help-49

[last slate]

so

ya

ok

ty

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[raven gorge]

Hello i need help with this
It's in Arabic but the question says

"The solution set of the inequality is ..... "

[raven gorge]

Talking about question 14 of course

Answers 2 and 5 are just "all are false" don't mind them

[heady plume]

$[x + n] = [x] + n\forall n \in \bZ$

[grand pond]

Arya

[heady plume]

@raven gorge

Where are you stuck after using this?

[raven gorge]

Honestly i didn't even use this
Just trying to settle a debate because solving it on desmos gave a different answer than the teacher gave

[heady plume]

[x] > 6 => [x] ≥ 7 => x ≥ 7 should be your answer solution

[heady plume]

Did you guys not get Option 3?

Oh right, your question doesn't specify [x] is the Greatest Integer Function... So yeah the question is up for debate ✓✓

[raven gorge]

Yes yes that's the answer we got

Also because Desmos doesn't recognize [x] as the greatest integer function

[raven gorge]

Yes exactly

[raven gorge]

It caused a debate

[heady plume]

Well, now you know what caused the debate

Confirm with teacher ✓

[raven gorge]

Yes tysm

[raven gorge]

I will be

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[cerulean temple]

why $f(x)=e^x$ is not continous at $x=\infty$?

[grand pond]

Slowaq

[runic hamlet]

for starters its not defined at x=infty

[cerulean temple]

how come?

its domain is R

[dawn dagger]

and what's R

[runic hamlet]

and infinity is not a real number

[cerulean temple]

hm yes thats true

so we can not say about any function from reals to reals that its continous at infty?

@runic hamlet

[runic hamlet]

yes

[cerulean temple]

alright thanks

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[buoyant yoke]

Im having some trouble understanding what $A$ is here. Is it a subset of $\N$?

[grand pond]

Bonk

[buoyant yoke]

So then when I need to use disjoint sets $A_i$ for showing $\sigma$-additivity I just get bigger and bigger subsets of $\N$?

[grand pond]

Bonk

[buoyant yoke]

Would this be valid?
Let $B=\bigcup_{i=1}^\infty A_i$ with $A_i$ disjoint. Then $\mu(B)=\sum_{n\in B} a_n=\sum_{n\in\bigcup_{i=1}^\infty A_i} a_n=\sum_{i=1}^\infty\sum_{n\in A_i} a_n=\sum_{i=1}^\infty\mu(A_i)$

[grand pond]

Bonk

[buoyant yoke]

Can I just do this?

[dawn dagger]

I am not sure but it seems reasonable

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[dawn dagger]

You think I knew Euler?

[heady plume]

4th option is nonsense, CD is no path.
1-3 have LADA, now BC is not optimal
so LADACBRAB -> 2nd option is optimal

Is it a test =_=

alr

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@paper valley Has your question been resolved?

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[wind oxide]

What's a good place to generate partial fractions for complicated things?

[wind oxide]

I have to partial fraction $\frac{x^{2m}}{1-(k-1)x^2 - kx^4}$

[grand pond]

rak³en

[buoyant yoke]

well first youd have to factor the denominator

[wind oxide]

should be $(1-kx^2)(1+x^2)$

[grand pond]

rak³en

[visual tiger]

do you want to factor 1-kx^2 further?

or can k < 0?

[wind oxide]

k can be anything

[visual tiger]

any real number

so it can also be -1

k=-1 case apart

[wind oxide]

well I'd like that but tbh all I need is a closed form for $\left| k \right| \leq 1$

[grand pond]

rak³en

[visual tiger]

k complex number is possible?

[wind oxide]

no, k is real

[visual tiger]

ok

[visual tiger]

so this will have to be apart

x^(2m) we can deal with later if division by denominator is not obvious enough

[wind oxide]

hm?

sorry I didnt get you

[visual tiger]

usually the partial fraction decomposition would go x^2m/(....) = integral part + .../... + .../...

but the integral part can be hard to compute

[wind oxide]

I see but how do I compute it?

[visual tiger]

if you want to compute it now

then I suggest maybe computing x^(2m) / (1+x^2)

and then dividing the quotient again by 1-kx^2

[wind oxide]

I see

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[visual tiger]

please don't ping helpers at random

[olive matrix]

also stop constantly closing and reopening help channels

also show a tiny bit of effort

also get a better screenshot

[thick epoch]

Can someone help me with this

[celest brook]

You can post the photo rotated so people can read it easier 🙂

[thick epoch]

My brain is so foggy right now I can barely speak

I did 4 assignments for social studies 3 for math and 2 for IT

today

Tomorrow I got a test

MY LIFE IS FUCKED

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

[heady plume]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[heady plume]

where are you even stuck.

!status

[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

[heady plume]

Lost? Do you know the first step to do these problems?

Did you mark the odd vertices?

.... they have odd number of neighbours?

Fr how are you doing Eulerization problems when you don't know what odd vertices are

Did you identify your start, end vertices?

for semi-eulerization, you need to make sure all your odd vertices are made into even except 2, which need to remain odd

the starting and ending locations

what you're doing by adding edges between all odd vertices

[olive matrix]

do you know what a graph is

[heady plume]

._>

.

here, clearly A is bogus because of that top diagonal edge, B is same because the top right corner edge connecting (1,4) to (2, 5). the proper "shortest path" would be (1, 4) to (1, 5) to (2, 5). Also D is nonsense because it connects (2, 5) to (4, 5) bypassing (3, 5)

C is the only one following all the steps here

HUH? Both use Fleury Algorithm

??? do you only want the answers to your questions?

don't understand what. Can you be specific?

