help-33

[cold pecan]

What just happened

[static quarry]

you really are better off not knowing

[cold pecan]

All good then

[slate yarrow]

you could guess from the reactions

[buoyant geyser]

Is math 100% logic?

[cold pecan]

It happens can't blame whoever he was . Maths can sometimes destroy the mind condition

But whoever he was should have relaxed

[buoyant geyser]

Yes I was doing this word problem and it wasn’t making any sense in my head so I used AI to understand the steps

[cold pecan]

Maths is all about patience, practice and logic

If you lose one them, it's over

[static quarry]

interesting that removal of the top message due to banning doesn't seem to auto-close the channel

[buoyant geyser]

I got this from someone. This is true right?

[static quarry]

for calculation-based math classes it's reasonably true

like non-proof based calculus

[buoyant geyser]

I see

[buoyant geyser]

What math would it not be true

[static quarry]

well proof-oriented math classes for one

[buoyant geyser]

Oh right

[static quarry]

also things like combinatorics often require clever thinking, and even if you do a lot of practice problems you can still easily be stumped by a new problem that requires a different trick

[buoyant geyser]

Yeah

[buoyant geyser]

For me why I struggle is because there’s so many variations

like

Word problems

[static quarry]

yea

[buoyant geyser]

Sometimes they make the word structure so confusing I don’t know which variables are which

[static quarry]

true, there are a lot of different ways to word the same problem

practice does help with word problems though

[buoyant geyser]

And English comprehension is my weakest ability my literal weakest

[buoyant geyser]

Yeah

[static quarry]

yea all i can really suggest there is to take your time and read it carefully

[buoyant geyser]

yeah

[static quarry]

easier said than done i know

[marsh citrus]

Channel closed

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[buoyant geyser]

This was one of them. I didn’t figure this out until AI showed me how to

[static quarry]

btw you can open your own channel if you want to continue this convo

this one is about to go away

[buoyant geyser]

How do I do that

[static quarry]

!help

[marsh citrus]

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[opaque frost]

how do i graph the lines

[opaque frost]

for c

[marsh citrus]

@opaque frost Has your question been resolved?

[zenith escarp]

Ok so I might be wrong here but bear with me

[opaque frost]

im all ears

[zenith escarp]

Do you know how to make a tangent function

[opaque frost]

not really

ik what it looks like but not how to get there

[zenith escarp]

Gimme a sec gotta pull up my equation

[opaque frost]

okayy

[zenith escarp]

Ok got it

So you know derivatives right?

[opaque frost]

yes

[zenith escarp]

First you need to find the derivative of the function

Once you do you have your slope function

[opaque frost]

-1/2(a)^3/2

?

[zenith escarp]

Yeah

But the power is negative

[opaque frost]

so would i have to move it to the top?

[zenith escarp]

Move what to the top?

[opaque frost]

the power

[zenith escarp]

Yeah

[opaque frost]

so that it can be un-negative

[zenith escarp]

Wait no

[true parcel]

the a part is in denominator

so its right

[zenith escarp]

Wait lemme do it rq

The slope formula aka derivative is just -0.5a^-1.5

Just input your x value into your "a" variable

In the derivative

[true parcel]

^ you will get the slope at that point

[zenith escarp]

Then you got slope

[true parcel]

once u have slope

y = mx+c

then find c using the given point

thats it

u have the tangent line at that point

[opaque frost]

-1/2

[true parcel]

yes

[zenith escarp]

Isn't it y = y1 + m(x -x1)

[true parcel]

that works too

[zenith escarp]

Then your y1 value is just your y value and your x1 value is just your x

[zenith escarp]

Yeah seems easier

[opaque frost]

so 1 =1 + m(1 - 1(1)) ?

[zenith escarp]

Anyways once you got this func just graph it

[zenith escarp]

Yeah

[true parcel]

what

u replaced everything with 1

[zenith escarp]

But your m = -0.5*1^-1.5

[zenith escarp]

That's one of her points

Oh wait no

Yeah he's right

[opaque frost]

huh?

[true parcel]

$y=1+(\frac{-1}{2})(x-1)$

[zenith escarp]

Y = 1 -0.5(x-1)

[elfin berry]

Astar777

[zenith escarp]

Why do you have 2 x values

Yeah

[opaque frost]

y = 3x/2 ?

[zenith escarp]

Where's this coming from?

[opaque frost]

solving the y = 1 + (-1/2)(x -1)

[true parcel]

u did it wrong

do it again

[zenith escarp]

You don't need to simplify first of all

[zenith escarp]

Why would you do it in the first place

Just leave equation as is

And use a graphing calc or smt

[true parcel]

idk, to get it in slope-intercept form?

[zenith escarp]

I mean you can just graph it as is

[true parcel]

i didnt say i would do it, im just sayin maybe thats why

[opaque frost]

would i do the same thing for (4, 1/2) and put them on the same graph

[true parcel]

yes

[zenith escarp]

Yeah

Idk gl

And gn

[opaque frost]

gn

would it be 4 + (-1/2) (x - 4)

?

[true parcel]

why is the slope still -1/2

-1/2 is the slope when x=1 (for point (1,1))

[opaque frost]

oh

[true parcel]

also why is that 4 doing on outside

[zenith escarp]

Gotta input the new x value

[opaque frost]

4

?

