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12:21 AM
does $-\Delta x + \ln(1+\Delta x)$ imply the total is less than 0
i.e. is $\Delta x > \ln(1+\Delta x)$ for all $\Delta x > 0$?
 
@Obliv yes
equality holds only for $\Delta x = 0$
 
Hint: First degree Taylor + remainder
 
12:37 AM
or use that $\ln$ is strictly concave
 
or Hint: Beginning of $1/(1+x) = $ …
 
1:06 AM
-2
Q: A property of morphisms about Euclidean spaces

portonIn 2019 I gave definition of "space" that encompasses all kinds of spaces in general topology (metric spaces and locales/frames and everything in between: topological spaces, uniform spaces, etc.) (This paragraph is just to give a background, this is not used in the question.) The definition of a...

Hmmm
its an interesting question. A bit unclear on what author means but looks creative
does anyone have an idea what is he talking about?
 
1:59 AM
thanks to me he corrected his post like 5 times already
seems like it wasn't just me being confused
 
2:17 AM
the defn does not make sense to me
 
yea
 
 
5 hours later…
8:19 AM
Hi chat!
I am trying to solve the simultaneous equations for $x,y,z,u$. The simultaneous equations are \begin{aligned*} \begin{equation*} \frac{a-x}{b} - z - u &= 0,\\ x - min(\beta u, c y) &=0,\\ x -y &= 0,\\ x-1 &= 0\\ \end{equation*}\end{aligned*}
$ \frac{a-x}{b} - z - u = 0,\\ x - min(\beta u, c y) =0,\\ x -y = 0,\\ x-1 = 0$
From last two equations, we have $x = 1, y =1$, then what is $u,z$ ?
For the second equation we will have $1 = min(\beta u, c.1)$
$a,b,\beta,c$ are constants
 
8:40 AM
hi guys
 
9:14 AM
Hiiigh
πŸŒΏπŸ€πŸ
 
@EE18 I don't really know about Halmos, but if you want to study set theory formally from the beginning, you can check out Enderton's textbook "Elements of Set Theory". I have personally read it and so far I had no problems with it.
 
 
1 hour later…
10:24 AM
@DanielDonnelly Hi :)
 
Hey mn
:D
Not so sure about my point now. Working on TPC at elementary level is going in circles!
 
How are you? I saw your email this morning.
 
Doing good
Hw are you?
Made $1500 last month, but this mnth no work
 
I could be better. I had to cancel plans for the day as I'm struggling with avolition :/
@DanielDonnelly That's cool :)
 
Oh not wanting to do stuff?
Same here
Maybe do new things for novelty feelings
 
10:27 AM
Yeah, but more clinical; it's a symptom of schizophrenia.
 
I have that too
They had me with BP, but then changed it
I'm not sure what I am
 
I want to do things. I just can’t bring myself to do them.
I'm sorry to hear that :(
 
Try serotonin - how much do you exercise?
A swift jog / walk will boost your sertonin then you'll want to do stuff even quit smoking e.g.
 
I don't exercise at all. I used to hit the gym twice a week before the pandemic though l
 
There you go that's always the reason, fr me too !
 
10:29 AM
My OT and I are working towards getting back into that habit.
I always feel bad cancelling plans over avolition. It feels like I'm just being lazy.
 
Don't worry about it too much. I'm probably worse than you
 
I was in great shape, too. I had a skinny guy six pack going on.
 
What I mean is, everybody can't fit this normal mold perfectly
 
Indeed :)
 
Oh you studied $\textbf{Ab}$s aka abelian grups?
lol
 
10:33 AM
Lol :)
 
Let's study together bro
we could make it a rule to do pushups or something for 5 minutes every 30 minutes
 
I'd like that!
 
You know I ran out of paper I write so much
 
I can hardly do one pushup lately though
 
here I am scribbling notes in an EMFT shaums outline book
I have carpel tunnel from both keyboard and majorly writing now
have to do yoga or something
@Shaun if you are interested in elementary number theory stuff like sums over divisors, I could teach that
It's strange, each integer is like a simplex
nvm β•°οΌˆβ€΅β–‘β€²οΌ‰β•―β•°οΌˆβ€΅β–‘β€²οΌ‰β•―
 
10:37 AM
Why? What's up?
 
