event you care about and you compare that to the total number ouf outcomes0

[novel lion]

uh

[novel lion]

what is this for

[oak fable]

idk

what is it?

[novel lion]

in third question you are picking 1 black and 1 white ball

[oak fable]

.

ooh

so

[novel lion]

the procedure is same as the previous questions

[oak fable]

2/18

[novel lion]

the difference is

the order matters, so

1. first white ball then black ball
2. first black ball then white ball

you sum these two for final probability

[oak fable]

uh

so like 1/9 - 1/9

[novel lion]

no

u have 9 balls

u picked 1 white ball

whats the probability

[spare seal]

so ... throwing out numbers isn't super helpful. it would be a lot easier to explain if you try to emulate the method we went through before for 1)

[oak fable]

5/9

[novel lion]

yes, now you have 8 balls left

now you pick 1 black ball

whats the probability of that

[oak fable]

4/8>?

[novel lion]

yes

so whats the probability of picking: first white ball then black ball

[oak fable]

5/9 - 4/8

?

[novel lion]

why minus

the actions arent independent, they're one after the other

[oak fable]

.

[novel lion]

so you multiply both

[novel lion]

thats not for this

and also sum, not difference

[oak fable]

oh alr

5/9 * 4/8

[novel lion]

yes, so 20/72

[spare seal]

btw when it comes to probabilities

P(X happens and Y happens) means you're going to be multiplying

P(X happens or Y happens) means you're going to sum

[novel lion]

now

this is for when u first pick white and then pick black

what about when u first pick black and then pick white

both of these are valid scenarios

so you have to consider both

[oak fable]

u get the difference

[novel lion]

why difference

[oak fable]

oh nvm

[novel lion]

both of these are independent, either 1st happens or 2nd

so you sum the probabilities of both

20/72 + 20/72

tahts the final probability

[oak fable]

so add or multiply

[novel lion]

here, add

because

[novel lion]

.

multiply is when actions arent independent

[oak fable]

so 40/72?

[novel lion]

.

[novel lion]

yes

[oak fable]

like 20/36 = 10/18 = 5/9

[novel lion]

thats by using choose function

thats different that what we did above

[spare seal]

(I mean, not technically true P(A and B) = P(A)P(B) if A and B are independent. Not to get too in the weeds of it)

[spare seal]

I think they just simplified 40/72

[oak fable]

ye i did

[novel lion]

oh nvm

my bad

[oak fable]

so its correct

[novel lion]

yeah thats correct

[oak fable]

?

[spare seal]

yep!

[oak fable]

oh alr

thanks guys

[spare seal]

so part 2) is just by part 1) just different numbers but same method

[oak fable]

okay

[novel lion]

welp

yeah right

should've worded it better

[spare seal]

yeah I thought you knew what you meant but it could be a bit confusing

I tried to avoid discussion of independence just so we could get through the question in a fairly intuitive way

[novel lion]

@oak fable remember this instead of what i said about independence

[oak fable]

oke

[novel lion]

!dome

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[novel lion]

.close

[midnight plank]

Channel closed

Closed by @novel lion

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• Show your work, and if possible, explain where you are stuck.
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[twilit field]

Let $x$ and $y$ be elements of a group $G$. Assume each of the elements $x$, $y$ and $xy$ have order $2$. Prove that the set $H ={1,x,y,xy}$ is a subgroup of $G$ and that it has order $4$
\textbf{I'm kind of confused at what the question is asking me to do, each element has order $2$ and there are 4 elements, so order 4?}

[grand pond]

What a wonderful world !

[runic hamlet]

show that H is a subgroup

show that the elements in H are distinct

[twilit field]

okie

thanks

[lyric charm]

do not confuse order of an element vs order of a group

[twilit field]

We start by showing $H$ is a subgroup. To do so we check for closure, presence of inverses, and presence of the identity
\begin{enumerate}
\item Identity element : 1 is present, there is an identity in the group
\item $x^2=1 \implies x=x^{-1},;y^2=1; (xy)^2=1 \implies xyxy=1 \implies xy=y^{-1}x^{-1}=(xy)^{-1}$\We thus have the presence of inverses in the set
\item closure: \textbf{ how do I do this efficiently}
\end{enumerate}
\textbf{ To prove elements are distinct I was thinking of using uniquness of inverses, but that feels circular}

[grand pond]

What a wonderful world !

[runic hamlet]

its 4 elements

just bruteforce closure

write out the multiplication table

[twilit field]

this is the table I got [
\begin{array}{c|cccc}
H & 1& x & y & xy \
\hline
1 & 1 & x & y & xy \
x & x & 1 & yx & xyx\
y & y & xy & 1 & x\
xy & xy & y & yxy & 1 \
\end{array}
]

I guess I have to deal with xyx and yxy

[runic hamlet]

check the table again

[grand pond]

What a wonderful world !

