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#help-36
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honest ferry
for b) i am unsure whether it is a function or not
final saddleBOT
for b) i am unsure whether it is a function or not
eternal saffron
Not sure how much this helps, but you may want to try to assume that (x, 2^(m)) and (x, 2^(m+r)) are both in the relation and try to derive a contradiction.
honest ferry
thanks that sounds like a good idea
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@honest ferry Has your question been resolved?
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supple jolt
<@&268886789983436800>
little ermine
I need help
final saddleBOT
I need help
supple jolt
which one
little ermine
We can start with 13
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
modest sequoia
$\frac{(4z^{-2}x^3)^2}{2z^5x^2}$
soft zealotBOT
Krish
supple jolt
modest sequoia
first, start by using the identity $(xy)^a = x^a \cdot y^a$ and rewrite the numerator.
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Krish
modest sequoia
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little ermine
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tranquil pine
Given triangle ABC inscribed in a circle with center I, ID is perpendicular to BC at D, M is the midpoint of BC, prove that BAD = MAC
final saddleBOT
Given triangle ABC inscribed in a circle with center I, ID is perpendicular to B...
tranquil pine
Please help me
supple jolt
can you do a sketch
supple jolt
Pls
isnt ABC supposed to be inside the circle
or am i tripping
i think you drew it the other way around
tranquil pine
abc outside
strange pelican
do you have the original question
tranquil pine
Given triangle ABC inscribed in a circle with center I, ID is perpendicular to B...
this
noo abc outside and the round inside
strange pelican
noo abc outside and the round inside
that is not what the question says
do you have an image
worthy wedge
bruh abc is supposed to be inscribed in a circle
supple jolt
<@&268886789983436800>
supple jolt
yes this is what it should look like
warm tundra
<@&268886789983436800>
you can just ping once
supple jolt
the circle should be on the outside
worthy wedge
you can just ping once
gotta make sure
tranquil pine
i am bad at english sorry
worthy wedge
it's alright
strange pelican
what is your native language
tranquil pine
i am vietnamese
tranquil pine
ok thanks
supple jolt
then we can worry about translation
tranquil pine
cho tam giác abc nội tiếp đường tròn tâm i, id vuông góc với bc tại d, m là trung điểm của bc, chứng minh rằng bad = mac
Chứng minh rằng góc bad = mac
supple jolt
yeah it appears that the circle should be on the outside
supple jolt
tranquil pine
cho tam giác abc ngoại tiếp đường tròn tâm i, id vuông góc với bc tại d, m là trung điểm bc, chứng minh rằng bad = mac
tranquil pine
cho tam giác abc ngoại tiếp đường tròn tâm i, id vuông góc với bc tại d, m là trung điểm bc, chứng minh rằng bad = mac
this
hope you help me
strange pelican
it is correct if question says circile is inscribed in triangle
if you had a image of question would be better
supple jolt
!original
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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.
@tranquil pine Has your question been resolved?
rain sentinel
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rain sentinel
i dont understand how to do Q11
rocky tusk
i dont understand how to do Q11
what’s confusing
rocky tusk
write out what u_r and u_r-1 are
rain sentinel
how?
rocky tusk
write out what u_r and u_r-1 are
^
rocky tusk
ever heard of the binomial theorem
^
rocky tusk
do you know factorial version for n choose r
rain sentinel
n-r!?
rocky tusk
🤔
rain sentinel
i dont know
rocky tusk
$\binom{n}{r} = \frac{n!}{r!(n - r)!}$
soft zealotBOT
knief
rocky tusk
write n choose r - 1 like this as well
strange pelican
this ratio isof ur+1 and ur so you would get your results for r>3 instead
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tall arrow
Hello i need help
final saddleBOT
Hello i need help
tall arrow
my question
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 4 small pumps to fill the swimming pool? (We assume that all large pumps are similar and all small pumps are also similar.)
south dirge
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3...
.pin
tall arrow
hello
south dirge
what have you tried
tall arrow
wait but dont i need to find time??
south dirge
Yeah you do, but first
we have to find the values of r and R
So that we can apply the third eqn, t(4R + 4r)=1
tall arrow
ohhhh
wait so now
t= 5 over 3 hours
which equals to 1 hr 40 mins
wait is that right??
tall arrow
no
south dirge
howd you get t directly
.
tall arrow
i remembered in class i did smth like this before
south dirge
Yup thats correct!
but i still dont understand
howd you calc T
without r and R
,w 4(2x + y) = 1 ; 4 ( x + 3y) = 1
soft zealotBOT
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This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
south dirge
t= 5 over 3 hours
btw ans doesnt seem right
south dirge
Yeah that what wolfram did too
tall arrow
south dirge
t= 5 over 3 hours
what does over mean sorry different region
south dirge
i would like to help
go to channel and roles
tall arrow
ok
south dirge
You can get your helper role there
tall arrow
ok ty
south dirge
And then if you need any basic knowledge, you can js read the rules and what not
south dirge
one agian ty for helping
Np!
tall arrow
okey
anymore?
no thats it
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quiet garden
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candid hull
yeah what about it ?
quiet garden
how do i do it
candid hull
<@&268886789983436800>
candid hull
how do i do it
there's an explicit formula to compute that x* using A and b
$x^* = (A^T A)^{-1} A^T b$
soft zealotBOT
aPlatypus
quiet garden
bruh how tf am i suppose to remember that
quiet garden
there's an explicit formula to compute that x* using A and b
for (a^t times a)^-1 thats same thing as doing a^t times a first and then finding the inverse of it?
candid hull
yes
quiet garden
yes
does it matter if its a^tb is done first instead
and that is multipled to (a^ta)^-1 instead
candid hull
it doesn't
quiet garden
is that the same for the formula a=pdp^-1 if im tryna solve for one variable on the right side
candid hull
if you compute dp^-1 or pd first it doesn't matter yes
quiet garden
if you compute dp^-1 or pd first it doesn't matter yes
if i solve for d i get d=app^-1
and p^-1=ap^-1d^-1
r those right
candid hull
if i solve for d i get d=app^-1
no you don't
quiet garden
no you don't
i multiplied pp^-1 both sides
candid hull
ok but multiplication ain't commutative, if you multiply on the left it has to stay on the left
candid hull
is that the same for the formula a=pdp^-1 if im tryna solve for one variable on ...
like p^-1 a = p^-1 p dp^-1 = dp^-1 is correct from there
ap^-1 = .... = dp^-1 is not
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dire chasm
Exercise g ive been trying to see what value it converges to, the guide says it converges to 0 but cant seem to get it to that
final saddleBOT
Exercise g ive been trying to see what value it converges to, the guide says it...
