help-42

[winter loom]

Where's that at

[blazing coyote]

ok

*oh

right

so it would be

$\frac{\left(\lambda_0\right)}{16\ \epsilon_0R}\ $

$\frac{\left(\lambda_0\right)}{16\ \epsilon_0R}$

[potent lotus]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

and this would be the y component too

[winter loom]

There is no y component.

[blazing coyote]

oh

[winter loom]

The y component cancels.

[blazing coyote]

so the other two quads have the same result I pressume

[winter loom]

That was the point of this image.

[blazing coyote]

yeah, forgot, sorry

[winter loom]

Yes it would, you don't even have to worry about it. You can just use the bounds 0 to pi. And the result will be what it will be.

[blazing coyote]

ah, thanks!

.close

[calm coral]

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[calm coral]

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[wind phoenix]

hi, are my answers correct?

[amber bolt]

there's no second correct answer

[wind phoenix]

okay

[wind phoenix]

then the first one?

only the first one

<@&286206848099549185>

<@&286206848099549185>

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[glad sinew]

can we go over these?

ok so a is true

[calm coral]

@glad sinew Has your question been resolved?

[glad sinew]

I'll send pictures of my work

but I think I got it

yea I gotta prove the other way for intersection

[dull storm]

For (b) and maybe (c), have you tried finding counterexample? Maybe try a function with many inputs go to the same output, like {(1,2), (2,2), ...}

[glad sinew]

ok

[dull storm]

Okie, lgtm for (a).

[glad sinew]

can you explain a proof for c)?

oh wait

atually i think its ok

ill send a picture after

for (<-)

if I already said y cannot be in f(B) even if x1!= x2, I can’t say there exist such x1 st y=f(x) in f(B)

so the last part is wrong..

I could say x1 in A intersect B..

@dull storm

can you check my answer for c?

[dull storm]

Something's wrong in (->): even if $x\notin B$, it's still possible that $y \in f(B)$.

[potent lotus]

poypoyan

[glad sinew]

how should I write out (->) then?

so I just state that its still possible

abd hence it is a contradiction to f(A)\f(B)?

how come this is not true

but,

this is true

I did the proof, but isnt inverse basically in the domain and its function in codomain...

[potent lotus]

poypoyan

[dull storm]

I'm sorry I have to write this again. May LaTeX is not that good rn.

You can do this for (c):

Here's a counter-example: Suppose $x_1 \neq x_2 \in A$ such that $f(x_1) = f(x_2) = c \in B$.
Let $H = {x_1, x_2}$ and $G = {x_1}$. Then $f(H \setminus G) = f({x_2}) = {c}$, but $f(H) \setminus f(G) = \ldots$.

[potent lotus]

poypoyan

[glad sinew]

but f(H)\f(G) = c still hold s true...

[dull storm]

Nope, because f(H) = f(G) = {c}.

[glad sinew]

ah

ok

[glad sinew]

injective is x != y imples f(x) != f(y) right?

yes

ok

[glad sinew]

how to prove?

[calm coral]

@glad sinew Has your question been resolved?

[stray tundra]

what does it mean by (p, q) = 1?

gcd of both being 1?

[calm coral]

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[wind phoenix]

could you help me please? i dont have much time.

[wind phoenix]

i have multiply -8, 8, and 9 with the same number

so that it gets infintely many solutions

[timid oracle]

make the second equation identical to the first to make it redundant

[wind phoenix]

i still dont know how to make them identical

2*(-4)

(-4) would be an option?

[timid oracle]

yeah so a(-4)=-8 and b(-4)=9

[wind phoenix]

okay

so -4 is the anwser

for a and b

[timid oracle]

no, a is 2 and b is -9/4

think about how you can multiply both sides of the second equation by -4 to get the first equation

[wind phoenix]

oh yeah

[wind phoenix]

i get it now

so 9/(-4)

is -9/4

?

[timid oracle]

that’s b

[wind phoenix]

yeah

[calm coral]

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[wind phoenix]

Is this correct?

[rough arrow]

I don't understand german, but I assume it wants you to have the vector a + (1/2)b from point P?

[wind phoenix]

Yes

[rough arrow]

Notice the direction of the (1/2)b vector you wrote, it has the opposite direction of b.

So you wrote a -(1/2)b

[wind phoenix]

Because it’s a plus

Oh

What

[final marsh]

Can you guys do 10th grade

Calc

And pre calc

[rough arrow]

Direction matters. If you go north, then going south isn't going north, that's going the opposite direction of north.

[wind phoenix]

!occupied

[calm coral]

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).

[final marsh]

Can you do some math

Like show me how to do it

[wind phoenix]

Like this?

[rough arrow]

Open a new channel. You'll find them in a section called "MATH HELP (AVAIABLE)" above this channel.

[rough arrow]

You're going in another direction.

[final marsh]

What language

German?

I cant read

But it looks like i did it back in 7th

[wind phoenix]

yeah but the b vector is minus, and i have to make it plus because it says +1/2b

[final marsh]

Just basic 10th grade math

plz help

[wind phoenix]

can you go to a different help channel please?

[wind phoenix]

here

[rough arrow]

Your b vector has its direction towards west-north. Then that's it. That is the positive direction of b, and you want that, (1/2)b is asked.