Oh, if you take $q_1^{k_1} \lt q_2^{k_2} \lt \dots q_n^{k_n}$ where $k_i \geq 1$ and our input number $X = \prod_{i = 1}^n q_i^{k_i}$, then the above inequalities are strict and give the "natural ordering" (of the simplex) such that the following map is a chain complex on certain $\Bbb{Z}$-modules of functions. Anyway, I could show you that some time. I don't know how to apply it though to ANYTHING!!!
 
I could use books as dumbbells:)
 
$q_i$ primes
You need a 40lb kettlebell. I got one
So you're studying matrix applications in group theory?
That goes deep
So I hear, I've onlly seen glimpses from Lang
 
Yes :) Mostly $SL(n,q)$.
 
Special linear group of $n\times n$ mats over $\Bbb{Z}/q$ the field?
 
10:43 AM
The project started out as studying linear algebraic groups. And yes :)
 
I know absolutely nothing about that :D
 
But with det 1.
 
But you get all sorts of interesting theorems when you do that
matrixify things
namely rings lol
 
Every finite group is a linear algebraic group.
 
what's that?
 
10:45 AM
But SL(n,q) and other groups are more direct in how that's so.
 
They nearly solved the graph isomorphism problem using permutation groups
What's weird is first they encode the graphs as strings and then solve the string isomorphism problem using groups
 
A l.a.g. is a group whose elements are solutions to polynomials, essentially.
 
A string is just "abababaccacaba"
 
There's a bunch of ways do define them.
They're app complicated.
*all
 
So the elements together solve a single polynomial?
 
10:47 AM
Yes :)
 
So strange!
 
The Zariski topology plays a role.
 
I know that
 
Topology isn't my forte :(
 
Closed sets are algebraic zeros
to familys of polys
 
10:49 AM
I could use an HPC soon.
 
O__O only made it through the first chapter of AG lol
What's HPC?
 
I did calculations on my laptop and they took ages, so that's why.
 
Oh, IC
You need to learn C++ or D
or C#
 
"High Performance Computer", almost a supercomputer.
 
something faster than Python / Sage
But I believe GAP is already implemented in C, so that's good to go
 
10:51 AM
Yeah, I use GAP.
 
I got a Getac rugged laptop with tiny touch screen and convertibility into tablet for $400
 
The amount of issues we've had getting the HPC to work in stupid.
 
At school?
 
Anyway, I'm computing a function Delta of SL(n,q).
Yeah.
They have two here.
 
Is it linux env or something?
 
10:52 AM
I'm using the less powerful one for now.
I don't know.
 
Write her in assembler using SSE instructions
 
I'll leave it to the IT team!
 
Maybe the HPC has virii
Someone made a supercomputer virus at school lol
 
Maybe!
 
But it was just a randomly buggy group theory program somene wrote that spawned the viral code into the machine
 
10:55 AM
They're pretty strict on security for it though.
The HPC version of GAP is in its alpha stage still. So maybe!
 
Self replicating & mutating programs wuld be cool, but it's very non-trivial
 
I might just use GAP instead, the ordinary version, but that's got the same issues on HPC so far anyway.
 
Buy online supercomputer HPC time
perhaps
then you can scale with what you require
 
I didn't know you could do that!
 
I don't know where, you'd have to look into it
 
10:58 AM
But with two at my disposal already, I should be fine!
 
But just general purpose "cloud" things such as Amazon AWS have pay per usage and you can upgrade hardware for dough
 
I'm going to get something to eat now. It's nearly 11:00 and I haven't had breakfast.
 