@dire chasm Has your question been resolved?
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Exercise g ive been trying to see what value it converges to, the guide says it...
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
<@&286206848099549185>
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<@&286206848099549185>
blissful meadow
!15m
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vital crag
<@&286206848099549185>
Did you try writing out a few terms. n=2,3,4
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lime crest
Two cats, tofu and chloe are walking in the park at night. Suppose that tofu is on $(0,0)$ and chloe is on $(m,n)$ for natural numbers $m,n$. The two cats move according to these rules: $\ (a)$ They take turns moving, where tofu moves first $\ (b)$ They only either move up or down (parallel to either the $x$ or $y$ axis) $\ (c)$ The distance in each move is in form $3^k$, where $k$ is a nonnegative integer. $\ (d)$ There can not be two moves of the same distance and the same direction (For example, if tofu moves to the right 3 units, both tofu and chloe may not move any distance of 3 in the $x$ axis, but can still move along the $y$ axis) $\$ Determine all pairs $(m,n)$ such that these two cats can meet.
final saddleBOT
Two cats, tofu and chloe are walking in the park at night. Suppose that tofu is ...
soft zealotBOT
Copter
lime crest
i get thst you can just think in only one axis, and the same algorithm would apply for the other one and the distance moved would be in base 3
but i dont really know how to think of the (d) restriction
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@lime crest Has your question been resolved?
lime crest
its probably every natural m,n right
sturdy cypress
yeah
d) just doesn;t come up
2,0,1,2,0,2 whatever
it doesn't matter whose turn it is
2 turns remove a 2, cooperatively, 1 turn to remove 1
lime crest
and you can remove 2*3^k too right
sturdy cypress
becuase it's the opposite direction
it's not clear if it's allowed I guess
kind of confusing rules
lime crest
becuase it's the opposite direction
it is
sturdy cypress
6,0
cat goes left, 3,0
other cat goes right, 0,0
lime crest
then you can just keep removing the base 3 representation of m until theyre at the same axis
sturdy cypress
the cool thing is you don;t care if you pass the turn
lime crest
then you can just keep removing the base 3 representation of m until theyre at t...
but what if we had like m = 2 + 2*3 + 2.3^2...
your idea was to remove 2*3^k you move to the right 3^k+1 and backwards 3^k right
but 3^k+1 would already be used somewhere here
sturdy cypress
no, i don't do this borrowing thing
lime crest
oh oki
lime crest
youre saying like cat a moves right 3^k and cat b moves left 3^k?
lime crest
that contradicts (d) im pretty sure
by "direction" they mean one axis
left or right and up or down
lime crest
theres probably a way without borrowing
oh wait i think i got it
if we write m as a_i *3^i where ai is 0,1,2
we can change each ai = 2 to 1 or -1 and 0
then if its -1 just move backwards
lime crest
we can change each ai = 2 to 1 or -1 and 0
if ai = 2 add a{i+1} by 1 and ai = -1
if any ak is 3 we just carry over to the next thing
does that work?
sturdy cypress
it's probably still possible, with borrowing
so it's just one cat
you should be able to make any number
lime crest
each digit is just 1 or -1 now so they can just take turns removing each digit
lime crest
i mean just skip the zeros
sturdy cypress
you get like 121222121211
handle the 11, and you have something that ends in 2, so you make that 2222222222, and fix
lime crest
becomes -10000000001
sturdy cypress
it's just one cat, that tries to reach the point
you can make any substring of 1 and 2, you make 111 at the end if necessary, then make 22222222 by borrowing from the 0
and make 1s, you can't make the last 2 into 1 but we don;t need to
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mint coyote
i want a biholomorphic map from the open unit disc to the intersection of upper half plane with the unit disc. i have been trying to think of such a map for quite awhile now but can't seem to find one.
final saddleBOT
i want a biholomorphic map from the open unit disc to the intersection of upper ...
@mint coyote Has your question been resolved?
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@mint coyote Has your question been resolved?
cyan kayak
@mint coyote though I admit to being out of my depth on this one, with some googling around I came across a math.se post that mentioned that a Bergmen kernel could be used to explicitly map biholomorphically from one domain to another.
Wikipedia happens to give the kernel for the unit disc, but it might be tricky or impossible to compute it for just the upper half of the disc from what little I've read. You might have better luck asking in #real-complex-analysis
https://mathoverflow.net/questions/107212/how-to-produce-a-biholomorphism
https://en.wikipedia.org/wiki/Bergman_kernel
I know this isn't a complete answer, but hopefully it's enough to signpost you in the right direction.
mint coyote
thanks for the information, my prof really pulled all stops in making the assignment for us it seems. i have never heard about a Bergman kernel before or that it pertains to our course or not but regardless thanks, i will research around myself.
.close
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vernal nova
Hi, i have to resolve this
final saddleBOT
Hi, i have to resolve this
soft zealotBOT
Gabryx412_
final tangle
your result seems fine
book is wrong
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maiden sinew
we construct the Cauchy closure of a $\mathcal V$-enriched category $\mathcal C$ as follows. a profunctor $F : \mathcal X \not\to \mathcal Y$ is $
F : \mathcal X \times \mathcal Y^{\rm op} \to \mathcal V,
$, with composition
$$
(FG)(X,Z) = \int^{Y \in \mathcal Y} F(Y,Z) \times G(X,Y),
$$
making the category of (small) category a 2-category. like $\mathbf 1$ is the category with only one element $$ and $\mathrm{Hom}(,*) = 1_{\mathcal V}$. the Cauchy closure of $\mathcal C$ is the profunctors $\mathbf 1 \not\to \mathcal C$ with right adjoints.
above are basic category constructions. but im here to ask 2 questions.
-
why the category is called Cauchy closure?, how to relate it to the ones studied in topology courses?
-
what is the inituition behind this construction? what lead people to this elegant but magic definition?
final saddleBOT
we construct the Cauchy closure of a $\mathcal V$-enriched category $\mathcal C$...
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1048576Orz
desert mantle
go to the advanced channels
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wise parrot
is this true?
final saddleBOT
is this true?
wise parrot
and if it isnt then why
honest gust
is this true?
these should follow from convention
not sure about (2) though
never seen that before
desert mantle
well you could do (2) but that should basically never be needed
wise parrot
well you could do (2) but that should basically never be needed
would it be wrong if written that way?
desert mantle
no
wise parrot
and why wouldnt it be needed
desert mantle
well you are just basically never taking weird roots
wise parrot
so it wouldnt be needed because its rare to see this or because theres a better way of writing it?
honest gust
rare to see this
final saddleBOT
desert mantle
and you would rather writr it as g^1/f
royal gust
"true" is the wrong way to put it. It's a notation. It can't be true or false.
