you should have 5te^(-t) @halcyon stratus
#help-38
dark sundial
dark sundial
@halcyon stratus
dark sundial
ok
in the 4th question you have to use a power series
and it will end after a few terms
probably 2 or 3 terms of the series
halcyon stratus
I’m back
dark sundial
great
halcyon stratus
Do u think we have time
dark sundial
How much time left?
halcyon stratus
2 hours more
dark sundial
What was the expression for y(t) in the 2nd?
halcyon stratus
halcyon stratus
I directly substituted infinity
dark sundial
but that will give you an indeterminate result
halcyon stratus
No worries
halcyon stratus
Thoughts on 3a
dark sundial
I belive he is using Alpha and Beta instead of A and B
or he expects you to do so
dark sundial
Yeah, and you need to find them
this contradicts my previous hypothesis
he is using A, B, Alpha and Beta
halcyon stratus
So basically we need to find equations in terms of alpha and Beta
dark sundial
yeah
dark sundial
seems to be correct
halcyon stratus
What about b
dark sundial
let me think
halcyon stratus
.
dark sundial
Do you understand that?
halcyon stratus
Where did the limit go
halcyon stratus
Right
dark sundial
applying it to both terms, separately, we get to that expression
I mutliplied the whole expression by e^(-t)
to isolate the exponential
in a way that I could make it go to 0, as t->infinity
halcyon stratus
So what are the conditions for alpha and beta
halcyon stratus
Basically the 1/e^4t will always become zero
No matter what conditions we give to alpha and beta
dark sundial
no
no
it will get become zero, but the other term won't
that will create a false statement
halcyon stratus
it will get become zero, but the other term won't
Yup that’s what I’m saying
The one term will always be zero
The other will have a condition
dark sundial
we need to get rid of both exponentials
in order to determine the values of alpha and beta
halcyon stratus
How do I confirm this solution in graph?
dark sundial
that make this y go to 0, as t becomes larger and larger
dark sundial
dark sundial
so 4*alpha=-beta
this makes the first term disappear
and you're left only with the second term
the decreasing exponential
dark sundial
let's do it
dark sundial
Well, I've been in undergrad for quite a long time, changed major from computer engineering, to physics, and gave up to go back to computer engineering.
since 2017 at university
at least the public ones are tuition free here in Brazil
and I like to teach
halcyon stratus
I’m doing computer science
Honours in computer science
Double minor in math and physics
halcyon stratus
Damn how do I do this
halcyon stratus
Bro I don’t understand a thing
halcyon stratus
Oh yeah I remember Taylor series
dark sundial
it's just equivalent to a taylor series
halcyon stratus
Lmao so he just used summation
dark sundial
yup
halcyon stratus
My question is only asking for order 1
halcyon stratus
I can’t do it bro
dark sundial
why?
halcyon stratus
How do we find the general solution
halcyon stratus
I asked ChatGPT
dark sundial
it is error prone
halcyon stratus
The solution kinda makes sense
To show that ( y = x ) is a solution to Legendre's equation of order 1 and find the general solution for (-1 < x < 1), let's start by substituting ( y = x ) into the Legendre's equation and then solve for the general solution.
Legendre's Equation of Order ( n ):
[ (1 - x^2)y'' - 2xy' + n(n + 1)y = 0 ]
For ( n = 1 ):
[ (1 - x^2)y'' - 2xy' + 1(1 + 1)y = 0 ]
[ (1 - x^2)y'' - 2xy' + 2y = 0 ]
Step 1: Verify ( y = x ) as a Solution:
Substitute ( y = x ) into the equation:
First derivative:
[ y' = 1 ]
Second derivative:
[ y'' = 0 ]
Substitute ( y, y', ) and ( y'' ) into the equation:
[ (1 - x^2)(0) - 2x(1) + 2(x) = 0 ]
[ 0 - 2x + 2x = 0 ]
[ 0 = 0 ]
This confirms that ( y = x ) is a solution.
Step 2: Find the General Solution:
To find the general solution for (-1 < x < 1), we need another linearly independent solution. Since ( y = x ) is a solution, let's use the method of reduction of order.
Reduction of Order:
Let ( y_1 = x ). We seek another solution of the form ( y = v(x) x ), where ( v(x) ) is an unknown function.