I've gt it
 
See you later :)
 
build your own photonic computer (analog) that just solves the one Delta Function cmputation
 
11:33 AM
I have some basic questions regarding a sentence in my measure notes;
> For example, if $N\subset \mathbb R$ is not a Borel measurable subset of $\mathbb R$, then $\{0\}\times N$ is not a Borel-measurable subset of $\mathbb R^2$ even though it is contained in the subset $\{0\}\times \mathbb R$ of zero Lebesgue measure on $\mathbb R^2$.
Why is $\{0\}\times N$ not a Borel-measurable subset of $\mathbb R^2$? Why does $\{0\}\times \mathbb R$ have zero Lebesgue measure on $\mathbb R^2$?
 
Read the FIRST word of the example.
"If".
The product of a non Borel measurable set and a Borel measurable set is not Borel measurable.
 
ok, thanks, that clarifies the first part
 
But the product of a Lebesgue measurable set and a Lebesgue null set has Lebesgue measure zero---in this case, the set is contained in a line segment, which had 2d Lebesgue measure zero.
 
11:49 AM
ok, I need to look this statement up further, i.e. that a product of a Lebesgue measurable set and a Lebesgue null set has Lebesgue measure zero
 
This follows from lower dimensional linear subspaces having Lebesgue measure zero
 
How do you define the two dimensional Lebesgue measure?
As I recall, the usual way to do this is to define a premeasure on $\mathbb{R}^2$ by assigning the measure $\lambda^2(A\times B) = \lambda(A)\lambda(B)$ for any measurable subsets $A,B \subseteq \mathbb{R}$.
Then construct the complete measure in the "usual" manner.
 
@XanderHenderson in my notes, a subset $N\subset \mathbb R^d$ has Lebesgue measure zero if and only if for every $\epsilon>0$ there is a countable collection $\{I_n : n\in\mathbb N\}$ of, not necessarily disjoint, $d$-dimensional rectangles such that $$N\subset \bigcup_{n=1}^\infty I_n, \quad \sum_{n=1}^\infty m(I_n)<\epsilon.$$
 
But here $N$ is not necessarily Lebesgue measurable
 
I'm using these notes, example 6.
Above, $m$ is the Lebesgue measure, sorry.
 
11:57 AM
@psie Each natural number $n$ is contained in the rectangle $[n-1/2, n+1/2] \times [-\varepsilon/2^{n+1}, \varepsilon/2^{n+1}]$.
(That is, each point $(n,0)$ in $\mathbb{N}\times\{0\}$).
 
Why $N\times \{0\}$ is not Borel measurable in $\mathbb{R}^2$, is that otherwise it would be Borel measurable in $\mathbb{R}\times \{0\}$. So $N$ would be Borel in $\mathbb{R}$.
 
12:28 PM
Oh, I just realized that I misread the question $N$, not $\mathbb{N}$. Doink. Still, enumerate the integers in some manner (e.g. $(0,1,-1,2,-2,3,-3,\dotsc)$). Then cover $N\times\{0\}$ by $[z_n-1/2,z_n+1/2]\times [-\varepsilon/2^{n+1},\varepsilon/2^{n+1}]$, where $z_n$ is the $n$-th integer in your enumeration.
 
12:44 PM
anyone here familiar with the term backward error from numerical analysis?
 
1:04 PM
thanks for the help πŸ‘
 
@ShaVuklia I am familiar with it, but I haven't thought about it since my masters degree (so it's been... uh... more than a decade).
 
let me give it a shot then :)
above is a definition of backward error
 
I'm trying to understand why the backward error is useful (and easier to compute)
I found one explanation in another book (which works with a slight modifcation of the definition)
 
My recollection is that the backward error is a measure of how much you can perturb the initial data before you aren't looking at the right solution any more. This is practical, since we actually have some control over how precise our measuring instruments are.
 
1:13 PM
they say that the forward error is hard to compute, because we don't know the true value of $f(x)$. however, for the backward error we still need to solve $\hat y=f(x+\Delta x)$ using the true function $f$ (instead of the algorithm), so wouldn't that be equally hard if not harder?
 
What is Figure 1.1? Have you looked at that?
 