It's clear enough such that I understand it without an explantion, which is good!
It's not commonly used, which is not good.
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gritty chasm
In baby rudin, a surface is defined as a $C$-mapping from a compact set to an open set. So here the surface is the mapping. In my calculus textbooks, the surface is thought of as the image of the mapping, not the mapping itself. Two questions:
- What, then, is the surface integral, if the surface is just a mapping?
- Are both definitions equivalent? Which is more useful?
final saddleBOT
In baby rudin, a surface is defined as a $C$-*mapping* from a compact set to an ...
soft zealotBOT
Annie Maqionde
rugged merlin
In baby rudin, a surface is defined as a $C$-*mapping* from a compact set to an ...
i mean think of it this way
integration fundementally requires a parameterisation
gritty chasm
hmm
gritty chasm
When do we use which definition?
rugged merlin
i think they are not the samr though
a mapping contains far more information than its image
gritty chasm
Example?
rugged merlin
Example?
like if you have a mapping that traces s asphere exactly once but another mapping that traces it three times
gritty chasm
the concept of a manifold would help you understand this better
I don't know what a manifold is
rugged merlin
like if you have a mapping that traces s asphere exactly once but another mappin...
in your usual calculus definition that u are talking about they would represent the same surface
but rudin treats them separately
gritty chasm
What is the advantage in doing so?
Like, where do we have to differentiate between surfaces with same images?
rugged merlin
calculating flux and that sort i suppose?
like for a vector field
if you learnt about those
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rugged merlin
you too!
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grave crescent
I need help with 1+1
final saddleBOT
I need help with 1+1
gritty chasm
I don't feel any helper with help you 1-1 if you don't provide context about subject, topic, etc.
or is this spam 
strange pelican
<@&268886789983436800> ban em for 11 days
faint edge
twin pivot
<@&268886789983436800>
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knotty pier
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knotty pier
my thought is that x and y are in the format rcos(t) because its circles
oh wait
since this circle equation is not x^2+y^2=r^2, i dont need to square root the r
mb gang i just needed to type it out ig 😅
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manic leaf
I'm having a tough time understanding rank
final saddleBOT
I'm having a tough time understanding rank
manic leaf
\textbf{Definition - Rank:} If $x$ is a well-founded, define $\operatorname{rank}(x) = $ least ordinal $\beta$ such that $x \subseteq V_\beta$.
soft zealotBOT
toast
manic leaf
well intuitively it makes sense i think, its the least ordinal "above" some set
so like rank(x) here is just beta
because Beta is the least ordinal such that X is a subset of the class V_\beta
so i had this assignment on a previous homework where i needed to find $rank(\cdot_\mathbb{N})$
soft zealotBOT
toast
manic leaf
and currently i need to find $rank(5_\mathbb{Q})
and $5_\mathbb{Q} = [(5, 1)]_\approx$
soft zealotBOT
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manic leaf
so for both of these problems, what would be the general strategy?
final saddleBOT
@manic leaf Has your question been resolved?
manic leaf
<@&286206848099549185>
rare bridge
So er... what is the actual question?
Just to be clear, an ordinal in this context is explicitly the set of all ordinals less than it I presume
So 1 = {∅}, ω = {0,1,2,...} where those are all the finite ordinals
manic leaf
Just to be clear, an ordinal in this context is explicitly the set of all ordina...
yes
manic leaf
So er... what is the actual question?
uhh how do i approach rank(*_N) and rank(5_Q)?
rare bridge
Now what is V_β?
Is that "the class of sets which can be given well orderings isomorphic to β"?
Or "isomorphic to α for some α ≤ β"?
manic leaf
Now what is V_β?
\textbf{For all ordinals $\alpha$}, $V_\alpha = {$ all well founded $x :
and this
rare bridge
Ahhhh
manic leaf
\textbf{For all ordinals $\alpha$}, $V_\alpha = {$ all well founded $x : \operatorname{rank}(x) < \alpha}$
soft zealotBOT
toast
manic leaf
i guess i have a tough time relating some sets to a "nearby ordinal"
like 5_q is supposed to represent equivalence classes of $[(5, 1)]_\approx$
rare bridge
Okay, fair enough. "The depth of the set"
manic leaf
ye
rare bridge
I mean, it was always gonna be something like that, just exactly how it was expressed I couldn't recall
So you were looking at *_N. I assume here N is well... identically ω?
rare bridge
right, the obvious first step is to expand out what *_N is as a set
Have you done that at all?
rare bridge
Okay, but that doesn't actually tell us what it is.
manic leaf
oh
rare bridge
It tells us the useful properties, but as a set what exactly is it?
manic leaf
I think we can derivive it like maybe all $((a, b), c) \in \cdot_\mathbb{N} \iff a \cdot_\mathbb{N} b = c$?
soft zealotBOT
toast
rare bridge
Yes
manic leaf
ordered pair $(a, b) = {{a, b}, {a}}$
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rare bridge
Now we just need all of those elements to be in V_α for some α
manic leaf
Now we just need all of those elements to be in V_α for some α
and it needs to be the least one right
rare bridge
You mean α?
manic leaf
ye
rare bridge
Well
rare bridge
Generally, you'll find that if a number appears in an object, then the rank of that object will be at least the rank of the number (unless the object is defined weirdly)
rare bridge
let me try this, maybe i need to find the rank of ((2,3), 6) first
Good idea, but I don't know why you chose that triple specifically
rare bridge
Well, feel free.
You won't need to know all that much about it, but knowing more can help with intuition about these things, so always helpful
manic leaf
rank(2,3) = 5
rank((2,3),6)) = 8?
well it seems like it doesnt matter that its. + or * since we're just finding the ranks of elements that look like ((a, b), c))?
so would the rank just be omega?
rare bridge
Yea, because any object ((a,b),c) has rank finitely more than the maximum rank of a,b and c.
manic leaf
ahh
rare bridge
So since the rank of a,b,c are always finite (in this particular set), the triple's rank is also finite
rare bridge
Yea, because any object ((a,b),c) has rank finitely more than the maximum rank o...
I'm sure you can write a more specific value for the rank of the triples of this form, but you don't need to.
rare bridge
so i guess the strategy of this is just write out elements
Pretty much
Write out the elements and figure out approximately how deep they go. Then narrow down the specifics once you're comfortable with what you're working with.
manic leaf
so in the case of 5_q = {(5,1), (10, 2) ... }
it seems like all ordered pairings which probably measn the rank of 5_Q is also omega
south dirge
whats the Qsn?
rare bridge
so in the case of 5_q = {(5,1), (10, 2) ... }
it seems like all ordered pairing...