[true parcel]

yes

[opaque frost]

-1/2(4)^3/2

[true parcel]

yes

[opaque frost]

so -4

[true parcel]

no

-1/16

,calc (-1)/((2)(4^(3/2)))

[elfin berry]

Result:

-0.0625

[true parcel]

yeah -1/16

[opaque frost]

so how would i plug it into the second equation 😭

[true parcel]

$y = y_1 + m(x-x_1)$

$x_1 ,,\text{and},, y_1$ are the coordinates of the given point

[elfin berry]

Astar777

[opaque frost]

1/2 = 1/2 + -1/16 (x-4)

[true parcel]

wait sorry i wrote the equation wrong 💀

[elfin berry]

Astar777

[true parcel]

fixed

[opaque frost]

y + 1/2 + -1/16 (x-4)

[true parcel]

its = at the starting

y = 1/2, not y + 1/2

[opaque frost]

oh mb

[true parcel]

rest is fine

[opaque frost]

then do i just put both in like desmos and graph it?

[true parcel]

yes

[opaque frost]

okay

i dont think i did it right 💀

[true parcel]

why are u graphing the derivative of the function

[opaque frost]

great question

😭

[true parcel]

[true parcel]

yeah

[opaque frost]

tysm 💀

[true parcel]

[marsh citrus]

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

[opaque frost]

.close

[marsh citrus]

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[marsh citrus]

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[cobalt zealot]

Hi

[cobalt zealot]

Can someone help me with this I'm not too sure where to start

[small vector]

you can use the grid to help kinda

[glass silo]

Remember the properties of the derivative, in relation to the original graph: f increases whenever f' is positive, decreases whenever f' is negative, stationary points of f are where f' is zero...

[small vector]

not exact obviously but e.x. the derivative is high for the leftmost part
because it goes up 6 squares in 1 square right

[marsh citrus]

@cobalt zealot Has your question been resolved?

[marsh citrus]

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[vapid rose]

Inside the triangle ABC there was such a point L that CL = AB and BAC + BLC = 180, a straight line parallel to BC and passing through point L, crosses AC at point K prove that AB = BK

[marsh citrus]

@vapid rose Has your question been resolved?

[vapid rose]

<@&286206848099549185>

[marsh citrus]

@vapid rose Has your question been resolved?

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@vapid rose Has your question been resolved?

[vapid rose]

.close

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[thorny stag]

how can one apply the sinus function to an inequality?

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[sinful thistle]

just apply sin to both sides

,tex .sum diff trig

[elfin berry]

ren

[thorny stag]

it doesn't affect the inequality itself?

[sinful thistle]

yeah but it helps

if it affects the inequality you're doing something wrong

[thorny stag]

what i mean by affecting it is just like when a decreasing function changes the order of the inequality

but the sinus is periodic

so idk

[shadow ibex]

$$ \frac{-3\pi}{2} < \frac{\pi}{2} \text{ TRUE} \
\sin{\left(\frac{-3\pi}{2}\right)} < \sin{\left(\frac{\pi}{2}\right)} \text{ FALSE}$$

[thorny stag]

i see

[elfin berry]

Melvin Eugene Punymier

[thorny stag]

so i can't apply it?

[shadow ibex]

I think you can apply it when the expressions on either side of the inequality symbol evaluate within 2pi units of each other

[thorny stag]

hmmm

what about (sin o f)(x), and f defined as f(x) = x*floor(pi/x)

[shadow ibex]

Restrict the domain of x such that -1/2 <= x < 1/2

...I guess maybe you can expand that range since that's the floor function

pi/x < 4

So I guess

-1/2 < x < pi/4
???

dunno, it's dark and I'm doing this in my head

Ugh

Lower input can't get less than -3, upper input can't reach 4

So you need

pi/x > -3,
pi/x < 4

[thorny stag]

whuuut

.close

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[spice coral]

I din understand how the statement 1 implies x=a

[mellow crag]

let's say x = a + 0.00000001 for example

then |x - a| < epsilon for all epsilon > 0

so 0.00000001 is smaller than all positive numbers

which is false

[spice coral]

Sorry i din get it

I mean how do we end up gettin x=a? Coz epsilon is some positive quantity in both the cases
So x shud always be tending to a, irrespective of whether epsilon is fixed or not

[mellow crag]

no so the first statement says that x is a number such that

whatever positive value of epsilon i give you

|x-a| will always be smaller than that

[mellow crag]

x is a 'fixed number'

so x can't 'tend towards' a

[spice coral]

There's this example given along

<@&286206848099549185>

[marsh citrus]

@spice coral Has your question been resolved?

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@spice coral Has your question been resolved?

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[agile forge]

i need help with this math problem

[agile forge]

bascicaly it says if 2a(a-5)+2b(b-1)+13+0 then calculate (a-1)^2005+(a-8b)^2005

please anybody

[bleak crystal]

mia sabse accha hu madharchiodo bhain ke laudo

[agile forge]

bro i need help not hindi😭

.close

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[wary kite]

which one

[swift bronze]

32

Mb my internet is lagging

[swift bronze]

How can I find using the limit thing for the tangent line

Can you help

<@&286206848099549185>

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[gloomy salmon]

so here they take x^2 out of the top and x out of the bottom

oh wait

nvm i see now lol

.close

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[jagged wharf]

Can someone help me with this Linear Algebra question?