[final marsh]

Like which one is which

[wind phoenix]

huh?

[rough arrow]

Shorten your b, and apply it.

[wind phoenix]

aah

west north is positive because it goes up

and not down?

but i think it is negative

because it is turned to the left side

[rough arrow]

Look at the arrow, that is the direction it's pointing.

[wind phoenix]

b goes to the left

it cannot be a postive vector

i thought i need to make b positive

[rough arrow]

https://www.youtube.com/watch?v=fNk_zzaMoSs&pp=ygULdmVjdG9yIG1hdGg%3D

[wind phoenix]

i know that i have to shorten it

[rough arrow]

Direction of its own matters, not the plane.

[rough arrow]

Watch this video, you'll probably get a more wider vision of vectors.

[wind phoenix]

i dont have that much time

._.

xd

[rough arrow]

It wouldn't be easier to explain with words how a vector works.

[white atlas]

in essence, you can't rotate your vector around without actual cause

you can only translate it (move it up, down, left, or right)

[wind phoenix]

https://tenor.com/view/milk-and-mocha-milk-mocha-milk-mocha-bear-milk-bear-mocha-bear-gif-1899012525858648612

kanna

[white atlas]

That should suffice to do your assignment

[wind phoenix]

This

doesnt make sense

or does?

and i should put this all to point P?

[white atlas]

your 1/2 * b is okay

but your resultant (a + 1/2 * b) isn't

wrong direction

[white atlas]

yes, that's better lol

[wind phoenix]

thank youu

[white atlas]

no worries-

[calm coral]

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[torpid vine]

https://cdn.discordapp.com/attachments/908078008609431674/1217196858187714622/image.png?ex=660325e9&is=65f0b0e9&hm=bc7c3c08805cdbecd570b6545e2a05772b110538d6f0edeb2d984341fb0da863&

[torpid vine]

Can someone help me figure out how squared logarithms work

[glacial lily]

Squared logarithms don't have a property so try to do normal algebraic manipulations

[torpid vine]

Okay

[glacial lily]

But the last term has an identity: $log_8x^2 = 2log_8x$

[potent lotus]

Oğuzhan

[glacial lily]

Only the $log_8^2x$ doesn't have a property

[potent lotus]

Oğuzhan

[torpid vine]

Oh yeah

I did that

Let me send a picture on how far I got

[glacial lily]

Alright

[torpid vine]

Here it is

,rotate

[torpid vine]

Oops

,rotate

[torpid vine]

,rotate

[torpid vine]

Okay

[glacial lily]

Do a substitution where y=log_8(x)

Then solve the inequality for y

[torpid vine]

Okay

,rotate

[torpid vine]

I got this so far

Is this right?

Well I don't know what to do after this

Also, am I allowed to cross multiply with inequalities

@glacial lily

<@&286206848099549185>

Sorry if you did not want to be pinged Oguzhan

I got it into one fraction

I am having trouble factoring the quadratic

I am stuck with this

[calm coral]

@torpid vine Has your question been resolved?

[calm coral]

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[heavy spoke]

If f is surjective and g is not surjective then g o f is not surjective

[heavy spoke]

Is this statement true or false?

[glass heart]

what do you think

you could try an example

[calm coral]

@heavy spoke Has your question been resolved?

[calm coral]

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[tepid oriole]

$f(x) = \begin{cases} \frac{kcosx}{\pi -2x}, &\text{ if x } \neq \frac{\pi}{2} \ 3, & \text{ if x = } \frac{\pi}{2} \end{cases}$

[tepid oriole]

MAN WHY DO I NEVER GET THESE RIGHT

[potent lotus]

Barbie

[tepid oriole]

yes this is it

find k

so that the function is continous

[calm coral]

@tepid oriole Has your question been resolved?

[tepid oriole]

<@&286206848099549185>

[remote mural]

note x not equal pi/2 to means greater than or less than pi/2

[tepid oriole]

yes i am aware of that

for some reason i have a feeling that this question is incorrect

[remote mural]

think the question is wrong

[tepid oriole]

yes i think so too

thanks

.close

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[scarlet saddle]

given f(x)=3^x-1 +2 as x approaches negative infinity

what happens to f(x)?

[scarlet saddle]

i dont know where to begin

but i would assume -1 or 0

[tepid oriole]

$f(x) = 3^{x-1} +2$

[potent lotus]

Barbie

[tepid oriole]

is this what you mean?

[scarlet saddle]

yea

sorry

[tepid oriole]

i think it just equates to 2

[scarlet saddle]

how?

im an idiot

yea it most likely does

lmao

how did u get it tho?

i plugged in

[tepid oriole]

i also plugged it in.

[scarlet saddle]

any other way?

[tepid oriole]

well there is one other

f(x) = 3^(x-1) + 2
= 3^x * (1/3) + 2
= (3^x / 3) + 2

lim[x → -∞] f(x) = lim[x → -∞] ((3^x / 3) + 2)
= lim[x → -∞] (3^x / 3) + lim[x → -∞] 2
= (lim[x → -∞] 3^x) / 3 + 2
= 0 / 3 + 2
= 0 + 2
= 2

[real sentinel]

if you remember 3^-1 = 1/3 so the more negative the x-1 term gets the smaller it becomes so at negative infinity 3^x-1 is basically 0

[scarlet saddle]

we didnt learn limits

[tepid oriole]

x approaches anything basically just means lim x-> whatever that is

but if you didn't just plug in.

it doesn't really matter.