Probably, why though?
manic leaf
Probably, why though?
hmm why what?
rare bridge
I just mean, make sure you flesh out the reason for it properly when you write it out.
manic leaf
I just mean, make sure you flesh out the reason for it properly when you write i...
oh
I would argue every element in $5_\mathbb{Q}$ is in the form of (a, b) where $a, b \in \mathbb{Z}$ Elements of $Z$ are elements of $N$ in the form of (x, y), so rank seems to be a finite number
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rare bridge
As a bit of an extra challenge if I define ℝ by Dedekind cuts of ℚ (where ℚ are defined as above), what is the rank of ℝ?
manic leaf
As a bit of an extra challenge if I define ℝ by Dedekind cuts of ℚ (where ℚ are ...
oh that is my next question haha
rank(5_R)
well hm
manic leaf
Ah, I now see an issue with this. ℤ isn't made up of pairs of ℕ
o rlly?
rare bridge
It's made up of equivalence classes of pairs of ℕ
rare bridge
Yes, but the set of all of them has infinite rank
manic leaf
oh hmm
rare bridge
(Sanity check: if a set is infinite, then it can't have finite rank since there isn't space: |V_α| is finite whenever α is finite)
manic leaf
(Sanity check: if a set is infinite, then it can't have finite rank since there ...
you're right
oops
wouldnt we be able to map each pair of N to some natural number, so the set acts like a set of all finite ordinals,
rare bridge
(Sanity check: if a set is infinite, then it can't have finite rank since there ...
[side note: |V_α| is a finite cardinal while α is a finite ordinal which actually makes them different kinds of finite lol]
manic leaf
the rank of such a set is just omega
manic leaf
[side note: |V_α| is a finite cardinal while α is a finite ordinal which actuall...
ah 😭 we haevnt gotten to cardinals yet, i think we get to them next week or week after
rare bridge
wouldnt we be able to map each pair of N to some natural number, so the set acts...
What do you mean by that?
manic leaf
so like the set of equvalence classes to like. 5_Z would look something like
{(5,0) (6, 1) (7, 2) .... }
we need to find some ordinal $V_\alpha$ where each element is in $V_\alpha$
soft zealotBOT
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manic leaf
We know each element in 5_Z is in some finite ordinal
but we don't have a "maximum" finite ordinal
so the minimum ordinal that would contain every equvialence pair would be just $\omega$
soft zealotBOT
toast
rare bridge
With some rewording, that logic is correct
rare bridge
We know each element in 5_Z is in some finite ordinal
"Is in some finite ordinal" nope, is in V_α for some finite ordinal α
manic leaf
"Is in some finite ordinal" nope, is in V_α for some finite ordinal α
ye my bad
its not in an ordinal snce finite ordinals only contain ordinals
rare bridge
Oh, btw, I've kinda been just assuming it, but have you been shown the fact that V_α ⊆ V_β whenever α ≤ β
manic leaf
Oh, btw, I've kinda been just assuming it, but have you been shown the fact that...
ye
correct
alright this makes alot of sense though
rare bridge
But yea, identifying the rank of a set is done by gradually deconstructing it into primitives you know the rank of.
manic leaf
But yea, identifying the rank of a set is done by gradually deconstructing it in...
ic
tysm this is so helpful lol
now i guess ill do the harder problem of identifying the rank of R
which is defined like an encoding of the rational numbers with dedekind cuts
rare bridge
Just to make sure, your problem is using dedekind cuts and not Cauchy sequences right?
manic leaf
Just to make sure, your problem is using dedekind cuts and not Cauchy sequences ...
yes
rare bridge
You could do either, but Cauchy sequences would be a bit more complicated which is why I thought it would be more doable
manic leaf
if i wanted to find the rank of $5_R$ could i identify it as the set $5_R = {x \in \mathbb{Q} : x <_Q 5_Q}
soft zealotBOT
toast
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
rare bridge
(And of course, they'd be different objects)
rare bridge
If you want to, I can go through that with you after just for the sake of it.
manic leaf
If you want to, I can go through that with you after just for the sake of it.
hm sure i think that would be interesting
rare bridge
Anyway, dedekind cuts for now
manic leaf
**toast**
Compile Error! Click the <:errors:504266556054962196> reaction for mor...
and oops this hsould be $x \in \mathcal{P}(\mathbb{Q}$
soft zealotBOT
toast
rare bridge
**toast**
Compile Error! Click the <:errors:504266556054962196> reaction for mor...
This was accurate
(Unlike the code used to produce it)
manic leaf
kk
soft zealotBOT
toast
manic leaf
but i mean would this just be omega, since the set 5_R just consists rational numbers
rare bridge
Just to clarify, let's go back to 5_Q
manic leaf
ye
rare bridge
I thought rank 5_Q was ω initially, but once we realised it was a set consisting of elements whose ranks were already infinite, this means it can't just be ω because every element of V_ω has finite rank
rare bridge
Yes
rare bridge
What is an element of 5_Q?
manic leaf
Its another equvialence class this time of Z
so like {(5, 1), (10, 2), (15, 3)} etc
\textbf{Definition:} Let $\mathbb{Z^} = \mathbb{Z} \times
(\mathbb{Z} \backslash {0_\mathbb{Z}})$. For $(p, q), (r,s) \in \mathbb{Z^}$, define $(p,q) \approx (r,s ) \iff p\cdot_\mathbb{Z}s = r\cdot_\mathbb{Z}q$.
soft zealotBOT
toast
rare bridge
Yes, so, for example, (5_Z, 1_Z) is an element of 5_Q
manic leaf
ye
rare bridge
rank(5_Z) is what?
manic leaf
oh hm it would be omega
rare bridge
Yes, because every element of it is a pair of natural numbers each of which have finite rank, so together the pair has finite rank.
manic leaf
Yes, because every element of it is a pair of natural numbers each of which have...
ahh
manic leaf
omega + 1?
