help-38

[lime sphinx]

in 3d space you can be sure of your geometric intuition and a picture like this make sense

[urban copper]

PM = M - P = (-2,1,7)-(-5,7,4) = (3,-6,3)
L' : X = k(3,-6,3) + (-2,1,7)
k € R

[lime sphinx]

in R^n n>3 you have to rely more on the abstract theory, intuition is harder to obtain

[urban copper]

ok

so can we verify

[lime sphinx]

yep

[urban copper]

L : X = k(-1,1,3) + (0,-1,1)

L' : X = k(3,-6,3) + (-2,1,7)

L' n L NON EMPTY INTERSECTION
P = (-5,7,4) € L'
L' perp L

(-5,7,4) = k(3,-6,3) + (-2,1,7)

i) -5 = 3k - 2
ii) 7 = -6k + 1
iii) 4 = 3k + 7

-3 = 3k ==> k = -1

it works

P is in L'

,w (3,-6,3).(-1,1,3)

[solid kiln]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

L n L' :

-x = 3y -2
x - 1 = -6y + 1
3x + 1 = 3y + 7

---

-1 = -3y - 1
0 = 3y

(x,y) = (2,0)

ok it works I think

L : X = k(-1,1,3) + (0,-1,1)
L' : X = k(3,-6,3) + (-2,1,7)

L n L' = 2(-1,1,3) + (0,-1,1) = (-2,2-1,6+1) = (-2,1,7)

L n L' = 0(3,-6,3) + (-2,1,7) = (-2,1,7)

ok thanks, tricky exercise, ty both

@lime sphinx @leaden nacelle ❤️

[lime sphinx]

victory

!rats

[trim jolt]

[urban copper]

.close

[trim jolt]

Channel closed

Closed by @urban copper

Use .reopen if this was a mistake.

[urban copper]

cool drawing mate

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[urban copper]

1. is interesting, we did an similar exercise in class, but already forgot how

oh I think I get it

,w (2,-1,-2)x(-2,2,-3)

[solid kiln]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

,w sqrt{49k + 100k + 4k} = 1

[solid kiln]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

haha

2. is the complicated one, idk

[clear cloud]

The intersection of three planes is neccessarly a point or nothing

So what im proposing

Is get the equation of the plane with k

[urban copper]

what?

the intersection of two planes is a line unless its the same plane, or unless its empty

[clear cloud]

I said three

[urban copper]

I am just saying what I know from basic theory

never seen a triple intersection of planes before

so I wouldnt know

[clear cloud]

Wait nvm

It can also be a line

[urban copper]

yes bro

because if the intersection of two planes is a line

and then the line is entirely contained in the other plane

[clear cloud]

Yep

[urban copper]

is a whole line the intersection of the three

:D

[clear cloud]

You can get the definition of the intersection of the two planes you know

[urban copper]

wdym

[clear cloud]

Like the intersection of pi1 and pi2

[urban copper]

?

can you elaborate on what you mean?

[clear cloud]

You want the intersection of these 3 to be empty right ?

You can first get the intersection of the two first one

See that its a line (else if its empty, the exercice is done) and get the eqn of the line

[urban copper]

ok

can we do it step by step

I kind of got lost

@clear cloud

we know the cartesian equation of the planes Pi1, Pi2

[clear cloud]

Yep

Set it as system of equation

And uh row reduce it (is that how its called ? With augmented matrix)

[urban copper]

Pi1: x - 2y + z = 4
Pi2: -2x - 6y + 3z = -3

Pi1 n Pi2:

i) x - 2y + z = 4
ii) -2x - 6y + 3z = -3

,w rref {{1,-2,1,4},{-2,-6,3,-3}}

[solid kiln]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

x = 3

y - z/2 = -1/2

y = -1/2 + z/2

y = (1/2).(z - 1)

[clear cloud]

Gotta eat, bro

[urban copper]

(x,y,z) = (3, (1/2).(z - 1), z)

[urban copper]

but bro

we were in the middle of something

[urban copper]

ping me when you are back bro

fuck my life bro

[trim jolt]

@urban copper Has your question been resolved?

[clear cloud]

Ok im back @urban copper

[clear cloud]

Ok and what about the directionnal vector of this

[trim jolt]

@urban copper Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[limpid dawn]

\textbf{Show:} \
If $f,g$ are analytic functions on some open connected set $G$ such that
[ f'(z) = g'(z) \quad \forall z \in G ]
then $f(z)=g(z)+c$ for some constant $c\in\R$.

[limpid dawn]

hi, i am a bit confused what to actually show, doesnt this simply follow

like isnt this immediate or am i missing something

[pseudo goblet]

Sort of

So far, as far as "analytic" is concerned, I think you would have been taught that you can safely say that "derivative = 0" implies "function is constant", right?

[limpid dawn]

yea that makes sense

[pseudo goblet]

So - here we've actually got f' = g'

[limpid dawn]

ok i actually think that's it

.-.

[pseudo goblet]

Walk me through the rest?

If f'(z) = g'(z), then ....

[limpid dawn]

then f'(z)-g'(z) = 0, now we integrate both sides getting f(z)-g(z)=C

[pseudo goblet]

yeee

[limpid dawn]

😭

[pseudo goblet]

Then you get to your conclusion

[limpid dawn]

ok thanks, i overthought this

.solved

[trim jolt]

Channel closed

Closed by @limpid dawn

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[olive totem]

I don't know how to proceed further I need some assistance

[trim jolt]

@olive totem Has your question been resolved?

[pseudo goblet]

You have a squared term which you're then square-rooting

...

Provided you can check which of 2root(6) - 5 and 5 - 2root(6) is positive...

[olive totem]

Ah yeah wait let me write it

I'm sorry for the question but I'm not sure I know what I'm writing

Something like this?

Ah this doesn't make sense tho

[heady furnace]

there is a square and a square root

[pseudo goblet]

You have a number, you're squaring it and square-rooting it

They cancel out*

*(if the bit inside the squared brackets is negative, it becomes positive)

compare how sqrt(10^2) = 10

[olive totem]

You're talking about √(2.√6-5)^2?