Substitute ( y = v(x) x ) into the Legendre's equation:
Derivatives:
[ y = vx ]
[ y' = v'x + v ]
[ y'' = v''x + 2v' ]
Substitute into the equation:
[ (1 - x^2)(v''x + 2v') - 2x(v'x + v) + 2(vx) = 0 ]
[ (1 - x^2)(v''x + 2v') - 2x^2v' - 2xv + 2vx = 0 ]
[ (1 - x^2)v''x + (1 - x^2)2v' - 2x^2v' = 0 ]
[ (1 - x^2)v''x + (2 - 2x^2)v' = 0 ]
[ x(1 - x^2)v'' + 2(1 - x^2)v' = 0 ]
Simplify and divide by ( x(1 - x^2) ):
[ v'' + \frac{2}{x} v' = 0 ]
This is a first-order equation in ( v' ):
Let ( u = v' ):
[ u' + \frac{2}{x} u = 0 ]
Solve using an integrating factor:
[ \mu(x) = e^{\int \frac{2}{x} dx} = e^{2 \ln |x|} = x^2 ]
[ (x^2 u)' = 0 ]
[ x^2 u = C ]
[ u = \frac{C}{x^2} ]
Since ( u = v' ):
[ v' = \frac{C}{x^2} ]
[ v = \int \frac{C}{x^2} dx = -\frac{C}{x} ]
So the second solution is:
[ y_2 = v(x) x = -\frac{C}{x} x = -C ]
solid kilnBOT
Flix 🇫🇷
LaTeX source sent via direct message.
```Compilation error:```! You can't use `macro parameter character #' in internal vertical mode.
l.51 #
## Legendre's Equation of Order \( n \):
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
If you're in the wrong mode, you might be able to
return to the right one by typing `I}' or `I$' or `I\par'.```
halcyon stratus
Can you write step by step solution and send me please
dark sundial
halcyon stratus
Thanks
halcyon stratus
I lost hope
dark sundial
ok
halcyon stratus
Are u willing to try
halcyon stratus
this video has the complete solution
His video is too long
halcyon stratus
Makes sense why he gave it as the last question
dark sundial
It's been some time since I took the calculus 4 course
halcyon stratus
It's been some time since I took the calculus 4 course
It’s okay you helped me a lot bro
dark sundial
it's been already answerd at math stack exchange
I would trust it more than chat gpt
dark sundial
If you're working while atteding university, that would explain a lot, like your difficulties, and the scarce time to solve the exercises.
halcyon stratus
I understand now
dark sundial
do it
dark sundial
yaaay
dark sundial
to do what?
halcyon stratus
to do what?
Basically we know y1 and I found y2. That’s the solution
dark sundial
yeah
dark sundial
none, I don't even watch football
halcyon stratus
You’re a nerd fr bro
dark sundial
It was my pleasure to help you
trim joltBOT
trim joltBOT
Available help channel!
wraith hinge
Solve the limit if $[\cdot]$ denotes the greatest integer function
$$\lim_{x\to \infty}x^{\frac{1}{[x]}}$$
solid kilnBOT
pun pun
cyan zinc
x^(1/x) is a continuous function over the reals so i don't think there's a reason that the continuous limit should be any different than the discrete
so just take the limit of x^(1/x)
wraith hinge
ok yea im so dumb i was thinking infinity raised to the power 0 is 0
😭
it should be 1
thanks for the help anyway
.close
trim joltBOT
trim joltBOT
Available help channel!
marsh forum
how does this imply the only way to write zero is if all elements are zero
marsh forum
like if I'm having two subspaces
can't one have the element (1,1...) and the other (-1,-1...)
left oriole
sure
but subspaces are closed under scalar multiplication
so if a subspace contains (1,1,...) it also contains (-1,-1,...) and vice versa
cosmic gate
hello
cosmic gate
was doing this question, and i was slightly confused about the second assertion (second dot point)
wait i got it lmao
you essentially equate two planes together
and after pulling everything to one side, you get a new plane equation that passes through the origin
.close
trim joltBOT
twilit latch
you essentially equate two planes together
wait can you elaborate on this
that's not what I got from reading it
cosmic gate
so i assumed that a has arbitray components a1, a2 and a3
and same thing with b
and the same thing with x
as a dot x = 0, we can do component wise multiplicaiton to get a1x1 + a2x2 + a3x3 = 0
same thing with b
so you get a1x1 + a2x2 + a3x3 = b1x1 + b2x2 + b3x3
twilit latch
can you reopen this
cosmic gate
and then pulling the b's to the other side, you get an equation for the plane
.reopen
trim joltBOT
✅
cosmic gate
ok where does it break down?