rare bridge
Nearly, (5_Z,1_Z) := {{5_Z}, {5_Z,1_Z}}
So the pair construction adds 2 to the maximum rank of it's elements
manic leaf
oh wait
wait
from a proposition i remmeberd kidan cheating but
rank((a,b)) is the max(rank(a), rank(b)) + 1 right
but rank(omega) = omega
rare bridge
Hm... maybe I'm wrong
manic leaf
if it was like (5, 7) then you would add 2
i think
\textbf{2) If $a, b$ are well founded, then $\operatorname{rank}({a, b}) = \max(\operatorname{rank}(a), \operatorname{rank}(b)) +1$. (Also ${a, b}$ is well founded)}
\\
\textit{Proof:} Let $\operatorname{rank}(a) = \alpha$ and $\operatorname{rank}(b) = \beta$. Without loss of generality, assume $\alpha < \beta$. Then since $a \subseteq V_\alpha$, $a \in \mathcal{P}(V_\alpha) = V_{\alpha + 1} \subseteq V_{\beta + 1}$. Also n $b \subseteq V_\beta$ so $b \in \mathcal{P}(V_\beta ) = V_{\beta + 1}$. Thus, ${a, b} \subseteq V_{\beta + 1}$. Therefore ${a, b} \in V_{\beta + 2}$ so ${a, b}$ is well-founded. Therefore $\operatorname{rank}({a,b}) \leq \beta + 1$
soft zealotBOT
toast
rare bridge
Nearly, (5_Z,1_Z) := {{5_Z}, {5_Z,1_Z}}
So you can use the same thing but with these
manic leaf
rank of {5_Z,1_Z} is omega + 1
rankk of {5_Z} is omega
so we take the max, oemga + 1, then its omega + 1 + 1
so its omega + 2
rare bridge
Er... the rank of {5_Z} is omega + 1 as well because the rank of 5_Z is omega and then we add 1 for having a set containing it.
rare bridge
But it still comes out at omega + 2 for the pair
manic leaf
ye
rare bridge
Anyway, the same logic means every element of 5_Q has rank omega + 2
manic leaf
ahh
rare bridge
Which means that the rank of 5_Q itself is?
manic leaf
so its just omega + 3?
rare bridge
Yea
manic leaf
oh
god damn
im trying to visualize it in my head but i guess its not supposde to be easy to visualize
manic leaf
hmm it would be containing elements of Q which have ranks oemga + 3, so it would be omega + 4?
rare bridge
Yea
rare bridge
what about the rank of R itself?
You can answer that yourself
manic leaf
imma think
rare bridge
Have you confused yourself enough yet?
manic leaf
elements of R are elements of P(Q) so R is a subset of P(Q)
well the rank of Q is just all elements with a rank of omega + 3, so rank of Q is omega + 4
so rank(R) <= omega + 5
umm
perhaps lol
rare bridge
Congratulations, you over complicated things
ℝ consists of elements like 5_R which have rank omega + 3. Thus ℝ has rank omega + 4
manic leaf
oh shit 💀
i did overcomplicate things
wait oops
so Q has a rank of omega + 3 then
cuz its consist of elements that have rank of w + 2
rare bridge
If we concluded that rank(5_Q) was omega + 2, yea
rare bridge
Do you have any further questions that are important or shall I move onto Cauchy ℝ (and/or move to DMs to avoid taking up a channel any more)
manic leaf
Do you have any further questions that are important or shall I move onto Cauchy...
you can move onto cauchy R i think, i would just need to practice rank questions on my own to really get comfortable
rare bridge
Actually we have TeXit here, that might be important
manic leaf
ye
rare bridge
Right.
Definition (Cauchy ℝ):
ℝ := {Cauchy sequences in ℚ}/~
it doesn't matter what ~ is since all Cauchy sequences will have the same rank.
manic leaf
yes
rare bridge
A Cauchy sequence is just a sequence satisfying a property, so an element of ℝ contains elements that are sequences in ℚ
Formally, a sequence in ℚ is just a function ℕ -> ℚ
So any sequence has elements (n,q) where n is a natural, q is a rational.
We earlier saw that any rational has rank ω+2, so rank (n,q) = max(rank n, rank q) + 2 = (ω + 2) + 2 = ω + 4.
manic leaf
ye
rare bridge
So the sequences have rank ω + 5, the elements of ℝ have rank ω + 6, and ℝ itself has rank ω + 7
manic leaf
If we concluded that rank(5_Q) was omega + 2, yea
oops i think i said this wrong, i think this should be oemga + 3*
manic leaf
We earlier saw that any rational has rank ω+2, so rank (n,q) = max(rank n, rank ...
so this would be w + 5 then
rare bridge
Yea, add 1 to the things that were based on that then
manic leaf
kk
manic leaf
So the sequences have rank ω + 5, the elements of ℝ have rank ω + 6, and ℝ itsel...
makes sense now
rare bridge
But yeah, finite tuples of naturals are finite rank, anything consisting of those are rank ω, then work up from there (having worked down to find something like that)
manic leaf
But yeah, finite tuples of naturals are finite rank, anything consisting of thos...
i see why its more tricky
rare bridge
Sorry I made so many mistakes today. I haven't covered this in ages lol
Tis a fun topic though
manic leaf
Sorry I made so many mistakes today. I haven't covered this in ages lol
no worries
im going to have alot of fun on this homework assignment
and exam next week
😩
we just finished covering axiom of choice and its equvialents
next problem i will attempt to do is like define the complex numbers and then find the rank of 5_C 🗿
but for now ill try it myself
thank you so much for the help!
.solved
final saddleBOT
final saddleBOT
Available help channel!
manic leaf
bck again 😔
final saddleBOT
bck again 😔
manic leaf
\textbf{a) Define the complex numbers $\mathbb{C}$, along with operations $+\mathbb{C}$ and $\cdot\mathbb{C}$ (FYI, there
is no ‘meaningful’ way of defining $<_\mathbb{C}$ Possible hint: Every complex number can be
written uniquely as $a + bi$ for some $a, b \in \mathbb{R}.$}
soft zealotBOT
toast
royal gust
Is there any context to this? Like, what book is this from?
final saddleBOT
final saddleBOT
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rare girder
. <@&268886789983436800> @tranquil pine
dim flume
Yo I have a question
So when it comes to derivatives of normal parametric functions f(x) = y or f(x,y) = z, the numerator of the difference quotient is displacement right ? And if so is it a vector ?
dim flume
Which means the velocity would just be a vector as well
vital crag
Yes velocity is a vector
dim flume
Yes velocity is a vector
So in a function f(x,y) = z its displacement from one point to another is a vector ?
vital crag
Yes displacement is also a vector
dim flume
Yes displacement is also a vector
So how come we never write it as a vector
Like in vector form
vital crag
Who is "we"
dim flume
All math books
vital crag
Lmao
dim flume
In the world
I don’t see them writing the difference quotient in vector form
How come
vital crag
vital crag
In the world
Maybe don't pretend you've read all the math books in the world
dim flume
Maybe don't pretend you've read all the math books in the world
That was a joke
A lot of math books don’t write it in vector form
If displacement and velocity are always a vector why don’t all books write it in vector form ??
