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12:01 AM
Anyways, I'm going to get back to working on this quotient algorithm and finding the greatest common integer among the set of integers that make up the interval defined by the quotient of the msb and lsb for all bits.
 
Had a quick question about framing a question in the right way to get the solution. The question asks me to find the angle that a stream makes with the horizontal at point $(\mathbf{a}, c)$ bold if intentional. if it flows in the $e_{2}$ direction at the point. Pertinent information to this is $z = f(\mathbf{x})$, $\mathbf{x} \in U \subset \mathbb{R}^{2}$. Also $f)\mathbf{a}) = c$ and $Df(\mathbf{a}) = (3, -4)$.

Should also add that the solution is $-\arctan(4)$.
My problem is I don't see how to "visualize" the stream against the horizontal properly.
this is perfect for a mathematica picture...........
 
are quantifiers basic, or is there a way to prove, e.g., $\exists x P(x) \iff \neg \forall x \neg P(x)$? it makes sense in natural language, but I'm wondering if there is some sort of logical calculus for this
 
1:05 AM
@dc3rd keyword directional derivative
 
@shintuku there's a tree-system in formal logic
oh cool thanks shintuku
 
hmm, so the directional derivative in this case is the vector $(3,-4)$.............so you saying directional derivative just now triggered the memory of you saying in one of your lectures of having to specify it as a unit vector....
 
i need a job where we do not record time in tenths of an hour. this is driving me crazy.
 
o.9
can someone explain this ansa to mi ?
1
A: Chance of getting a date

PotatoUse the fact that expectation is linear. A single coin flip has an expected value of $1/100$th of a date. To get an expected value of 1 date, flip the coin 100 times.

 
@dc3rd huh?
 
1:13 AM
never mind the second part about unit vector...I thought about what I had said...
 
No, the first…
Expectation is linear. $E(nX) = n E(X)$.
 
@leslietownes. Which is better zoom or in class?
 
o.9
that allows us to conclude for $100$ there is $1$ expected date
 
@TedShifrin. Expectation is linear transformation?
 
o.9
Does that mean $100$ is the expected number of attempts we need to get our first date.
 
1:17 AM
avra, in class. zoom is OK for presentation but not for conversation.
 
Go read, $E(X+Y) = E(X)+E(Y)$ whether or not independent random variables.
 
which is what class ought to be.
 
o.9
I agree that the expected value is $1$ when we have $100$ attempts.
But that isn't really what is being asked
 
It isn’t?
 
o.9
No we want the expected number of attempts till the first success
 
1:19 AM
No.
 
the linearity of expectation might go on my list of top 10 profound thoughts.
 
@o.9. What about this answer please?
Use the fact that expectation is linear. A single coin flip has an expected value of 1/100th of a date. To get an expected value of 1 date, flip the coin 100 times.
It's from the same page above
It seems logical
Since each date is independent, so we should multiply 1/100 100 times
In case date are dependent (don't think this is true), so we sum 1/100 100 times
 
Oh... I see what you meant Ted $D(f(a)) v = (3, -4) (0,1)^t$ you know what I mean in terms of the directional derivative...pardon the formatting
 
@dc3rd. This is directional deritivative in the direction of $v$
 
yes. with $v = e_2$
 
1:23 AM
(0,1) is the unit vector and (3, -4) is the vector along it
@leslietownes. They might continue the closure, which means mooore zooms
@leslietownes. COVID cases are striking again
 
yeah, it's not good. my office had a plan to return to the office after labor day. they are reconsidering it.
 
o.9
I thought you got some days to rest after labour
to take care of the child
 
@leslietownes. It's serious! The virus has polarized itself to a new form
 
it doesn't help that i work in orange county, which is a hotbed of people who disbelieve in the virus. even if we're all vaxxed, the people delivering stuff to the office or visiting might not be.
 