[pseudo goblet]

yes

That is a thing you're squaring then square-rooting

[olive totem]

What exactly do u mean the square root disappears, the ^2 is inside of it, meaning a formula in the equation inside

[pseudo goblet]

Look here

[olive totem]

Oh my gyatt I see it now 👎👎👎👎👎

Sigh

I'm tweaking rn

[heady furnace]

$\sqrt{a^2} = (\sqrt{a})^2 = \abs{a}$

[solid kiln]

Evotushon

[pseudo goblet]

dear lord

[pseudo goblet]

(i'd be careful about that middle one, but the LHS and RHS are right yes)

[olive totem]

Sorry for wasting your time lol

[heady furnace]

yeah speaking generally I wouldn't throw around absolute values left and right but for the last one it is needed

[pseudo goblet]

yh

[olive totem]

Closing now

[pseudo goblet]

And in this case too, since 2root(6) < 5

[olive totem]

Yeah I noticed this too

Okay bye

.close

[trim jolt]

Channel closed

Closed by @olive totem

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[vagrant kraken]

ok a very easy way of dealing with this problem

a parabola has zeros -3 and 7, what does that imply?

[terse python]

that it's like (x-3)(x-7)

?

[vagrant kraken]

yes exactly

however

it needs to pass through (5,36)

what changes can you make to y=(x-3)(x-7) so that is does not change the zeros and it also can STRETCH

also it should be +3

y=(x+3)(x-7)

[terse python]

so like 36=a(5+3)(5+7)

?

[vagrant kraken]

yes

not +7

[terse python]

-7

[vagrant kraken]

the zeros are -3 and 7

that means that those number + a number should equal to 0

yeah you would have a(x+3)(x-7)

[terse python]

how do i write it in standard form?

[vagrant kraken]

it only asked you to write it in y=a(x-s)(x-t)

[terse python]

theres another question

after this

[trim jolt]

@terse python Has your question been resolved?

[trim jolt]

@terse python Has your question been resolved?

[fierce rain]

Go on

What's the other q

[trim jolt]

@terse python Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[mystic veldt]

is this not wrong

shouldn't $\int d\vec{r}=L$

[solid kiln]

yajat

[mystic veldt]

nvm

.close

[trim jolt]

Channel closed

Closed by @mystic veldt

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[upper zinc]

no of solutions of summation n=1 to n=6 of x to the power n / n factorial = -1

[slender lion]

,, \sum_{n=1}^{6}\frac{x^n}{n!}=-1 ?

[solid kiln]

Gregory

[upper zinc]

yes

[marsh forum]

Well, 0, I think?

[upper zinc]

cant it be a negative value

[marsh forum]

This is part of the taylor expansion of e^x

I'm not sure though

I mean some terms will ne gative

[upper zinc]

i was getting opp signs for x=1 and x=2

[marsh forum]

but the positive terms should overpower the negatives ones

,w \sum_{n=1}^{6}\frac{x^n}{n!}=-1 solve for x

[solid kiln]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by math_rocks

[marsh forum]

,w solve x+ x^2/2!+x^3/3!+x^4/4!+x^4/5!+x^6/6!=-1

see

no real solutions

[slender lion]

this doesnt include the first term so it it negative for part of it but yeah not quite to -1

[solid kiln]

Results provided by WolframAlpha

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

Requested by math_rocks

[upper zinc]

would there be any way to show this mathematically

[slender lion]

take the derivative and find its minimum then evaluate it at the minimum to show that it is > -1

that of course requires finding 0 of 5th order polynomial so idk

[marsh forum]

or use the fact that this is part of taylor expansion of e^x, which is always positive

[upper zinc]

yeah ur right

thnx

!close

.close

[trim jolt]

Channel closed

Closed by @upper zinc

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[manic lagoon]

What is a tangent?

[manic lagoon]

To a circle?

[trim lichen]

a straight line which has exactly one point in common with the circle

[manic lagoon]

Ques. Prove that a tangent is perpendicular to the line joining the point of tangency to the centre of the circle.

[trim lichen]

do you want to present your own proof attempt or do you want to get an idea tossed your way?

[manic lagoon]

Wait.

I am presenting my proof.

Proof:

Consider a circle with a tangent CD drawn, touching the circle at B. The centre of the circle is A.
The assumption is that the circle is only in contact with a single point of the tangent, which is B. Or that the whole circle lies above the tangent.

If AB is perpendicular to CD, we're done.

If not, then let AG be the perpendicular from the centre to the tangent. Since triangle ABG is a right triangle, AB > AG. Then there must lie a point F such that AF = AB = radius. This means the circle must pass through F. Clearly, F is a point which is on the other side of the tangent. Therefore, the circle has a point below the tangent, and this contradicts the assumption (a certainty). This contradiction will always arise if BG is non-zero (when GF is positive). So BG must be zero. We're done.

I will immediately go see the textbook proof after this.

My proof is based on the fact (which has just been proved by me) that the perpendicular from A to CD coincides with the point of tangency.

I forgot to ask the question.

Q. Is my proof correct?

<@&286206848099549185>

[trim lichen]

yeah looks correct ish

you have to use in one way or another that the hypotenuse of a right triangle is longer than either leg

though i will say that you will find people more willing to read your proof if you follow the convention of calling the center of a circle O

[manic lagoon]

Okay.

[manic lagoon]

ok

Thanks.

.close

[trim jolt]

Channel closed

Closed by @manic lagoon

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[maiden falcon]

The numbers 1, 2, 3, . . . , 100 are written down on each of the cards A, B, and C. One
number is selected at random from each of the cards. Find the probability that the
numbers so selected can be the measures (in cm) of the three sides of a right-angled
triangle.

help pls

[empty prairie]

Use the pythagorean triplets

[outer kiln]

^^ those are the possible sides

[maiden falcon]

is there a known way to find all pythagorean triplets in a given range

[native shuttle]

Why do I feel like I have seen this question somewhere

[outer kiln]

Im not sure if there’s any fast way to find them you’d probably just have to count

[native shuttle]

There was a way if I remember

[maiden falcon]

u do realise

😭

[native shuttle]

diptava

[maiden falcon]

i think so, yeah

it was in a mock test i was taking

[native shuttle]

Yep seen this question before

[maiden falcon]

be careful using that abbreviation tho

💀

[native shuttle]

ik

[maiden falcon]

if anyone knows the procedure please do help

@dapper swift maybe check this out

[native shuttle]

ann might know

[trim jolt]

@maiden falcon Has your question been resolved?

[dapper swift]

https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

yes, you're looking for Euclid's formula

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[glad patio]

is his right?