[scarlet saddle]

this one makes sense

thx tho barbie

.close

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[calm coral]

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[dusky tendon]

Are there any shortcuts to knowing if something is a permutation or combination?

[amber bolt]

no

[dusky tendon]

so the only way is to read it and analyze :/ there aren't any keywords or anything?

[civic dirge]

These questions can be extremely free and varied, sadly.

[dusky tendon]

Oh, well thank you for that.

.close

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[solid steppe]

yo

[solid steppe]

can someone help me with a question

[quick karma]

[solid steppe]

so

i need to invert C

and C= A-2I

so my question is how do i get C ( i dont understand what 2I does)

[normal wadi]

I is the identity matrix

[solid steppe]

yes i know that

but when its A-2I

where does it change?

[normal wadi]

wdym

[solid steppe]

yea in this case C=A-2I

[normal wadi]

that right part just says that the C matrix is equal to the difference of A and 2*I

[solid steppe]

yes

the thing is i dont understand how to do it

[normal wadi]

were you taught how to add and subtract matrices?

[solid steppe]

yes

[normal wadi]

then what exactly don't you get about the operation here

[solid steppe]

the thing is i learned it but i have forgot it ( i am currently preparing for a retest)

[normal wadi]

https://youtu.be/QXUbFzEd3Ww?si=M9onE81aO9UvCDxk

incredibly trivial stuff

and should be the least of your concerns if you're diving into matrix inversion

[solid steppe]

i know this stuff

[normal wadi]

you just said you forgot it?

are you talking about inverting matrices or matrix addition

[solid steppe]

not what the video speaks about

i am talking about

when

it says A-2I

the 2I

what does that indicate?

[normal wadi]

do you know what the identity matrix is?

[solid steppe]

u mean A A^-1=I

?

[normal wadi]

yes but what is it exactly

for a 3x3

[solid steppe]

[100 010 001]

[normal wadi]

yes

now what happens when you multiply this matrix, I, by -2?

[solid steppe]

uhhhh

omg

lol

so its just that easy lol?

[normal wadi]

yes that's why I was so confused

[solid steppe]

[-2 0 0 ,0 -2 0 ,0 0 -2]

[normal wadi]

yes

now what happens when you add this result to the A matrix?

[solid steppe]

its just minus

the A

so we get

[2 2 1, 0 1 -2, 3 2 3]

[normal wadi]

yes

that is the C matrix

[solid steppe]

yea

i think when it comes to math

i always make harder then it should be

but thx for the help

i have 1 more question

if u can help me it would be the world

for me

[normal wadi]

yes

[solid steppe]

so this question

i have solved a)

and stuck on b

[normal wadi]

idk what this says man

needs translation

[solid steppe]

so it says The plane Π contains the points (−1, −3, 2) and (1, 0, 1) and is parallel to
lelt with the vector

a)give an equation for Π.

b) Calculate the distance from the point (10, 1, 1) to Π.

[normal wadi]

I believe the first thing is to construct an additional vector so that you can utilize the cross product

[solid steppe]

yea answer is they add p_0= [0 0 3]

but idk how they do that

[normal wadi]

not sure how they got that vector from the 2 points because neither addition nor subtraction of one point to another turns p_0 into that

[solid steppe]

here is how the professor did

[normal wadi]

oh my bad I was looking at part a

[solid steppe]

okay but

how do they get the p_0?

[normal wadi]

I'm checking

ok so I got a plane equation 2x - y + z = 3

[solid steppe]

yes same as me

uhh mb i should have told u what i got on a)

[normal wadi]

let me test something out, I believe you dont need that specific point that your professor gave

I believe you can use either given point from the problem; let me try it first

[solid steppe]

oki

[normal wadi]

I get what your prof is doing but idk how they got the point, can you translate that second image

[solid steppe]

it does not say anything

[normal wadi]

"punkten p0 = ... ligger i planet"

etc

[solid steppe]

just on the right side it say the point p0 is in the plane

[normal wadi]

what about the stuff before the final step

[solid steppe]

thats all

he gave us

[normal wadi]

??

[solid steppe]

or u mean a)?

[solid steppe]

the shortest distance is given by

[normal wadi]

got it

you can use any point that lies on the plane using the formula they provided

I would use one of the two points that was given in the problem

[solid steppe]

so like (1 0 1)?

[normal wadi]

yes

[solid steppe]

uhhh okay

does that always apply like that?

or can there be a question where u need to find out?