@leslietownes. If they reopened it, it might be catastrophic
 
1:27 AM
o.9 i got four weeks after my child was born. if i were a girl it would have been six.
our relatively small business has already lost two people to the coronavirus and it has been hard to replace them.
 
o.9
would you also get six if it was your wife delivering the child and you were a girl?
 
yes, i think so. there are probably legal issues with the agreement.
 
o.9
pog
 
almost anything i have ever gotten at work was due to the gay rights movement, and even if it wasn't the right thing to do, i would support it for that reason alone.
i think the smart thing to do if you run a large business is to make everything gender-neutral.
it's just easier and you don't get into the weeds of people's lives.
once you say 6 weeks for this and 8 weeks for that, you are asking for trouble.
i'm an attorney but i'm not offering legal advice.
i'm just commenting on what's out there.
 
o.9
why not?
I do not agree with the answer I linked previously :/
 
1:33 AM
my now-wife and i were 'domestic partners' in iowa for a few years. thank you gay rights movement although we are hetero. it let her get on my health insurance. big thing in the USA. huge thing. hard to explain to people in civilized countries.
 
o.9
Glad you were able to do that :)
 
insurance is a weird creature in the USA. it is state-specific so what you get in iowa might not be what you can get in maine.
 
o.9
that answer also only works when $p$ is the reciprocal of a positive integer
 
haha
In one country, woman take 3 months
Now they shrink it to 4 weeks
 
o.9
in which country?
 
1:38 AM
@leslietownes. It seems it's standardized !
 
companies do uniformize their policies to the extent they are able.
 
Russian
 
o.9
they don't standardize pay tho :'(
they try to scam you and you have to negotiate it
 
uniformization is generally a good idea. it's the concept of insurance.
 
o.9
fuckin' pain
but I guess they save a lot of money that way
 
1:40 AM
Yep!
Here in the US is $$
 
o.9
what does that mean?
 
So people would like to come back to work ASAP ASAP SSSS
 
o.9
that cash is king?
 
It's very expensive
 
o.9
oh
 
1:41 AM
So you can not take unpaid vacation for 3 months
 
o.9
I remember when I went to CA when I was 12 with my parents we had to bargain for a ton of stuff
for the house
cellphone plans, matresses, cars, car insurance, laptops
 
it's called freedom
 
comrades, it is time to say we've had enough
 
@o.9. Yes. Here it's FREE as @leslietownes with BUTS
 
freedom of choice is what you got, freedom from choice is what you want
i'm going to spend the next 3 hours approximating devo riffs
 
o.9
1:44 AM
You can also buy stuff from regular folk in Mexico
well, I guess salesman aren't regular folk
Salesmen are professionals
I don't really know what I'm saying
 
You have been practicing math for life and they have been bargining for life. Do $E(n)$
 
o.9
I guess I just don't like bargaining with professionals
 
if only distribution and retail was the mandate of centralized public agency
 
o.9
no that would suck
but you can get people who have actual jobs to also do the sales
instead of having people that just talk shit for a living
 
my wife just discouraged my daughter from pulling something out of the refrigerator by saying "no, that's a wellness shot" which to me seems peak west coast
 
1:48 AM
fine comrade, if you enjoy being sold things by people who are trained to make you want things you aren't sure you need, suit yourself
 
i'm an american, i buy what i'm sold
 
true
 
o.9
I guess many contractors are not salesman but will also overcharge you if they think they can get away with it.
knowing the price of many things can be very helpfull I guess.
 
@o.9. 100%
NO MERCY
Unfortunately.
You have to be circumspect
 
woah these logical trees are so damn useful
why haven't I seen these before
 
1:56 AM
@shintuku. Which one ?
 
MM
this is expressions tree?
What course?
 
learning first-order logic
 
*expression tree?
WOW!
I did not expect that?
What course?
 
i think people see this in an introduction to logic class
 
1:59 AM
Oh! I thought introductory to graph theory
or compiler theory
 
the above tree shows the following argument is valid:
$P \iff Q$
$Q \iff R$
$\therefore P \iff R$
which, well, is obvious but it's nice to have a sort of calculus for these things
 
@leslietownes. Have you seen these in any math class before?
Is $(\log(n)) ^ {\frac{n}{3}}$ same as $\log (n^{\frac{n}{3}})$? I guess there is a typo here?
 
2:26 AM
i've seen those trees, not in any class i was in but in general, yes.
 