[twilit field]

We start by showing $H$ is a subgroup. To do so we check for closure, presence of inverses, and presence of the identity
\begin{enumerate}
\item Identity element : 1 is present, there is an identity in the group
\item $x^2=1 \implies x=x^{-1},;y^2=1; (xy)^2=1 \implies xyxy=1 \implies xy=y^{-1}x^{-1}=(xy)^{-1}$\We thus have the presence of inverses in the set
\item closure: We Prove this in the multiplication table attached below
\end{enumerate}
\begin{array}{c|cccc}
H & 1& x & y & xy \
\hline
1 & 1 & x & y & xy \
x & x & 1 & yx & xyx\
y & y & xy & 1 & x\
xy & xy & y & yxy & 1 \
\end{array}
]
However, $(xyx) = y^{-1}x^{-1}x = y^{-1}=y$
\and $yxy = yy^{-1}x^{-1} = x$
\ $xy=y^{-1}x^{-1} = yx$
\
thus the table is
\begin{array}{c|cccc}
H & 1& x & y & xy \
\hline /[
1 & 1 & x & y & xy \
x & x & 1 & xy & y\
y & y & xy & 1 & x\
xy & xy & y & x& 1 \
\end{array}]
\textbf{ To prove elements are distinct I was thinking of using uniqueness of inverses, but that feels circular}

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

what's wrong with my TeX

😭

anyway, it's closed

[runic hamlet]

ignoring the tex you somehow managed to not see the nice thing about multiplication in this group

look at the equality xy = y^-1 x^-1 closer again

[twilit field]

the inverse is the elment itself

It's abelian

[runic hamlet]

yes

so therefore fuck all annoying stuff afterwards

and finally, what happens if for example x=y ?

[lethal path]

I love Cayley tables

[twilit field]

The group ends up having 2 elemets

[runic hamlet]

which would be a problem

so why can it not happen

[twilit field]

hmm

[twilit field]

the elements are distinct?

[runic hamlet]

why are they distinct

you are only given that x,y,xy have order 2

the question said nothing about the elements being distinct

you have to show that

[twilit field]

hm

if x=y, then xy would have order

1

xy would have order 1, yea

[runic hamlet]

yes

[twilit field]

okay

thanks

I think that completes it

Thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tidal turret]

3. Let ( T ) and ( H ) be the subspaces of ( \mathbb{R}^4 ):

[
T = {x \in \mathbb{R}^4 \mid 2x_1 - x_2 + x_3 = 0; x_1 - 2x_4 = 0}
]
[
H = {x \in \mathbb{R}^4 \mid x_1 - 2x_2 + 2x_3 - x_4 = 0}
]

Find, if it exists, a subspace ( S \subset \mathbb{R}^4 ) such that ( S \oplus (T^\perp \cap H) = H ).

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[west pulsar]

I'm learning Spanish, though an english math discord group!

[tidal turret]

are u from america?

[visual tiger]

so what did you try?

[west pulsar]

no, from HK, not important at all.

Go back to the question, though

[tidal turret]

status 1 I think

[visual tiger]

did you try and compute the subspaces that are mentioned but not explicitely computed?

for starters

[tidal turret]

you mean find a basis

[visual tiger]

(although there is an easy way to find a suitable S very fast)

[visual tiger]

why not

[tidal turret]

how? hints not spoilers

[visual tiger]

do you know about the property $(A\oplus B) \cap C = (A\cap C) \oplus (B\cap C)$?

[grand pond]

rafilou is not not born in 2003

[carmine sigil]

Conceptually, S will be whatever space was removed from H by the intersection operation. There should be such an S. And if we look at the dimensions of each space involved we see that it's probably 2 dimensional.

T: 2, Tperp: 4-2 = 2, H: 3, Tperp cap H: 1 or 2, but generally 1. So S is 3-1 = 2.

[fallen sparrow]

Hi yes excuse me but what does the + inside a circle symbol represent?

[carmine sigil]

It's the direct sum operation

[visual tiger]

direct sum, sum of two subspaces with trivial intersection

[fallen sparrow]

Could you give an example

[carmine sigil]

If you need an example please open your own help thread

[fallen sparrow]

Ok sorry

[tidal turret]

<(1,0)> (+) <(0,1)> = R2, no?

[carmine sigil]

This is true

[tidal turret]

i dont know that, I guess I need to find a basis then

[tidal turret]

how did you figured T is dim 2? 4 unknowns 2 equations?

[carmine sigil]

Exactly

Well, it could be 3 dimensions if the equations are redundant, but it's clear that they are not in this case.

(x4 only appears in one equation)

[tidal turret]

yeah

but I got lost on how you found that dim Tperp Cap H is generally 2

dim Tperp = 2

dim H = 3

but how is dim Tperp cap H generally 2?

anyways, I still need to do compute the basis of T and H, no?

because idk $(A\oplus B) \cap C = (A\cap C) \oplus (B\cap C)$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

I would need to prove that in order to use it

[carmine sigil]

My bad, it's generally 1 but can be 2, not the other way around

[tidal turret]

yes, now its more understandable

[carmine sigil]

Which makes S generally 2 dimensional

[tidal turret]

what happens if Tperp Cap H is dim 2 tho? Tperp is entirely contained in H

[carmine sigil]

Yup

In which case S is equal to T cap H

[tidal turret]

are you saying if dim S is 2, and Tperp is entirely contained in H

S (+) (Tperp n H) = H

dim(S) + dim(Tperp n H) = dim(H)
dim(S) + dim(R^2) = dim(R^3)

[tidal turret]

ohh, three equations, 4 unknowns, thats why you say S is dim(R)

only if these hyperplane cartesian equations are not redundant

[carmine sigil]

Sorry, let's take a step back. If we have Tperp in H, then S is only 1 dimensional, not 2.