[normal wadi]

as long as you're picking a point that satisfies this plane equation it will work I'm pretty sure

[normal wadi]

but yes if you have some layout like this, the equation for [distance from point to plane] that your prof provided should always work

[solid steppe]

Uhhh okay i see

thank you so much for the time and effort

❤️

[normal wadi]

np

[solid steppe]

have a good night

or morning depens where u are lol

.close

[calm coral]

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[wind phoenix]

can someone help me with this? I am supposed to be asleep and i dont have much time

[wind phoenix]

.close

[calm coral]

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[torpid hinge]

Can someone just give me a hint for this problem (other than the one already given)? Or maybe a way to actually apply the hint that was given. Preferably just the first step or smth

[pure kayak]

difference of two squares

wait im high

i just imagined a whole different equation

[torpid hinge]

Lol dw it happens

I think I'm actually supposed to use that tho, I just don't know how to get to that point

[fathom solstice]

try putting it in the form (y+1)^2 for some y

its gonna be a couple of terms of (y+1)^2 for different y

[torpid hinge]

Wait wdym by this

Oh nvm I think I got it

[fathom solstice]

add and subtract some terms to have a polynomial that is of the form (y+1)^2 on the left side that uses the x^2 term

x^4 term*

[torpid hinge]

Ty for the help

.close

[calm coral]

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[fathom solstice]

youre welcome

[exotic falcon]

...

[calm coral]

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[heady jungle]

so I'm finding dr/dt

nvm height was dumb and in inches

.close

[calm coral]

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[desert talon]

can someone just lmk if I'm doing this right?

[desert talon]

I've been absent with a cold and I can't take the test later because this is the last week of the quarter

[empty quail]

you can't round the number when you start multiplying

because you get a different answer

[desert talon]

oh shoot

[empty quail]

idk where did you get sin 40 from

[desert talon]

where did I write that ;-;

hol;d up

gimme a moment

[empty quail]

okie

[desert talon]

I couldn't find sin(40 so I just didnt round and it doesnt feel right

[empty quail]

oh nvm yea

I see it is sine(theta) times 40

[desert talon]

sorry for my hand writing xd

[empty quail]

you should do it this way

because you see the circle part is confusing

[desert talon]

Ill try to be more clear in the furture 👍

[empty quail]

[desert talon]

so im on the right track?

[empty quail]

did you multiple sin 71 by 39.6 to get number like 40.165?

I gave you a clear step to restart again by multiplication

[desert talon]

oh shoot

alr gimme another moment xd

[empty quail]

no problem

[desert talon]

something like this?

then i divide both sides by 38.2?

[empty quail]

yep

after that you can get your answer

[desert talon]

is this correct?

YAAAY

alr lemme continue to solve for the triangle then ill send you another screenshot

[empty quail]

it is getting too late for me

I need to gts

[desert talon]

ooh alr alr

gn

hopefully someone else comes to my rescue

Now I would use the new angle I just in the law of cosine to find the last side?

[calm coral]

@desert talon Has your question been resolved?

[desert talon]

<@&286206848099549185> ?

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@desert talon Has your question been resolved?

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@desert talon Has your question been resolved?

[desert talon]

.close

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[warm warren]

So I’m trying to understand this but I can’t see where this magical covector is coming from

[warm warren]

The claim is that a linear map is a (1, 1)-tensor

And he’s showing here that the notion of 𝚽 being a linear map that takes vectors to vectors is equivalent (in the finite dimensional case) to taking a vector to the dual of the dual space

He also claims that this output, that lives in the dual of the dual space, T( • , v), is something that can eat a covector, which maps vectors to reals

So suppose I have a linear map 𝚽

I can feed this map some vector v

What covector do I feed it to make it go to ℝ?

Because on the bottom left he writes that the T constructed from the data of 𝚽 maps a covector and vector to ℝ

By feeding the covector: 𝚽(vector) which gives a vector, which the covector clearly maps to ℝ by the definition of a covector

But I don’t get where this particular covector (denoted varphi) comes from

[ancient thistle]

[ancient thistle]

like the bottom left corner?

[warm warren]

Ye

[ancient thistle]

its coming from V*

[warm warren]

Well yes it’s a covector it comes from V*

But where do I get this covector from

Say I have a linear map 𝚽 and a vector v

[ancient thistle]

its just some element of V*

you dont get it

[warm warren]

How do I make this covector

[ancient thistle]

you dont make it

its any element of V*

T(φ, v) = φ(phi(v))

phi is the other phi

[warm warren]

So my 𝚽 can eat a covector and a vector and give me something in ℝ in a bilinear fashion

[ancient thistle]

isnt phi a map from V to V

[warm warren]

Ok my T_𝚽 can eat a covector and a vector and give me something in ℝ in a bilinear fashion

[ancient thistle]

yeah

,, \phi : V \to V, \quad T_\phi : V^* \by V \to \R

[warm warren]

So from a linear map, I can create a (1,1)-tensor

Ah, and the top part is saying from a (1,1)-tensor I can create a linear map

Then the (1,1)-tensor and the linear map must be the same thing

[ancient thistle]

theres a correspondence between the two

[warm warren]

Right

He also says that (V*)* = V

Should that be = or isomorphic

[ancient thistle]

iso

[warm warren]

Ah

So if I have a (1,1)-tensor and I give it a vector

Now it’s something that can eat a covector and spit out a real number

And this thing, has an isomorphic mapping to some vector v in V

So this thing is the T( • , v) object, which lives in (V*)*, and has a corresponding v in V by some isomorphism

And this corresponding v is what I’d have gotten if I just fed my original vector into 𝚽

[warm warren]

But then it depends on the isomorphism I chose at this step no?

