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12:01 AM
I'm trying to show that f and f' are cont. on [a,b] and f'' exists on (a,b) then there exists c in (a,b) s.t. f(b) = f(a) + f'(a)*(b-a) + f''(c)/2*(b-a)^2. any ideas? Ted helped me last night and was saying to use cauchy MVT on R(x)=f(x)-f(a)-f'(a)(x-a) and g(x)=(x-a)^2 but I'm still not seeing how the third term results from considering those functions
 
how do I centre an equation in latex for an answer?
 
@kyle: Tell me what Cauchy MVT tells you.
@JakeRose double dollar signs will do fine
 
12:16 AM
@TedShifrin It says that if R and g are cont. on [a,b] and R and g are diff. on (a,b), and g(a)=/=g(b) then there exists a c in (a,b) s.t. R'(c)/g'(c)=R(b)-R(a)/g(b)-g(a)
 
Hi Ted.
 
OK, so what are $R(a)$ and $g(a)$?
hi @anakhro
 
And hi the rest of you, I don't know any of you though. :(
 
They're both 0
 
So we have $R(b)/G(b)=R'(c)/g'(c)$. Can you apply it again, @kyle?
 
12:18 AM
Between two points in (a,b)?
 
where does $1/z$ map $y > -5$?
 
No, no. What are $R'(a)$ and $g'(a)$?
Where does the mapping map $y=5$, @JoeShmo?
 
@TedShifrin math question: if you had a nickle for every time someone made an "$x$: A Geometric Approach" joke, how rich would you be?
 
I'm rather over it, @anakhro.
But I realize that's my best legacy.
 
the circle that sits right below the horizontal, touching the origin
 
12:20 AM
You have a legacy of being a kind person here, at least.
 
of radius and center.. im not sure
oh
no no
 
Below the horizontal?
 
silly me
although im still not sure what the center are
 
well, g'(a) = 0 too...
 
Well, figure this out, @JoeShmo.
 
12:22 AM
below the horizonal, according to the picture im staring at
 
Yes, @kyle, good, and what about $R'(a)$? Look at the definition of $R$.
How do you end up with negative imaginary part, @JoeShmo?
 
it seems R'(a)=0 too?
 
Aha. Now finish, @kyle.
 
Rolle's?
 
No, no. We're doing Cauchy MVT again, remember?
 
12:23 AM
oh, I see. one sec then.
 
We shouldn't have used $c$ already, btw.
 
because we want an f''(c)?
 
Right. So save the letter $c$ until the last step :)
 
$y=5$ gets mapped to $-5(u^2+v^2)-v =0$
 
Oh, ugh. If you know it's a circle, you should find it immediately.
And what you just wrote is not an equation of anything.
 
12:28 AM
edited :0) although i should probably deny deny deny
 
I'm still not seeing it. I have R'(b)/g'(b)=R''(c)/g''(c) where R'(b) = f'(b) - f'(a) and g'(b)=2(b-a), R''(c)=f''(x). Should I combine both results somewhere?
 
You shouldn't have $b$ in there, @kyle. That was not an eligible letter.
Remember, we started with $R(b)/g(b) = R'(\xi)/g'(\xi)$ for some $\xi$.
 
should find who what immediately? $y>-5$ gets mapped to $\{y \le 0\}\cup$ circle above the horizontal
 
You know better than that, @JoeShmo.
 
agreed, and we reasoned that g'(a)=R'(b)=0... so I figured I can just apply cauchy again to get something with the second D
 
12:31 AM
oy vey
 
woops, g'(a)=R'(a)=0* typo
 
So $\dfrac{R(b)}{g(b)} = \dfrac{R'(\xi)-R'(a)}{g'(\xi)-g'(a)}$. Now finish @kyle.
 
ok. lets see. $\{y \ge 0\}$ gets mapped to $\{y \le 0\}$?
 
smacks @JoeShmo
Oh wait. I'm smacking too quickly.
 
!!!!!!!!!!!!!!!!!!!!
 