And equal to T cap H

[tidal turret]

by Tperp in H you mean only one of them is in H or both of the basis vectors of Tperp (entirely contained)

[carmine sigil]

Tperp in H means that Tperp is a subspace of H

Entirely contained

But this is not the common case

[tidal turret]

ok, yes in that case, dim(Tperp n H) = 2
and thus, dim(S) = 1

[tidal turret]

ok

the other case is dim(Tperp n H) = 1

[carmine sigil]

In which case we have S = (Tperp n H)perp n H

[tidal turret]

In which case we have $S = (T^{\perp} \cap H)^{\perp} \cap H$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

this u mean?

[carmine sigil]

Yup

[tidal turret]

no idea mate

[carmine sigil]

So we want S to be the part of H not in Tperp n H

We can think of the subspace of R4 independent of Tperp n H, this is (Tperp n H)perp

Then we want the part of this that is in H

(Tperp n H)perp n H

And this is S

Satisfying your requirements by construction

[tidal turret]

you lost me here, you mean

$(T^{\perp} \cap H) \oplus (T^{\perp} \cap H)^\perp = \mathbb{R}^4$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[carmine sigil]

If V is any subspace of R4 then V oplus Vperp = R4

[tidal turret]

yes

but we dont know the dimension of the intersection between the orthogonal complement of T and H

because it can be 1 or 2

dim(H) = 3
dim(Tperp) = 2

[carmine sigil]

Doesn't matter, the above is true no matter what, it's just simpler in the exceptional case

[tidal turret]

nah I think I get what you mean

because (Tperp n H)perp is linearly independent to (Tperp n H)

and we need S to be contained in H

S is exactly (Tperp n H)perp n H

but we still dunno the dimension of the intersections

but at least we found S

[carmine sigil]

Doesn't matter, you figure this out as you perform the work

You'll know as soon as you find Tperp n H

[tidal turret]

ok

Tperp = <(2,-1,1,0), (1,0,0,-2)>?

[carmine sigil]

I haven't put pen to paper, and unfortunately, I don't think I can right now, I need to go afk soon

[tidal turret]

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

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

is ok, I think I am on the right track, I appreciate the help

[tidal turret]

you dont need pen and paper tho, I noticed from the cartesian equation of the hyperplane, the normal of a plane is the vector that is orthogonal to all the vectors that are living in the plane

but this dimension counting, we did was really helpful for my brain, ngl

(x1,x2,x3,x4) = a(2,-1,1,0) + b(-1,0,0,2)

(x1,x2,x3,x4) = (2a-b, -a, a, 2b)

H = {x1 - 2x2 + 2x3 - x4 = 0}

(2a-b)-2(-a)+2(a)-2b = 0

2a-b+2a+2a-2b = 0

a(2+2+2) + b(-1-2) = 0

6a -3b = 0

6a = 3b

2a = b

a(2,-1,1,0) + b(-1,0,0,2) = (2a-b, -a, a, 2b)

(2a-b, -a, a, 2b) = (2a-2a,-a,a,4a) = a(0,-1,1,4) ==> Tperp n H = <(0,-1,1,4)>

dim(Tperp n H) = 1

(Tperp n H)perp = { -x2 + x3 + 4x4 = 0}

,w nullspace{{1,-2,2,-1},{0,-1,1,4}}

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

S = <(9,4,0,1),(0,1,1,0)>

I guess we can throw a bunch of orthogonal complements at the problem XD

there was multiple ways to attack this problem

let me verify if my S meets the condition

S = {(9a, 4a+b, b, a) | a,b € R}
H = {x1 - 2x2 + 2x3 - x4 = 0}
9a -2(4a+b) + 2b - a = 0
9a -8a -2b + 2b -a = 0
a(9-8-1) + b(-2+2) = 0
9-8-1 = 0
-2+2 = 0
so yes, S is entirely contained in H

aka $S \subseteq H$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

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

Tperp = {(2a+b,-a,a,-2b) : a,b € R}

H = {x1 - 2x2 + 2x3 - x4 = 0}

2a + b -2(-a) + 2(a) - (-2b) = 0

2a + b + 2a + 2a + 2b = 0

a(2+2+2) + b(1+2) = 0

6a + 3b = 0

6a = -3b

3a = -b

b = -3a

Tperp n H = {(2a-3a,-a,a,-2(-3a)) : a,b € R}
Tperp n H = {(-a,-a,a,6a) : a,b € R}

Tperp n H = <(-1,-1,1,6)>

S = <(9,4,0,1),(0,1,1,0)>

,w rank {{9,4,0,1},{0,1,1,0},{-1,-1,1,6}}

[grand pond]

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[tidal turret]

ok so this S works

.close

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[crude totem]

Hello. I have the function f(x)=3x^2 - 2x + 1. I have to calculate the tangents equation in the point x=2. i already have an answer but i am not sure if i calculated the slope of the line correct.