[oblique current]

Yeah but you wanna think of the map the other way, you send v to T(-,v) and this will turn out to be an iso

[ancient thistle]

the iso is canonical

its just evaluating at the vector you get

[oblique current]

Which is presumably why they're equals sign pilled

[warm warren]

How do I map Hom(Hom(V, ℝ), ℝ) to V

That’s the (V*)* to V right?

[ancient thistle]

you usually don't

you're gonna need some kind of finiteness of basis argument here cuz its just straight up not true when V is infinite dimensional

[oblique current]

You have to figure out that everything in the left vector space can be written as T(-,v)

In finite dim^

[warm warren]

Well he did restrict to finite dimensional cases here

[oblique current]

But this we can just get from considering the map the other way like above and show its an iso

[ancient thistle]

you can probably pick bases and do linear algebra stuff to say that the map is uniquely determined by its values on the bases

and that'll give you the map back from V** to V

[warm warren]

I’m trying to think perhaps say 𝚽 is a matrix right nice and concrete

[oblique current]

Well then you definitely do have to pick bases

[warm warren]

What sort of information am I getting out of the matrix such that I can feed it a vector and a covector to get a real number

[ancient thistle]

well

when you have a matrix its kinda dumb right

like

[warm warren]

Because T_𝚽 is saying we extract the information out of 𝚽

What specifically what information

[ancient thistle]

if you have the matrix [phi] and the vector [v]

then [phi][v] is the image of v under phi

[warm warren]

Sure

[ancient thistle]

now you can just take a row vector [w]^t

and [w]^t [phi][v] is a real number

[warm warren]

Indeed

[ancient thistle]

but then like

[warm warren]

That is indeed true

[ancient thistle]

the V* x V -> R map is just

(w, v) -> [w]^t [phi][v]

[warm warren]

Ah

So in matrix land the covector is just the row vector of length equal to the dimension of V

Right because it is the thing that sends vectors to real numbers linearly

Hey that’s just the dot product

Kinda

So from the identity linear map, I can make the dot product

That’s the (1,1)-tensor that corresponds to the dot product

[oblique current]

Once you choose (orthonormal?) bases yeah

[ancient thistle]

just pick a new dot product

[warm warren]

Oh right the dot product uses the basis as a definition

Hey but I don’t need it I can just slap some (w, v) into I(identity linear map)

And I get something that works like the dot product

[oblique current]

Your T is essentially an abstraction of the dot product yeah

But its just the obvious thing you would do to get a number if you have a vector and a function from vectors to R: apply that function to your vector

[warm warren]

Ok so if I have a function that takes a covector and a vector

Naturally I just feed the vector into the covector

But suppose I wanted to change it up a bit and include a linear map somewhere in there

Then I just feed the image of the vector into the covector

[oblique current]

Yeah you can "twist" the map by some iso endo of V

[warm warren]

But how do I know the image of the vector, is something that the covector can eat?

[ancient thistle]

in some sense the isomorphism between V and V** is trivial when you talk about actual matrices

[warm warren]

What if my linear map lost dimensions

[ancient thistle]

[v] is also something in V** because you can do [w]^t [v]

[oblique current]

It's need to be a map from V to V

[warm warren]

Oh

Ah

That’s quite a strong restriction

So then not all linear maps are (1,1)-tensors

Only the ones from V to V

[oblique current]

Otherwise like you say your covector can no longer in general eat something in some other non isomorphic space W

[ancient thistle]

when you write (1,1)-tensors you kinda already mean like

V* x V -> R or something

but really like

Hom(V, W) and V* x W can be identified in some sense for finite dimensional spaces

[warm warren]

I don’t get that

😦

[ancient thistle]

oh god it's gotta be tensor product

i can't type this on my keyboard

[warm warren]

For the V* and W?

[ancient thistle]

yeah

like

[warm warren]

I tried thinking about that but I didn’t get anywhere either

[ancient thistle]

if you have an element [ \sum_i \varphi^i \tensor w_i \in V^* \tensor W ]

[warm warren]

That’s a basis isn’t it

[ancient thistle]

no it's just a general element

something in the tensor product is a finite sum of pure tensors

which are those phi x w thingos

you can associate to it a map in $\Hom(V, W)$ by doing something quite similar [ T(v) = \sum_i \varphi^i(v) , w _i ]

[warm warren]

Are tensor product like a product topology

[ancient thistle]

just feed the phis the v and multiply by the corresponding w

no not quite

but also kinda

[warm warren]

You don’t just append the W part onto V?

[ancient thistle]

no

that would be more like a direct sum

[warm warren]

Ok I’m reading Wikipedia and it says

Tensor product of V and W is a vector space with a bilinear map

[ancient thistle]

that's kinda not helpful

[warm warren]

It maps (v, w) to v tensor product w

In a bilinear fashion

I changed up the words a bit

[ancient thistle]

the bilinear map is ${-} \tensor {-}$

[ancient thistle]

it's kinda tautological

[warm warren]

Right ok nice

So Hom(V, W) is just a vector space of all linear maps from V to W

And V* tensor product W is

Something

So it’s like you take a row vector length dimV, and a column vector length dimW

Ah you can make a matrix out of that

That’s the linear map!