12:34 AM
<--- smacks self
 
:)
 
My apologies.
 
the smaller $y$ is, the larger the radius under the horizontal
 
So we should get a circle passing through $0$ and $-i/5$, right?
 
and the radius grows without bounds
 
12:35 AM
And orthogonal to the $y$-axis, of course. So that determines it.
You were right originally about a little circle under the horizontal axis. My apologies.
 
and orthogonal to the horizontal
ok
 
No, it's tangent to the $x$-axis.
 
yes yes
tangent* to the horizontal
 
So the part above $y=5$ clearly maps to ...
 
a circle below the horizontal, tangent to it at $0$
i said that earlier
 
12:36 AM
I'm back to $y>5$ now.
 
but R'(xi)/g'(xi)= f'(xi)-f'(a)/2(xi-a)
 
which means that $\{y > -5\}$ gets mapped to what i said
 
Why are you doing that, @kyle?
 
because that's the RHS since R'(a)=g'(a)=0
 
But we want to apply Cauchy MVT again.
@JoeShmo Typo? I'm lost.
 
12:39 AM
Maybe you'd like to comment on it
 
Then R'(b)/g'(b)=R''(c)/g''(c) ?
for some c in (a,b)
since R'(a)=g'(a)=0
 
@Balarka: How did you address his interest in Laurent?
Right, @kyle. Now finish!
 
$\{y>-5\} = \{y\ge0\}\cup\{0 > y > -5\}$ gets mapped to $\{y\le0\}\cup$ another disk above the horizontal this time, once again tangent to it at $0$
 
I thought the original question was $\{y>5\}$, @JoeShmo.
 
@TedShifrin I dunno what the jet space would be for meromorphic functions. If it's not defined at a point the jet is meaningless
 
12:42 AM
$-5$
 
There are all sorts of sheaf-type things with principal parts, though.
 
Maybe you can look at things with fiber being a space of formal power series instead of polynomial functions
 
But g''(x)=0
 
But that sounds annoying
 
yes. I put $-5$ up there
 
12:42 AM
So you told me the wrong question to start, @JoeShmo. OK.
 
nope
 
Oh, you didn't. Hell with me.
OK, so, yeah, you're right that you want to break it into two parts. I don't think you're right about the disk.
@kylecampbell Say what?!
 
I yield f'(b)-f'(a)/2(b-a)=f''(c)/0
 
why? the strip gets mapped to the disk above no?
 
Aw come on, man.
Where does $0$ map, @JoeShmo?
 
12:45 AM
back to itself
 
Say what?
 
$y=0$ presumably?
 
I think this time you get a smack.
3
 
$\infty$
 
Aha.
So that has to be a limit of points to which points in the strip $-5<y<0$ map.
This room is driving me to drink. It's time for a few martinis.
 
12:48 AM
hmm.. ok
no wonder youre not following the questions
 
Watch it, boy.
 
@kyle: Have you reconsidered the second derivative of $(x-a)^2$?
 
wow, oops
 
That and $1/0=0$ is driving me to martinis.
 
12:51 AM
it's just 2 then, so f'(b)=f'(a)+f''(c)(b-a)
 
Come on. Just put things together carefully.
Go back to the equation I displayed ages ago and finish correctly.
 
hmm
you're referring to R'(b)/g'(b)=R''(c)/g''(c)?
 
i thought you meant $y=0$
which i still contend gets mapped to $y=0$ under $1/z$
 
OK ... I'm done multitasking. Bye.
 
no hold on
 
12:56 AM
aw man.. well thanks
 
this is almost a palindrome
depending on the number of martinis youve had
oh. is it everything BUT the disk whose boundary passes through $\{(0,0),(0, i/5)\}$
ted's gone :(
 
he's tired of us
 
can't blame him
 
lol
nope
 
one can only put up with so many stupid questions in one day
its $\{(0,0),(0,1/5)\}$* btw
 
1:26 AM
man, I'm still just getting f'(b)=f'(a)+f''(c)(b-a) in the first, and in the second I get f(b)-f(a)-f'(a)(b-a)/(b-a)^2=f'(xi)-f'(a)/2(xi-a). If only I could let xi=b.
rather in the second, then the first
 
whats the question?
 
I'm trying to show that if f and f' are cont on [a,b] and f'' exists on (a,b) then there exists a c in (a,b) s.t. f(b)=f(a) + f'(a)*(b-a) + f''(c)/2*(b-a)^2. Ted recommended considering cauchy MVT with R(x)=f(x) - f(a) - f'(a)*(x-a) and g(x)=(x-a)^2
I see some progress but I'm not quite there yet...
 
right. just iteratively apply MVT?
 