[dusky thunder]

take 6 to the power

not the coefficient

[whole coral]

This isn’t equivalent to what you were given either

[dusky thunder]

(also you arent dividing by log 2 when you cancel them out, that is the wrong way to think about it)

you are raising them to the power of 2

aka applying the inverse

[red loom]

although it is correct to assert that:

[\log_2(3p-1)-2\log_2(q)=\log_2\left(\frac{3p-1}{q^2}\right),\quad p,q>0]

[solid kiln]

PajamaMamaLlama

[red loom]

hello chartbit!

[whole coral]

Hiiiii

[glad patio]

so this?

[whole coral]

This bit here is not quite correctly written

[whole coral]

You should have the right hand side of here

The subsequent steps are sus, even allowing for that

[trim jolt]

@glad patio Has your question been resolved?

Channel closed

Closed by @glad patio

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[lusty pond]

Could anyone help me spot the error in my work? I keep getting -1.5. Here is the associated graph

[trim jolt]

@lusty pond Has your question been resolved?

[trim jolt]

@lusty pond Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[empty tulip]

Can someone check this please

also can mph be written as mh^-1?

[trim lichen]

you shouldn't write miles as m

m is for meters

maybe abbreviate miles as mi instead

[empty tulip]

right okay

[viral merlin]

just add units for 6b) , and yeah everything else looks good

[trim jolt]

@empty tulip Has your question been resolved?

Channel closed

Closed by @empty tulip

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[marsh forum]

Trying to integrate this using green;s theorm

[marsh forum]

so $\iint (2xy^3 - x )dA$

Is that right so far

[worldly wing]

[solid kiln]

What a wonderful world !

[worldly wing]

looks good

[marsh forum]

so I then have $\int_{0}^{1} \int_{0}^{2x} (2xy^3-x)dydx$

[worldly wing]

uhh

that doesn't look right

[marsh forum]

what's wrong

the curve joinign (0,0) to (1,2) is y=2x

[worldly wing]

why are you integrating from 2x to 1?

should be 0 to 2x

[marsh forum]

the lines in fubini's theorm enter from y=2x and exit from x=1

wait

oh

my bad

yea

[solid kiln]

What a wonderful world !

[worldly wing]

should be easy from here

[marsh forum]

yea, I'll leave it here, I'll just do one more green's theorm question

$\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} -3(x^2+y^2) dy dx$

[solid kiln]

What a wonderful world !

[marsh forum]

this feels wrong

[worldly wing]

why?

[marsh forum]

look at the order

[worldly wing]

what

[lament reef]

it looks right to me 🤔

[marsh forum]

sorry, basically using fubini's right

So the "lines" enter from $-\sqrt{4-y^2}$

[solid kiln]

What a wonderful world !

[marsh forum]

,w plot \sqrt{4-x^2}$

[solid kiln]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by math_rocks

[worldly wing]

the lines?

[marsh forum]

wait, I'll send a picture

oh

got my mistake

makes sense now

nvm

thanks

[marsh forum]

this can be re-written as $\int_{0}^{2 \pi} \int_{0}^{2} -3r^3$

[lament reef]

don't forget your jacobian

[solid kiln]

What a wonderful world !

[marsh forum]

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[zenith stream]

Does anyone have anything on explaining annihilators ?

I have hw due about determining if an annihilator of a vector space is subspace of it but I can’t find anything on annihilatiors themselves

[whole coral]

(did they not even give you the definition? )

[clear cloud]

If so thats not very kind of them

[zenith stream]

It’s in German but if you mean like the definition of the annihilatior yes, but I don’t know how to work on it

Jesus that’s unreadable hold on

[clear cloud]

Its good

[whole coral]

And you're doing (i) basically?

[clear cloud]

Unterraum = subspace if im not too rusty

[zenith stream]

Well I need to do all of it

[zenith stream]

I think so!

I’ve been looking online for any kind of explanation but I can’t find anything that tells me where I’m even supposed to start :(

[whole coral]

Well, how about you start with us

[zenith stream]

[clear cloud]

What are the questions in your hw ?

[whole coral]

Do you know how to show something in general is a subspace of a larger vector space?

[zenith stream]

I just finished watching some lectures on it and made notes so I think so?

[zenith stream]

Hold on I’ll translate it

[whole coral]

Awwww, that's cool, can you state it by any chance? (if not, then don't worry!)

[clear cloud]

Dw about this

Chartbit understand german aswell

[whole coral]

I don't, but I can figure out what they're asking

[zenith stream]

“V is a K-Vector space (K meaning Body here) and U is a subspace of V. The ammount
[here the definition]
is de annihilatior of U. Show:
i.) U0 is a subspace of V*

[whole coral]

(@limpid dawn may I pingeth you )

[clear cloud]

Call the avengers

[clear cloud]

Oh indeed, it was already the hw, sorry

This looks like a special case of orthogonal vector space

[limpid dawn]

Was ist V*

[clear cloud]

Algebraic dual

[limpid dawn]

hab gegooglet

[zenith stream]

Shit sorry my parents are calling me for dinner I’m really sorry I’ll be back asap like 15min

[limpid dawn]

set of all linear maps from V to a field K

[clear cloud]

Yeah

gegooglet is pretty funny to say

[limpid dawn]

body is funny to say, it's field actually

[clear cloud]

Körper 🗿

[limpid dawn]

für UVR musst du drei Sachen zeigen

1. Neutrale Element ∈ U⁰
2. a,b ∈ U⁰ folgt a+b ∈ U⁰
3. λ ∈ K, a ∈ U⁰ folgt λa ∈ U⁰

[clear cloud]

bzw ?

[limpid dawn]

beziehungsweise

Du musst hier wissen und benutzen, dass U ein UVR ist und dass f eine lineare Abbildung ist

Du musst dann deren Eigenschaften ausnutzen

[trim jolt]

@zenith stream Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[limpid dawn]

.reopen

[trim jolt]

✅

[whole coral]

Saved it

(also, if you're curious, there's also another one #help-3 )

[limpid dawn]

i am outside

[whole coral]

Awwww

Hope you're having fun at least

[limpid dawn]

i haveewe

good wwather

[whole coral]

Been a long 15mins

[trim jolt]

@zenith stream Has your question been resolved?

[clear cloud]

its important to take time to eat

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[wet current]

I can’t figure out these last two questions

[wet current]

<@&286206848099549185>

[lament reef]

what have you tried?