Ok but I lose all sense of control when I’m forced to give up the notion of a matrix, a row vector and a column vector

[ancient thistle]

its kina like uh

,, \sum_i \mat[b]{|\ w_i \ |} \mat[b]{- & \varphi_i & -}

ew

[ancient thistle]

it looks so gross

anyway

now when you get a vector

its gonna be like

,, \sum_i \mat[b]{|\ w_i \ |} \underbrace {\mat[b]{- & \varphi_i & -} \mat[b]{| \ v \ |}}_{\text{some number}}

[ancient thistle]

so if you think of the w_is as the basis of W

that will somehow be able to give you any vector in W

[warm warren]

What is this

[ancient thistle]

my shitty representation of what [ \sum_i \varphi^i \tensor w_i ] is supposed to be

[warm warren]

Ok so this is what anything on the right side looks like

[warm warren]

Right side of this statement

[ancient thistle]

yeah

and then like

you can see what the linear map looks like

from V to W

[warm warren]

And supposedly it represents a linear map because that’s what Hom(V, W) contains

[ancient thistle]

well heres how the linear map works

[warm warren]

So varphi_i is kinda telling me how much of my v gets turned a particular basis vector of w_i

[ancient thistle]

yea

perhaps what you want is something more like
[ \phi : V \to W, \qquad T : W^* \by V \to \R ]

[warm warren]

Why is W the starred one now

😭

[ancient thistle]

well the way im thinking of it is like

[warm warren]

WAIT NO HE DID SAY SOMETHING LIKE THAT

[ancient thistle]

T is really a map V -> (W* -> R)

which is like

V -> W**

[warm warren]

something about “throw something on the right to the left, and give it a star”

[ancient thistle]

and W** is meant to be W

[warm warren]

“And works the other way too”

[ancient thistle]

[warm warren]

[ancient thistle]

yeah so

Hom(V, W) becomes V* x W (tensor)

[warm warren]

Where is this from

[ancient thistle]

like

T : W* x V -> R

but instead

you can just give it a v in V

and you're left with a w in W* to give

[warm warren]

Ok suppose T is such a map

If I give it a v, then it now looks like T(v) : W* -> R

[ancient thistle]

T(-, v) : W* -> R

[warm warren]

Ah but this T(v) is in (W*)*

[ancient thistle]

yeah

so thats an element of W under the isomorphism

[warm warren]

The?

[ancient thistle]

canonical one

[warm warren]

I don’t know what’s the canonical isomorphism

[ancient thistle]

its the substitution map

v -> (phi -> phi(v))

aka the do nothing map on column vectors

[v] -> [v]

since [v] is a column vector, and its also a cocovector since if you take a row vector [w]^t, you can do [w]^t [v]

[warm warren]

So you’re saying a cocovector takes covectors to R

And in particular, [v] is a cocovector because it takes the covector [w]^t to R by [w]^t [v]

[ancient thistle]

yeah

[warm warren]

Ok sure

So then if I have T( • , v), this cocovector is canonically the vector v

And if I fed the cocovector T( • , v) some covector w

[ancient thistle]

well here you actually want to apply phi first

[warm warren]

Then I get out w(v)

Ok this is the T( • , v) not constructed from anything

[ancient thistle]

yeah okay

[warm warren]

But suppose I wanted to construct a particular T from a linear map

Then the cocovector T_𝚽( • , v) is canonically the vector 𝚽(v) which lives in V

Hmm, but I don’t know that 𝚽 is full rank, what if the image of 𝚽 is not all of V

Then if 𝚽(v) = 𝚽(w), for some v,w ∈ V, I have that T_𝚽( • , v) = T_𝚽( • , w)

Ok sure I just have less cocovectors

And clearly the different cocovectors I have depend on the 𝚽 so it seems like at least some of the information of 𝚽 is indeed getting passed on to T

If I feed my cocovector a covector, I just apply the covector to my 𝚽(v)

And that’s why he write 𝚽(varphi(v))

Ok

And the idea is that a linear map 𝚽 bring v in V to 𝚽(v) in V

But this 𝚽(v) has the capability to be fed into a covector

So it is in a sense a cocovector

So I get out something that can take a covector and get into R

So $V\overset{\phi}{\to}V\overset{v^\in V^}{\to}\mathbb R$

[potent lotus]

Frosst

[warm warren]

So a linear map is like

Unfinished

I can still feed the result into a covector

Ok this is quite clear now, assuming I’ve not got the wrong idea in my head

Alright I will now roam freely with this new delusional idea

Thanks snow and zac

.close

[calm coral]

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[tranquil basin]

can someone explain me this

[tranquil basin]

the content in the brackets is (x^2 + 1)^2 right? but why did x^2 change its sign

[zealous hawk]

add and subtract x^2

[alpine stone]

What do you mean by changing its sign?

x^2 was rewritten as 2x^2 - x^2

[tranquil basin]

why

[alpine stone]

So that you would have x^4, 2x^2 and 1 in the sum; Those add up to (x^2 + 1)^2

[zealous hawk]

it's a trick to factorize x^4+x^2+1

we add x^2 so it becomes x^4 + 2x^2 + 1, then to balance it we subtract the x^2 we added

[tranquil basin]

to keep it equal?