[jagged wharf]

This is my work so far

[sharp vessel]

span(W) is not V

span(W) is just W

W is a subspace

it already contains all linear combinations of vectors in it

and i don't understand your usage of span: in (ii) you are comparing dim and spans of things using < and >

like, u wrote spanW <= dimV

what does that mean?

If A is a set of vectors, then span(A) is the set all of linear combinations of vectors from A

if A is a subspace of V, and not just a set, then span(A) = A since A already contains all linear combinations of vectors from it

dim(V) is the dimension of V, and it is a natural number (since V is finitely generated)

you cant compare it with a set

[jagged wharf]

i see what you mean, idk im kinda lost on this question ngl

sorry

[sharp vessel]

do you know that you can take any set of linearly independent vectors, and complete it to a basis of V?

[jagged wharf]

a basis of V implies it spans to all the vector space right? whereas the independent vectors dont need to satisfy that condition

[sharp vessel]

a basis is a linearly independent spanning set

[sharp vessel]

yes

[jagged wharf]

yeah like in R3 you can start with (1,0,0) and (0,1,0) which are LI but adding (0,0,1) makes it a basis, right?

[sharp vessel]

u should view linear independence and spanning as two different properties
a basis is a set that satisfies BOTH of those

[sharp vessel]

yes

in general, you can start with a vector

and keep adding vectors

you can keep the set linearly independent

but once you pass dim(V) vectors

for example if you have dim(V) + 1 vectors

it cant be linearly independent anymore

and the opposite is true for spanning sets:
any spanning set must have size of AT LEAST dim(V). anything less and it cant span V

dim(V) is exactly the number of vectors for which you can make it both linearly independent and span V

[jagged wharf]

okay that makes sense

so for (i) how do i even start? bc it is given that V is finite dimensional and that W is a subspace of V, then W cannot be infinite dimensional because if the dim(W) > dim(V), W cant be a subspace of V anymore

[sharp vessel]

?

[jagged wharf]

yeah i know that

[sharp vessel]

remember how the dimension is defined:
first, you define what a basis is. then, you show any basis of V must have the same number of vectors.
once you know that the size of a basis is fixed, you can call that the dimension of the vector space.
its defined as the number of vectors in some basis (doesnt matter which, that number is constant)

so show that a basis of W must have size smaller than a basis of V

if you can a basis B of W, what can you do to get a basis of V?

hint: i already told you

[jagged wharf]

well i can expand it to get a basis of V I understand that, but how do i show that one dimension is less than the other? do i just go like this:

[sharp vessel]

Let B = {w_1, w_2, ..., w_k} be a basis of W

you can complete it to a basis of V, by adding the vectors {w_k+1, ..., w_n}

so B' = {w_1, ..., w_k, w_k+1, ..., w_n} is a basis of V

now, since you added vectors, the dimension of V must be at least the dimension of W

either you added none,

or u added some positive amount

this proof is really trivial

a basis of W can be completed to a basis of V since its a linearly independent set

[jagged wharf]

okay but thats for like part (iii)

[jagged wharf]

is this wrong for (i)?

[sharp vessel]

oh wth

how do u know m <= n

thats false

[jagged wharf]

because its a subspace

so it has to be at least less than V

[sharp vessel]

u mean n <= m

u wrote n is dim(W)

[jagged wharf]

oh wait yeah yeah thats what i mean

[sharp vessel]

ok but thats what ur trying to prove

[jagged wharf]

but it is given that it is a subspace of V

[sharp vessel]

yes

that just means its closed under linear combinations and that its a subset of V

[jagged wharf]

and isnt that by definition, a smaller dimension_

[sharp vessel]

no

thats what ur proving

[jagged wharf]

okay so what I did is wrong because of my assumption that n <= m

[sharp vessel]

yes

[jagged wharf]

how is this (you can complete it to a basis of V, by adding the vectors {w_k+1, ..., w_n}) "allowed" but saying that it has to be of equal or smaller size than V as it is a subspace "not allowed", I know that im proving that one has smaller dimension than the other but why cant i use that fact, isnt that a definition of a subset/subspace? thats why it is subspace/set, right?

[sharp vessel]

ok wait

I guess just assume dim(W) > dim(V) and get a contradiction

a basis of W must be linearly independent, but it's also linearly dependent since it has more vectors than dim(V)

[jagged wharf]

doesnt this also work: As V is finite dimensional and W is a subspace of V, then W cannot be infinite dimensional because if the dim(W) > dim(V), W cant be a subspace of V anymore

[sharp vessel]

ok why dim(W) > dim(V) implies W cant be a subspace of V?

[jagged wharf]

because that would imply that there are elements in W that are not in V

[sharp vessel]

why?

dimension is size of basis

this question is asking you to prove something very obvious and intuitive, and thats exactly why you cant make careless assumptions

it's all about unpacking definitions

[jagged wharf]

okay. i understand. would you mind going over why this is allowed?

[sharp vessel]

i mean on a second thought its kinda cheating

just do a proof contradiction

[jagged wharf]

how is it cheating

[sharp vessel]

ehhh

i mean

maybe its not

[jagged wharf]

...

[sharp vessel]

like the theorem im relying on says, that if V is finitely dimensional and you have a set of linearly independent vectors

then you can complete it to a basis of V

but i guess you aren't supposed to use it

even tho its proven very early on

[sharp vessel]

a proof by contradiction like this is better i think

[jagged wharf]

wouldnt it be dependent if it has more vectors than dim W?