[wet current]

i have tried to calculate the area but failed

[lament reef]

okay, well each of those tiny squares has an area of 9

how many squares are there

[wet current]

there are 42 i think

[maiden falcon]

youre right about that, so just multiply and get your answer 🙂

[wet current]

nice i got the orange one done

[maiden falcon]

yess, you can similarly get the purple one done as well 🙂

[trim jolt]

@wet current Has your question been resolved?

Channel closed

Closed by @wet current

Use .reopen if this was a mistake.

[zenith stream]

Sorry, had to clean the kitchen

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[zenith stream]

Oh ok ty I could t figure out why it was

[zenith stream]

Muss ich das irgendwie beweisen? Oder wie kann ich zeigen dass es eine abb ist?

[limpid dawn]

nein dass weiß man

ein element aus U^0 bedeutet dass es ein element von V* (Menge der linearen abbildungen) mit der eigenschaft dass halt auf das neutrale Element immer abgebildet wird

[zenith stream]

Ohhhhhh

Ok dass macht viel mehr Sinn

[zenith stream]

Ok also ich muss zeigen das U^0 ein UVR ist durch Beweis der drei Bedingungen von f?

[limpid dawn]

mithilfe der eigenschaften von f (ist ja eine lineare abbildung) zeigst du die 3 uvr axiome

[zenith stream]

Ok… ich glaube ich habe nur irgendwo ein Beispiel vob solchen Aufgaben gespeichert.. Danke dir! Versuche erstmal das zu lösen

[limpid dawn]

wir können es zsm machen wenn du magst

[zenith stream]

Im welchem Sinne :0

[limpid dawn]

ja hier

[zenith stream]

Ah ok hole nur mein pc auf händy ist beschissen

[zenith stream]

Will dir auch nicht Zeit nehmen fyi

[limpid dawn]

wie du willst

[zenith stream]

Im Sinne wenn du etwas wichtiges hast will ich jetzt niecht das du meinend wegen das verschiebst aber ich wäre dankbar für die Hilfe

[limpid dawn]

hätte ich was wichtiges, wäre ich nicht hier

[zenith stream]

ok.. also erst beweisen dass 0 in U^0 enthält ist

?

[limpid dawn]

ja

frage, ist 0 in U?

[zenith stream]

ein 0 vektor der zur menge dazugehören muss?

[limpid dawn]

ja, der gehört dazu, weil U ja ein UVR ist

das sollte man wissen/erwähnen

[zenith stream]

ein neutrales element dann?

[limpid dawn]

ja

[zenith stream]

ok.. um ganz ehrlich zu sein ist mein größtes problem gerade ist u^0 definition verstehen ich versuche mir das selbst gerade zu erklären sorry, gib mir eine sekunde

[limpid dawn]

U^0 ist die Menge aller linearen Abbildungen V -> K mit der Eigenschaft, dass sie auf 0 abbilden

also von all den linearen abbildungen in V* ist U^0 eine teil menge mit elementen die nur auf 0 abbilden

man will jetzt zeigen, dass sie ein uvr ist

also dass man da so rechnen kann

[zenith stream]

ist V* das gleiche wie V->K...?????

[limpid dawn]

ja L(V,K) so habe ich es verstanden

[zenith stream]

sorry, wenn dumme fragen, sitze an dem thema schon ganzen tag

[limpid dawn]

kein problem lol

[zenith stream]

was bedueted nochmal auf 0 abbilden..

im sinne wie rec hnet man sowas

[limpid dawn]

du setzt was in die abbildung ein, und da kommt 0 raus

[zenith stream]

OH'

[limpid dawn]

ja

[zenith stream]

ok ok also, hold on, ich schreibe es mir kurz ins notitztbuch

[limpid dawn]

also abbildung, also auf etwas wird abgebildet

du gibst dem was, der gibt dir was zurück

[zenith stream]

ja stimmt, habe vergessen, abbildungen waren letztes jahr bei mir chnell gemacht, sehe gerade das war ein fehler von ir longterm

ok also U^0 sind elemnte in V* die abbildungen auf 0 geben.. ok ich habs endlich kappiert, hoffe ich

[limpid dawn]

elemente in U^0 ... aber ja

[zenith stream]

wait

[limpid dawn]

und diese elementen (abbildungen) bilden auf 0 ab

[zenith stream]

ok uhh jetzte irgendiwe uvr beweisen.. du hasst nicht zufälliger weise irgendein beispiel erkläreng oder weist wo eine ist nach der ich mich richten kann..????

[limpid dawn]

ich habe es selber versucht zu beweisen, deswegen habe ich angeboten dir zu helfen

[zenith stream]

ah ok danke

[limpid dawn]

fangen wir an mit (1)

[zenith stream]

sorry nochmal, unser prof erklkärt nicht so und versuche alles selber zu lernen aber ist schwer

[limpid dawn]

wir wissen dass 0 in U, weil U ein UVR ist von V. Außerdem wissen wir dass f in V* ist, das bedeutet also dass es eine lineare abb. von V auf K ist

welche eigenschaft haben wir für lineare abbildungen

[zenith stream]

uhm, hier oder prinzipiell

also esy ist ene funkttion?

ja?

hier im sinne

[limpid dawn]

angenommen f ist eine linear abb. welchen wert hat f(0)?

[zenith stream]

0..

[limpid dawn]

ja, wegen der homogenität gilt ja f(0) = f(0v) = 0f(v) = 0

jetzt wo wir das alles wissen

was passiert mit f(0_u)

?

[zenith stream]

das gleiche..?

also im sinne

sekunde

[limpid dawn]

moment mir ist gerade was aufgefallen man

wir müssen 0 in V* betrachen also die nullabbildung

nicht das element 😮‍💨 mb

aber umso einfacher

erfüllt die null abbildung die bedingung?

[zenith stream]

oh ok

[zenith stream]

also es hat ein nullelement auf jeden fall..

[limpid dawn]

hu

[zenith stream]

oder? fuckoing hell im dumb

[whole coral]

(I was about to comment about whether you were considering the zero map, can't say I understand the conversation but guessing you spotted that(?) )

[zenith stream]

nehme koffein hold on

[limpid dawn]

yeessss

i am sorry

[zenith stream]

gehirn functioniert nicht

[limpid dawn]

ok wir fangen neu an

[zenith stream]

im sorry

:(

[limpid dawn]

(1) wir wollen zeigen dass die Nullabbildung n(u) in V* auch in U^0 ist

jetzt musst du dich fragen, gilt denn n(u) = 0 für alle u?