"I don't understand this. " Am I supposed to be reading your mind?

You don't know Eulerization or Semi-Eulerization. You don't know to identify odd vertices

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[heady plume]

Thanks. Bye

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[granite cipher]

I don’t fully get the terms symmetric anti symmetric and transitive

[granite cipher]

Can someone maybe explain them?

[grim vector]

Symmetric relation is like you can swap the side

x = y <=> y = x

= is a symetric relation

For antisymetric

[quartz otter]

< and >

these are anti symmetric then

[grim vector]

<= and >=

[quartz otter]

but i honestly don't know what's the meaning of transitive

[grim vector]

Cuz < and > if you have x > y and y < x its not possible

Transitive is for like

[quartz otter]

yeah aren't we looking for that

[grim vector]

x,y,z if x >= y > z then x > z

[quartz otter]

if it's not possible when you switch them the sign is anti symmetric

[grim vector]

No we want that it implies x=y

[grim vector]

And for transitive

Hum

Its like if im taller than someone

[quartz otter]

who said that lol

[quartz otter]

got it

[grim vector]

Idk antisymetric relation is defined like this

[quartz otter]

ok but you defined symmetric like that at first

x = y is symmetric because no one cares if you switch their position

and x > y is anti-symmetric because it makes a lot of difference when you switch their positions

[grim vector]

No antisymetric doesn't mean quite this

[granite cipher]

Can u explain

Say a>=2 for all a in Z

[quartz otter]

didn't understand that

[granite cipher]

It’s not reflexive

[quartz otter]

give me an example

[grim vector]

= is not reflexive ?

[granite cipher]

No

[grim vector]

Why so

[granite cipher]

It doesn’t hold for all a

[grim vector]

That a<=a ?

[granite cipher]

Like if a is less than 2

[grim vector]

Yes

It still hold

[quartz otter]

2 >= a

a <= 2

see it's reflexive

[granite cipher]

Look at c

[quartz otter]

uhh incomprehensible definitions again

i can't understand anything as a person who has no idea about this definition

and

i read it

try to understand it

but i can't

that's how flipping complex it is

[granite cipher]

We should call the helpers

<@&286206848099549185>

[quartz otter]

just

i'm not familiar with these kind of representations

i should be learning their meaning first

[granite cipher]

Yea that’s okay bro thanks for trying ;))

[quartz otter]

then i can understand

[quartz otter]

like what does 0 R 0 mean for example

lmao

[granite cipher]

0 in relation with 0

aRa

[odd solar]

@granite cipher what was the original question

[granite cipher]

It was

Can someone explain the properties of antisymmetric symmetric and transitive

The example was a>=2 for all a in Z

[granite cipher]

I tried to understand them but they aren’t making sense

[odd solar]

this is a bit of a strange relation because instead of being a relation between pairs in a set, it's a relation between one of the pairs and 2

[granite cipher]

Yea

C

[odd solar]

so given a pair from Z, (a, b), you have aRb -> a >= 2

start with symmetric, do you know the definition?

[granite cipher]

Yea so if aRb is true and bRa is true then it is symmetric

It’s that right?

[odd solar]

i believe it's not restricted to just if it's true. The result of the relation just has to be the same when the pair is flipped

[granite cipher]

Ah okay

What if I took a as 2

Then wouldn’t it be symmetric

[odd solar]

yes, but it needs to be symmetric for all a and b in Z

[granite cipher]

Ahhh

For all okay

[odd solar]

so the relation would be symmetric if it's a relation on { 2 }

[granite cipher]

In that case it can’t be symmetric

For all values in Z

[odd solar]

but if you can find a single counter example for a,b in the set the relation is defined on, then the property doesn't hold

make sense?

[granite cipher]

Yea that makes sense

[granite cipher]

Is that for all reflexive symmetric antisymmetric and transitive?

[odd solar]

yes

for this R it's simple to find a pair that is fails, a and b just need to be on either side of 2

[granite cipher]

Yea

[odd solar]

that way it will fail for one, but pass for the other

and thus not be symmetric

[granite cipher]

Symmetric makes sense

[odd solar]

now antisymmetric

definition?

[granite cipher]

It’s when

aRb and bRa implies a=b

But how does that work I’m a little confused

[odd solar]

it means that for all a,b in the set, you can choose a pair (a,b) such that the relation is symmetric only when a = b

so what would you want to ensure when looking for a counter example?

[granite cipher]

Let me think

If a was 3 and b was -3

Would that work?

[odd solar]

test it

[granite cipher]

Okay

3 is greater than 2 so passes but -3 fails the relation

[odd solar]

so is it symmetric for that pair?

[granite cipher]

It’s not symmetric for that pair

[odd solar]

then it's not a valid counter example

you want a counter example for which the relation is symmetric and a =/= b

[granite cipher]

So do an and b have to be equal?

[odd solar]

for the relation to be asymmetric on the pair, yes they must be equal

[granite cipher]

Oh

[odd solar]

for the relation to be asymmetric on the set, aRb must be symmetric ONLY when a = b

so you can't have R be symmetric for a pair (a,b) where a =/= b

because then there exists a pair (a,b) such that a =/= b for which R is symmetric, which contradicts asymmetry

[granite cipher]

Let me process that

can you clarify a=/=b

[odd solar]

a and b not equal

e.g. (3, 4)

[granite cipher]

Okay

[granite cipher]

Oh find a pair where a and b are not equal

But a and b are symmetric

That would contradict asymmetry

Like a is 5 and b is 6

Then

They pass the relation

But a doesn’t equal b

That would be a counter example right?