@XanderHenderson that is sensible, but the definition already works with the measured $\hat y$, so it's not as if we know beforehand whether $\hat y$ is close to $f(x)$
figure 1.1 explains what's going on formally
 
Yeah, I figured it was going to be that picture.
 
but to me it doesn't clarify why the backward error is useful (or easier to compute)
 
You compute $\hat{y}$, right?
That is the thing your algorithm gives you?
 
1:16 PM
yes, that's our numerical result
one interpretation I can get behind:
if the backward error is big, then we would have to deviate a lot from $x$ to obtain the numerical solution exactly
and I suppose if $f$ is continuous, and our backward error is small, then we know that the actual solution is close to the real value?
 
No, I don't think that is the right way of thinking about it.
 
I am glad I mentioned it then, because I would like to know the right way :)
 
It is more about how much you can perturb the input, and still get the same output, which may or may not be close to the "true" solution.
Like you compute $\hat{y}$ using $x$. But $\hat{y}$ isn't actually $f(x)$. It is actually $f(x + \Delta x)$ for some error term $\Delta x$. The backward error is the smallest size of that error term $\Delta x$.
 
I assume small $\Delta x$ is good?
which would then indicate that we can apply some perturbation? (I don't see that implication)
 
Yeah, in general, you want your errors to be small.
No, it isn't about "applying" a perturbation.
It is about understanding how much the data can be perturbed until things go wonky.
If $f$ is sufficiently nice (e.g. $C^1$), then small perturbations in the initial data should lead to small perturbations in the output. The backward error is, in some sense, giving a bound on this (smaller is better).
My very vague recollection is that the forward error can be estimated in terms of the backward error and the condition number...?
 
1:25 PM
yea, that works by definition
 
But, like I said, it's been more than a decade.
 
I'm afraid still don't see the implication: if $\Delta x$ such that $f(x+\Delta x)=\hat y$ is small, then "the data can be perturbed without things going wonky"
it seems like there is a tacit assumption there that $\hat y$ is already pretty good
but if we knew that, we would be done (since backwards error is presented as a more computation friendly thing to check than forwards error)
so we don't know if $\hat y$ is good, and therefore I don't see how $\Delta x$ being small tells us anything
(if we already knew that $\hat y$ is good, then we would know that a small perturbation still yields a good result, so then it makes sense)
but maybe I shouldn't see them as conditions with the same end in mind
 
Look at it this way: you have a ruler which measures in cm. You obtain some measurement $x$ cm, and get a result $\hat{y}$. You know that $\hat{y}$ isn't the actual value of $f(x \text{ mm})$, but you know that the backward error is 0.5 mm. So this means that the $\hat{y} = f(x+\Delta x)$, where $\Delta x$ is smaller than 0.5 mm. Your ruler measures in cm, so you are probably close to the actual value.
And, again, my vague recollection is that you can bound the error in $\hat{y}$ using the condition number and the backwards error.
 
@XanderHenderson hm, to generalise this: if we know that $f$ is not sensitive to small errors, then a small $\Delta x$ indicates that we found a good approximation
I think the first point mentioned here is what you just illustrated
we generally have "noise"
so if our backward error has the same magnitude as the noise
then we shouldn't expect our algorithm to distinguish $f(x)$ and $f(x+\Delta x)$ anyways
 
1:48 PM
I wonder sometimes where folks get their energy to track 4, year old questions to give little lectures math.stackexchange.com/a/3690071/27978
 
The first point in the last screenshot I shared is now clear to me (and I can get behind it), and I suppose the second point about perturbation is what you were talking about Xander. I'll have a look in the book where they elaborate the point on perturbation further
 
@copper.hat not sure what Anne is trying to do here
 
,@Jakobian I have no idea, seems a bit pointless:-)
 
@copper.hat Yeah, that's not really okay.
 
gotta admit, she got our attention
@XanderHenderson this seems to be the perturbation argument (in the book from the last screenshot)
 
2:03 PM
The term on the left looks like the relative error in the output, being bound by the product of the condition number and the backward error.
 
yes
 
Yeah, the second quote.
Isn't that exactly what I said before? :P
 
the condition number is some cases (like evaluating $f$ for $f$ $C^2$) is easy to compute
@XanderHenderson yes, but I wasn't aware that the condition number was sometimes easy to compute
to me the condition number was just by definition "relative forward error"/"relative backward error"
@XanderHenderson hence my reaction to this with "yea, that works by definition"
I'm starting to see now why it is useful
 
It is a theoretical tool, which lets you prove bounds. Super useful.
 