[zenith stream]

n(u)?

[limpid dawn]

ja so habe ich mal die null abbildung genannt

[zenith stream]

ja??

[limpid dawn]

ja

so ist die ja definiert

[zenith stream]

ok..

[limpid dawn]

(mathematiker würden sagen trivial)

ok das zu (1), also ist n(u) auch in U^0

klar

so

jetzt (2)

[zenith stream]

fage, ist n(u) f(u)????????? nein oder?

[limpid dawn]

f(u) ist einfach so ein name

so ein namenshalter

[zenith stream]

ok,war mir gerade nikcht sicher

[limpid dawn]

platzhalter macht das sinn?

[zenith stream]

ja, war einfach nicht sic her ob es ein n(u) gab das ich übersehen habe und habe die frage 30 mal niochmal gelesen

sorry

[limpid dawn]

ich habs hier n genannt

damit wir wissen von wo wir reden

[zenith stream]

ok

[limpid dawn]

\textbf{(2). Zu zeigen: } Seien $f,g \in U^0$. Dann ist auch $f+g \in U^0$.

[zenith stream]

2 muss ich zeigen dass die summe zweier vektoren wieder in U^0 sind?

[limpid dawn]

jaa

[zenith stream]

oh waqit hier kann man mit latex schreiben???

[limpid dawn]

ja

[zenith stream]

yoooo what geil

[limpid dawn]

habs dir anscheinend vorenthalten haha

[zenith stream]

das die summe von f und g in U^0 sein müssen..? also

hold on

[limpid dawn]

genau das willst du ja zeigen

[limpid dawn]

das haben wir, ok?

[zenith stream]

mhm (also wie schnell schnriebst du in latex omg 😭 )

[limpid dawn]

was gilt dann unbedingt für f(u) und g(u)

[zenith stream]

ich habe das gefühl ich bin da komplett falsch, aber müssen die beiden.. 0 ergeben???

[limpid dawn]

jaaaa

[zenith stream]

[limpid dawn]

für alle u ergben die 0

[zenith stream]

ohhh ok

[limpid dawn]

das steht hier ja

also du weißt f(u) = 0 und g(u) = 0 für alle u

[zenith stream]

mhm

[limpid dawn]

zweifel?

[zenith stream]

soll ich den haben????

und jetzt beweisen dass das produkt aus skalar und Vektor in U^0 vorhanden sind.... also 0? weil multipliziert mit 0..

[limpid dawn]

wir sind nicht fertig

wir fangen jetzt an

[zenith stream]

uh oh

[limpid dawn]

wir haben nur gesammelt an informationen die wir benutzen

werden

f+g in U^0 willst du zeigen

[zenith stream]

mhm

[limpid dawn]

wie ist vektoraddition mit abbildungen definiert?

[zenith stream]

uhh... hold on machen die abbildungen ein unterschied von einer normalen vektoraddition..?

[limpid dawn]

in einem vektorraum nennt man bezeichnet man die elemente als vektoren

das ist vlt bisschen seltsam weil das in der schule anders beigebracht wurde

aber so ist es in der linearen algebra

und auf einem vektorraum gibt es zumal eine vektoraddition also wie man die sogennanten elemente (vektoren) zusammen addiert (muss nicht das klassische addieren sein, was man kennt)

[zenith stream]

v + w ∈ U, ∀v, w ∈ U?

[limpid dawn]

das willst du zeigen hier

nur für U^0

[zenith stream]

ja..

durch f(v + w) = f(v) + f(w), ∀ v, w ∈ V???

[limpid dawn]

nein

du addierst abbildungen

also fängst du mit (f+g)(u) an

[zenith stream]

ist das nicht =fu+gu

[limpid dawn]

yep so ists ja definiert

f(u)+g(u)

SO

und was ergibt das, welche Annahme trafen wir über f(u) und g(u)

[zenith stream]

f(u)=0 und g(u)=0??

[limpid dawn]

ja

also erfüllt f+g auch die bedingung, dass es auf 0 abbildet, für alle u

damit wärst du durch

Du benutzt doch das, was schon wahr ist

Wir haben gesagt f und g sind elemente in U^0, dann gilt f(u) = 0 und g(u) = 0 für alle u

das darfst du da benutzen

[zenith stream]

ok.. und für skalarprodukt
(λ · f)(u) = λ · f(u)=0?

[limpid dawn]

ja weil f(u) ein element von U^0 ist

[zenith stream]

yey :)

[limpid dawn]

jetzt noch bisschen alles vernünftiger aufschreiben und das wars

[zenith stream]

https://tenor.com/view/kiss-gif-11816971814746635421

danke T-T

[limpid dawn]

kd!

[whole coral]

(minor comment from me lurking but as you know f(u) = 0 for all u in U, you may want to add in that lambda * f(u) is lambda * 0, which then works out to be 0 )

[zenith stream]

ich verstehe nicht was sogar mein problem jetzte war, fühl mich so dumm, ist in sich einfach..

[zenith stream]

OH ok ty!!

[limpid dawn]

i actually thought about whether to mention it, should have

[limpid dawn]

willkommen in mathe bzw. lineare algebra

[whole coral]

Makes things clearer, and shows how things follow on

[zenith stream]

ill try to write it prpperly, and do my best to do the erst... tbh my biggest issue have been definitions

[limpid dawn]

yess die musst dud dir einprägen und benutzen

alles baut darauf auf

[zenith stream]

im starting to understand why the biology majors chose biology to avoid math

my biochem class genuienly easier than beginners math

[limpid dawn]

ah biology

the study of

what

[zenith stream]

silly small things that will kill you somehow

[limpid dawn]

smells like definitions, if you dont know them

[zenith stream]

bio majors memorising all of math so tghey never have to deal witgh it again

[limpid dawn]

die anderen aufgaben sehen krasser aus

[zenith stream]

i shoukld get on that grind

[limpid dawn]

i se

[zenith stream]

ich habe bis 8 uhr morgen, chat how cooked am i

[limpid dawn]

wie bis 8 uhr?

oder ab 8 uhr

[zenith stream]

this is all of it but my part is 2.2

[limpid dawn]

(ii)

[zenith stream]

gotta hanbd it in tmw

[limpid dawn]

ah

übungsblatt oder?