[odd solar]

yea that's perfect

real simple

any other qs about asymmetry?

[granite cipher]

I get asymmetry now thank you

Just transitive left

[odd solar]

definition

[granite cipher]

It means

If aRb and bRc then aRc

[odd solar]

and again, that's for all a,b,c

in the set R is defined on

[granite cipher]

Okay

So for counter example aRb bRc but not(aRc)?

[odd solar]

yes

this one usually takes more thinking to get a counter example

[granite cipher]

Yea I’m going to try some pairs

What’s the most efficient way for these types of problems

[odd solar]

hmm

I don't really know

I haven't done these problems since first semester of uni which was 3 years ago

[granite cipher]

Hahaha I’m in first semester

I hate this module honestly

[odd solar]

I did too

[granite cipher]

I don’t mind probability or stats or calculus but this sucks

[odd solar]

it's worth while to get decent at though since it really helps you learn to read sets and quantifiers etc

[granite cipher]

Yea that’s true

[odd solar]

which comes up in a lot of maths

[granite cipher]

Yea

[odd solar]

I took a quick peek at what the answer sheet said for transitivity

can you make sense of it?

[granite cipher]

Looking at it

Does aRb here mean a >=2 and b>=2?

[odd solar]

R acts only on the first of (a,b)

so aRb is a >= 2

[granite cipher]

Oh

So a >= 2

We know that

And b>=2 as well

Oh

It doesn’t matter what c is because

[granite cipher]

R will only depend on A

And as a >=2

Then automatically

aRc doesn’t fail the relation

Correct?

[odd solar]

yeah, that logic is sound

[granite cipher]

Awesome

I have a question

I’d probably think these are false

And find counter examples

How can you know it’s true or false efficiently or does that take some guess work

[odd solar]

what do you mean by 'these'

[granite cipher]

Like these properties

Some you can see more easily

But like for example transitive

You cant see that at the start

[odd solar]

just have to take a moment to think about it

[granite cipher]

Okay

[odd solar]

there's not single method for it since it depends on how the relation is defined

[granite cipher]

True

Thank you so much for the help man

[odd solar]

no problem, hope it helped

[granite cipher]

It did very much

Have a good one

[odd solar]

you too

[granite cipher]

.close

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[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

(f(0,1,-1))_B = (1,a,1)
(f(1,1,1))_B = (-1,3,0)
(f(0,0,1))_B = (b,1,2)

,w inverse {{0,1,0},{1,1,0},{-1,1,1}}

[grand pond]

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[tidal turret]

,w {{0,1,0},{1,1,0},{-1,1,1}} * {{1, -1, b},{a,3,1},{1,0,2}} * {{-1,1,0},{1,0,0},{-2,1,1}}

[grand pond]

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[tidal turret]

,w monomorphism

[grand pond]

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[tidal turret]

dim(ker)=0

,w det {{1,-1,b},{a,3,1},{1,0,2}} = 0

[grand pond]

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[tidal turret]

,w det {{1-a,a+1,1},{-a-2(b+1)+1,a + b + 2, b + 1 },{-a-2(3-b)+4, a-b+3, 3-b}} = 0

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[tidal turret]

f(1,0,0) = (1-a,a+1, 1)
f(0,1,0) = (-a-2b -2 +1, a+b+2, b + 1)
f(0,0,1) = (-a -6 + 2b + 4, a-b + 3, 3-b)

f(2,3,1) = (4,3,6)

(4,3,5) = 2(1-a, a+1, 1) + 3(-a-2b-2+1, a + b + 2, b +1) + 5(-a-6+2b+4, a-b +3, 3-b)

4 = 2(1-a) + 3(-a-2b-2+1) + 5(-a-6+2b+4)
3 = 2(a+1) + 3(a+b+2) + 5(a-b+3)
5 = 2 + 3(b+1) + 5(3-b)

fuck this is not the way

I am doing something wrong

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[tulip token]

what are you doinng

just solve det(M) = 0, write (2, 3, 1) and (4, 3, 6) in terms of B, and solve for a and b by writing M(whatever (2, 3, 1) is in B) = (whatever (4, 3, 6) is in B)

[tidal turret]

thats for the first part, what about the monomorphism part though?

[tidal turret]

why det(M)=0? that will give me a and b values such that the matrix has a linear dependent row or column

[tidal turret]

monomorphism means the dim(ker) = 0

[tidal turret]

but solving when det (M) = 0 gives me the a,b values when dim(ker) >= 1

also, det(M)=0
gives 2a -3b + 5 = 0

what about it?

in any case we would need to satisfy
2a-3b+5 ≠ 0 for f to be a monomorphism ( dim(ker) = 0)

well I can find (2,3,1) in B coords though

[tulip token]

tuviste tiempo para escribir 4 mensajes pero no tuviste suficiente tiempo para leer la pregunta

y f no es monomorfismo

as for the rest of it, I encourage you to read and think about it a little more

[tidal turret]

ohhh

ty

(2,3,1) = a(0,1,-1) + b(1,1,1) + c(0,0,1)
2 = b
3 = a + b
1 = -a + b + c

3 = a +2
a = 1

1 = -1 + 2 + c

1 + 1 - 2 = c

c = 0

(2,3,1) = a(0,1,-1) + b(1,1,1) + c(0,0,1)
(a,b,c)=(1,2,0)