[sharp vessel]

?

I don't understand your question

B has exactly dim(W) vectors

if B is a basis of W, i mean

Assume dim(W) > dim(V). Let B be a basis of W.
B is linearly independent since it's a basis, but also linearly dependent since it has more than dim(V) vectors

Contradiction

[jagged wharf]

why more than dim (V) tho

i thought it had to be more than dim W

[sharp vessel]

the number of vectors in B is by definition dim(W)

since we said B is a basis of W

[jagged wharf]

yeah i understand that but where does V come in

[sharp vessel]

last part here

We said B is a basis of W. That also means its some set of vectors in V, which must be linearly independent if |B| > dim(V)

but because we said its a basis (of W), its also linearly independent

we're getting a contradiction

[jagged wharf]

so you are using the fact that W is a subspace of V

[sharp vessel]

yes

now to prove dim(V) = dim(W) iff V = W:
one direction is trivial - if V = W then obviously dim(V) = dim(W)
for the other direction, just prove that a basis of V is also a basis of W, and that means dim(V) = dim(W) since they have a basis of the same size

[jagged wharf]

okay but i dont think i can just write that it is obvious that if V=W then dim(V)=dim(W)

[sharp vessel]

why

they are the same vector space

literally the same thing

replace the letter V with the letter W

dont overthink it xd

[jagged wharf]

i mean, idk... it does say "prove"

[sharp vessel]

one of the directions is trivial

you are still proving the other one

its fine

this is common in some proofs

[jagged wharf]

so for the other direction, if dim (V) = dim (W), then basis of W = Basis of V, and therefore V=W?

[sharp vessel]

well, prove the basis thing

but yes

i mean

no

[jagged wharf]

huh

[sharp vessel]

wait

i mean

not exactly

[jagged wharf]

im lost

[sharp vessel]

ok yes

xd

if B is a basis of V and W

then V = span(B) = W

[jagged wharf]

so: if dim (V) = dim (W), then basis of W = Basis of V, V = span(B) = W, and therefore V=W

[sharp vessel]

yes

prove that they have the same basis tho

take a any basis of V, and prove its a basis of W

[jagged wharf]

wait so i know that the number of vectors must be the same because the dimensions are the same, but how do i show that the elements are the same, that the basis itself is the same

[sharp vessel]

remember that a basis is a set of linearly independent vectors which span the space

let B be a basis of W

prove to me, that B is also a basis of V

this isnt hard to do

[sharp vessel]

also remember that any set of n = dim(V) linearly independent vectors is a basis

from the three conditions:

1. the set spans V
2. the set is linearly independent
3. the set has dim(V) vectors

any two of these is enough to show the set is a basis

[sharp vessel]

thats not very accurate

[jagged wharf]

what can i do to fix it

[sharp vessel]

use this

[sharp vessel]

.

[jagged wharf]

well it has to span V because they have the same dimension and its like saying that B spans all R^3 and because V and W are dim R^3, the span has to be the same for both, and it has to be lin independent as it is the basis of W

.close

[marsh citrus]

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[burnt totem]

i have no idea how to solve question3, i guess i have to use the inequality from question 2 but even i tried to change x for x^2, the integral is disgusting

[marsh citrus]

@burnt totem Has your question been resolved?

[marsh citrus]

@burnt totem Has your question been resolved?

[glacial hedge]

4 other people agree this is a hard problem btw
Maybe eventually someone will come who can do this

[marsh citrus]

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[vague zinc]

This can be done in different ways. Just replace t=x^2/n and you get Beta-function which can be easily counted through Gamma-function. In case you know nothing about these just make several integrations by parts.

[marsh citrus]

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[real warren]

Hi I'm stuck on a pre-cal inverse functions question, its very simple but I just cant seem to wrap my head around it, I know the answer but if someone could give me their perspective on how they solved it that would be really amazing!!. The question is "Determine the restrictions on each function in order for its inverse to be a function. f(x) = x^2.

[glacial portal]

inverse function is just swap x and y in position at the function and isolate the x

[real warren]

oh its juat asking the restrictions, so x>=1

so the answer for this question is x>=0

but I just don't know how I would find that for the restriction

[glacial portal]

no wait

[glacial portal]

is x>0

or x<0

depend on which function you use

because it will be 2 functions

[real warren]

oh wait

the answer says x>0 or x<0

[glacial portal]

yeah bc you need to past the vertical line test to become a function

you can just stat one of the function or both

[real warren]

oh

[glacial portal]

and write restrictions below

[real warren]

alright thank you!

I think i understand now<3

[glacial portal]

np

[real warren]

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[static quarry]

define "mound shape"

[marsh citrus]

@solemn loom Has your question been resolved?

[sleek lake]

sounds symmetric yeah

it could be like this

nah i guess it could be worse

like this

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[balmy ruin]

[harsh falcon]

!status

[marsh citrus]

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

[balmy ruin]

Uh 1 & 4 please

[harsh falcon]

uhm

that contradicts

[balmy ruin]

too scared to sumbit

🦶

[harsh falcon]

x axis is your input, y axis the output
the range of a function is the set of y-values that the function outputs

[balmy ruin]

Dumber pls

[wraith sun]

It's very much wrong

[balmy ruin]

:(

[balmy ruin]

What would be the correct then :v

[wraith sun]

Range means, minimum possible values to maximum possible value

Of y axis, yes

[harsh falcon]

here i marked it

[balmy ruin]

ohh

so

[harsh falcon]

can sometimes be the union of two disjoint sets as well, if the function is not continuous

[balmy ruin]

So what would be the answer..?