[zenith stream]

im ina group with two engineering masters and last week we got a bad grade bc of me so im paranoid about making mistakes rn, honestly was scared af to ask for help here

[zenith stream]

tutorium

[limpid dawn]

okok

Ja also (i) haben wir richtig, du musst halt schauen, dass du den Beweis richtig schreibst, also dass er logisch ist

[zenith stream]

yessir

[limpid dawn]

Ok also für die (ii) a)

[whole coral]

(btw you're always welcome to ask for help here now, we're quite friendly, I would like to think )

[zenith stream]

tbh i think like a year or more ago i asked for help in general and got like someone message me mean stuff so i kinda was too afraid to try again but im glad i didm you guys r awesome :)

[limpid dawn]

Wenn ich mir das anschaue, die linke Menge bedeutet
[ (U_1+U_2)^0 \coloneqq {f \in V^{\star} \mid f(u) = 0, \quad \forall u \in U_1+U_2} ]

[zenith stream]

ih nein, muss ich das davor definieren..

ok macht sinn

[limpid dawn]

nein, das ist schon definiert, ich wollte nur herausfinden wie genau, damit man ja iwie anfangen kann

[zenith stream]

ah ok weis tbh jetzt auch nicht ib mann das etzte einfach ebweisen soll oder ausrechnen oder wie

[zinc shard]

Seems like i cant lurk here no more

Off to anohter channel

[limpid dawn]

,, U_1^0 \cap U_2^0 \coloneqq { f \in V^{\star} \mid f(u_1) = 0 \text{ und } f(u_2) = 0, \quad \forall u_1 \in U_1, \forall u_2 \in U_2 }

[limpid dawn]

does this make sense chartbit?

[zenith stream]

$f(u_1)=0$ weil $f(u)=0$ ?

[solid kiln]

Inoghmia

[limpid dawn]

was wir machen können ist zeigen, dass
[ A = B \lr \forall a \in A : a \in B \text{ und } \forall b \in B : b \in A ]

[limpid dawn]

also jedes element in (U1+U2)^0 ist auch im Schnitt und umgekehrt

[zenith stream]

ohh ok

[limpid dawn]

Also fangen wir an mit der einen seite

also im prinzip you show that both are subsets of each other

[zenith stream]

was ist nochmal der unetrschied zwischen inem teilraum und einer teilmenge? ist es nur weil es uvr sind?

[limpid dawn]

teilraum meint man uvr

[zenith stream]

ah ok

uh also hm

[limpid dawn]

Frage wie ist f definiert?

[limpid dawn]

Kannst du mir sagen was für f gilt?

[zenith stream]

uhh sekunde...

[zenith stream]

warte steht da nicht alles schon was für f gilt?

[limpid dawn]

ich will es in deinen worten was konkret für f gilt

[zenith stream]

f muss auf allen vektoren von u1 und u2 0 annehmen ..?

[limpid dawn]

konkret weißt du f(u) = 0 für alle u in U1+U2

wenn u in U1+U2 was gilt dann für u nach Definition der direkten summe?

[zenith stream]

u1 +u2 müssen 0 ergeben aber ich folge glaube nicht was das jetzt für u bedueted..

[limpid dawn]

u ist doch die summe von einem element aus U1 und U2

zum Beispiel

[zenith stream]

ja

[limpid dawn]

u = u1+u2

f(u) = f(u1+u2)

[zenith stream]

ja

[limpid dawn]

mach weiter

f ist lineare abb was kann man machen weiter

[willow urchin]

are you doing 2.2.ii?

[zenith stream]

yeah hes basiacly tutoring me while my brain turns to nush bc i need to hand it in by tmw

[zenith stream]

ganz ehrlich, ich weis nicht...also ich weis mit linearen abb kann man rechnen aber ich bin mir jetzte nicht sicher was man hier weiter machen kann

[willow urchin]

i see

for (a) it might be easier to do a double subset proof rather than trying to write a chain of set equalities

[limpid dawn]

thats what we were actually aiming for

[willow urchin]

mb idk german

[limpid dawn]

maybe i got lost somewhere

[zenith stream]

can you give me a minute, im gonna make some coffe

[limpid dawn]

1 am

[zenith stream]

im struggling to form thoughts

[willow urchin]

did u already prove LHS subset RHS?

[limpid dawn]

thats what we do rn

[willow urchin]

https://cdn.discordapp.com/attachments/908078008609431674/1366903472104149123/587715738484211754.png?ex=6812a38f&is=6811520f&hm=612420285394ce88b5d512d19aac2203be2693e6a3b81d25a51785bea958365f&

ah this

[limpid dawn]

yea

[zenith stream]

yeah, but i currently have issue remebering words, ill survive tzmw on 2h or smth

[willow urchin]

let $f\in(U_1+U_2)^0$ then aim to show $f\in U_1^0$ and $f\in U_2^0$

[solid kiln]

ロケットジャンプ

[willow urchin]

yes?

[limpid dawn]

yes

i mean you are the pro but yes

[willow urchin]

ofc u know but what about inoghmia 🙂

[limpid dawn]

oh what about her?

i just met her

oh i see lol mb

[willow urchin]

chart needs a hobby

ig we wait for inoghmia to caffeinate

well adonis and chart can wait. i have a meeting soon

[limpid dawn]

i think we were actually doing well so far

[whole coral]

I'm literally just watching videos and chilling

[zenith stream]

hi im back

sorry for the wait

[limpid dawn]

hey willst du auf Deutsch oder Englisch weiter

[zenith stream]

at this point im just trying my best to understand anything so that i can word it nicely later in latex, ill worry about fully understaning remebering and practicing later

[zenith stream]

mathe wörter deutsch aberv alles andere egal

[willow urchin]

so far nothing about concepts, just writing goals

claim: $(U_1+U_2)^0\ss U_1^0\cap U_2^0$

[solid kiln]

ロケットジャンプ

[zenith stream]

last week my group flopped bc of me so im just trying my best to not dissapoint them again

[willow urchin]

to prove this, let $f\in(U_1+U_2)^0$ then aim to show $f\in U_1^0$ and $f\in U_2^0$

[solid kiln]

ロケットジャンプ

[zenith stream]

ok

[willow urchin]

agree?

i didnt invoke any concepts. i just set up the goal

[zenith stream]

yeah i thik i saw an example similar to this

[willow urchin]

now we bring in concepts

why $f\in U_1^0$?