twilit latch
you are using 0-0=0 to justify why this is a plane
did you use the fact that one of the vectors is 0?
we are discussing the second bullet point right?
twilit latch
the 2nd is if "exactly one of a or b is 0"
cosmic gate
oh frick
im really sorry
i meant the third bullet point
very sorry for the confusion
twilit latch
even in that case this reasoning is flawed
twilit latch
so you get a1x1 + a2x2 + a3x3 = b1x1 + b2x2 + b3x3
you are essentially reducing your two equations of contraint into one equation of constraint
you need to use the fact that they are parallel
cosmic gate
hmm i guess you could reformulate the second equation where you replace the b's with lamba * a's
wanton rune
i think you can simplify this down using inner product distributivity property
twilit latch
hmm i guess you could reformulate the second equation where you replace the b's ...
do you want to try again to see if you can come up with a good explanation
wanton rune
this makes proving parallel case extremely nice :3
twilit latch
tag me if you want me to look over it
cosmic gate
ok i think i got it
so we have the first equation, a dot x = 0
now we consider the second equation
b dot x = 0
since they are parallel, we can say that b = lambda dot a where lamba is an element of the real field
thus we can reforulate the equation to be lambda * a dot x = 0
we can divide both sids by lambda to get our previous equation a dot x = 0 which is a plane that passes through the origin
@twilit latch would this be correct reasoning?
cosmic gate
very cool
i sometimes get formulate my proofs incorrectly so this was very educational
one more question
regarding the last dot point
i thought that W must only be the origin
but the solutions imply that its a line passing through the origin
twilit latch
so we want to think about this in two important ways
geometrically
and also algebraically
what does a dot x = 0 tell about how vectors a and x are related
cosmic gate
they must be orthogonal
twilit latch
right
cosmic gate
oh meaning that a and b must be parallel
twilit latch
nope
that would be true in R^2
in R^3 we have a three dimensional space
you want to think about problems like these are in terms of dimension/rank
so what is the span of two lineally independent vectors
what is the rank of that vector space generated by 2 linearly independent vectors
cosmic gate
so what is the span of two lineally independent vectors
should it be 0?
cosmic gate
doesn't it refer to the linear combinations of the vectors a and b
cosmic gate
oh shit i got confused by the definition of linearly independent vectors having some constants that make them 0
anyways, the span can be represented by a 3 by 2 matrix
twilit latch
of what rank
cosmic gate
....rank 2
cosmic gate
i had to google search this
but apparently it tells us the dimension of the vector space generated
twilit latch
ok I'm going to answer this so we can move on
rank 2 geometrically is a plane
rank 1 is a line
point is asking what is orthogonal to a plane
(it is more than 1 point)
twilit latch
no
cosmic gate
ah yeah i was thinking about the vectors in the plane
twilit latch
unless we have more than 3 dimensions
then it's possible
R^n can have at most n linealy independent vectors
ok algebraically
every unique linear equation
adds 1 dimension of constraint
ie your solutions will have 1 less dimension than the vector space they live in per unique equation
cosmic gate
sorry i don't really understand this notion of constraint
twilit latch
let's go back to high school level examples
how many unique equations do you need to solve for two variables in R^2
cosmic gate
2
twilit latch
how many equations do you need to specify a line in R^2
cosmic gate
1 right?
cosmic gate
1 equation
cosmic gate
is this in R2?
twilit latch
you put no constraint on the possible solutions
cosmic gate
like the plane equation ax1 + bx2 + cx3 = 0?
twilit latch
which actually contains no ifnormation
twilit latch
like the plane equation ax1 + bx2 + cx3 = 0?
this is R^3
cosmic gate
okay i thought we were operating in R2 my bad
cosmic gate
okay
twilit latch
you are kind of seeing what is happening, no?
in R^3 we need 1 equation for a plane
in R^2 we need 0 equeations for a plane
In R^3 we need 2 equations for a line
in R^2 we need 1 equation for a line
in R we need 0 equations for a line
so before we have any equations contraining what our point can be
it can be anything in the vector space
since we know nothing about it
cosmic gate
but there are two equations, meaning that it has to be a line?