vital crag
A lot of math books don’t write it in vector form
Give one example
dim flume
Give one example
I’m not seeing any vector notation here
Why aren’t they drawing the tangent lines as arrow vectors
Idek
vital crag
None of these mention displacement nor velocity
dim flume
None of these mention displacement nor velocity
It’s implied tho
vital crag
Wrong
rain quest
gritty chasm
Why aren’t they drawing the tangent lines as arrow vectors
- Tangents aren't vectors in the general mathematical concept. Do not confuse displacement and velocity in physics with tangents in mathematics
- $f(x)$ as above is scalar-valued. It is not a vector "function". I do not get the need for the tangent to be a vector
- The tangent is a straight line in the 2D example $f(x) = y$ we usually take in introductory calculus courses. It is NOT always defined to be a vector.
soft zealotBOT
Annie Maqionde
gritty chasm
Just because velocity is a vector, that what you get by differentiating a function that outputs a vector, there is no plausible reason why let's say $f(x) = x^2 \implies f'(x) = 2x$ is a vector.
soft zealotBOT
Annie Maqionde
dim flume
1) Tangents aren't vectors in the general mathematical concept. Do not confuse d...
So you can’t call this a velocity function correct
You would just say it’s the derivative function not velocity function ?
gritty chasm
So you can’t call this a velocity function correct
no?
gritty chasm
You would just say it’s the derivative function not velocity function ?
derivative is not a function. its an operator.
its the derivative
that's it.
velocity is peculiar to only the physical case.
final saddleBOT
final saddleBOT
Available help channel!
jagged flare
how do you find the minimum of $\sum^{n}{i=1}\frac{a_i}{a_i+a{i+1}}$, with $a_{n+1}=a_1$, and $a_1,\dots,a_n$ are positive reals?
final saddleBOT
how do you find the minimum of $\sum^{n}_{i=1}\frac{a_i}{a_i+a_{i+1}}$, with $a_...
soft zealotBOT
ihave<skissue>
jagged flare
n=1 is 1/2, n=2 is 1
but for n=3 you can do some bullshit like wlog a_1=1, a_i=eps^i for i≠1 and as eps aproaches 0 then the sum approaches 1
but i cant get the equality, i cant find it
unless theres some infimum bull but i dont know how that works
loud sundial
but i cant get the equality, i cant find it
you won't
loud sundial
how do you know you wont
contradiction suffices
Let $x_i=\frac{a_i}{a_i+a_{i+1}}$ and note that $x_1 x_2 x_3=(1-x_1)(1-x_2)(1-x_3)$
soft zealotBOT
Civil Service Pigeon
loud sundial
so what, like x_1+x_2+x_3=1+x_1x_2+x_2x_3+x_3x_1
covert tree
how do you find the minimum of $\sum^{n}_{i=1}\frac{a_i}{a_i+a_{i+1}}$, with $a_...
a_{n+1} is used here in the sum, and u r treating it as a_1 right?
oh wait im blind
i thought u said an=a1 lol
loud sundial
<:pandahmm:858834883760750602>
I should say this is because I wrote it differently
this might be right
but I'm lazy
jagged flare
huh
loud sundial
but I'm lazy
,texsp ||If the sum equals $1$, then $x_1+x_2+x_3=1$. Then
$$x_1 x_2 x_3=(1-x_1)(1-x_2)(1-x_3)=(x_2+x_3)(x_3+x_1)(x_1+x_2).$$
But by the AM-GM inequality, the right hand side is at least $8x_1 x_2 x_3$. So we have
$$x_1 x_2 x_3 \geq 8x_1 x_2 x_3.$$
$x_1 x_2 x_3$ is positive, contradiction.||
soft zealotBOT
Civil Service Pigeon
loud sundial
huh
x_1+x_2+x_3=1+x_1x_2+x_2x_3+x_3x_1
this might be relevant but my brain is too off to see the connection
I'm in my "I did something and it works so I quit" mood
jagged flare
> x_1+x_2+x_3=1+x_1x_2+x_2x_3+x_3x_1
this might be relevant but my brain is too ...
i mean x_1x_2+x_2x_3+x_3x_1>=0 with equality at atleast two being 0 (cause it has to be positive), but that implies two of a_i being 0, and /0 contradiction
loud sundial
mmm ur orz if you say it works then it probably works
jagged flare
ur orzer tho
loud sundial
ur orzer tho
me dum
jagged flare
**Civil Service Pigeon**
oh wait this is prolly generalizable?
jagged flare
actually this only proves that the sum cant be 1
whats stopping it to be less than 1 and basically skipping over it
or is it sum ivt thing
loud sundial
whats stopping it to be less than 1 and basically skipping over it
make sequence 
jagged flare
make sequence <:happy:908558103992225822>
whar
jagged flare
im no orz i cant think
covert tree
but for n=3 you can do some bullshit like wlog a_1=1, a_i=eps^i for i≠1 and as e...
is this not approaching 1/2?
jagged flare
no?
covert tree
oh wait no im stupid
loud sundial
im no orz i cant think
but for n=3 you can do some bullshit like wlog a_1=1, a_i=eps^i for i≠1 and as eps aproaches 0 then the sum approaches 1
The idea of considering an exponential sequence is very relevant, but don't send eps to zero.
covert tree
😅
covert tree
> but for n=3 you can do some bullshit like wlog a_1=1, a_i=eps^i for i≠1 and as...
example this works for all n i believe
or eps^{n-i+1} works as eps->0
prob just a typo
jagged flare
wait hold on
why is x_1x_2x_3=(1-x_1)(1-x_2)(1-x_3)
wont the variables be flipped
loud sundial
$$\prod^{n}_{i=1} \frac{1-x_i}{x_i}=\frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_1}{a_n}=1 \implies \prod (1-x_i)=\prod x_i$$
soft zealotBOT
Civil Service Pigeon
loud sundial
because $$x_i=\frac{a_i}{a_i+a_{i+1}} \implies 1-x_i=\frac{a_{i+1}}{a_i+a_{i+1}}$$
soft zealotBOT
Civil Service Pigeon
jagged flare
fr
loud sundial
I thought that $1-x_i=x_{i+1}$ for a min lol
soft zealotBOT
Civil Service Pigeon
loud sundial
short term memory ahh
jagged flare
carb is the real orz
loud sundial
carb is the real orz
🎁
soft zealotBOT
Civil Service Pigeon
covert tree
WLOG assume that $a_n$ is the largest entry. If $a_i$ is not an increasing sequence, then you can show that there exists at least $2$ entries for which the $\frac{a_i}{a_i+a_{i+1}}\geq \frac{1}{2}$. And this means that the sum must be greater than $1$. Now we consider the case where $a_i$ is increasing, then it is obvious that $\frac{a_n}{a_n+a_1}+\frac{a_{n-1}}{a_{n-1}+a_n}\geq 1$ using the fact that $a_1\leq a_{n-1}$ and you also yield a sum greater than 1
jagged flare
**Civil Service Pigeon**
dont throw trash in my help channel
loud sundial
dont throw trash in my help channel
lol
covert tree
WLOG assume that $a_n$ is the largest entry. If $a_i$ is not an increasing seque...