(2,3,1)=(0,1,-1) + (2,2,2)

(4,3,6) = a(0,1,-1)+b(1,1,1)+c(0,0,1)
4 = b
3 = a+b
6 = -a + b + c

3 = 4 + a
a = 3-4 = -1

6= -(-1) + 4 + c
6 - 1 - 4 = c

c = 1

(a,b,c)=(-1,4,1)

(4,3,6) = (0,-1,1)+(4,4,4)+(0,0,1)

(4,3,6) = (-1,4,1)_B

(2,3,1) = (1,2,0)_B

M . (-1,4,1) = (1,2,0)

M_BB(f) . (-1,4,1) = (1,2,0)

,w {{1, -1, b},{a,3,1},{1,0,2}} * {{-1},{4},{1}} = {{1},{2},{0}}

[grand pond]

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[tidal turret]

,w {{1, -1, b},{a,3,1},{1,0,2}} * {{-1},{4},{1}}

[grand pond]

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[tidal turret]

(b-5, 13-a, 1) = (-1,4,1)

b- 5 = -1
13-a = 4
1 = 1

b = -1 + 5 = 4
13 - 4 = a = 9

2a -3b + 5 = 0

(a,b) = (9,4)

2(9) -3(4) + 5 = 0

18 - 12 + 5 ≠ 0

what?

[tidal turret]

I managed to find a,b such that f(2,3,1)=(4,3,6)
but when solving det M = 0 I get infinitely many possibilities for a and b

det(M)=0 gives
2a -3b + 5 = 0
2a = 3b - 5
a = (3b-5)/2
(a,b) = ((3b-5)/2, b)

what am I even doing

[tulip token]

it's asking you to find the values of a and b for which both conditions are true

stop writing random things and read the question

[tidal turret]

after finding (2,3,1) and (4,3,6) expressed in base B and performing
M.(2,3,1)=(4,3,6)
I got (a,b)=(9,4)
but if I plug (a,b)=(9,4) the equation
that I got when doing det(M)=0
2a-3b+5=0
does not holdd

I mean not performing M.(2,3,1) = (4,3,6)
but (2,3,1) in bass B and (4,3,6) in base B

[tidal turret]

don't delete your message doe, I didn't managed to read what you wroteee

I was talking on the phone with my grandma

[tidal turret]

both simultaneously? I thought they meant separately

[tidal turret]

holy mother of math bro

I am missing something, I just dunno what

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

I just noticed the mistake

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[wind oxide]

yo wsp

[wind oxide]

new integral time

eh actually

before that

[blissful totem]

I love how casual it is

[wind oxide]

🙂

gimme 1 sec

okay yeah

[blissful totem]

Bro is gonna pull out the most egregious integral

[wind oxide]

wait nvm

I amma just write the integrals

[mossy spindle]

💀

[wind oxide]

$I(k) = \int_0^1 \frac{ \ln u}{(1-ku) \sqrt{u(1-u^2)}} du$. I need to find $I(1) - I(-1)$

[long dagger]

Interesting

[grand pond]

rak³en

[mossy spindle]

ncert misc prob vibes 💀

[long dagger]

Use reduction

[wind oxide]

can yall wait? lemme type the question out.

[long dagger]

oh wait

just solve both integrals

I(1) and I(-1)

[wind oxide]

Next I noted that for $J_k (\alpha) = I(k) = \int_0^1 \frac{ u^{\alpha} }{ (1-ku) \sqrt{u(1-u^2)} } du, I(k) = J'_k(0)$

[long dagger]

Are we going for a Feynman here

[wind oxide]

yes.

[long dagger]

J_k(alpha) isn't equal to I(k) though

[wind oxide]

it derivative evaluated at 0 is

was an acciden @long dagger , mb

Given, $$I(k) = \int_0^1 \frac{ \ln u}{(1-ku) \sqrt{u(1-u^2)}} du $$
I need to find $I(1) - I(-1)$.
To do this, I noted that for $J_k (\alpha) = \int_0^1 \frac{ u^{\alpha} }{ (1-ku) \sqrt{u(1-u^2)} } du$, $$I(k) = \frac{d}{d \alpha} J_k( \alpha)$$
So now I tried to solve for $J_k(\alpha)$. But I wanna ask, cant I actually solve for $J_1(\alpha)$ and $J_{-1}(\alpha)$ (assuming that it is easier) and then differentiate, so that I directly get $I(1)$ and $I(-1)$?

Also for those who say simply find $I(1)$ and $I(-1)$, I couldn't seem to do it. If you do though, I'd be happy to see a solution.

[grand pond]

rak³en

[midnight plank]

@wind oxide Has your question been resolved?

[wind oxide]

<@&286206848099549185>

[last slate]

This ain't pre university maths

It's unique definite integral

I've never seen this

@wind oxide

Does this come in exam

[wind oxide]

I have no idea

I found an integral on MSE, question got removed, got stuck on it, never moved on

I am in 10th grade 💀

<@&286206848099549185>

[wind oxide]

question statement.

[last slate]

I'm also in 10th grade

Stop wasting ur time on this

[urban widget]

😔

[last slate]

First clear basics

[urban widget]

Question looks interesting

[last slate]

That bro doesn't even know 1+1 and it's doing msc

This is not fair for the server lol

[urban widget]

Msc?

[urban widget]

Your trying to use ftcp1 to solve this 💀

You just don’t want to solve the integral huh

[wind oxide]

excuse me? my choice

[wind oxide]

i tried that for enough time that I no longer want to.

[wind oxide]

and what do you know about me?