twilit latch
for point 4 yes
so if the vector has n dimensions
and you have k unique constraints/linear equations
then the solutions are n-k dimensional subspace
does this make sense
each equation/contraint if it is unique eliminates a dimension
cosmic gate
yeah i can kinda see the logic
twilit latch
actually for equations of the from a dot x=0 this just rank of the null space
you put a b whatever you have that is being dotted with x in a matrix
and then calculate the rank of that matrix
have you heard of rank nullity theorem
cosmic gate
no
twilit latch
have you looked at null spaces at all?
twilit latch
indeed
cosmic gate
isn't that the kernel
twilit latch
they are the same thing
twilit latch
something about the solutions to x for Ax = 0
so you would put a and b as the rows of A
which is just the equation you had a dot x= b dot x=0 in matrix form when solving for the nullspace
twilit latch
damn there is so much to learn lmao
linear algebra is very cool
still my favorite part of math
twilit latch
feel very dumb right now lol
one they it will all click
it is kind of like that
just takes some more exposure
learning linear algebra is non-linear lol
cosmic gate
lmao
btw, if you don't mind, could you explain why planes with dimensions higher than 3 can be orthogonal
or if that is too time consuming, a resource will do
just something to itch my brain cause i distinctly rememver learning something very similar to it
twilit latch
2+2=4
ok I will elaborate
but that is literally what it is
a plane takes needs two dimensions
to have two orthogonal ones you need at least 2+2=4 dimensions
cosmic gate
hmm i remember that in R4 you can have the planes intersect at one point
twilit latch
yes this is it
cosmic gate
hmm that makes sense
twilit latch
if you want an explicit example the plane spanned by [1 0 0 0]^T and [0 1 0 0]^T is orthogonal to the one spanned by [0 0 1 0]^T and [0 0 0 1]^T
cosmic gate
in R3, would it be correct that two planes just always intersect
but in R4, you can have them never intersecting
twilit latch
in R3, would it be correct that two planes just always intersect
planes can be parallel, but would no longer be vector subspaces
like the plane z=1 and z=2
twilit latch
good idea
cosmic gate
thanks for all your help arsam really appreciate all the help
twilit latch
you're welcome!
cosmic gate
.close
trim joltBOT
trim joltBOT
Available help channel!
cosmic gate
so what is the span of two lineally independent vectors
sorry arsam for reopening this, i just was confused why we are looking at the span
cosmic gate
.reopen
twilit latch
no worries
by knowing where your points aren't, you find out where they are
we were looking at what object the solution vectors had to be orthogonal to
if x is orthogonal to a and b
it is orthogonal to the span of a and b
cosmic gate
so essentially, if x is orthogonal to both a and b, it must be orthogonal to their plane
cosmic gate
so we can the plane of a and b being lambda times a + mu times b
twilit latch
so essentially, if x is orthogonal to both a and b, it must be orthogonal to the...
of course assuming a and b are linearly independent and actually span a plane
cosmic gate
so we can the plane of a and b being lambda times a + mu times b
times everything by x, we get lamda * a dot x + mu * b dot x
as a dot x and b dot x are 0, the sum must also be 0 meaning that x and the span are orthogonal
twilit latch
perfect
twilit latch
np
cosmic gate
.close
trim joltBOT
trim joltBOT
Available help channel!
random niche
$\int_1^{2} x \ln x dx \$
here would I start with the antiderivative of ln x is x ln x - x + c
solid kilnBOT
Tomi
clear cloud
Nah integrate by parts with u=lnx and v'=x
livid thunder
Nah integrate by parts with u=lnx and v'=x
u would get xlnx-x
random niche
shouldnt it be u = x v'= ln x dx
livid thunder
oh i didnt see the question properly
clear cloud
u would get xlnx-x
How ?
livid thunder
shouldnt it be u = x v'= ln x dx
typically u let dv equal the easiest thing to integrate
clear cloud
Since you find antiderivative of x its x^2/2
And you always want to take the derivative of ln
random niche
typically u let dv equal the easiest thing to integrate
but like when working with definite integrals and integration by parts, it is the same thing the only difference is that I would need to evaluate it between the boundaries.
clear cloud
Dont you have to jusitfy the antiderivative of ln ?