😭

[harsh falcon]

thats not the goal here

the goal is you achieving the answer

going by how we defined the range, what would you say would be the answer?

[balmy ruin]

Ehh um

-2 & 10?

[harsh falcon]

what made you get to that answer?

[balmy ruin]

I guessed

[harsh falcon]

in other words, what is yyour reasoning

you shouldn't

[wraith sun]

Range means, minimum possible values to maximum possible value, ** that's actually possible to get from the graph **

[balmy ruin]

oh okay

[wraith sun]

Guesswork won't help you in calculus

[balmy ruin]

😔

Glad I won’t be taking calculus

[wraith sun]

Oh this isn't calculus?

[balmy ruin]

no

im a freshmen, this is algebra 1

……..

[wraith sun]

Oh, we are taught this, with full precalc, calc1, calc2 in the "senior year" in 6 months

[balmy ruin]

sigh

I hate math

Can u still explain how to solve for andwer pls

[still temple]

?

[balmy ruin]

it’s okay I got it lol

[marsh citrus]

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[slim cypress]

how do you simplify this further

[digital robin]

is this boolean algebra?

[slim cypress]

yes

[lofty gyro]

i think i know, but just to check, is the answer ∅

if not, i should be wrong during my thinking process

[slim cypress]

uhh is that even possible in boolean algebra

[lofty gyro]

haha, sorry man my bad then

[slim cypress]

np

[lofty gyro]

lemme rethink

[digital robin]

i completely forgot how to do this

In Boolean algebra, the sum operation is idempotent right?

[slim cypress]

x + x = x if thats what you mean

these are the theorems i was given

[lofty gyro]

i splitted A'B'
into A'B'C +A'B'C'

and then do some operations to the latter terms
i.e.
A'B'C+(A'B'+A'B+AB')C'

would this help

[slim cypress]

uhh let me think

hmm idk, because where would you go from there

idk if theres a way to simplify A'B'+A'B+AB'

[lofty gyro]

yea

you can split A'B' into A'B' + A'B'

[grave dove]

i dont think this can be simplified any more

[lofty gyro]

interesting

[grave dove]

i think its the MSOP

[slim cypress]

i put it in a calculator and it says this

im not sure how they got this

[grave dove]

yeah, what lol

generate the truth tables of the original and the "simplified"

bet they wont be the same :d

[slim cypress]

ill try lol

[grave dove]

original expression

[grave dove]

holy

oh my gosh

[grave dove]

but like how do you go from AB+A' to B+A'

no wayyy

[slim cypress]

yeah i have no idea lol

[grave dove]

that tool is on another level

[slim cypress]

theyre not even equivalent unless i wrote it wrong lol

[grave dove]

the + is v not /\

[slim cypress]

oh

oops

[grave dove]

lets see

[grave dove]

fair enough

its not possible to go from AB+A' to B+A' using the theorems

the tool just happens to know another function which has the same truth table

thats how i would explain it

[slim cypress]

oh okay

ill just not simplify any further then lol

thanks

[grave dove]

if i were on exam and i ended up with this

i would stop right here, nothing to do anymore

[slim cypress]

👍

[grave dove]

its crazy oh my god

[grave dove]

thank you too!

[slim cypress]

.close

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[high edge]

how do i determine the new integral boundaries for a non trivial jacobian transform?

[high edge]

What if I have $\int_{0}^{2}\int_{0}^{1-\frac{x}{2}}(x+y) \mathrm{ d}y\mathrm{d}x \text{ with } u=x+y \text{ and } v=x-2y$ for example?

[elfin berry]

Jill ♡

[main idol]

Calculate the Jacobian and do the substitution

[high edge]

yea but how do i find the new boundaries?

[main idol]

Solve for x and y in terms of u and v

[high edge]

im still confused

i got $\int_{0}^{2}\int_{0}^{u}\frac{u}{3}\mathrm{d}u\mathrm{d}v$ which probably isnt correct

[elfin berry]

Jill ♡

[high edge]

idk what im doing wrong

[marsh citrus]

@high edge Has your question been resolved?

[main idol]

Do an example first. 14.7.4 here
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_212_Calculus_III/Chapter_14%3A_Multiple_Integration/14.7%3A_Change_of_Variables_in_Multiple_Integrals_(Jacobians)

[high edge]

Oke thanks

.close

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[lyric kelp]

is it

i figured no but i forgot how i figured it out

[glass silo]

I'll make you figure out why

[sand fable]

[glass silo]

Oh

Well, we've said that A must be invertible, can you name what its inverse is?

[lyric kelp]

if i have to guess

is it cuz A is an identity matrix

and therefore

it is invertible

[glass silo]

A doesn't need to be the identity matrix

What does it mean to be invertible?

[lyric kelp]

have a determinant?

[glass silo]

I mean, that may be one way you have it defined-

Really, $A$ is invertible if you have another matrix $B$ such that $AB = BA = I$ (with all the appropriate dimension stuff of course)

[elfin berry]

@glass silo

[glass silo]

But we can use determinants if you want (we don't need to though!)