[zealous hawk]

which becomes (x^4 + 2x^2 + 1) - x^2

yes

and now you can see it's a a^2 - b^2 form

cuz that left side quantity is square of x^2+1

[tranquil basin]

so the step between those 2 was (x^4 + x^2 + x^2 +1) + x^2 and then to keep it equal it comes down to (x^4 +2x^2 +1) - x^2

[zealous hawk]

yeah

[tranquil basin]

which is (x^4 + x^2 + 1) which is (x^2 + 1)^2

[zealous hawk]

yea

it's done to factorize the thing

[tranquil basin]

i understand it now thanks for elaborating

.close

[calm coral]

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[tranquil basin]

wait

isnt x^4 + x^2 +1 after subracting it what i started at :/

[zealous hawk]

yeah we dont subtract it again obviously

this is what you have

now use (a+b)(a-b) formula

[tranquil basin]

ohhhhh i was subtracting it wrong

now it should be clear

[calm coral]

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[woven wigeon]

How can i simplify this easily

[calm coral]

@woven wigeon Has your question been resolved?

[white atlas]

Break it into their respective prime factors

15 = 5 * 3
6 = 2 * 3
10 = 5 * 2
21 = 7 * 3```

Then exponentiate each appropriately and then simplify

[calm coral]

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[calm coral]

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[lucid knot]

i used a calculator, and i think its right

i have 2 questions

1. is it even correct?

2. how would you solve this without a calculator?

[glacial lily]

You need a calculator.

[lucid knot]

ok that makes me feel better

[glacial lily]

x > ln(sqrt(63/7))

[lucid knot]

thats ln (3)

1.09861

[glacial lily]

Yeah and whatever your calculator gives is the answer

[glacial lily]

If that's it then your answer is correct

[lucid knot]

😌

i got concerned for a minute

thank you!

.close

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[grim vale]

convert 0.281 to a fraction (81 is recurring)

[full niche]

just let x = the number

and try cancelling out the recurring part

[calm coral]

@grim vale Has your question been resolved?

[calm coral]

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[gray briar]

I am having a hard time understanding how to evaluate these functions I have tried to teach myself and haven't made any sense of these.

[gray briar]

what does the graph serve in understanding how to evaluate the functions?

[blissful phoenix]

The graph shows you which values the functions take at different values of x

[gray briar]

so at -1 f(x) is at 0 and g(x) -2 in terms of the y axis or is this not relevant?

[blissful phoenix]

Correct. f(-1) = 0, and g(-2) = 0

So for the first question, (f+g)(-1) = f(-1) + g(-1), can you tell me what that is by looking at the graph?

[gray briar]

Im sorry I am confused.. Am i looking at the difference of the to graphs? If so the difference is -2

[blissful phoenix]

Correct. (f+g)(-1) = f(-1) + g(-1) = 0 + (-2) = -2

[gray briar]

so then i am using the (-1) as a reference to see on the graph of where they are talking about and then seeing what the difference is. are we setting the function equal to zero to find the answer?

In part two the answer would then be 0 as there is no difference?

Also how does the function (f+g) , (g-f), ((g)(f)), (g/f) affect evaluating?

[blissful phoenix]

Let's make sure we know what a function is first. A function f(x) takes a value x and maps it to the y-axis in this case (see y = f(x)). So, for example, if x=-1, then f(x) = f(-1) = 0, as you can see on the picture

The first question states: (f+g)(-1) = ?, and you're supposed to answer what ? is. (f+g)(-1) is a syntactic way of writing f(-1) + g(-1), i.e., it's the same thing.

The second question states: (g-f)(-2) = ?, which is the same thing as g(-2) - f(-2) = ?

You have correctly answered that the second question is g(-2) - f(-2) = -1 - (-1) = 0

Does it make sense?

[gray briar]

sorta.. so in i could view however they put g and f in the syntactic way as basically saying = and not be hyper focused on it as its going to not effect what i am seeing on the graph? cause (g/f) (1) would be 1/1 right?

(g/f)(1) is -4 in that case correct?

[blissful phoenix]

Make it easier on yourself by writing out the whole thing. (g/f)(1) = g(1)/f(1) = ?

[gray briar]

So then like this correct?

[blissful phoenix]

yes

Next, what is g(1), and what is f(1) ?

[blissful phoenix]

Bravo. Correct

[gray briar]

ahhh I understand this now the function is saying the rule that you are going to use to evaluate but also where on the graph you will be looking on the graph and the once you find the (f) and (g) you can then use the values from the graph to get the solution

correct?

[blissful phoenix]

Correct.

[gray briar]

So then this how this one is set up?

[blissful phoenix]

First evaluate f(-4), then take the answer you get and plug it into the g function

So

• let f(-4) = z
• look at the graph to figure out the value of z
• when you have the value of z, look at the graph to find the value of g(z)

[gray briar]

so (f(-4)) = -3

[blissful phoenix]

Correct. You have figured out steps 1 and 2. Next, do the third step, i.e., evaluate g(-3)

[gray briar]

Wouldn't that mean then that (-1(-3)) as g(-4)=-1 so then the solution would be 3

[blissful phoenix]

Where are you getting (-1(-3)) from?