2:18 PM
In other news, the finance people at the college can't do basic arithmetic, and it scares me.
In a recent meeting, they said something to the effect of "Faculty have gotten 5% pay raises in each of the last five years, for a total of a 25% pay raise" (note: these numbers are completely made up.
$1.05^5 \approx 1.28$, so that is a 28% pay raise (the magnitude of the actual mistake made by these folk is closer to a 5% error).
I'm totally okay with them making this mistake, as it benefits me, but the same error has lead them to under report CPI over the same period, and that has compounded much more to the detriment of faculty.
 
that's wild
 
Of course, the administration is also on a kick to pay everyone at the college a minimum wage of \$20/hr (which is going to cost us about half a million dollars, out of an annual budget of about \$40 million, in a year when we are looking at a $2-3 million shortfall due to the legislature being a bunch of idiots). And they want to make summer classes tuition-free.
I am fairly well convinced that our current president is looking to go somewhere else, and is trying to do things which look good on a resume ("Hey! I increased the minimum wage at the college to \$20/hr! I gave students free classes! Aren't I the best?!"), while making no real concessions to the long-term viability of the college.
It is very frustrating.
 
2:35 PM
@XanderHenderson fail them
 
@Jakobian They aren't in any of my classes. :(
 
Oh its not the students?
 
@Jakobian No. It is the finance people at the college. Like, the Vice President of Finance.
I guess her actual title is "Vice President of Administrative Services / CFO".
 
She's not made for the job
 
3:11 PM
Well, one of our professors rick-rolled the entire class, including me.
 
@ε†₯ηŽ‹Hades That doesn't seem comparable. Your professor seems like a class act.
 
3:25 PM
say a finite number of isometric surfaces intersect in a 'nice' way. How do you generally pass from having four individual metrics to one global metric on the union of the surfaces s.t. the metric becomes singular at the smooth curves of intersection?
 
3:42 PM
@XanderHenderson No it's just that I find it hard to believe I got tricked so easily. I am relatively adept at avoiding rickrolls.
 
3:58 PM
@ε†₯ηŽ‹Hades I don't believe you can be good at that
 
4:09 PM
Can someone give an example of $\partial A\not\subset\partial(int(A))$ Where $A \subset X$ for some topological space $X$?
 
4:20 PM
nvm just the rationals
 
@SoumikMukherjee $\partial(A) = \overline{A}\setminus \text{int}(A)$ while $\partial(\text{int}(A)) = \overline{\text{int}(A)}\setminus \text{int}(A)$
 
Your question boils down to $\overline{A}\neq \overline{\text{int}(A)}$
any set which is closed but not regular closed works
A subset S {\displaystyle S} of a topological space X {\displaystyle X} is called a regular open set if it is equal to the interior of its closure; expressed symbolically, if Int ⁑ ( S ¯ ) = S {\displaystyle \operatorname {Int} ({\overline {S}})=S} or, equivalently, if βˆ‚ ( S ¯...
This is also related to this answer of mine where I show that if $A, B\subseteq \mathbb{R}^n$ are homeomorphic and $A$ is regular closed, then $\overline{B} = \overline{\text{int}(B)}$
In fact I prove an equivalence: $A\subseteq \mathbb{R}^n$ is homeomorphic to regular closed subset of $\mathbb{R}^n$, if and only if its homeomorphic to a closed subset of $\mathbb{R}^n$ and $\overline{A} = \overline{\text{int}(A)}$
 
@Jakobian Thanks
 
4:47 PM
@Jakobian Nah, I've had a lot of training since I spent a lot of time on Discord and other online forums.
It's about as easy as identifying: "If it's too good to be true, it probably is."
 