I applied it twice yes
 
you could also check out how taylor's expansion is derived
the error bound
 
1:39 AM
yeah, I remember it's the remainder but I'm trying to just show it with cauchy
 
cauchy what?
 
cauchy MVT
it seems as though I have to combine the results for R(b)/g(b)=R'(c)/g'(c) and R'(b)/g'(b)=R''(c)/g''(c)
but I don't see how, and I guess I should call them c_1 and c_2
 
2:09 AM
I figured it out another way by defining another function where it =0 at the endpoints and applying Rolle's twice, but I still want to know how Ted did it with Cauchy MVT :(
 
2:32 AM
Man, I just missed Ted. I have a proof to show him :\
 
 
1 hour later…
3:43 AM
20 hours ago, by Silent
35 mins ago, by Silent
how to show in Young's inequality, if equality occurs, then $a^p=b^q$? @TedShifrin
 
Hello
Guys I have a question
is the ball in topology the same as the neighborhood of a point?
 
4:12 AM
@CaptainAmerica16 Generally not in America. However, if I remember right, "positive" meaning $\geq 0$ is/was common in France.
 
4:28 AM
@mathsresearcher depending on the context used, yes. mathworld.wolfram.com/OpenBall.html
 
4:38 AM
Guys, I have another question
consider Let \textbf{a} $\in$ R. If there exists an \textbf{x} $\in$ R such that $\forall$ $\epsilon>$0, \textbf{x} $\in$ B(\textbf{a},$\epsilon$) then $x=a$.
why is it that x after the epsilon>0
does not have a dependency on epsilon?
is it because its uniqueness was specified in the beginning?
@TedShifrin can you help please?
or is it that B(\textbf{a},ϵ) has the dependency on epsilon?
 
What context?
@mathsresearcher It doesn't have dependency on epsilon because you picked an x before considering the various epsilons
 
I have another question
how would I interpret
a ball
when talking about convergent sequences
in R^2
intervals?
or can I interpret them as literal balls?
 
4:54 AM
8
Q: What is the difference between open ball and neighborhood in real analysis?

John HassI'm learning real analysis. Open ball: The collection of points $x \in X$ satisfying $|x - x_{0}| < r$ is called the open ball of radius $r$ centered at $x_{0}$ Neighborhood: A neighborhood of $x_{0} \in X$ is an open ball of radius r > 0 in $X$ that is centered at $x_{0}$ I'm usin...

A ball in R^2 is going to be an open disk. Not an interval, 'cause an interval is a one-dimensional thing.
 
so how would I interpret
a ball
when discussing the convergence of
the sequence (1/n)
 
... over R?
So not R^2?
 
oh i see my confusion
i thought just because when graphing sequences
they are supposed to be in R^2
 
A ball in R is an open interval. A ball in R^2 is a open disk. A ball in R^3 is an open sphere, etc.
 
yeah, i understand it now
 
5:40 AM
If I have two functions in x and y which are continuous then is the product of them continuous function of two variables
 
5:57 AM
Can anyone give some idea to proceed in this prob. - In how many ways $11$ items can be distributed among $3$ peoples such that, number of items received by any two person is more than the number of items received by the third person.
 
6:07 AM
It can be done case by case though.. I want a general approach, so that it works for big numbers. Like 13 people and 100 item etc.
 
6:45 AM
Well, first thing I think is "How many items does one person need to receive in order to make it impossible for every other person to have more than them?"
 
7:02 AM
if f is a diff. function on (a,b) and we have x,z,y in (a,b) with x<z<y can we conclude by MVT that there exists c_1 in (x,z) and c_2 in (z,y) such that f(z)-f(x)/z-x=f'(c_1) and f(y)-f(z)/y-z=f'(c_2)? I mean, can we partition a known diff. set into smaller subsets and make the same conclusion conclusion from MVT?
 
 
2 hours later…
9:03 AM
The difference between two consecutive cubes is $$(n+1)^3-n^3=n^3+ 3 n^2+3 n+1-n^3= 3 n^2+3 n+1$$ We have that

- if $n=0$ then the difference is equal to $1$.

- if $n\geq 1$ then $3 n^2+3 n+1\geq 3 +3 +1=7$

What do we have if $n\leq -1$ ?
 
 
1 hour later…
10:27 AM
@apnorton, @kylecampbell, you may like this, that is, $x^p$ continuous for any real power.
 
10:51 AM
@Fargle Really? Does that not make things confusing communicating some things with mathematicians outside of France?
 
@MaryStar Complete the square
Alternatively, compute $(n+1)^3-n^3$ for various values of $n$
(from $-5$ to $5$, say)
and see if you notice anything
 
I wonder, if this will help reboot continuum analysis...
 