[urban widget]

Hm f’(x)=F(x) where f(x) is integral

Or sorry

Differentiating the integral

Just gets you f(x)

And then you might be able to evaluate it using the ftcp2?

F(b)-F(a)

The issue is we don’t know what k is

[wind oxide]

dont really get you, I dont think I need to FTC here

[urban widget]

How would you solve it then

[delicate arrow]

I hate math

[wind oxide]

like I wrote? FTC says that F(b)-F(a) is the integral right

(p2)

[urban widget]

Mhm

[wind oxide]

and p1 is F'(x) = f(x)

[urban widget]

Yeah

[wind oxide]

Well that is a) for a single variable and I have multiple

b) that is not my question.

[urban widget]

You in multivarible…?

Stokes or green thereom then?

[wind oxide]

technically yeah, k and alpha cus

[wind oxide]

i dont know how to apply those honestly

<@&286206848099549185>

[last slate]

What if we try chatgpt

[wind oxide]

SHUT UP.

[robust bear]

Where does the problem starts because it's mess in this chat

[wind oxide]

.

[wind oxide]

<@&286206848099549185>

Given, $$I(k) = \int_0^1 \frac{ \ln u}{(1-ku) \sqrt{u(1-u^2)}} du $$
I need to find $I(1) - I(-1)$.
To do this, I noted that for $J_k (\alpha) = \int_0^1 \frac{ u^{\alpha} }{ (1-ku) \sqrt{u(1-u^2)} } du$, $$I(k) = \frac{d}{d \alpha} J_k( \alpha)$$
So now I tried to solve for $J_k(\alpha)$. But I wanna ask, cant I actually solve for $J1(\alpha)$ and $J{-1}(\alpha)$ (assuming that it is easier) and then differentiate, so that I directly get $I(1)$ and $I(-1)$?

Also for those who say simply find $I(1)$ and $I(-1)$, I couldn't seem to do it. If you do though, I'd be happy to see a solution.

[grand pond]

rak³en

[urban widget]

I would ask this in multivarible or calculus

[wind oxide]

hmm

[wind oxide]

<@&286206848099549185>

[willow obsidian]

Are you this person who is talking with me?

@wind oxide

[wind oxide]

what? yes i guess?

[wind oxide]

no, i aam not

harper != itzkraken.

[willow obsidian]

Okay

[zealous schooner]

!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.

[zealous schooner]

what is the original integral you are trying to solve

not this supposed in between step you're sharing

[grand pond]

rak³en

[zealous schooner]

okay so you turned this into $$-\int_{-1}^1\int_0^1\frac{u\log u}{(1+yu)\sqrt{u(1-u^2)}}\dd{u}\dd{y}$$

[grand pond]

kheerii

[zealous schooner]

you're missing a factor of u here then

[wind oxide]

hm?

wait really?

[wind oxide]

okay i did not realise that

[wind oxide]

oh god i've been solving the wrong thing now that is crazy

[zealous schooner]

which is why !xy exists

[wind oxide]

sighs

[zealous schooner]

your integral is invariant under the substitution t=(1-u)/(1+u)

[wind oxide]

my bad, I guess

[wind oxide]

meaning..? it looks the same?

[zealous schooner]

yes

where did you get this integral?

[wind oxide]

mse

[midnight plank]

@wind oxide Has your question been resolved?

[wind oxide]

i'll look into this later, i guess

.close

[midnight plank]

Channel closed

Closed by @wind oxide

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[midnight plank]

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[tidal turret]

how do I find the nullspace of this matrix

[midnight plank]

@tidal turret Has your question been resolved?

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[tidal turret]

.reopen

[midnight plank]

✅

[odd solar]

I thought you already completed this question

[tidal turret]

sorttt off

[honest urchin]

I don't think the basis matters.

Try doing basis changes like in the Gaussian algorithm to simplify the matrix.

[tidal turret]

basis changes?

[honest urchin]

yeah

[tidal turret]

can you elaborate

[honest urchin]

Do you know the Gaussian solution approach?

[tidal turret]

no?

[honest urchin]

Oh

[tidal turret]

I wanted to find the nullspace of the matrix first, because using the determinant = 0 at an arbitrary submatrix is not the standard procedure

[honest urchin]

There are some number of operations you can do to a matrix that will not change the dimension of the kernel or the image.

For example you can interchange rows and sum one row to another with some factor. That corresponds to changes of the range's basis.

Likewise you can do it with columns.

[tidal turret]

row operations do not change dim kernel bro

[honest urchin]

Yes? That's what i am saying too.

[tidal turret]

changes in the ranges basis?

[tidal turret]

maybe i am minsinterpreting your words then, can you elaborate a bit more

I can agree that giving the basis B is not necessary though

[honest urchin]

Basically I am just trying to argue why this is true.

Also why the same is true with column operations.

[tidal turret]

yeah I agree that row/column operations do not change the dim of the kernel, it only makes it easier to see which value of a to choose for getting the dim ker = 2

[honest urchin]

b?

[tidal turret]

edited

[honest urchin]

Alright, that is what I would do.

After arriving at a simpler matrix the dimension of the kernel should be easy to spot.

[honest urchin]

looks like they did just that

[tidal turret]

it is not though

[tidal turret]

like we need to find a such that dim ker = 2

[honest urchin]

why is the matrix one wider

are we now looking for dim ker = 3?

[tidal turret]

we want dim ker = 2 which values of a make the kernel dim 2

[honest urchin]

for this matrix?