Or it is allowed to take it out of your cap like a known thing ?
random niche
Or it is allowed to take it out of your cap like a known thing ?
what? i dont know what you are saying
im trying to solve that integral
that i sent
clear cloud
Yeah but you want to take v' as lnx
split chasm
it works both ways here
(not always the case)
but the way you're approaching it is more complex
livid thunder
if u let dv=lnx then you'll end up with xlnx-x as ur 2nd integral, which is a recursion which is fine but idk why u would complicate it
random niche
okay but I end up having these 4 variables, dv = x dx, v = x^2/2, u = ln x, du= 1\x dx
clear cloud
okay but I end up having these 4 variables, dv = x dx, v = x^2/2, u = ln x, du= ...
Yeah
Put it in formula
split chasm
dv = x dx
but yeh
random niche
Put it in formula
$\ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \frac{1}{x} dx$
solid kilnBOT
Tomi
clear cloud
Ye
brittle barn
and now this is trivial
random niche
Ye
$= \frac{\ln x ^2}{2} - \frac{1}{2} \int x^2 \frac{1}{x} dx \$
$ =\frac{\ln x ^2}{2} - \frac{1}{2} \cdot \frac{x^3}{3} \cdot \ln x \$
solid kilnBOT
Tomi
random niche
Ye
is this correct to this point? now I just have to simplify and evaluate it between the bounds.
split chasm
no
random niche
oh that is another ibp there
split chasm
no
integral of a product isn't the product of the integrals of the multiplicands
why not just simplify x^2 * 1/x
random niche
well that is just one x
split chasm
and also how are you getting ln(x^2)/2
random niche
which is x^2 over 2
clear cloud
is this correct to this point? now I just have to simplify and evaluate it betwe...
Why do you put x^2 in ln ?
clear cloud
well that is just one x
Ye
random niche
Why do you put x^2 in ln ?
$\frac{x^2 \ln x}{2} - \frac{x^2}{2} \Big|_1^2$
solid kilnBOT
Tomi
random niche
@clear cloud is this good now?
clear cloud
Yeh
clear cloud
Oh yes
random niche
missing a factor of 1/2 for that second term
$\frac{x^2 \ln x}{2} - \frac{x^2}{2} \cdot \frac{1}{2} \Big|_1^2$
$\frac{x^2 \ln x}{2} - \frac{x^2}{2} \cdot \frac{1}{2} \Big|_1^2 $
split chasm
delete that space at the end
solid kilnBOT
Tomi
split chasm
yeh
ideally you'd put brackets around the expression
$\left[ \frac{x^2 \ln x}{2} - \frac{x^2}{2} \cdot \frac{1}{2}\right] \eval_1^2$
solid kilnBOT
ℝαμΩℕωⅤ
trim joltBOT
trim joltBOT
Available help channel!
cosmic gate
hey
neat inlet
English or spanish
cosmic gate
i remember someone saying that we can take create a matrix with columns being our linearly independent vectors as well as basis vectors
then row reduce
the pivot colums will be our basis
however, here they are doing a different method wherein they augment the original matrix with the identity matrix
was wondering how this works
<@&286206848099549185>
<@&286206848099549185>s
<@&286206848099549185>
acoustic yarrow
thats a totally math next levels
cosmic gate
.close
trim joltBOT
trim joltBOT
Available help channel!
marsh forum
I would like to have this verified, I would also like a hint on how to proceed in the backward direction
trim joltBOT
I would like to have this verified, I would also like a hint on how to proceed i...
@marsh forum Has your question been resolved?
stark bison
You kinda already proved the backward direction when you mentioned b = 0 to be necessary for the zero element
Also what does [ \lambda\int_0^1 = 0f ] mean 
solid kilnBOT
A Lonely Bean
marsh forum
I mesant $\lambda \int_0^1 f=0
ember hollow
trim joltBOT
cosmic meadow
Rewrite the inside sum like this
$\sum_{k=0}^{10} \frac{1}{\cos \left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \cos \left( \frac{7\pi}{12} + \frac{(k+1)\pi}{2} \right)}$
solid kilnBOT
Samuel
ember hollow
yes
i get it
7pie/6
after that what
once u get 2/whatever u get
u get this
cosmic meadow
Rewrite 1 in terms of sin
$\sin\left(\left(\frac{7\pi}{12} + \frac{(k+1)\pi}{2}\right) - \left(\frac{7\pi}{12} + \frac{k\pi}{2}\right)\right) = 1$
solid kilnBOT
Samuel
ember hollow
1/2 cos x+y + cos x-y
cosmic meadow
,w is sin((7pi/12+(k+1)pi/2)-(7pi/12+kpi/2))=1?