11:25 AM
@AkivaWeinberger,
yesterday, by Silent
35 mins ago, by Silent
how to show in Young's inequality, if equality occurs, then $a^p=b^q$? @TedShifrin
 
11:45 AM
Do you mean the following?
$$(n+1)^3-n^3=n^3+ 3 n^2+3 n+1-n^3= 3 n^2+3 n+1=3 n^2+6 n+3-3n-2=3 (n^2+2 n+1)-3n-2=3 (n+1)^2-3n-2$$
Or can we just say that the differences for the negative ones are the same as for the positive ones? @AkivaWeinberger
 
I meant $3n^2+3n+1=3(n+\frac12)^2+\frac14$
but we can in fact say that the differences for the negative ones are the same as for the positive ones because $(n+1)^3-n^3$ doesn't change when you substitute $-n-1$ for $n$
 
12:27 PM
Raising both sides to the $1/(pq)$ power, we get that $a^p=b^q$ implies $a^{1/q}=b^{1/p}$, and this implies $a^{1/p+1/q}=b^{2/p}$ @Silent
 
12:40 PM
Ok, working on it. Thank u
 
1:38 PM
@AkivaWeinberger
 
@AkivaWeinberger No, no! I wanted to know: If $a b = \dfrac {a^p} p + \dfrac{b^q} q$, then how to derive $a^p=b^q$. I think you have helped me with converse, which is fine to me.
 
Oh
Hm, so we have $ab-b^q=\dfrac{a^p}p+\dfrac{b^q}q-\left(\dfrac{b^q}p-\dfrac{b^q}q\right)$
$ab-b^q=\dfrac{a^p-b^q}p$
Don't think that goes anywhere
 
2:03 PM
ok
 
2:39 PM
@blue_eyed_...
 
2:49 PM
what's with the huge number of homework questions in the chat
Makes me want to drink vodka
and then smack them all with a Dedekind finite set
 
3:16 PM
@AkivaWeinberger Ok! Thank you!! :-)
 
3
Q: Perko pair - What's the handedness of these pictures?

Akiva WeinbergerIn 1974, a paper titled On the Classification of Knots appeared showing that the knots $10_{161}$ and $10_{162}$ in Dale Rolfsen's knot table were actually the same knot. He included this picture, showing how to deform one into the other: From that point on, $10_{161}$ and $10_{162}$ became kn...

Finally finished this
^Perko knot
 
3:32 PM
What conditions do you need for L2 integrability to imply L1 integrability?
Or is there no relationship?
Thinking of an expected value measure
 
what will be the answer of a and b? I am not in the situation of showing my work or even trying it.is there anyone to lend me a hand on it?
also I am asking for a clarification.
 
is there a name for a structure which contains parallel values, for example, an array which contains values that mirror their indexes?
0->0,1->1,2->2 etc...
or is this so mundane that there is not a concept or name associated with it
forget it, figured it out.
 
4:05 PM
Hey does anybody know if the function $\frac{1}{m^2+|k|^2}$ for $m > 0$ has a easy inverse Fourier transform for $k \in \Bbb R^n$? I know the solution for $n=1$ is $\frac{1}{2m} e^{-m |x|}$. Is the answer the same for the general case?
 
@AlessandroCodenotti: Care to help me with a logic question?
 
Depends on the question
 
Sure.
Before that let's agree on the following.
We have different kind of Logics. Like Predicate Logic, Modal Logic, Temporal Logic, Paraconsistent Logic. Right @AlessandroCodenotti?
 
Yes, but I must warn you I only know about the first
 
@AlessandroCodenotti That's not an issue. My question is 'simple': What is the definition of a "Logic"?
In other words, why do we call all of them a "Logic"?
 
4:11 PM
I think a logic is made up of its syntax and semantic, but I've never really dealt with other logics apart from FOL so I don't really know how much structure do you need to have something you can call "a logic"
 
-1
Q: Prove that the straight lines whose direction cosines are given by the relations

blue_eyed_...Prove that the straight lines whose direction cosines are given by the relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are parallel if $\sqrt {af} \pm \sqrt {bg} \pm \sqrt {ch}=0$. My Attempt: Given: $$al+bm+cn=0..........(1)$$ $$fmn+gnl+hlm=0........(2)$$ From equation $(1)$, $$l=-\dfrac {bm+cn}{...

Anyone?
 