[tidal turret]

like

like the kernel of the linear transformation

but like

if we multiply by change of basis matrix to get Matrix representation of f with input basis standard basis and output basis standard basis the kernel is the same

[tidal turret]

ye

because kernel of M_EB(f) is same as kernel M_EE(f)

[honest urchin]

It's not the same asnwer as to your question sicne there is another column

[tidal turret]

we are tying to find the nullspace

the dimension of the nullspace = the dimension of the kernel

[honest urchin]

if there is another column, the null space is one dimension bigger

[honest urchin]

Alright...

[tidal turret]

I removed the last column

this one is the reduced matrix

[honest urchin]

if a = 1 the second column degenerates (unless the last two columns would degenerate to just dim im = 1, but you can check that)

there is no other solution since you can't make the last two columns never change the last coordinate of the range.

[tidal turret]

when a = 1, Dim(Im)=3 and dim(ker)=1

<@&268886789983436800>

[honest urchin]

<@&268886789983436800>

[honest urchin]

huh, you are right, I think in that case there is no solution.

[tidal turret]

for this matrix, plug a = 1, check the rank

,w rank {{1,1,-2,1},{0,0,-3,1},{0,0,3,-1}}

[grand pond]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

what?

[honest urchin]

oh

[tidal turret]

dim(Im)=2

[honest urchin]

I shouldn't say stuff like that without computing myself :X

[tidal turret]

dont worry I am confused as hell myself aswell

anyways

[tidal turret]

only solution is a = 1
?

[honest urchin]

yeah

Do you grasp why?

[tidal turret]

not really

strong intuition?

[honest urchin]

if a isn't 1

[honest urchin]

with this

[honest urchin]

We are trying to solve M v = (x,y,z) with any x,y,z to prove the image has dimension 3

you can put any value in the third component of the range by choosing some factor in v=(0,0,r,0) or v=(0,0,0,r).

by suming the second column using v = previous v + (0,q,0,0) you can put any coordinate in y.

and likely you can do the same with the x.

[honest urchin]

Due to the shape of the matrix

this will not disturb the z coordinate again

hence you can produce any value as long as the diagonal isn't 0 anywhere

To say it another way, if a != 1

discard either the third or the fourth column whcihever isn't 0 in the end

if anything this shrinks the image

then for this matrix the determinant is not 0 since the determinant is the product of the diagonal entries in thsi case.

[tidal turret]

but the matrix is non square, which determinant

[honest urchin]

Hence dim im = 3 in the shrunken matrix, henceforth the original matrix has dim im > 3 and therefore 3.

[honest urchin]

see:

[honest urchin]

^

if you shrink the input space by considering a subspace

the image can stay the same or shrink

it can not get bigger

if we already get dim im = 3 with the smaller matrix

we know the larger one is at least dim im = 3

[tidal turret]

x1 + x2 - 2x3 + ax4 = 0
(a-1)x2 -3x3 + x4 = 0
(a+2)x3 - ax4 = 0

[tidal turret]

yeah Rank = 3 if a = -2 or a ≠ 1

only possibility is a = 1

what about the basis for the image

[honest urchin]

plug in a = 1 and work it out

[tidal turret]

that is why they give basis B

we need to multiply by a change of basis matrix

or plug a = 1 and rref and then translate those B coordinates into standard basis

[honest urchin]

yes

[midnight plank]

@tidal turret Has your question been resolved?

[honest urchin]

Is this still open?

[tidal turret]

what about the basis for the image

M_BB(f) . (v)_B = (f(v))_B

[honest urchin]

if that is M, then { Mv | v in R³ } = R x R x {0}

[tidal turret]

wdym

[honest urchin]

if we use the standard basis for first attempt

then the range is R x R x { 0 }

in other notation { (x,y,0) with x and y any real }

[honest urchin]

(here I mean the cartesian product with x, and the reals with R)

[tidal turret]

cartesian prod?

[honest urchin]

https://en.wikipedia.org/wiki/Cartesian_product?useskin=vector

It is pretty common, you should get to know it.

[honest urchin]

for the question at hand

you can also think about it this way first.

It's just notation

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

wait a second

x1 + x2 -2x3 + x4 = 0
-3x3 + x4 = 0

x4 = 3x3

x1 + x2 -2x3 + 3x3 = 0

x1 + x2 + x3 = 0

x1 = -x2 -x3

(-x2-x3,x2,x3,3x3) = x2(-1,1,0,0)+x3(-1,0,1,3)

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

it was solved in #linear-algebra

finally we can put an end to this

.solved !!

[midnight plank]

Channel closed

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[midnight plank]

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

1 2, 1 3, 1 4, 1 5
2 1, 2 3, 2 4, 2 5
3 1, 3 2, 3 4, 3 5
4 3, 4 2 , 4 1, 4 5
5 1, 5 2, 5 3, 5 4
14/20 = 7/10 = 0.7
is there any other way to solve this ?

[rose trout]

The product is even if one of the two numbers you select is even.

[abstract wind]

yes
even * even = even
even * odd = even

[rose trout]

So what is the probability that one of the numbers you pick is even?

[next rover]

first number is even: (2/5)(4/4)
first number is odd: (3/5)(2/4)

you can add it together because it covers every case

[abstract wind]

i don't get it

1 2 3 4 5
first number is even: 2,4
first number is odd : 1,3,5

[next rover]

when the first number is even, any remaining number works

so there's (4/4) chance it works

[obsidian wedge]

simply because even x odd would still be even

[next rover]

yeah

[abstract wind]

yes so u mean 2/5 * 4/4 ?