I am finding myself a bit confused with the notion of the domain and range of a (set-theoretic) relation and am hoping someone can help me straighten things out
Halmos begins by defining the Cartesian product of two sets, constructing it explicitly via specification from $P(P(A \cup B)$
He then says that the "reverse direction" is important: that is, given any set whose elements are all ordered pairs, we want to know (and we subsequently show) that there exist sets $A,B$ such that $R \subset A \times B$
No problem so far. He then goes on to say that "it is often desirable to take $A,B$ as small as possible". And so then we define e.g. $A:=\{a \in \cup \cup R : \exists b \in B((a,b) \in R) \}$, the domain of the relation
I had naively thought, since he spoke in formal english, that $dom R =:A$ and $ran R =:B$ as constructed would be such that $A \times B = R$, but my question is about confirming that that suspicion is wrong and that in general all we can have is $R \subset A \times B$ (i.e. it's in general proper)
A second question (again, my sense is the answer is no) is whether there is any construction such that, in general, I can get $R = A\times B$ given only $R$
Maybe to summarize, if someone can confirm this claim: given $R$, the sense in which Halmos uses "smallest" is that for any $A,B$ s.t. $R \subset A \times B$, we have $dom\,R \subset A$ and $ran\,R \subset B$
 
5:09 PM
@ε†₯ηŽ‹Hades you're imagining things
sorry to tell you that
@EE18 I don't see how that would be confusing
 
you are a much better mathematician than I (to the extent that I can even use that word for me), so that is not surprising :)
 
thanks for the compliment but thats not my point
 
given that i am self-studying i am generally grateful when i can reach out for some confirmation of something i'm having trouble with
 
you did learn what domain and range is in high school or earlier
other than writing it in symbols this really isn't that different
don't treat it like you're learning something new, you're learning things you already know, but in a different way
 
agreed, hence my earlier mistake was realized (embarrassingly only in a later chapter) but i wanted to confirm
my last comment above in particular (my "claim")
 
5:15 PM
(Graphs of...) Most functions $A\to B$ aren't equal to $A\times B$
If they were then we already get restrictions: $B$ can have at most one element
 
i guess that's an immediate counterexample to my second question above, thank you
 
@EE18 Smallest here would just mean that $R\subseteq A\times B$ and if $R\subseteq C\times D$ then $A\subseteq C$ and $B\subseteq D$
 
Yuck, I got the inclusion directions mixed up
thank you
 
you know any group theory say?
very basic group theory
 
the basic definitions
 
5:19 PM
yeah okay, so think of it this way
You have a group $G$ and you have some set $A\subseteq G$
what would be subgroup generated by $A$?
you're trying to take smallest subgroup containing $A$
okay now back to set theory, this is the same
you're trying to contain $R$ in the smallest square
 
Got it, but $R$ itself may not be "square"
hence my earlier error
 
and an arbitrary subset $A\subseteq G$ doesn't have to be a subgroup
 
touche
thanks jakobian
 
you just have this closure operator that to a subset assigns the subgroup its generated by
 
may I also ask, one speaks of a relation in $X$ when R is such that $domR,ranR \subset X$. If the $\subset$ is an equality, is there a special name given? I have a vague recollection of "on" being used but maybe that's wrong
Halmos doesn't mention anything
 
5:22 PM
and here you have closure operator that to each relation assigns the smallest square that contains it, the square is being generated by $R$
@EE18 when $\text{ran}(R) = X$ then $R$ is often called surjective, mostly for functions though
as for $\text{dom}(R) = X$, well, there was some name but I'll have to check
 
fair enough, thanks as always Jakobian!
 