[gray briar]

wait..

Is it saying g(-3)? Since g(z)=-3

Or am I just confusing myself more?

[blissful phoenix]

Yes, it seems you are thinking about it wrong.

g(f(x)) is a two-step process. First, you evaluate f(x). Then, you take the answer you got from f(x) and plug it into g

So you've correctly figured out that f(-4) = -3

The next step is to evaluate g(-3)

then you have the answer

[gray briar]

So then g(-3) is -1

ahhhh

[blissful phoenix]

correct

Thus, g(f(-4)) = g(-3) = -1

[gray briar]

That makes more sense now

[blissful phoenix]

👍

[gray briar]

Thank you so much I was so lost and didn't understand any of this you definetly helped me understand how to solve these correctly

[blissful phoenix]

Glad to help!

[gray briar]

last question on this

As far as ((g)(f)) would be the same as ((f)(g)) as its just multiplication but just switching around g and f right?

[blissful phoenix]

Yes

[gray briar]

Thank you

ops okay for sure last question

is this not multiplication ?

[blissful phoenix]

f ∘ g is usually denoted as function composition, so (f ∘ g)(0) = f(g(0))

and (g ∘ f)(-3) = g(f(-3))

This should be defined wherever you are learning this from

[gray briar]

... Now I am more confused

Oh is it saying it similar to how g(f(-4)) just written different?

[blissful phoenix]

It is the same but written differently, yes

[gray briar]

Thank you sooo much 🫶

That should be all !

.close

[calm coral]

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[calm coral]

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[jagged surge]

Hey I need help soling this fraction. I would appreciate if you showed how you solved it.

[jagged surge]

Im not sure where to start

[remote mural]

hm

can u simplify the numerator more?

if so what will u get?

[jagged surge]

the numerator? Im sorry im not fluent in english

[remote mural]

the top part

over the divide line

[jagged surge]

hm

im not sure how

we can probably do soemthing with the "b" to start

no we cant, the other Bs multiplied

[remote mural]

Understand this?

[jagged surge]

how did the 2 from the first line turn into a 4 in the second?

[remote mural]

Because 2 multiplied by 2 is 4

2 x a square x 2 x b

[jagged surge]

yeah, but why did that mulitply and not the others

[remote mural]

So 4 x a square x b

[remote mural]

They did

[jagged surge]

oh lright

just to make it match with the other side?

im sorry, I dont understadn where the second B went in the second line

[jagged surge]

hello? If you went away you could tell me, so I could ask someone else

.close

[calm coral]

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[proud osprey]

how do you find the tanget line

[swift laurel]

what kind of operation do you use to find tangent lines in other contexts?

[proud osprey]

y - y0 = m (x - x0)

[swift laurel]

so in other contexts, you find a tangent line based on a point and a direction. what kind of operation gives you the direction of a tangent line?

[proud osprey]

first derivative?

i mean the point is literally like r(pi/4)

[inner zealot]

Yes.

[proud osprey]

uh you dont need the second derivative anywhere right

[inner zealot]

Not to calculate the tangent vector, no.

[proud osprey]

whats the point?

because r(pi/4) is ust one value

and this needs x0 and y0

oh wait

sorry im stupid

bruh

its a vector

sorry im really tired

.close

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[remote mural]

hi

[remote mural]

i was wondering

if there is an alternative more elegant proof than i already did

i will state the problem

translation is show that the number c = .... is a perfect square

now my proof consists of taking an natural number(any number works) and seeing the general rule

complementary combinatorics

so you get

lets say n = 2

you get

C(5,1) * C(5,2) * C(5,3) * C(5,4) * C(5,5)

now

heres where things get nice

wait actually

maybe no

t

yeah it is

so

C(5,1) is 5, C(5,2) is 10 , C(5,3) is 10, C(5,4) is 5 and the other one is 1 which doesnt matter

now

10 * 10 * 5 * 5 = 2500 sqrt(2500) = 50

so perfect square

now from that

i just generalize for C(2n+1,k)

i showed that the rule still holds because using complementary formulas ( C(n,k) = C(n,n-k))

that's how i proved it

now i was wondering if there is a more elegant way of doing it

you COULD also

do

something like this

let me upload it wait

yeah something like this works good

and still a proof

since the first is equal to the one at the end

and there are 2n combinatorics in total

then there will be n pairs

so that will result in perfect square

that's another good proof i think

what are your thoughts?

[calm coral]

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[twilit bramble]

Is the answer to this c?

[calm coral]

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[fluid edge]

Am i correct?

[exotic falcon]

yes

no

[fluid edge]

dude its the same question

[exotic falcon]

lmao ik so why ask again

you were correct the first time when I said yes

[fluid edge]

i just wanted to clarify

Any way is this correct

[kind wagon]

yes

[fluid edge]

nice thx

[calm coral]

@fluid edge Has your question been resolved?

[calm coral]

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[lucid knot]

Joint Variation: One variable increases (or decreases) along with two or more others, all in the same way. More of one thing means more of another (and vice versa).
Combined Variation: One variable is affected by a mix of increases and decreases from other variables. An increase in one thing might be countered by a decrease in another.