[obsidian wedge]

our objective is to get an even result

[next rover]

yes

multiplication

[abstract wind]

8/20 + 6/20 = 14/20

hmm how can u know x * 4/4 and x * 2/4 ?

2 1,2 3, 2 4,2 5

ahh nvm cause even * odd or even still even

for x * 2/4
cause odd only can be even by odd * even so only 2 option

this is might be algebra ?

[subtle zinc]

yes

nahhhh but them pronouns tho 💀

get what

what

im talking about ur pronouns

no

the other one

[abstract wind]

so u guys already masterize pre algebra before learning algebra ?

[subtle zinc]

masterize 💀

[abstract wind]

so what is x and y in algebra ?

[rose trout]

I just witnessed a

[rose trout]

What do you mean by that?

[subtle zinc]

anything u want

depends on context

[abstract wind]

i mean usually we using x in the equation...

[rose trout]

It's only convention that we use x, or y. You can use any letter you want, doesn't even need to be Latin.
They are variables, so they serve as placeholders for values.

[midnight plank]

@abstract wind Has your question been resolved?

[abstract wind]

ok

.close

[midnight plank]

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Closed by @abstract wind

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

I dont get 8b

How do I do it

[midnight plank]

@granite cipher Has your question been resolved?

[west iron]

think about what's required for cos(x)=cos(y) and sin(x)=sin(y) to both be true

what conditions on x and y are needed

[granite cipher]

X=y?

[west iron]

that is sufficient but not necessary

here's an example

x=0, y=2pi

[granite cipher]

Ah

So

Y=x+2pi

[honest urchin]

how about y=x+pi

[granite cipher]

That would not work?

Because

Oh wait

Yea x+pi

[honest urchin]

sin(0) = 0 = sin(pi), yeah

but but cos(0) = 1 != = -1 = cos(pi)

[granite cipher]

Ahhhh

So it has to be 2pi

[honest urchin]

(also it's not true for all sines, but there would be another symmetry)

[granite cipher]

Okay

[honest urchin]

,wolf graph sin(x)

[grand pond]

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Requested by myxogast

[honest urchin]

if it was just for sin, there is a mirror symmetry around pi/2

[granite cipher]

I see the mirror

But it's sin and cos here so ir doesn't apply right

[honest urchin]

yeah but you need to convince the judge

[granite cipher]

Oh

So what do I say

[midnight plank]

@granite cipher Has your question been resolved?

[granite cipher]

.close

[midnight plank]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[somber canopy]

Can anyone give me a summation question with infinite sums to do, doesn't necessarily have to be infinite geometric, could involve factorials, squares or smth like that

[viral dagger]

let there be a function $P(n)$ such that $P(n)$ is the greatest prime thats not larger than $n$, and let there be another function $N(n)$ such that $N(n)$ is the smallest prime thats greater than $n$. If $n+1$ is prime, prove that
$$\sum^{n}_{i=2}\frac{1}{P(i)N(i)}=\frac{n-1}{2n+2}$$

[grand pond]

skissue.in.a.teacup

[somber canopy]

Do you have any idea

how much I hate prime numbers

people simply give them more credit that they deserve

[viral dagger]

[somber canopy]

also

it doesn't have infinite sum

[viral dagger]

ok heres one w infinite sum but is much easier

[somber canopy]

i dont mean

ridiculous leve

[viral dagger]

$\sum^{\infty}_{n=1}\frac{1}{n^4+n^2+1}$

[somber canopy]

give me something that i can solve

but find hard

mhm

see

much better

I have no idea how to solve this

but

I feel

much better

seeing this

[grand pond]

skissue.in.a.teacup

[somber canopy]

let me do this

I think I can solve it

[viral dagger]

but interesting

[somber canopy]

no

prime are boring

how tf a prime calculator was discovered is beyond me

equation i mean

@viral dagger

give me a hint

[viral dagger]

factor it

[somber canopy]

u cant

[viral dagger]

you can

[olive yew]

Notice that

[somber canopy]

wait u can

damn

ok thx

it kinda looked similar to

x² + x + 1

and u cant factor that

[olive yew]

@viral dagger ?? Dawg

[viral dagger]

then do fraction decomposition

[somber canopy]

mb

ik

how it works

[viral dagger]

:3

[somber canopy]

nope I was right

it cant be factor

without imaginary terms

@viral dagger

[olive yew]

Notice that n^4+n^2+1 = n^4+2n^2+1-n^2

[somber canopy]

yeh

[viral dagger]

do smth with it

[somber canopy]

$n^4 + 2n^2 + 1 - n^2$

[grand pond]

kronium_

[viral dagger]

then what

[somber canopy]

$n^2(n^2 +2) + (1-n^2)$

[grand pond]

kronium_

[viral dagger]

no not that

$n^4+2n+1\quad -n^2$

[grand pond]

skissue.in.a.teacup

[somber canopy]

thats what I did

[olive yew]

Does a^2+2a+1

Ring a bell

[somber canopy]

yes

$(n+1)^2\quad -n^2$

[viral dagger]

no

[olive yew]

Uhhhh nearly

[somber canopy]

yes

[olive yew]

Nearly

[somber canopy]

YES

I SEE IT

oh right

mb

no bells 😦

😭

[olive yew]

So imagine if instead of (a+1)^2

Which became a^2+2a+1

[somber canopy]

$(a^2 + 1)^2$

[grand pond]

kronium_

[viral dagger]

yes

[olive yew]

[viral dagger]

so $(n^2+1)^2-n^2

[viral dagger]

any ideas?