[lyric kelp]

ah yes this ofc

actually how i thought was

cant haev a zero col or row

therefore A has to be invertible

for A^3 to be identity

😭

[glass silo]

I mean, that A^3 can't, or that A can't?

[lyric kelp]

A

A must not have a zero col or row

[glass silo]

Well, that isn't an incorrect statement, but bear in mind that not having a zero row or column isn't enough for invertibility...

e.g. the matrix $\pmqty{1 & 1 \ 2 & 2}$ is not invertible but has no zero rows/columns

[elfin berry]

@glass silo

[glass silo]

Let me give a hint: The fact that $A^3$ is the identity is saying that $A^3 = I$...

[elfin berry]

@glass silo

[marsh citrus]

@lyric kelp Has your question been resolved?

[lyric kelp]

ahhh ty !

i got it

[marsh citrus]

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[lyric kelp]

had to do the multiplication

[marsh citrus]

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[timber imp]

How do we know ab and Ed parallel

[marsh citrus]

@timber imp Has your question been resolved?

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@timber imp Has your question been resolved?

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@timber imp Has your question been resolved?

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[sonic stag]

I'm currently stuck on 8b and don't really know how to approach this question or what to do

[marsh citrus]

@sonic stag Has your question been resolved?

[marsh citrus]

@sonic stag Has your question been resolved?

[vital oracle]

50/36 = 1.38888...
so (b) is 1

[marsh citrus]

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[vague sigil]

can someone please help with this question, part iii. the answer is 103

[lofty gyro]

have you done part i and ii?
please show your work

[cloud field]

<@&268886789983436800>

[vague sigil]

okay but it’s pretty messy

[lofty gyro]

it's alright 😊

[lofty gyro]

reading

[vague sigil]

i understand how its a right angle, bc 53 + 37

i just have no idea how to do the bearing

and i doubt the 2nd compass is correct

[lofty gyro]

are you allowed to use sine cosine?

[vague sigil]

yes

i also tried the sine rule: sin53/16 = sin/7

but it got me this rlly off decimal

[lofty gyro]

you can use cosine rule

brb 10mins, (feel free to ask others meanwhile)😁

[vague sigil]

ohhh

first would i need to find what one side equals

ill try that

[marsh citrus]

@vague sigil Has your question been resolved?

[lofty gyro]

back

[lofty gyro]

anything done so far? 😄

do ping me if you need further help for this question
@vague sigil

[vague sigil]

no I still got it wrong:(

i really dont get it

[lofty gyro]

oh okay, let's try together

[vague sigil]

yup

[lofty gyro]

did you get this red one?

[lofty gyro]

and have you learnt this?

if not, we can use another method

[vague sigil]

ohh, no i did it in a very wrong way

that makes much more sense

[vague sigil]

yes i have

[lofty gyro]

that's great!

[vague sigil]

isn’t it 2ba?

nvmm

i got 66

[lofty gyro]

btw, i got this:

oh, i just found that you can also use sine law also, omg.... there are so many methods, that's why we can't think of them

,w arccos(7/sqrt(16^2+7^2))/pi*180

[elfin berry]

Results provided by WolframAlpha

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

Requested by biscuityxd

[lofty gyro]

is it this? 66.37062226934318335785168353975...

[vague sigil]

the text book says the answer is 103

[lofty gyro]

if it's correct,we can find theta, which is the final step

to be exact, we need to find the bearing of the pigeon from home.

so it will be 37+66.37062226934318335785168353975...
which is around 103°

[vague sigil]

why do we add 37?

[lofty gyro]

since we have 37 and the angle we found is the angle in thr triangle at H

[lofty gyro]

green angle plus the orange one = bearing

[vague sigil]

Is the orange 66?

[lofty gyro]

yes!

that's correct!

(approximately of course)

[vague sigil]

ohhhh that makes so much sense

omg thank you so much for all your help

[lofty gyro]

you're welcome!

[marsh citrus]

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@vague sigil Has your question been resolved?

[marsh citrus]

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[vernal forge]

i have this

[vernal forge]

how do i do a?

i would do it by a comparison with 1/n but suppose i can't do that and just go off of the definition

in the previous exercise it involved essentially inverting the sequence to get N in terms of eps but i can't really invert n! so idk what to do

[glass silo]

How did you plan to compare?

[vernal forge]

$\frac{1}{n!} < \frac{1}{n}$

[elfin berry]

artemetra

[vernal forge]

for n>=3

[vernal forge]

is that not good enough

cuz the N you get with 1/n is definitely larger (given a fixed epsilon) than the one you get for 1/n! but it still satisfies the conditions in the definition

[glass silo]

It is good enough in fact you could make the inequality non-strict for any strictly positive n

[vernal forge]

is there a way without the comparison?

[devout mauve]

the key about analysis is doing comparisons

[vernal forge]

fair enough

[devout mauve]

this is a feature not a bug

the key in epsilonics is being able to make good comparisons

[vernal forge]

okay okay

[devout mauve]

that dont destroy your behavior but make your terms easier to deal with

[vernal forge]

okay and for part b

i was thinking of comparing it with $\left(\frac{5}{6}\right)^n$

[elfin berry]

artemetra

[vernal forge]

but i am not sure if we can use the fact that $n! > 6^n$ for all sufficiently large $n$

[elfin berry]

artemetra

[devout mauve]

prove it by induction

[vernal forge]

cuz we are essentially proving that $n! > 5^n$ for all suff. large n

[elfin berry]

artemetra

[vernal forge]

fair enough

[devout mauve]

well just > is not enough

n^2+1 > n^2 but their quotient does not go to 0

[vernal forge]

oh yeah

i don't have a formal notion of an "asymptotically faster growth"

but i get your argument

thanks

[devout mauve]

try showing only n! > 5^n and then just using that and comparison

for the 5/6 you would also need to show that that converges to 0

so actually as an exercise just do both

[marsh citrus]

@vernal forge Has your question been resolved?