In mathematics, a binary relation R βŠ† X×Y between two sets X and Y is total (or left total) if the source set X equals the domain {x : there is a y with xRy }. Conversely, R is called right total if Y equals the range {y : there is an x with xRy }. When f: X → Y is a function, the domain of f is all of X, hence f is a total relation. On the other hand, if f is a partial function, then the domain may be a proper subset of X, in which case f is not a total relation. "A binary relation is said to be total with respect to a universe of discourse just in case everything in that universe of discourse...
if $\text{dom}(R) = X$ then its called a total relation
You might also try to visualize the smallest square $A\times B$ that contains $R$
for example draw $[0, \infty)\times [0, \infty)$ or something, and try to see that if you choose some subset, what do you have to add to make it a square
say, you have points $(2, 1)$ and $(1, 2)$
but if this is confusing to you then maybe you should consider grabbing an easier to understand book
 
Nope it's been smooth sailing so far (will no doubt get harder once I wade into ordinals which I've never seen before). Just a bad hiccup this morning
and yes I should have known re: total order! just like with the reals...
to your point, sometimes it's just the phrasing in formal set-theoretic language of things i already know which leaves me slower than i otherwise would be
 
yeah you should definitely think about those things like nothing new then
because it isn't new. Its just a more formal way of writing stuff you know already
malapropism popping up again
I swear its something with my brain
 
6:11 PM
perhaps just a natural consequence of fatigue
also, similarly spelled and pronounced words come down to noticing the difference between one or two letters πŸ”€
 
6:30 PM
so yes, it is something in our brains in between having a thought and processing it into words
πŸ§ πŸ€”πŸ“– and vice versa
 
7:26 PM
What are words?
 
@user85795 nah I wasn't tired. Also about it being something in our brains, sure. It happens to everyone. I'm just worried about the frequency in which it happens when it comes to me
 
@TedShifrin Strings of letters, but that's just a theory.
 
@robjohn Ah.
 
I guess I hadn't refreshed this page in a while. My gravatar still had a flag in it until I just refreshed.
 
7:45 PM
hey, folks. quick question. i have a problem statement that I do not get. simple trigonometry. it goes like this: "An object P is situated 345 m above a level plane. Two people, A and B, are standing on the plane, A in a direction south-west of P, and B due south of P. The angles of elevation of P as observed at A and B are 34 and 26 degrees respectively. Find the distance between A and B".
what does "due south" mean in this case? directly underneath P? then how come there's an angle of elevation? furthermore, how come the angle of elevation of A is greater than B's, since I assume A is farther from P than B?
 
South is behind.
West is to the left.
 
Right. It's standard to have the + $x$-axis be EAST, and the + $y$-axis be NORTH. You can figure the rest out.
Southwest means the direction at angle $5\pi/4$ (measured from the positive $x$-axis as usual).
I would suggest that you put P at the origin.
 
Or on the line $y=x$ in the 3rd quadrant
 
Draw a 3 dimensional sketch.
 
Why 3-D?
 
7:50 PM
let me wrap my tiny lil' head around your suggestions. i didn't think this to be 3d. it's a simple trig problem. i barely know sin and cosine rules, and this chapter is right after it.
 
Oh, I missed it that P is 345 m above the plane. Forget my earlier comment about the origin.
 
> what does "due south" mean in this case? directly underneath P?
 
They actually mean angles of elevation from the $xy$-plane to P.
No, the directions are in the plane, relative to the projection of P into the plane (which is where the origin should go).
 
Label your sketch carefully with the directions @nevvermind
::think::
 
@nevvermind It is indeed a trig problem.
 
7:53 PM
i did this initially, but it's wrong, I think
 
No, that's not right.
 
cool. lemme have another think.
thanks, all
 
You will need to use the law of cosines to solve this problem.
You do need a 3-D picture.
 
And careful thinking about directions
 
i see. let me try that. i couldn't wrap my head around "behind". but I think it now makes sense. mind unboggled.
 
7:55 PM
:-)
 
Basic question. In measure theory, one says that two functions are equal almost everywhere if they are equal except on a set of measure zero. Does this notion hold for inequalities too? Suppose one writes $\int_{\mathbb R} |f|d\mu <\infty$. Does this mean the integral is finite a.e.? The reason for the question is that in my notes they say "finite a.e." in regards to an integral, which I have not seen before.
I should maybe add, it's in regards to Fubini's theorem, where we have iterated integrals.
 
@psie it holds for any property
that depends on a real parameter
here the integral doesn't depend on parameters so "finite a.e." makes no sense
 
@nevvermind This should help you visualize.
 
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