@AlessandroCodenotti But why do you call FOL a "Logic"?
Is it just an arbitrary choice for an arbitrarily selected formal system?
 
hmm I've never thought about it. I guess there is a proper definition of what "a logic" is when you study different kind of logics and FOL fulfills it
 
@AlessandroCodenotti My thoughts exactly.
Any idea @LeakyNun?
For example, it seems clear to me that for each Logic there must exist a unique set whose elements we will call wff.
It should also have a notion of "follows from" but exactly defining it seems to be the main challenge.
Anything else that you thing should be an integral part of a "Logic" based on your intuitive conception of logic @AlessandroCodenotti?
 
All you talked about so far falls in the syntax part, I think the semantic should also be part of the logic. That is how to interpret which sentences are true and false in a model
 
4:24 PM
So we need to have notions of "follows from". For now it seems to be at least two. One for the syntax and one for the semantics.
This seems to suggest that we can identify a Logic with the following pair. $\langle L,\vdash,\vDash\rangle$ where $L$ is a non-empty set whose elements we will call well-formed formulas and $\vdash$ and $\vDash$ are two binary relations.
But, of course, this is still very imprecise.
Hi @MatheinBoulomenos.
 
Hi @user170039
 
5:25 PM
When can you deduce L1 integrability from L2? I'm thinking about random variables and variance
Or do you have to check L2 and L1 seperately always?
 
@GFauxPas if $X$ is a probability space, then you can use Jensen's inequality applied to the convex function $x \mapsto |x|^{p/q}$ for $p>q$ to show that $\|f\|_{L^p} \leq \|f\|_{L^q}$ for every measurable function $f$ on $X$
in detail: $\|f\|_{L^p}^p= \int_X |f|^p=\int_X (|f|^q)^{p/q} \leq (\int_X |f|^q)^{p/q}=\|f\|_{L^q}^p$, so taking the $p$-th root gets result
by rescaling, you can also get a modified result for each finite measure space
 
Ooo
Thanks!
 
but if the measure space is not finite, this doesn't work and indeed there won't be any inclusions between $L^p$ spaces in general
 
But Lp and Lq always imply Lr for p < r <q?
 
yeah
I meant if you don't take an intersection
 
5:39 PM
Thanks!
 
5:50 PM
Wait, you sure you have the direction of ineq correct there
?
 
oh right
Jensen's inequality goes the other direction
sorry
so $\|f\|_{L^p} \geq \|f\|_{L^q}$
 
Thougt something was whacky but nbd
Cool beans
Thats a nice inequality for moments then
 
6:38 PM
if sum A is bigger then sum B then if i raise each summand to some power greater then 1 does the inequality still hold?
and each term is bigger then 1
 
6:58 PM
Evening all
 
@ÍgjøgnumMeg evening
 
Hey @Mathein, how's it going?
 
pretty well, thanks
and yourself?
 
Yeahhh.. just preparing to leave the country in 2 months
lol
 
7:15 PM
you'll be in good company here, lots of guys are doing algebra/NT here
 
I prepared 2 days before, you're already in the German mindset!
 
7:31 PM
hahaha
I'm just trying to find a flat that doesn't cost 99999€ per month
atm I'm looking in Schwetzingen
 
@ÍgjøgnumMeg Good luck, if it is like in Bonn the situation is dire
 
according to someone who moved from Bonn to Heidelberg, Heidelberg is worse than Bonn when it comes to finding a flat
 
Sad, apparently the area is quite "flat" topographically
so I don't mind living further out
 
Schwetzingen is a nice place
there's the philosopher's way in Heidelberg, that's not flat
 
Lol I've seen that, I've also looked at Ziegelhausen which is up a mountain way isn't it?
 
7:40 PM
yeah
 
Could enjoy that.. lol
 
 
1 hour later…
8:55 PM
Hi. $X$ is a normed space over $\Bbb C$. given $x_1,\dots,x_n$ i need to prove that $x_1,\dots,x_n$ are independent iff for all $\alpha_1,\dots,\alpha_n \in \Bbb C $ there exists $\phi \in X^*$ s.t $\phi(x_i) = \alpha_i$
Suppose the x'is are independent , then define $Y = span(\{x_i\})$ . and given $\alpha_1,\dots,\alpha_n$ define $\phi(x_i) = \alpha_i$ and this way we have a linear functional on $Y$
its bounded (right? ) so we get that we can expend $\phi$ to all $X$ by Han-Banach. where did i need the fact that the x'is are independent?
 