[somber canopy]

yes

yes ideas

$a^2 - b^2 = (a+b)(a-b)$

[grand pond]

kronium_

[viral dagger]

mhm

[somber canopy]

$(n^2+1)^2 - n^2 = (n^2 + n + 1)(n^2 - n + 1)$

[grand pond]

kronium_

[viral dagger]

mhm

[somber canopy]

WAIT

i dont understand

if it was factorable this entire time

why didn;t the computer factor it

when i put it in

[viral dagger]

skill issue ig

[viral dagger]

thats so far??

oh

skill issue

[olive yew]

[viral dagger]

(the computer probably only finds first power factors, ie x-1, 2x+3, ect)

[somber canopy]

what fughing good is it

I can do that myself

[olive yew]

Is that mathpapa

[somber canopy]

yes

now leave me be

[viral dagger]

meanie

[somber canopy]

I must attain

utter concentration

to perform

decomposition

[honest urchin]

,wolf factor x^4+x²+1

[grand pond]

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Requested by myxogast

[honest urchin]

People should use Wolfy more often.

[viral dagger]

wolfy =w=

[olive yew]

Peak calculator

[midnight plank]

@somber canopy Has your question been resolved?

[somber canopy]

@viral dagger

after partial frac decomp

[viral dagger]

[somber canopy]

,w factor n^2 - n + 1

[grand pond]

Results provided by WolframAlpha

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Requested by kronium_

[somber canopy]

ok

HOW DO U SOLVE THIS NOW

skissue

can u do this and show me

[viral dagger]

uhhh

your cooked (i typed in the wrong question LMAO)

there should be an n in the numerator

so n/n^4+n^2+1

[midnight plank]

@somber canopy Has your question been resolved?

[midnight plank]

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[ionic magnet]

A square is inscribed in an isosceles right triangle such that two of its vertices lie on the hypotenuse and the other two lie on the legs. The side of the square is 5. Find the hypotenuse of the triangle.

[ionic magnet]

[quiet parcel]

have you tried to sketch what you're given in the text?

[ionic magnet]

yes

[quiet parcel]

May I see?

[ionic magnet]

😢

[quiet parcel]

alright now this may not work

but what i would do

since its an isosceles right triangle

and the legs are equal in size

the hypotenuse is x * sqrt(2), the diagonal of the square you get when you draw another right triangle

here let me sketch

[ionic magnet]

why hyp is x* sqrt(2)?

[ionic magnet]

ok

[ionic magnet]

why sqrt 2

[quiet parcel]

i just think you gotta manipulate the squares and stuff a lil to find x

[ionic magnet]

where we get it

[quiet parcel]

x^2 + x^2 = c^2

2x^2 = c^2

c = x sqrt(2)

[ionic magnet]

oh

what would u do

any ideas?

i mean with squares but what

[quiet parcel]

i think

in these situations

when you have an isosceles right triangle

and a square positioned like that

the legs are divided into three parts

which means

(x/3)^2 + (x/3)^2 = 5^2

2(x/3)^2 = 25

x/3 = 5/sqrt(2)

x * sqrt(2) = 5 * 3

x = 15/sqrt(2) = 15sqrt(2)/2

so therefore

c = x sqrt(2) = 15sqrt(2)/2 * sqrt(2) = 15 * 2 / 2 = 15

[ionic magnet]

wdym the legs are divided into three parts

can you draw it?

[ionic magnet]

аааа

.close

[midnight plank]

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[midnight plank]

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[mint ravine]

Can someone please help me on this question

[mint ravine]

The red part is my work so far

[buoyant yoke]

AB is not 11.2

[mint ravine]

wdym the area is 112

[buoyant yoke]

you are given the diagonal

not the height

area is base * height

base is CD/AB, height is AM

[mint ravine]

..close

.close

[midnight plank]

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[midnight plank]

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[safe onyx]

System of equations:
Find F in terms of $m, g, \mu_s$ (not $N$)
$$F cos\theta - \mu_s N = 0$$
$$F sin\theta + N - mg = 0$$

[grand pond]

smeagol

[safe onyx]

I first isolated N in the second equation:

$$N = -Fsin\theta + mg$$

Then used N in the first equation

[buoyant yoke]

thats incorrect

[silent compass]

The signs seem wrong

[grand pond]

smeagol

[silent compass]

That's correct

[safe onyx]

ty

[grim vector]

in the first you mean ?

[safe onyx]

yeah sorry

[grim vector]

ok

[safe onyx]

$$F cos\theta - \mu_s (mg -F sin\theta) = 0$$

[grand pond]

smeagol

[safe onyx]

then distributed the mu_s then moved F sin/cos to the other side

$$F cos\theta - \mu_smg + \mu_s F sin\theta) = 0$$

[silent compass]

The sin term should be with a plus sign

[grand pond]

smeagol

[safe onyx]

$$ - \mu_s mg = - F cos\theta - \mu_s F sin\theta$$

[grand pond]

smeagol

[safe onyx]

then factored out F

[grim vector]

would be easier to send mu_s mg on the other side but well it works

[safe onyx]

oh right

[grim vector]

the less minus sign to handle

the more you happy

[safe onyx]

$$F cos\theta + \mu_s F sin\theta = \mu_smg $$

this is better

[grand pond]

smeagol

[safe onyx]

$$F ( cos\theta + \mu_s sin\theta ) = \mu_smg $$

[grand pond]

smeagol

[safe onyx]

Factoring F