I think this works to show that the sum is greater than 1
soft zealotBOT
qwertytrewq
loud sundial
You could also just do
,texsp ||Let $M>1$ and $a_i=M^i$ for $i=1 \dots, n$. Then the sum is
$$\frac{M}{M+M^2}+\frac{M^2}{M^2+M^3}+\cdots+\frac{M^{n-1}}{M^{n-1}+M^n}+\frac{M^n}{M^n+M}=(n-1) \left(\frac{1}{1+M} \right)+\frac{M^{n-1}}{M^{n-1}+1}$$
which tends to $0+1=1$.||
soft zealotBOT
Civil Service Pigeon
covert tree
,texsp ||Let $M>1$ and $a_i=M^i$ for $i=1 \dots, n$. Then the sum is
$$\frac{M}{...
yeah this gives a possible infimum, combined with what I wrote it shows that 1 is indeed the infimum
jagged flare
**Civil Service Pigeon**
yeah do this for the general stuff then muirhead cause anything majorizes (1,1,1,...,1) (?)
loud sundial
yeah do this for the general stuff then muirhead cause anything majorizes (1,1,1...
Does Muirhead work
jagged flare
is there a sort of ivt for multivariable
loud sundial
is there a sort of ivt for multivariable
yeah
jagged flare
you get (binom coeff)×symsum
jagged flare
yeah
how does it work
like in the bounds, if we fix every other variable except one, the function is continuous, so we can do ivt, and this is for any single variable, so can we do this repeatedly to get ivt for the whole thing
loud sundial
you get (binom coeff)×symsum
Don't you only have "half" of a symmetric sum? Ex. We have $a_1^2 a_2$ but not $a_1 a_2^2$. The intuition about majorising is probably useful though.
soft zealotBOT
Civil Service Pigeon
loud sundial
like in the bounds, if we fix every other variable except one, the function is c...
Partial continuity isn't full continuity though. Also make sure that your domain is connected
jagged flare
Partial continuity isn't full continuity though. Also make sure that your domain...
yeah no idea what that means
jagged flare
**Civil Service Pigeon**
we do though?
loud sundial
we do though?
oh
jagged flare
,w expand (a+b)(b+c)(c+a)
soft zealotBOT
loud sundial
lol i meant to pop in and out for 1 being impossible and I still gotta finish my fuud so I'm washed
jagged flare
,w expand (a+b+c)(a+b+d)(a+c+d)(b+c+d)
soft zealotBOT
loud sundial
<:blobcry:971797574916907028> what did you thought i was talking about
$\sum \frac{a_i}{a_i+a_{i+1}}$ lol
soft zealotBOT
Civil Service Pigeon
jagged flare
loud sundial
I was trying to bs some shit with the numerator and denominator
but ur too orz to do something dumb like that
hence why I was like
am I missing smt
covert tree
how does it work
Imo the natural generalization: any continuous function $f:\mathbb R^n\to \mathbb R$ and two points $x=(x_1,\dots,x_n)$,$y=(y_1,\ldots, y_n)$ in $\mathbb R^n$, and for any $f(x)<a<f(y)$ there exists a point in $(x_1,y_1)\times (x_2,y_2)\times\cdots\times (x_n,y_n)\subseteq \mathbb R^n$, say $z$, such that $f(z)=a$
soft zealotBOT
qwertytrewq
cerulean epoch
Let \(n \geq 3\). We have numbers \(t_1, \ldots, t_n\) s.t.\ \(\sum t_i = 0\) and wish to show that
\[ \sum \frac{1}{1 + e^{-t_i}} > 1. \]
Denote \(f(t) \coloneqq 1/(1 + e^{-t})\) for shorthand. Assume there exist \(t_i\) such that \(\sum f(t_i) \leq 1\) for a contradiction. Since \(f\) is monotonic and positive, and \(f(0) = 1/2\), we must have exactly one \(t_i \geq 0\). Call this \(t_j\). Then the other \(t_i\)s must be in \([-t_j, 0]\)---hence these must have
\[ f(t_i) \geq \frac{1}{1 + e^{t_j}} = 1 - f(t_j) \]
which is a contradiction.
jagged flare
but ur too orz to do something dumb like that
nuh if it even works i wouldnt suggest if cause i i wouldnt even undersrand ir
covert tree
Imo the natural generalization: any continuous function $f:\mathbb R^n\to \mathb...
But this works for any path connected subset of $\mathbb R^n$ instead of the hyperrectangle
jagged flare
what the whar
soft zealotBOT
qwertytrewq
loud sundial
Anyway my lower tech generalisation was to sps $\sum x_n=1$, then
$$\sum_{j \neq i} x_j \geq (n-1) \left(\prod_{j \neq i} x_j \right)^{\frac{1}{n-1}} \implies \prod^{n}{i=1} (1-x_i) \geq (n-1)^n \prod^{n}{i=1} \left(\prod_{j \neq i} x_j \right)^{\frac{1}{n-1}}=(n-1)^n \prod x_k$$
and so
$$\prod x_i \geq (n-1)^n \prod x_i$$
which is bs
soft zealotBOT
Civil Service Pigeon
covert tree
**qwertytrewq**
if u wanna do n case doesnt this just work
if you want to show it has to be >1
loud sundial
**Civil Service Pigeon**
we were going along with generalising this
covert tree
Ah
jagged flare
**Civil Service Pigeon**
ok yeah this would prolly be better
covert tree
i c i c
jagged flare
cause muirhead i need to prove the coeffs are symmetric and such
jagged flare
is R^n (the domain right???) connected?
covert tree
is R^n (the domain right???) connected?
yes
jagged flare
wait so can we apply ivt then
covert tree
connected means that you don't have two disjoint open set seperating it
covert tree
wait so can we apply ivt then
no by path connected is a generalization of invervals for whuch you pick you point that satisfy ivt
jagged flare
ok I go back to eating dinner
byee
covert tree
no by path connected is a generalization of invervals for whuch you pick you poi...
so im changing the "[x,y]" in " for all x,y and continuous function $f:\mathbb R\to \mathbb R$, and any $f(x)<b<f(y)$, there exists $z\in [x,y]$ such that $f(z)=b$" to a path connected region.