Because you cannot always define such a $\phi$
For instance, if $\sum \lambda_ix_i =0$, then $\sum \lambda_i \alpha_i = 0$
@Liad
 
i guess something with linearity of $\phi$ won't work? (im trying to see what's the problem in your example..)
 
Remember you want that true for all $\alpha_i$ and you can fix the $\lambda_i$
 
ah. so we can take $\alpha_j = 1$ for $\lambda_j \ne 0$ and the rest zero to get a contradiction
so that's the other direction.
 
9:11 PM
Hi!!
 
Hi!
 
-1
Q: A function on $X$ is continuous iff its restriction to each element of an open cover is continuous

shimaaProblem If $\{ B_\lambda : \lambda\in\Delta\}$ is any collection of open subsets of $X$ whose union is $X$, a function on $X$ is continuous iff its restriction to each $B_\lambda$ is continuous. Context This is not true for general covers. For example, $X=(0,2)$ is the union of $(0,1)$ a...

why we took $U$ open in $Y$? I thought $U$ was an open cover of $X$
Heyo@apnorton
 
$\{U_i \}$ is an open cover of $X$, but $U$ is some open subset of $Y$.
There's a little bit of confusing notation in that answer due to variable reuse, imo
 
9:29 PM
yup, is it correct that $\cup_{i=1}^{n} U_{i} = U$ @apnorton
 
No
 
supports @apnorton's claim that the notation sucks: Let's use $V\subset Y$ for an arbitrary open subset of $Y$.
 
oh ok!
then we look at $f^{-1}(V)$ ?
Now how we can connect this to the open covers $U_{i}$ of $X$
 
@Baymax: In general, it's good to use letters that give some indication of where things live. If $f\colon X\to Y$, use $y\in Y$ and $x\in X$ and $y=f(x)$. Use different letters for subsets of $X$ and $W$, etc.
 
Hm, I see
 
9:37 PM
$f^{-1}(V) = \bigcup \big(f^{-1}(V)\cap U_i\big)$.
 
ohh
Now how can i say that $f^{-1}(V)$ is open?
@TedShifrin
 
Hi demonic @Alessandro
@Baymax: If you're assuming $f$ is continuous, then it is.
But the equation I wrote holds for any set.
 
I'm guessing there's no nicer name than the following, but
 
no we only have the info that $f$ is continuous restricted to each of the open covers and we need to prove that $f$ is continuous
@TedShifrin
 
9:46 PM
Suppose I have n-dimensional Euclidean space and some affine subspace of it, and I restrict it to the first orthant. So I'm interested in the nonnegative part of said affine subspace.
 
OK, What is $(f|_{U_i})^{-1}(V)$?
 
Is there a better name than "nonnegative part of an affine subspace"?
 
is it $U_{i}$ ?
@TedShifrin
or some subset of $U_{i}$ ?
 
What subset?
 
open subset?
 
9:50 PM
Hint: Look at the formula I wrote above
 
$f^{-1}(V) \cap U_{i}$ ?
 
Yup.
 
UYeah
 
Hello, i just got here, what are y'all talking about?
 
9:54 PM
A topology related question
 
@Semiclassical I would refer to as the portion in the first orthant. Perhaps coin a short name and stick with it in your writing.
 
@TedShifrin but after that how to proceed?
 
Union of open sets is open.
 
Use a + as a subscript maybe
 
9:55 PM
said subset is certainly convex (affine subspaces are convex, and the first orthant is convex, so their intersection is convex)
 
you mean take union $\cup f^{-1}(V) \cap U_{i}$
@TedShifrin
 
oh damn, this looks like complicated stuff
 
@Baymax: Exactly as I wrote it above.
 
And from what I know of my problem I'm pretty confident it'll always be a bounded convex set
Main reason I'm looking for known names is to see if there's a better algorithm to run
 
how do we know that $f^{-1}(V)$ is open ?
we hv to prove $f$ is continuous
 
9:57 PM
"nonnegative vertex enumeration", maybe
 
only knowing that f restricted to each open set $U_{i}$ is continuous
 
We established that $(f|_{U_i})^{-1}(V) = f^{-1}(V)\cap U_i$, remember?
You need to prove that.
 
OH GOTCHA
!!
 
might see if there's anyone in our math department who knows this stuff like the back of their hand
(think I know the first person I'll ask, though it's probably too boring for him)
 
sO $\cup(f|_{U_{i}}^{-1}(V) = \cup(f^{-1}(V) \cap U_{i})$
 
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