[rancid vigil]

Show your work

[crude totem]

1 sec

[rancid vigil]

f'(x0) seems to be wrong

The limit should be to 2 tho

Not 9

[crude totem]

Oh sorry it is 2 but that doesnt change anything i think

[rancid vigil]

It does

[crude totem]

oh yes

wait let me calculate

[crude totem]

idk why i said that

it gives 10 not 31

[rancid vigil]

Yep

[crude totem]

thank you very much

[rancid vigil]

Np

Are you allowed to derivate the function ?

It's easier than calculating the limit

[crude totem]

you mean with a rule of derivation?

[rancid vigil]

Yep

[crude totem]

i am but i didnt think of that

[rancid vigil]

$f'(x)=6x-2$

[grand pond]

<rajel />

[rancid vigil]

f'(2) = 10

Easily

[crude totem]

so thats the same with 3x+4 like what i got earlier?

[rancid vigil]

Yep

I would just do this method if I'm allowed

[crude totem]

but how is 6x-2 equal to 3x-4

[rancid vigil]

The other one involves more calculus

[rancid vigil]

They aren't equal

Calculating the limit would give you different values for different x

While derivating give you the expression that works for every x

[crude totem]

so 6x-2 works for every x but 3x-4 only works with certain values?

[rancid vigil]

only works with x being 2 , yes

Btw it's 3x+4

[crude totem]

oh yeah mispelled

well alright thank you so much

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[slate berry]

What should I do next?🥲

[slate berry]

Thx

[fallen sparrow]

Are you not able to understand what to do after a^2 - ab - b^2 = 0?

[midnight plank]

@slate berry Has your question been resolved?

[slate berry]

Yes

[fallen sparrow]

Hmm well does the equation resemble something close to a quadratic in any sense?

[slate berry]

I tried but failed 😞

[fallen sparrow]

I see

Try to make it into a quadratic

Divide by b^2

You should see a nice quadratic

Got it?

[slate berry]

Like this?

Sorry I might be a bit confused

[fallow scarab]

,rotate

[fallow scarab]

you simplified the middle term incorrectly

[slate berry]

My bad I was blind 😶

[slate berry]

I solve it

Thank you so much guys🙏 😊

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[tidal turret]

$(A\oplus B) \cap C = (A\cap C) \oplus (B\cap C)$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

how does this property work?

for A,B,C finite dimensional vector spaces

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[fresh sparrow]

wdym

[tidal turret]

is it true?

[fresh sparrow]

i think so

[tidal turret]

how?

[fresh sparrow]

just treat oplus and intersection as regular set operations

[fresh sparrow]

what of you had union instead of oplus

[tidal turret]

union of subspaces is not closed under addition

[fresh sparrow]

i know

[tidal turret]

so?

[fresh sparrow]

i am asking more like if you just had sets

[tidal turret]

you mean basis extension

[fresh sparrow]

no

$(A\cup B)\cap C = (A\cap C)\cup(B\cap C)$

[grand pond]

artemetra

[fresh sparrow]

this is true if A,B,C are just sets

not vector spaces

would you agree?

[tidal turret]

but its subspaces not sets. wtf

[fresh sparrow]

i know

but

it's basically the same argument

[tidal turret]

can u do a venn diagram

[fresh sparrow]

i can't rn, sorry

[tidal turret]

i am not well versed in set theory, I wouldn't know

[tidal turret]

how would I sketch a venn diagram out of this

which one is bigger?

[tidal turret]

in the case the sum is direct, then the sets are disjoint

[tidal turret]

the property is not true

but I cant provide an counterexample

[midnight plank]

@tidal turret Has your question been resolved?

[shadow schooner]

R^2 A = {(x,0) | x in R}, B = {(0,y) | y in R} and C = {(x,x) | x in R} seems to do the trick

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

what?

[tidal turret]

the property is fake mate, is not true all the time

[tidal turret]

A = <(1,0)>
B = <(0,1)>
C = <(1,1)>?

A (+) B = <(1,0)> (+) <(0,1)>
A(+)B = <(1,0),(0,1)>

(A(+)B)∩C = <(1,1)>

(A∩C) = {0}
(A∩B)={0}

[tidal turret]

whats the direct sum between two zero vectors

<(1,1)> ≠ {0} (+) {0}

<(1,1)> ≠ 0

[tidal turret]

ok, true, C is a line that passes through the origin

[tidal turret]

nice mate

.close

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[merry breach]

Just a meta question for anyone familiar with quantum computing, is it required to have quantum physics understanding

[runic hamlet]

surely at least to some degree

but not like you need a phd in physics

[merry breach]

Oh, but you need to say have taken a course in qm at least?