soft zealotBOT
qwertytrewq
covert tree
or, simply, a path from x to y
jagged flare
what does a path mean here
covert tree
what does a path mean here
continuous function from $t:[0,1]\to \mathbb R^n$ such that $t(0)=x, t(1)=y$
soft zealotBOT
qwertytrewq
covert tree
you can think of this as traveling a path on $\mathbb R^n$ on the time interval $[0,1]$
soft zealotBOT
qwertytrewq
covert tree
such that you begin at $x$, and end at $y$
soft zealotBOT
qwertytrewq
jagged flare
is a path like a generalization of an interval
covert tree
is a path like a generalization of an interval
ummm you couuuulddd think of it like that, it's more or less like drawing a line on R^n
jagged flare
the line has to be straight?
like the thing im really asking is can it just straight up do loops?
covert tree
the line has to be straight?
oh no no no, that's not what I mean,
u can curve or whatever
jagged flare
mmm
covert tree
it's like one random stroke with your pencil
cerulean epoch
you can do loops
tbh i think the most natural generalisation is that continuous functions take connected sets to connected sets
jagged flare
so like, f is continuous and connected from R^n->R^n, if f(x)<b<f(y), then theres z on the path from x to y so f(z)=b
covert tree
here is an example: take any $f:\mathbb R\to \mathbb R$ to be continuous, and let $t:[0,1]\to \mathbb R^2$ to be defined as $t(s)=(s,f(s))$. Then the curve $t$ basically draws out a graph of $f$ on $[0,1]$
soft zealotBOT
qwertytrewq
covert tree
so like, f is continuous and connected from R^n->R^n, if f(x)<b<f(y), then there...
not from R^n -> R^n.
R^n -> R
jagged flare
covert tree
how r u ordering R^n in the first place to say f(x)<b<f(y)?
jagged flare
oh yea
covert tree
if you say $a<b$ iff every coordinate is less
soft zealotBOT
qwertytrewq
covert tree
then this is a partial relation
and the statement would be false with counterexamples in $\mathbb R^3$ i think
soft zealotBOT
qwertytrewq
jagged flare
then this is a partial relation
what does that mean
covert tree
what does that mean
this is intuitive it's not nothing formal i meant like it is not the case that a<b or a=b or b>a
if you define it that way
also bad word to use
i meant this is a partial ordering
partial order is formal and it means that not every two element are comparable
jagged flare
ohh
covert tree
like if you have a tree, then the relation "ancestor of" is a partial order
because it is possible for two nodes to not be ancestors of eachother
jagged flare
yeah i think ive seen that before but dont remember the name
covert tree
but it still satisfy everything we want for an ordering
jagged flare
so like, f is continuous and connected from R^n->R, if f(x)<b<f(y), then theres z on the path from x to y so f(z)=b
covert tree
that is a>b and b>c then a>c stuff like that
covert tree
so like, f is continuous and connected from R^n->R, if f(x)<b<f(y), then theres ...
yup
any path from x to y works
jagged flare
wow then wouldnt it be really general
covert tree
wow then wouldnt it be really general
yeah
because you can just compose the path with the function and you get a function from an interval to R i think
jagged flare
*this* is a path?
i think this ezample is for R^2->R (did i get it? :<)
covert tree
*this* is a path?
yeah
jagged flare
wow ok
covert tree
the function $t:[0,1]\to \mathbb R^n$ is what we call a parametrization
soft zealotBOT
qwertytrewq
covert tree
because sure, you can alk on the path, but at what speed are you traversing the path
that, t might be able to talk about (although keep in mind that this function might not even be differentiable)
jagged flare
okay okay
covert tree
this is the immediate thing that I think i could generalize
but it honestly just reduce to the usual case quite easily
but afaik IVT is not a topic at all in multivar calc, probably due to this underwhelming generalization
i can't think of others ways one would generalize
you could also think of it like this: "if V is connnected and f is continuous, then f(V) is also connected"
what's connected in R, only intervals work (this is a pretty difficult proof for now), so f(interval) iis a interval too
jagged flare
if we were to apply it to our problem
f:=(the sum)
f is a function fron R^n->R
the domain R^n is connected
by contradiction, let there exist y such that f(y)<1. let x be some variable such that f(x)>1. by the ivt f(x)>1>f(y), and there exists z on the path from x to y such thay f(z)=1. however, we proved that 1 is not in the range of f, this contradiction, and there is no y st f(y)<1?
jagged flare
but afaik IVT is not a topic at all in multivar calc, probably due to this under...
aw thats dissapointing
covert tree
aw thats dissapointing
However, there is a generalization for metric spaces
more interesting, and it includes the boring case for R^n
covert tree
you could also think of it like this: "if V is connnected and f is continuous, t...
this is the statement i believe
covert tree
if we were to apply it to our problem
f:=(the sum)
f is a function fron R^n->R
t...
Ah i see... that works i think
covert tree
tho i have no idea what metric spaces are :<
$(M,d)$ where $M$ is a set, $d:M^2\to \mathbb R$, $d$ tells you how far away two points are, given two points $x,y $, $d(x,y)$ tells you their distance
soft zealotBOT
qwertytrewq
covert tree
of course, we want the distances to satisfy some conditions like $d(x,y)=d(y,x)$ (distance from x to y is the same the other way), $d(x,x)=0$ (distance from x to x is 0), and $y\neq x$ impplies $d(x,y)>0$ (distance is positive between distinct points).
soft zealotBOT
qwertytrewq
covert tree
and $d(x,y)+d(y,z)\geq d(x,z)$ which is the triangle inequality
soft zealotBOT
qwertytrewq
jagged flare
Ah i see... that works i think
hopefully it does
covert tree
hopefully it does <:blobcry:971797574916907028>
you can prove it using the normal ivt if you don't want this generalization
jagged flare
can you?
covert tree
Let's use ur notation
covert tree
if we were to apply it to our problem
f:=(the sum)
f is a function fron R^n->R
t...
let $g:[0,1]\to \mathbb R^n$ be a path from $x$ to $y$, then $f\circ g$ is a continuous function from $[0,1]\to \mathbb R$. Since $f\circ g(0)=f(x)$ and $f\circ g(1)=f(y)$, by IVT there exists a point $z'\in [0,1]$ such that $f\circ g(z')=1$ and thus picking $z=g(z')$ suffice
soft zealotBOT
qwertytrewq
covert tree
i guess the only nuance is that you need to show composition of continuous function is continuous.
covert tree
ohh wait do you need to do this for the multivar ivt?
yeah but it is a quite standard $\epsilon \delta$ bash iirc
soft zealotBOT
qwertytrewq
covert tree
wait didnt you post something earlier
yeah
jagged flare
**qwertytrewq**
yea this