[solid kiln]

ロケットジャンプ

[zenith stream]

because U1 and U2 are both of U^0..?

are theyß

?

[willow urchin]

to show $f\in U_1^0$ we must show $f(u)=0$ for all $u\in U_1$

[solid kiln]

ロケットジャンプ

[willow urchin]

ok gotta leave this to adonis. hope the setup helps

[limpid dawn]

guess i can go now

[zenith stream]

pleas e dont leave me-

unless u have to sleep

or do smth

i dont want to take up your time if oyu dont ha any to spare

[limpid dawn]

hmm okay

Assume $f \in (U_1+U_2)^0$ then we know $f(u) = 0$ for all $u \in U_1+U_2$.

[limpid dawn]

verstehst du?

(chartbit if we go wrong pls interrupt)

[zenith stream]

i think?

[limpid dawn]

ok

jetzt fragst du dich, was bedeutet es für u in U1+U2 zu sein

[zenith stream]

ja, also was bedueted das fur $f \in (U_1+U_2)^0$

[limpid dawn]

nein u

[solid kiln]

Inoghmia

[zinc shard]

Oh, so it’s german

[limpid dawn]

ja das habe ich hier geschrieben

[zinc shard]

[limpid dawn]

wir wissen f(u) = 0 für alle u in U1+U2

jetzt frage ich wwas bedeutet u in U1+U2 konkret

[zinc shard]

Sorry for disturbance, i go sleep now

[zenith stream]

meinet das als eine frage

[limpid dawn]

hier

das bedeutet es

wie ganz vorher nur dass u aus einer anderen menge kommt

[zenith stream]

u ist in U1 und U2 vorhandene..? ist es eine teilmenge..?

[limpid dawn]

nicht unbedingt und u ist keine Menge sondern element

[zenith stream]

[limpid dawn]

definition!!

[zenith stream]

oh ok

[limpid dawn]

,, U_1+U_2 \coloneqq { u_1+u_2 \mid u_1 \in U_1, : u_2 \in U_2 }

[zenith stream]

oh stimmt man macht ja dann u2, u1.. hab vergessen

[limpid dawn]

u ist in U1 + U2 also müssen doch solche u1 u2 existieren

aus denen u sich zsm setzt

[zenith stream]

habe das gemeint bin eben nur.. dumm, muss an definitionen arbeiten..

[limpid dawn]

ok kein problem ist auch spät

,, u \in U_1+U_2 \ri (\exists u_1 \in U_1) , (\exists u_2 \in U_2) \text{ such that } u = u_1+u_2

[limpid dawn]

klar?

[zenith stream]

wird das nicht durch ie vorherige definition von dir erklärt?

[limpid dawn]

ja

[zenith stream]

oh ok

[limpid dawn]

aber bei eine beweis schreiben wir das auf

die definition schreibe ich für dich

die sollte man im kopf haben

[zenith stream]

danke

[limpid dawn]

ok jetzt wo wir oberes wissen, setzen wir mal ein in f

mach mal

[zenith stream]

als in das f(u)=0?????

[limpid dawn]

ja

was kommt raus

wenn man u neu setzt

[zenith stream]

...0?

[limpid dawn]

f(u) = f(u1+u2)

einfach für u das einsetzen was wir gerade hergeleitet haben

[zenith stream]

ich glaube ich muss das mir auf ein zettel schreiben

oh

[limpid dawn]

jetzt nutz die linearität von f aus

[zenith stream]

ergibt sich dann nicht u1+u2=u1+u2 ???? 😭

[limpid dawn]

wat

[zenith stream]

i need to kill myself at this point

[limpid dawn]

are you good?

kinda worried

[zenith stream]

im honesstly not sure my vision is kinda bury

[limpid dawn]

literally or metaphorically?

[zenith stream]

literally

[whole coral]

That's a no then

How much sleep did you have the night before?

[zenith stream]

i think i havent slept for abiut 28h im insomniac

[whole coral]

That is very concerning

You must be really really exhausted

[zenith stream]

i took 400 mg caffeine im awake just, might have side effects, happens sometinmes, i thjink i need to take an hiourand wait till they wear off

[whole coral]

"side effects"

[whole coral]

I do hope they wear off but even at that, just the sound of that

When do you think you'd be able to rest by a decent amount?

[zenith stream]

probs tmw ill sleep in the afzernoon before work

or pass out in class my prof is chilkl like thagt, lets me sleep

[whole coral]

Awwww, well I do hope you get a decent (and uninterrupted!) amount of sleep soon

[whole coral]

For a moment when I saw this, I thought you may have needed to do more, but at least it's just 2.2 you need

I think they should be doable (the ones y'all haven't yet done, of course!)

[zenith stream]

ty man

[zenith stream]

oh no dude if i had to do everything i think id just cry

[whole coral]

I would cry too, and I'm not even the one doing them

[zenith stream]

idk, ill try to sleep now, caffeien sometimes has an olpposite effect on me of making me sleepy, ill wake up later and contnue in the morning

ive been at math for like 10h rn so mabe its just my brain overlaodng

[whole coral]

That is a long time to be working on something for sure, you definitely need to take a break, and get some rest

[whole coral]

And yea, sometimes it does be like that when do you think you might wake up?

[zenith stream]

u right, imma take a shower, eat, wait till my head isnt pulsing anymore and my vision evens out

[zenith stream]

youtube 1h timer, headphones in and set on full volume is my go to combo

[whole coral]

Yep, you should! If you aren't feeling great then better to wait until it gets better at least

[trim jolt]

@zenith stream Has your question been resolved?

Channel closed

Closed by @zenith stream

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[zenith hill]

Arg(-1+i) = 3π/4 and Arg (-1-i) = -3π/4. Is it because the result should be in -π and π?

[ebon rune]

yes

,w range of arg function

anyway, yes itll only output a number between -pi and pi

[zenith hill]

okay thank you i just needed to clarify it hehe

.close

[trim jolt]

Channel closed

Closed by @zenith hill

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[jolly basin]

can someone help me on the clarification for why we don't use standard error here? (ping me pls if u respond)

[jolly basin]

ik the transformation part

mu=3990

sigma=146.64

im aware z scores only use standard deviation

but like conceptually how would i know that i guess is the question

ik standard deviation is the variation of a single sample

and standard error is the variation of sample statistics from the true population parameter

if i used standard error would that be like saying we're trying to find if a sample with mu=4200 is significantly different from the population parameters?