A semesters worth

[runic hamlet]

from listening to a few others it doesnt seem like you need more than the basic ideas. like why you can do certain gates or something. but that was just my impression

I'm not doing quantum computing myself

[merry breach]

Oh alright, thx for the insight

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[south heath]

is this valid/sufficient working for this matrix question please?

[runic hamlet]

I havent checked for mistakes but the working looks good

[south heath]

great ty

could anyone help with this one?

im fully lost

i dont know what to do or check

[fallen sparrow]

Try to correlate with cayley Hamilton theorem and cube roots of unity

[south heath]

i have no idea what those are

ive searched up cube roots of unity which talks about complex numbers which is fine, its just it isnt really covered withinb the course matter and ide like to stick within it

so is there a simpler/other way?

[fallen sparrow]

Yeah

If you wanna find those values

Manipulate the given equation by multiplying A

And A inverse a bunch of times

[fallen sparrow]

No problem i just mentioned those to give a connection of sorts

[south heath]

per matrix rules ifi multiply the left A^2 by say the inverse of A, does that equal A^-1*A^2 = A^-2A - A^-1 I?

[fallen sparrow]

I don't understand what you mean can you write it down properly

[south heath]

for example

[fallen sparrow]

Yes

You're right there

[south heath]

oo, so would this prove one expression the A^-1? one

in this case

[fallen sparrow]

Yep

[south heath]

and then i assume for the first 2 i just do it with more powers

[fallen sparrow]

Yes

Don't assume

Solve it

[south heath]

ye

[fallen sparrow]

You'll get a relation

Independent of n i believe

[south heath]

hmmmmm like this?

sus~7

[fallen sparrow]

You can't use fractional indices in matrices

[south heath]

mmm oki

[fallen sparrow]

Keep multiplying by A

And substitute A^2 from initial equation

[south heath]

surely it is this?

and if so am i missing any working or justification

[fallen sparrow]

So close

Yet so far

It's correct no doubt

[south heath]

lol

[fallen sparrow]

But the answer is independent

Of n

Simplify the expression you've obtained further

[south heath]

i theorise its smth to do with the square of A-I always gives a standard value but icant figure it out

[fallen sparrow]

Yes precisely

Look at the initial equation

Multiply with A

Now observe

[south heath]

ahah i just keep coming back to a loop

like if i try look at the original equation then it just brings it back to what the A^x value

it is

[fallen sparrow]

[south heath]

when n is something

[fallen sparrow]

Substitute the value of A^2 from the previous equation

[south heath]

wait sorry which equation theres like too many now

[fallen sparrow]

A^3 = A^2 - A

A^2 = A - I

So A^3 = ?

[south heath]

= I

lol

[fallen sparrow]

Negative I

Not I

[south heath]

oh eya

mb

yes

[fallen sparrow]

So A^6 = I

So the A^6n term becomes I

[south heath]

i see

[south heath]

hence this??

[fallen sparrow]

You can just write A^3 as - I to make things easier 🙂

[south heath]

oh yea lmfaooooo

this question is actually crazy tho bru

aight thank you

[fallen sparrow]

Noice

You can further skmplify this btw

[south heath]

oh god

give me a sec

[fallen sparrow]

[south heath]

would this be the simplest most answer

[fallen sparrow]

I'm afraid to bring this to your notice... but we can further simplify the second answer into 1 term

[south heath]

= -A^2?

[fallen sparrow]

Yep

You got it

Nice

[south heath]

can it be argued that this is not simpler than I - A or is it just cuz ig you dont have any I value

just A

also thx

[fallen sparrow]

No id say more simplified would be less number of terms

[fallen sparrow]

Np

[south heath]

yea either way now i can point out both

[midnight plank]

@south heath Has your question been resolved?

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[south heath]

.close

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[sour sierra]

Hi! I got this question in exam I am not sure if I got it correct so I wanna check

[sour sierra]

There is 9 balls labelled 1-9,

Lee take 2 ball at random

She stated the chance the sum of that two ball is even is higher than the chance of the product of that two ball being even

Show that she is incorrect

[sour sierra]

,rotate

[lyric charm]

for even * odd you need to consider the possibility of drawing an even then an odd vs. an odd then an even

so it will give you 40/72 and not 20/72

[sour sierra]

Ohhh

ugh forgot to do that

Thank you! Have a good day

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[molten bay]

Limit x tends to infinity (sinx/x)^(1/x) should I apply

e^1/xlog(sinx/x)

[molten bay]

Expansion of sinx

[fallen sparrow]

Not needed

Assume Limit as L and take log both sides

Would be simpler

[dreamy lichen]

what you wrote works as well, its basically the same thing, but dont expand sin

[molten bay]

Then?

L'hopital?

[dreamy lichen]

that'd work, or you could try to make it into the form ln(1+u)/u

[fallen sparrow]

Yeah but after that you gotta have to take Log or convert into e^ form right?