[vernal forge]

i see

okay makes sense

thank you

[marsh citrus]

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[abstract dust]

Can you please give me problems about evaluating functions

[marsh citrus]

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[neat elk]

each year money is despoited in an account which pays an annual interest rate of r. the balance is multiplied bya growth factor x =1+4. at the start of year 1, 500 is deposited, then 200, then 600. write an expression for the value of the account in terms of the growth factor x

[marsh citrus]

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.

[neat elk]

.close

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[night lion]

my approach was

[night lion]

taking x outside

and do by parts twice

is there any other simple way u can think of?

[lethal bridge]

u-sub and then partial fractions might work

[night lion]

what sub

[zenith escarp]

You make a certain part of the function equal a variable usually its a "u"

Makes things simple sometimes

Likes here the 1-x^3 could be set to equal u to use the integral formulas easier and then simplifying

[night lion]

that would give me -2x^2 dx= dt

how will that help

.close

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[glass perch]

Find a system of infinitely many linear independent functions

[sacred rivet]

There is no finite basis that can span the entire space of functions from (0,1) to R tho?

[mellow crag]

if there is no finite basis then it's not finite-dimensional

and if it's not finite dimensional then it has to be infinite dimensional

[sacred rivet]

So this explains why it's infinite-dimensional?

I'm not sure if what I said above is accurate tho

[glass perch]

And if you find infinitely many linear independent functions (0,1)-> R we cannot have a finite basis

[sacred rivet]

Could we also say that F(0,1) is infinite-dimensional because it contains all polynomials (along with many other types of functions)?

[glass perch]

If you have already shown that the set of all polynomials are linear independent then yes.

If you don't, there are far better examples

[mellow crag]

a really easy set of examples are just

f(x) = 0 for all x other than x=0.5

etc.

[sacred rivet]

So that just leads back to this.

[devout mauve]

ehh, polynomials are great. also apply for other classic function spaces

not sure which ones you want to go for but the next one I would think of do not apply to most nice spaces

[glass perch]

Yes, but here it is overkill. Especially if you don't have the result at hand

[devout mauve]

trivial result about nonzero polynomials having finitely many roots

[sacred rivet]

Is it sufficient to just say that it's infinite-dimensional because no finite basis can span the entire space of functions from (0,1) to R then?

[devout mauve]

for completeness, the next ones I would think off are ||indicator functions|| but those dont apply to C^k or L^p spaces

[glass perch]

Yes, you're right. That's a far more efficient way of showing it then the one I was aware of. I'd just said what LY gave.

[devout mauve]

btw should be set of all (monic) monomials instead of set of all polynomials

[mellow crag]

whilst that is true, if a problem asks you to show that something is infinite-dimensional you'll basically have to do what we've done here

[sacred rivet]

Gotcha

I was also wondering about the vector space R itself with standard operations.

Would the dimension for this simply be 1? Because any real number can be written as a scalar multiple of the single basis vector 1, which spans R. Thus, since the basis for R has one element, the dimension of R is 1 as well?

[mellow crag]

yes but always remember that when we talk about vector spaces

we're talking about a vector space with respect to a scalar field

[mellow crag]

R is a R vector space, in which case it has dimension 1

but it's also a Q vector space, in which case it has infinite dimension

[sacred rivet]

Oh, so it's only 1-dimensional because in this case I was told R itself, with standard operations is a vector space over R and since any real number can be written as a scalar multiple of 1, it's 1-dimensional?

[royal sable]

yes

[sacred rivet]

In this case could R also be considered as R^1 which is one-dimensional because the standard basis consists of 1 vector?

[mellow crag]

yeah

[sacred rivet]

Gotcha, thanks a lot guys!

.close

[marsh citrus]

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[devout mauve]

which is pretty much the same argument because the standard basis vector would be 1 again. or (1) if you want to write it as a vector

[marsh citrus]

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[strong dagger]

So I got this to rref but i’m confused on the basis part. I mapped the matrix to a row vector so would my basis be the first two rows?

[strong dagger]

and then map it back to the original 3 x 2 matrix?

I believe i got it

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[marsh citrus]

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[half dagger]

How do I find the angle here?

[marsh citrus]

@half dagger Has your question been resolved?

[royal sable]

Find r' and use any formula for the angles. For example the one that involves scalar product

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[topaz rapids]

I need help with a

[topaz rapids]

This is my attempt but I got stuck

[marsh citrus]

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[marsh citrus]

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[topaz rapids]

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[sacred crown]

how do you plot |z|=Rez+2

[main idol]

use z = x + iy and |z| = \sqrt{x^2 + y^2}

solve for y in terms of x

[sacred crown]

ok

like a half sideways parabola

[main idol]

,w plot y = sqrt((x+2)^2 - x^2)

[elfin berry]

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[main idol]

there are two branches, but yea

actually is this a conic

,w plot y^2 = 4x + 2

[elfin berry]

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[main idol]

no you're right it's a parabola

i guess parabolas are conics too

alright i'm done waffling. sorry for waffling

[sacred crown]

no worries!

thank you lol

for smth like |z-1|+|z+1|=7

idk if I should rewrite z as x+yi

hm wait

im gonna try and see

[static quarry]

standard idea when you have the sum of two abs vals, move one of them to the other side

then square

[sacred crown]

I didn’t square but this is same?