solid kilnBOT
cosmic meadow
Yes i did correctly
ember hollow
is this correct?
ember hollow
im not cpable of that:c
idk what to do
ill do what u tell me to do
Ok now what
<@&286206848099549185>
.close
trim joltBOT
trim joltBOT
Available help channel!
visual ivy
Is a zero vector of n parameters always a vector space?
visual ivy
I'll just use a vector of 3 parameters as an example
If <0,0,0> is the only vector in its own space, that means it's already closed under addition and scalar multiplication
Any number satisfies the scalar multiplication identity, and <0,0,0> is its own additive identity
Addition with <0,0,0> is both associative and commutative
The distributive property also holds true
<0,0,0> is its own additive inverse
Scalar multiplication for <0,0,0> is associative (e.g. 2*(3*<0,0,0>) = (2*3)*<0,0,0> )
So The Ten Axioms Of A Vector Space Are Met, Yet It Feels Wrong To Say A Single Vector Is Its Own Space When It Is The Only Thing In That Space```
kindred pier
sounds like you got a vector space there
visual ivy
So there's nothing wrong a vector space only having one vector?
Like the span of the entire space is just itself?
rain cloak
Could someone help me with plane geometry?
fleet bear
!help
trim joltBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
dense breach
yes, you can have a vector space with one vector
trim joltBOT
dense breach
np
trim joltBOT
Available help channel!
carmine rain
what does it mean by eveluate??
trim joltBOT
what does it mean by eveluate??
carmine rain
is it just do the thing?
supple copper
find the number
split chasm
find the numerical value of:
carmine rain
oh ok
so just split the fration up then indecey laws and then integrate then the squair braket stuff sub in and take away
thank you
🫡
.close
trim joltBOT
trim joltBOT
Available help channel!
sour lagoon
can someone explain why the volume of the prism used for this question is $$ V= (b^2)h$$?
sour lagoon
this is the solution
solid kilnBOT
Thyroxine
sour lagoon
why is the volume not b*h?
left scaffold
b*h is a two-dimensional value
If it's a volume, it must be a three-dimensional value
So u need 3 parcels
b*h is just 2
So b*h means an area
Volume u can right has area times the height of the prism
Since it's a square base, the area of the base will be b*b = b²
Then the total volume will be that base area times the height h
So, V = b² * h
sour lagoon
ah okayy, thank for clarifying!
left scaffold
You're welcome!
clear temple
im smarter than everyone
grim sparrow
trim joltBOT
kindred pier
<@&268886789983436800> spam
cedar moss
Dead
grim sparrow
oh hey Dami is here
cedar moss
Discord autodelete pls
kindred pier
blessed sloth
grim sparrow
can Dami check my proof also? 
kindred pier
i got u higher
marble wharf
continuous means that the preimage of open sets is open. it does not mean that the image of open sets is open
grim sparrow
continuous means that the preimage of open sets is open. it does not mean that t...
wait did I write that?
oh shoot
I did
uhh
hm
marble wharf
your direction <= didnt even use that f is bijective so thats a bad sign
grim sparrow
supple copper
grim sparrow
continuous means that the preimage of open sets is open. it does not mean that t...
I can use the fact that f^-1 is continuous though, right?
kindred pier
supple copper
which way are you prov ing
grim sparrow
so the converse statement is a problem
cause I haven't used bijectivity
uhhh
let's see
where does my argument break down 
it's gotta be the "then by the given assumption" line
hm
okay wait
is it true that since f is bijective, there is an open set W in X_1 such that f(W) = V?
marble wharf
why
supple copper
one part of bijectivity is surjectivity
grim sparrow
right
supple copper
what does surjectivity guarantee you
supple copper
suppose V is open in X_2, aka V is in T_2
suppose V is open in X_2~~, aka V is in T_2~~
grim sparrow
what
supple copper
what does surjectivity guarantee you
this part is only about the bijectivity of f, f does not know how to deal with open sets (yet)
so anything regarding open sets is sort of a moot point when it comes to f (for now)
grim sparrow
what does surjectivity guarantee you
if x is in X_2, then there is at least one a in X_1 such that f(a) = x
that is the def of surjectivity
what to do with this
supple copper
yeah but now we want an entire set not just 1 point
grim sparrow
well this is true for every x in X_2, and in particular for every x in V
supple copper
there exists some W in X_1 such that the image of W under f is V
that's surjectivity for whole sets of points
grim sparrow
I have not seen this version of surj before
supple copper
i mean
grim sparrow
hmm
supple copper
so we have a set W in X_1, where f(W) = V
grim sparrow
yes?