[dreamy lichen]

e^1/xlog(sinx/x)
This already is in e^ form

you can just move the limit into the exponent basically

exp is a pretty nice function, so it works

[fallen sparrow]

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

yes then?

e^1/xln(sinx/x)

How can I convert without expansion of it

1+u?

no idea please explain properly not getting idea of your words

[fallen sparrow]

What is lim u tending to 0 of ln(1+u)/u

[molten bay]

1/(1+u)

=1@fallen sparrow

[fallen sparrow]

Right

Now express the term inside ln to be of the form 1 + u

[molten bay]

I have no idea@fallen sparrow

[fallen sparrow]

Uhh

[molten bay]

ln(1+u)

[fallen sparrow]

Since sinx/x tends to 1 as x tends to 0

[molten bay]

our limit is infinity+

[fallen sparrow]

Oh

Wait a minute

[molten bay]

Okh

[molten bay]

<@&268886789983436800>

[peak lantern]

<@&268886789983436800>

[molten bay]

.close

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[spark nebula]

$n^2 = a + b \times n + c \times n \times (n-1)$

[grand pond]

rulzer.

[spark nebula]

where does this come from ?

[dusty portal]

!xy

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[spark nebula]

you want the whole problem ?

ok what they want is

to calculate this sum

first we use the e^x sum

which is

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

tf

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

[grand pond]

rulzer.

[spark nebula]

okay

$\sum_{n=0}^{\infty}\frac{1}{n!} = e$

[grand pond]

rulzer.

[spark nebula]

then we have to use the $n^2 = a + b \times n + c \times n \times (n-1)$

[grand pond]

rulzer.

[spark nebula]

to decompose the n^2 then multiply it by e

then we get the sum but the problem is where they got this $n^2 = a + b \times n + c \times n \times (n-1)$

[grand pond]

rulzer.

[dusty portal]

I don’t think we need to use this to be honest

We can just use clever manipulation

$\frac{n^2}{n!}=\frac{n}{(n-1)!}$

[grand pond]

;(

[midnight plank]

@spark nebula Has your question been resolved?

[midnight plank]

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[grim sierra]

$$\int 0 , dx = \int = C$$
$$\int = C \implies \int x^3 , dx = Cx^3 , dx$$

[grand pond]

MARETU

[grim sierra]

can someone disprove this

[fallow scarab]

wut

is this math

[grim sierra]

∫ 0 dx = ∫
but ∫ 0 dx = C

so ∫ = C

∫(x^3 dx) = C(x^3 dx)

idk how to show this is wrong

[fallow scarab]

this is nonsense

[grim sierra]

how

oh

0 * dx = 0

not an empty integral 🙁

alr thank you.

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[steel coyote]

can someone help me with this

[steel coyote]

im used to having a vector < , > with two coordinates to find distance traveled

this is the formula but its one equation so idk how to do it

[ruby eagle]

i think since the question only says "up and down" that is means 1 dimensional movement

[steel coyote]

alr so how do i solve that

[fallow scarab]

just say v(t) = x(t)

and y(t) = 0

[steel coyote]

oh ok

ok so is the thing i did wrong is that i didnt square v(t)

https://cdn.discordapp.com/attachments/1018703621841494076/1372683989877395638/image.png?ex=6827ab16&is=68265996&hm=42c1b66872117551a3263e8d55ee50a0d6a02726fa2b6364cc64b4291d29a1fa&

and didnt square root it

[fallow scarab]

$\sqrt{v^2} = |v|$

[grand pond]

riemann

[steel coyote]

oh

but then what did i do wrong

[fallow scarab]

|v| is not always equal to v

[steel coyote]

so i have to do the integral of 0 to 3 of the absolute value of v(t)

does that change these positive then?

or is it just the end result

[fallow scarab]

nope

[steel coyote]

im confused on how to plug that into a calculator or do it by hand

[fallow scarab]

you'll have to find the zeros of v(t) between t = 0 and t=3

[steel coyote]

so the x intercepts? then what

[fallow scarab]

,tex .abs def

[grand pond]

riemann

[fallow scarab]

use that to re-write |v(t)|

[steel coyote]

im confused

[steel coyote]

i have three x intercepts how do i rewrite it based on that

[fallow scarab]

plug in v for x

[fallow scarab]

to the right side here

[fallow scarab]

looks like there's only one x intercept between 0 and 3

[steel coyote]

ohh

ok so what do ou mean plug in v for x

[fallow scarab]

|v| = -v if v < 0 etc.

[steel coyote]

what is v ??

[steel coyote]

ok

so

this point

do i plug it in?

im so confusd how does |v| = -v

[fallow scarab]

you need to read this more carefully

surely you've seen things like |-3| = 3

[steel coyote]

is the x the x interecept?

[fallow scarab]

so why is this confusing

[steel coyote]

what?

[fallow scarab]

?

[steel coyote]

if the absolute value of v equals - v if v < 0

bru whats V whats X

[fallow scarab]

this is v

[steel coyote]

https://cdn.discordapp.com/attachments/1018703621841494076/1372688001427046420/image.png?ex=6827aed2&is=68265d52&hm=62f3ad175173f8f3d60c25245d2632106ddb5d7476653166771c388e75d25f6b&

[steel coyote]

yeah so im taking the absolute value

idk how to

[fallow scarab]

plug this v(t) into the definition |v(t)|

|v| = -v if v < 0 and v if v >= 0

[steel coyote]

and so do we plug in
https://cdn.discordapp.com/attachments/1018703621841494076/1372688001427046420/image.png?ex=6827aed2&is=68265d52&hm=62f3ad175173f8f3d60c25245d2632106ddb5d7476653166771c388e75d25f6b&

this

for the t?