[static quarry]

if only it were that easy...
you're missing the cross term on the RHS

[sacred crown]

right

idkk hm

[static quarry]

your last line should be:
$$(x-1)^2 + y^2 = 7^2 - 14|(x+yi) + 1| + (x+1)^2 + y^2$$

[elfin berry]

Bungo

[static quarry]

that middle term starting with the 14 is what you missed

also note the sign of the last two terms

assuming that you agree with this, rearrange to get the remaining abs val term on one side and everything else on the other side

and square again

also note that you can do a fair amount of simplification before that

[marsh citrus]

@sacred crown Has your question been resolved?

[static quarry]

can multiply out the (x-1)^2 and (x+1)^2, and simplify more

[sacred crown]

righttt

ok

I mean now I can plot it on desmos, but I wouldn’t have known the shape myself

[static quarry]

you can simplify the stuff inside the square root

it's still gonna be of the form sqrt(some quadratic) though

[sacred crown]

thank you!

[marsh citrus]

@sacred crown Has your question been resolved?

[marsh citrus]

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[terse vine]

This is for inverse function and its a super easy question but im just confused

[terse vine]

why is it not:

x/-1 = y

im confused howh the inverse is the same

This is for inverse function and its a super easy question but im just confused

[grave dove]

x/-1 = -x

[terse vine]

ya but if u divide by -1 it remove the - beside the x?

bc -1/-1 = 1

so im confused why the inverse is the same as before

[grave dove]

lets explain the process with another function

maybe it'll be clearer

[terse vine]

yes plssss tysm

[grave dove]

$y=\frac{1}{x+2}$

[elfin berry]

Emily

[grave dove]

the first step to invert this function is to make x the subject

like write it as x = ...

$x+2=\frac{1}{y} \ x = \frac{1}{y} - 2$

[elfin berry]

Emily

[grave dove]

are you following Mila

[terse vine]

yes

[grave dove]

are you okay with that!

[terse vine]

wait why cant u just switch y and x

make it x = 1 / (y + 2)

thats how we were learning it

[grave dove]

yeah, both work

[terse vine]

can we do this one?

[grave dove]

i usually write x in terms of y, then switch

[terse vine]

😭

[grave dove]

yeah, it will be the same

$x+2=\frac{1}{y} \ x = \frac{1}{y} - 2 \$ switching x and y $\implies y=\frac{1}{x}-2$

[elfin berry]

Emily

[terse vine]

mhm

[grave dove]

you want to start by switching, then write y in terms of x?

$y=\frac{1}{x+2} \$ switching x and y $\implies x=\frac{1}{y+2} \ y+2 = \frac{1}{x} \ y = \frac{1}{x}-2$

[elfin berry]

Emily

[grave dove]

see, they both work

[terse vine]

ok

[grave dove]

now back to our original question

y = -x

using your method, we start by switching x and y

it becomes: x = -y

[terse vine]

right

[grave dove]

then multiply both sides by -1

-x = y

or, y = -x

so it remains the same

for real :p

follow the steps so you dont get lost

[terse vine]

okok got it

thank u

[grave dove]

1. take the original function, switch x and y

2. write y in terms of x

that will be your inverse function

[terse vine]

mhm

ok tys m

tysm

im gonna relook at this and copy it down my notes :)

tyy

ill close the chanel so other ppl get a chance to ask questions too lol

.close

[marsh citrus]

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[opal egret]

can you help with this

[main idol]

#❓how-to-get-help

[marsh citrus]

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[stark field]

guys does someone understand what this X^T means here

[main idol]

transpose

[stark field]

This only definition i cant comprehend

[main idol]

most math books just write $X \cdot Y$ instead of $X^T Y$

[elfin berry]

riemann

[stark field]

hmmm that is insane

And the part "assigns a real number to each pair of vectors X,Y" --> this means product of corespondent elements of vectors assigns to R^n some number or will equate to some number?

if i undertod corect

[main idol]

i don't know how abstract you wanna get, but dot products assume you have a field (in this case R) and the product of two numbers (x1 * y1, call this z1) is again itself a real number, and the sum of two real numbers (z1 + z2) is again a real number

i recommend being less abstract and just deal with numbers

[stark field]

[main idol]

have you seen this before?
https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/dot-cross-products/v/vector-dot-product-and-vector-length

[stark field]

hmm i watch now very nice

but im going insane rn going hard math

studying 24/7

if i pass i will make memorial for this discord 👍

[main idol]

[marsh citrus]

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[silver veldt]

struggling to process how i’m gonna graph this. forgive me for my stupidity i’m just genuinely at a loss

[marsh citrus]

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[marsh citrus]

@silver veldt Has your question been resolved?

[plain wing]

for x less than 0 draw 1/4x-3

for x in 0,4 draw y=2

for x greater than 4 draw x^2