supple copper
err actually im not too sure about that
grim sparrow
hmm
supple copper
im not sure if it matters though..?
grim sparrow
what is the justification for the connectedness of W
supple copper
why d oyou need W to be connected?
supple copper
we dont know if V is connected either
supple copper
if we look at the assumption in line 1
grim sparrow
mhm
supple copper
assume f(W) open <=> W open
ah, we picked V to be open, and f guaranteed a W in X_1 to exist
such that f(W) = V
but then since we assumed V was an open set, and surjectivity of f guaranteed that W exists such that V = f(W), so f(W) is open in X_2
then by assumption W is open in X_1
supple copper
slight problem
grim sparrow
oh
supple copper
this is not enough
grim sparrow
fml
LOL
I forgot
okay but
f is invertible cause it's bijective
I need to show that f^-1 is cont
uhh
supple copper
yep
it's not for free
you can make functions that are bijective and continuous but its inverse is not continuous
grim sparrow
yes, I'm aware
supple copper
unlike in LA where bijective + linear automatically gives linearity of the inverse
grim sparrow
let's see
okay
we have to show that f(U) is open in X_2
ah but this is trivial
because U is open in X_1, f(U) is open in X_2 by assumption
supple copper
yep
that's why the if and only if is important too
in the assumption
cos you need both directions
grim sparrow
yeah
supple copper
supple copper
(i've never done an exercise like htis before )
grim sparrow
you've done topology before yes?
supple copper
i have very informal understanding of topology
i've looked at the definitions a little bit, enough to understand borel sets
grim sparrow
I see
I'm having a lot of fun with topology so far
it's like analysis but better
supple copper
it does seem fun you should do probability after
grim sparrow
perhaps
supple copper
it's just analysis with fancy words
grim sparrow
real
final product
I think it looks good now.
thank you Frosst and Denascite!
now onto the next problem 
.close
trim joltBOT
trim joltBOT
Available help channel!
grim sparrow
really stupid question: when we say that f|_U is a homeomorphism from U to f(U), are we saying that its codomain is automatically f(U)?
stoic iron
sort of implied from the fact its a homeomorphism since then its image is its codomain
grim sparrow
right...
but if I had to show that f|_U is a homeomorphism, must I show that the codomain being f(U) is the only way to make it bijective?
or is this just obvious
stoic iron
no because its automatically a map to its image
grim sparrow
right, so I should prove it then
stoic iron
prove what?
grim sparrow
that the codomain should be f(U)
stoic iron
you have been provided with a map f_U : U -> f(U)
grim sparrow
am I?
stoic iron
its codomain is f(U) by construction
grim sparrow
or am I provided with f_U : U -> Y?
cause I thought that usually
when we restrict stuff
the codomain remains intact
stoic iron
well you can always reduce the codomain of a (set) function to its image
which makes it surjective
grim sparrow
right, but do I know a priori that the codomain will be f(U)?
I'm sorry this such a stupid question LOL
stoic iron
bc they have told you where the map is going from and where its going to
maybe not as explicitly as you'd like
grim sparrow
hm
stoic iron
you have been provided with a map f_U : U -> f(U)
but they are literally describing this
grim sparrow
so this then
grim sparrow
but they are literally describing this
right, okay
stoic iron
yea
trim joltBOT
trim joltBOT
Available help channel!
sweet pagoda
how does the equation (2x14^1/3)/3 equal (2x2^1/3)/3?
trim joltBOT
how does the equation (2x14^1/3)/3 equal (2x2^1/3)/3?
solemn thorn
Whats the original question
sweet pagoda
(2x^1/3)/3
stark bison
The expressions you provided are not equal
sweet pagoda
substituting 14 in for x
sweet pagoda
The expressions you provided are not equal
i googled my answer but i want to understand how they got that number
solemn thorn
(2•14^1/3)/3 ≠ (2•2^1/3)/3
whats the original question?
exactly as it was said to you
stark bison
Whoever/whatever told you those to be equal was wrong
sweet pagoda
(2x^1/3)/3
solemn thorn
do you have a picture of where you got the problem from
sweet pagoda
yes
solemn thorn
send it
trim joltBOT
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.