[fallow scarab]

not sure what you're asking

[steel coyote]

how do we know if its v < 0

[fallow scarab]

the graph tells you

you're plotting v aren't you?

[fallow scarab]

in here

[steel coyote]

this the graph so how can i tell

but theres only one point from t= 0 to t=3

[fallow scarab]

a function y = f(x) is negative if y is below the x axis

[fallow scarab]

if this is v(t) against t, then v < 0 means v below the t axis

[steel coyote]

how

[fallow scarab]

what is confusing about that

[steel coyote]

everything what

[fallow scarab]

if you understood this, it's identical

[steel coyote]

but how is it below the t axis

where

some parts are above

some parts are below

[fallow scarab]

i really don't know what you're confused by

[fallow scarab]

the graph of v is below the t axis yes

right

what is your confusion

[steel coyote]

bro how is the graph below the t axis

the graph of v some parts are below some parts are above

are you looking at just the t=0 to t=3

?

[fallow scarab]

what is "it" in this question

[fallow scarab]

yes that's what's given in the problem

[steel coyote]

yeah so that part is below the x axis

or the t axis

[fallow scarab]

correct the part you circled is below the t axis

[steel coyote]

and some parts are above

[fallow scarab]

i don't understand what you're confused by

[steel coyote]

so the integration of the parts below the x axis are going to just be positive

?

Okay let me break it down

https://cdn.discordapp.com/attachments/1018703621841494076/1372689835432874124/image.png?ex=6827b088&is=68265f08&hm=d89e75adc6fb6fc1f5cbd909328e9215bac5cf5e921290db15b3a9a02df80871&

we have this function

we want to find the absolute value of it

so you said look at the v on the graph from t=0 to t=3

this section right here

so now what we know that from 0-0.53 is negative and the rest 0.53-3 is positive

so how do we rewrite the equation or the integral

[fallow scarab]

you have to rewrite |v| using the definition i gave way up above

[steel coyote]

https://cdn.discordapp.com/attachments/1018703621841494076/1372686420602519552/261933205387477002.png?ex=6827ad59&is=68265bd9&hm=69308e92ee9e2b9f5eaa324906d864ae3f7c9f0e746c04f8859d4e12c519db34&

how do i use this to rewrite the equation

[fallow scarab]

plug in your v(t)

[fallow scarab]

the right side

[steel coyote]

.close

[midnight plank]

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[midnight plank]

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[sharp fiber]

I might be able to help

Dm me

[midnight plank]

@narrow viper Has your question been resolved?

[simple field]

not sure this is allowed here (🤓)
<@&268886789983436800>

[formal blade]

Yeah, please don't use the server/help channela for this.

[midnight plank]

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[twilit field]

Prove that the intersection $K \cap H$ of subgroups of a group $G$ is a subgroup of $H$,
\
Proof:
\
As $K$ and $H$ are both subgroups, it follows that $1$ is in both, and thus in the intersection. We now check for closure, each set is individually closed, thus we only have to check the case wherein the elements are from different sets. Let $x \in K, y \in H$. We then wish to show $x+y \in K \cap H$. To the contrary let $ x+y =z; z \notin K \cap H$. \textbf{ my concern is how do I deal with this is z \notin $K \cup H$}

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

I was thinking of a proof by contradiction

but if $Z \in U \setminus ( K \cup H)$, I can't really do much, can I

[grand pond]

What a wonderful world !

[twilit field]

where U is the universal set here

[nova yoke]

wait, why are you considering x in K and y in H

isn't the goal to show that the intersection is a subgroup?

[twilit field]

yea

[sharp coral]

i think you are confusing this with the sum of vector spaces?

[twilit field]

hmm, mayhaps

yea

I am

[nova yoke]

so take two elements that are in their intersection (so they are in both K and H)

[twilit field]

right, and then closure follows almost immediately

[nova yoke]

yea

[twilit field]

as does the presence of inverses

cool

[nova yoke]

in fact it's true for arbitrary intersections of subgroups (finitely many, infinitely many, even uncountably many)

by basically the same proof

[twilit field]

Thanks!

[nova yoke]

yw

[twilit field]

.close

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[tidal turret]

can I get some help with 1)

:(

[polar star]

could you translate it

[tidal turret]

yeah, is loading

beep beep boop boop

1. Let (S = \langle (3, 0, -1, 1); (1, 2, -1, 0) \rangle),
   [ T = { x \in \mathbb{R}^4 / x_1 + x_2 = 0 } \quad \text{and} \quad H = { x \in \mathbb{R}^4 / x_2 + x_3 + kx_4 = 0 }]
   be subspaces in (\mathbb{R}^4).

   Find, if possible, (k \in \mathbb{R}) and a subspace (W) of (\mathbb{R}^4) such that ((S \cap T) \oplus W = H).