[trim jolt]

@jolly basin Has your question been resolved?

[jolly basin]

.close

[trim jolt]

Channel closed

Closed by @jolly basin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[marsh forum]

So I need the region common to x^2+y^2≤1 and z=x+3

Well, I suppose I could start by paramatrizing the cylinder

(cos(t),sin(t), h)

they intersect along the curve $( rcos(t), rsin(t), 3+ rcos(t)); 0≤r≤1; 0≤t≤2π$

[solid kiln]

What a wonderful world !

[marsh forum]

Would that be right

[trim jolt]

@marsh forum Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[warm trench]

Hello, can anyone explain me what a finite cover is and how to identify them from open sets? Thanks

[trim jolt]

@warm trench Has your question been resolved?

[onyx mist]

A cover is a set of sets, such that the union over the cover contains the whole set

If the cover has finite elements, then it is called a finite cover

[warm trench]

does a cover need to be bounded for it to be a finite cover?

[onyx mist]

Yes

In fact, boundedness is a requisite for compactness

[warm trench]

interesting

[onyx mist]

In any metric, a set is compact iff it is closed and bounded

Hiene borel

[warm trench]

okok, for example i have the set A, set A has the covers

• (-\inf, 0]
• (-1,1)
• (1/k, 3-1/k)_k=1^\inf
  i would assume the first one is not a finite cover, the second one is a finite cover

[onyx mist]

Hang on

[warm trench]

im still unsure how i should categorise the third one

[onyx mist]

(-\inf, 0] and [0, \inf) is a finite cover

its only two elements

[warm trench]

but theyre intervals

[onyx mist]

But it is not the case that every cover is finite

I can give an infinite cover

[warm trench]

yeah im a bit confused at this one

[onyx mist]

Infinitely many sets = infinite cover

These are two sets

A cover is a set of subsets of the space

It is not a subset of the space

How many elements the set has determines the size of the cover

You just covered all of R in two sets
(-\infty, 0] and [0, \infty)

That cover has a size of two

[warm trench]

uh

hold on

i need to process this

so uh

im still unsure what a size is

[onyx mist]

Im confusing you lol

[warm trench]

lol

im forsure not taking analysis

cover is a set of subsets, so if i have (-1,2), the cover should be {(-1,1),(0,2)}

[onyx mist]

So, a cover of X is a set $A$ such that $$X\subseteq\bigcup_{i\in A}i$$

[solid kiln]

Cycadellic

[warm trench]

uh

this is pretty different from the definition i found

or maybe its just the wording

[onyx mist]

How did your definition word it

This is just the general topological definition

[warm trench]

${u_\alpha}\alpha\in S$ is a cover of $A$ if $$A\subseteq\bigcup{\alpha\in S}u_\alpha$$

[onyx mist]

Thats the same thing

But with lazier notation

[warm trench]

ah

its easier for me to process

[solid kiln]

Potus

[onyx mist]

We just say alpha instead of u_alpha

[warm trench]

alr

[onyx mist]

A cover is finite iff the size of the set is finite

In other words, there is a natural number

The white are a cover on the red circle

[warm trench]

i see it now

[onyx mist]

This is a finite cover, because its has 7 sets

[warm trench]

ohhhh

i starting to understand it now

[onyx mist]

And its not that its bounded i misunderstood

[warm trench]

lole

[onyx mist]

We call the elements the covering sets

If every covering set is open, it is called an open cover

[warm trench]

and for a set to be compact, every cover of A must be open and finite?

[onyx mist]

A set is defined to be compact iff every open cover has a finite subcover

[warm trench]

ah, so it only applies for the open covers of a set A

[onyx mist]

We define a subcover to be a set of covering sets such that, it covers a subset

[warm trench]

uh

[onyx mist]

Hm?

It must cover a subset of the particular “open cover” and also be a cover of the compact set in question

[warm trench]

ah

[onyx mist]

Then its a subcover

If every covering set is open, its an open subcover

[warm trench]

so, a cover that already covers a subset of an open cover and the compact set

is a subcover

[onyx mist]

Yeah

Just think of it as a subset of sorts

Theres actually a general word for this, a “subobject”

[warm trench]

i havent gotten to that part

[onyx mist]

Thats not in analysis lol

[warm trench]

lol

[onyx mist]

Thats more grad algebra and category theory especially

[warm trench]

yeah i cannot differentiate between analysis and algebra

[onyx mist]

Analysis is an algebra

[warm trench]

but not all algebra is analysis?

[onyx mist]

Its just studying continuity on a connected field

Yeah

Mayne

You can do every algebra up to vector spaces on them

[warm trench]

alr alr back on topic

[onyx mist]

Compactness

[warm trench]

so, what properties should the subcover possess for it to be classified as a finite subcover?

[onyx mist]

Just if i can cover it in finitely many sets

[warm trench]

ah ic

so $$(\frac{1}{k}, 2-\frac{1}{k})_{k=1}^\inf$$ is not finite

ffs

you get the idea

[onyx mist]

its \frac{ }{ }

[warm trench]

ah

[solid kiln]

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

[warm trench]

there we go

[onyx mist]

Yes

Except

Yeah

[warm trench]

if you union it with a finite set?

[onyx mist]

Covers are closed under unions

[warm trench]

ah

[onyx mist]

Wdym exactly

We have multiple sets to choose different types of objects of at this point

[warm trench]

yeah nvm

would (-2,1) be compact if the cover is {(-\inf,0), (-1,1/2),(0,1),(0.1,9)}?

i am inclined it believe it is, but there isnt a finite subcover

or mb im wrong

[onyx mist]

Okay so (-2,1)
Consider your open cover
We want to find a finite subcover

[warm trench]

ideally, a finite subcover must contain the compact set in consideration

but there isnt, so (-2,1) is not compact

[onyx mist]

Youre overthinking

[warm trench]

😭

[onyx mist]

The whole cover is finite

So theres your finite subcover

[warm trench]

uh

oh

[onyx mist]

Lets look at R

Say $${(a-2,a+2)}_{i}$$