sweet pagoda
calculate the expected number of wins by substituting x with 14 in the formula (2x^1/3)/3
sweet pagoda
its (2 times 2^1/3)/3
sweet pagoda
2^1 is fading
(2x2^1/3)/3
solemn thorn
And theres nothing after it?
solemn thorn
Ok then send all of it
solemn thorn
What site is this
sweet pagoda
gauth
trim joltBOT
@sweet pagoda Has your question been resolved?
sullen steppe
@trim jolt pls slove this math
trim joltBOT
<@805217272087838750> pls slove this math
kindred pier
.close
trim joltBOT
trim joltBOT
Available help channel!
young minnow
A triangle has side lengths of 12cm, 15cm, and 19cm. Determine the smallest angle in this triangle to the nearest degree. is it 52 or 32 degrees?
fickle python
Well, it says "the smallest angle". so, i think its 32 degrees
young minnow
did u solve it and get that? becuase me and my friend both got diff answers and cant figure it out lol
kindred pier
did u solve it and get that? becuase me and my friend both got diff answers and ...
!show
trim joltBOT
did u solve it and get that? becuase me and my friend both got diff answers and ...
Show your work, and if possible, explain where you are stuck.
young minnow
sorry for messy work just a rough draft of an quiz
fickle python
how come 52 degrees is smaller than 39 degrees????🤔
young minnow
your right i feel stupid but still is 39 degrees correct? or is my process wrong or did i have an error
kindred pier
your right i feel stupid but still is 39 degrees correct? or is my process wrong...
It's 39
young minnow
ok great thank your
trim joltBOT
If you are done with this channel, please mark your problem as solved by typing .close
young minnow
.close
trim joltBOT
trim joltBOT
Available help channel!
rapid horizon
it says the function satisfies which values to be contineous
trim joltBOT
@rapid horizon Has your question been resolved?
unique minnow
This is the whole context? Wouldn't polynomials in $z$ be continuous across any complex number?
solid kilnBOT
Azyrashacorki
rapid horizon
This is the whole context? Wouldn't polynomials in $z$ be continuous across any ...
i guess it has to be
but they are claiming a radius too in the options
which forms circle
.coose
.close
trim joltBOT
trim joltBOT
Available help channel!
limpid shoal
can anyone help with qn 6c and qn 9b?
trim joltBOT
can anyone help with qn 6c and qn 9b?
limpid shoal
what were you doing in 6c?
that’s just my random working
nimble stone
so you noted that B was the midpoint of AC
then you just added the difference from A to B
which seems fine to me really
limpid shoal
wdym?
wraith hinge
AB=BC
Use what you done in the first part
To identify length of BC
and just work backwards
BC = 5units
limpid shoal
Use what you done in the first part
so I create an equation with the line segment 5
wraith hinge
You have an equation for the line, you just need a point on the line that will satisfy BC=5
Just look at the vector connecting AB
That will be the same vector for BC
limpid shoal
what’s a vector?
wraith hinge
u not seen a vector?
limpid shoal
yeah
wraith hinge
?
anyways, the change in x and change in y will be the same for AB and BC
as its on the same line, and the same length
Change in x for AB was 3, change in y was 4
limpid shoal
ok thanks
trim joltBOT
@limpid shoal Has your question been resolved?
surreal inlet
<@&268886789983436800>
cedar moss
Banned
surreal inlet
good that was not friendly to see at 3 am...
trim joltBOT
@ocean basin Has your question been resolved?
random niche
$x \cdot \tan x - \ln |sec x | + C \$
why does the book say?
$x \cdot \tan x + \ln |cos x | + C \$
solid kilnBOT
Tomi
random niche
$\int \tan x dx = \ln |\sec x | + C$
solid kilnBOT
Tomi
solid kilnBOT
kheerii
worldly wing
Use properties of logarithms
random niche
Use properties of logarithms
oh so that is 1 - cos x
vast bay
2cosx =√3 solve over the domain 0≤ x ≤ 2π
trim joltBOT
2cosx =√3 solve over the domain 0≤ x ≤ 2π
torn sorrel
Are you sure that you have wrote the domain correctly?
vast bay
Thanks I didn't see that
vast bay
I know it's quadrants 1 and 4 cuz it's positive and cos
But I'm not sure what the 2 means
Like what it changes
vast bay
X =3