2. Let (H_1 = { x \in \mathbb{R}^4 / x_1 + a x_2 + b x_3 = 0 }),
   (H_2 = { x \in \mathbb{R}^4 / 2x_1 - x_2 + x_3 + x_4 = 0 }), and
   (S = \langle (1, 0, 1, 4); (4, 2, 0, 3) \rangle) be subspaces in (\mathbb{R}^4).

   Find the values of (a) and (b), and a subspace (T \subset H_2) such that (S \oplus T = H_1).

[grand pond]

renato

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

guys can I get some help with 1)?

at least with finding the dimension

[midnight plank]

@tidal turret Has your question been resolved?

[hard pewter]

find the basis of T in order to find the basis and hence dimension of S intersection T, which will help you to deduce possibilities for dimensions of H and W

[tidal turret]

for 1)?

[hard pewter]

yes

[tidal turret]

idk maybe if you follow but

W is exactly $(S \cap T)^{\perp} \cap H$

[grand pond]

renato

[tidal turret]

you follow?

[hard pewter]

hmm not really

i mean i dont follow really

[tidal turret]

ok, lets do your approach

[tidal turret]

I found basis of T

[tidal turret]

I still need to find the dimension of the SnT

[hard pewter]

to find S intersect T, let any arbitrary element in S be equal to alpha * v1 + beta * v2, where v1 and v2 are in basis of S

[tidal turret]

we didnt needed to find a basis for T

[hard pewter]

and then x1+x2 =0

[hard pewter]

yeah mb

[hard pewter]

this will give you the common elements

or S inters T

[tidal turret]

will be cool if we find k first before intersection

[hard pewter]

S, not H

[tidal turret]

oh, im stupid

mb haha

[tidal turret]

OK

one moment

[hard pewter]

this should give you alpha+beta=0

[hard pewter]

yh so dim is 1

[tidal turret]

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

[hard pewter]

now for H, if k = 0 then dimH is obviously 3
and if k!=0 then dimH is ..

ig still 3?

[tidal turret]

dim(R) + dim(W) = dim(H)
dim(H) = 3
1 + dim(W) = 3
dim(W) = 2

[hard pewter]

yes still 3

[tidal turret]

k does not determine the dimension

[hard pewter]

so dim W is 2

[hard pewter]

yh

[tidal turret]

idk why tho, denascite and rafilou2003 always say k = 0 is still dim 3

like it doesnt determine the dimension, idk why

we could plug k = 0 and check if dim(H) is still 3

[hard pewter]

yeah u can check it out mathematically as well but to get an intuition,

[hard pewter]

it will obv

it will just become like T here

[hard pewter]

we only have one relation

between 4 variables

[tidal turret]

yeah

[hard pewter]

cant reduce the indepedent variables to less than 3

no matter what k is

anyway

[tidal turret]

yeah lets continue, dim(W) = 2

do you know how to find k?

,, (S \cap T) + W = H \implies \begin{cases} (S \cap T) \subset H \ W \subset H \end{cases}

[grand pond]

renato

[hard pewter]

i am getting the feeling that there is no definite value of k maybe

because for a value K, i will get some basis of H. And i can extend basis of SnT to that basis of H

[tidal turret]

wtf?

did you read the latex I sent

[tidal turret]

W should contain the zero vector so if (SnT) + W = H then (SnT) ⊂ H and W ⊂ H

also SnT needs to contain zero vector otherwise is not a subspace

[tidal turret]

so?

[hard pewter]

yh this makes sense

[hard pewter]

it does already

[hard pewter]

nvm i was cooking some garbage lmao

[tidal turret]

is okay, this exercises are nasty

if we plug the vector of SnT into cartesian equation of hyperplane of H we get k

H = {x2 + x3 + kx4 = 0}

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

2 + 0 -k = 0

k = 2

[hard pewter]

yh makes sense

[tidal turret]

H = {x2 + x3 + 2x4 = 0}

[hard pewter]

now we just find W by finding the elements that we need to extend basis of SnT with to get basis of H

that will be one possible W

or just the orthogonal complement can do

either way is fine ig

[tidal turret]

XD

[tidal turret]

which one you prefer?

[hard pewter]

i prefer first one because it highlights that W is not unique

[tidal turret]

wait a moment

what did you said?

[hard pewter]

i said W is not unique

[tidal turret]

"find the elements that we need to extend basis of S with to get basis of H"

[hard pewter]

not S

SnT

mb

[tidal turret]

oh, ok

yeah, that is a good description of the task

:)

[tidal turret]

H is dim 3

[hard pewter]

yes so we need 2 elements, which justifies that dim W is 2

[tidal turret]

also since k = 2, we are guaranteed that SnT lives in H

[tidal turret]

yes exactly

we need two vectors in H that make a basis dim 3 with SnT

[hard pewter]

yes

i can literally go with 1,0,0,0 and 0,0,1,0 lmao

they work

[tidal turret]

,w nullspace {{0,1,1,k}} where k = 2

[tidal turret]

H = {x2 + x3 + 2x4 = 0}

no

(0,0,1,0) not in H