@TedShifrin Sorry to hear that. Hope it goes away quickly.
 
2:50 AM
@Avra Those are definitely not the same
 
o.9
I don't understand how that expression can be $\mathcal O(n)$
 
3:23 AM
what expression?
 
o.9
Either of them
2
Q: Growth rate of $(\log{n})^{3n}$ and $(n)^{\frac{n}{3}}$

AvraTo find the growth rate of $(\log{n})^{3n}$ and $(n)^{\frac{n}{3}}$, I took the log of both: $(\log{n})^{3n}$: $$\log (\log{n})^{3n} = 3n \log{\log{n}} \in O(n \log n)$$ but the answer I have says it's $O(n)$. $(n)^{\frac{n}{3}}$: $$\log (n)^{\frac{n}{3}} =\frac{n}{3} \log{n} \in O(n \log n) ...

This must be trolling right? math.stackexchange.com/questions/4215481/…
 
> The diagonals of a parallelograms are given by the vectors $\vec A=3\vec {i} + \vec {j} + 2\vec {k}$ and $\vec{B}=\vec {i} - 3\vec {j} + 4\vec {k}$. Find the area of the parallelogram.
$$| \vec A|=\sqrt{14}\\| \vec B|=\sqrt{26}$$
Area=$\frac12 d_1 \times d_2=\frac12\times \sqrt{14}\times \sqrt{26}=\sqrt {91}$
Is this wrong?
 
$(a-b)\times(a+b)=2a\times b$,
the area is $\frac12$ the cross product of the diagonals
 
That gives $5\sqrt 3$
 
yep
 
3:37 AM
But what is wrong in the above approach? :-(
 
you need more $\sin$
 
I don't see what you are doing above. explain
@copper.hat more sin... so hedonistic!
 
that's me to a tee :-)
always looking for the indulgent angle
 
I calculated the magnitude of diagonals (which gives the length) and used the formula :Area of parallelogram=1/2 product of diagonals.
 
not the product, the cross product
 
3:39 AM
hence the need for $\sin $ :-)
 
as copper says.
in a rhombus, you would take the straight product.
or a kite
In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Kite quadrilaterals are named for the wind-blown, flying kites, which often have this shape and which are in turn named for a bird. Kites are also known as deltoids, but the word "deltoid" may also refer to a deltoid curve, an unrelated geometric object. A kite, as defined above, may be either convex or concave, but...
Note that the diagonals are perpendicular
so the sin is $1$
full sin
 
Oh! So stupid of me. Thanks :-)
 
I can not figure out how small to make the inner radius of an annulus to make a holed cross-cap from a Mobius strip.
Can someone help me with the cross-cap please?
Im cutting the along the radius and putting the ends on like an intersection, but in every picture i find of the cross-cap there is just 1 loop for a boundary at the posterior side or behind the intersection. My paper has a hole there in a figure 8 and at the front of the intersection....
 
4:09 AM
I dont think i was self intersecting the inner radius. That would mean that I've just got to lengthen some parts of the figure 8 if im right.... Im still looking for how that part is 1 loop though!!
 
I've never used chat before so let me know if I'm doing this wrong. I am wondering if anyone agrees or disagrees with the following idea: Whenever you have a real symmetric matrix $A$, you should always think of $A$ as representing a quadratic form ($x \mapsto x^T A x$ ) rather than as representing a linear transformation ($x \mapsto Ax$).
 
What I read is 'chatting' and 'asking questions' have different guidelines -_-
 
it is helpful to think of the two things together.
particularly if A is unbounded.
in many applications the linear operator is of central importance and the quadratic form is secondary.
 
For example, the Hessian $Hf(x)$ is a good example of a symmetric matrix, and it appears as a quadratic form in the Taylor expansion of $f$. It seems that it is more illuminating to think of $Hf(x)$ as representing this quadratic form rather than as representing a linear transformation.
Is there a good example where a symmetric matrix $A$ arises in a context where it's more illuminating to think of $A$ as representing a linear transformation?
 
i agree there but would not extend this vibe to all linear operators.
 
4:16 AM
not even to all symmetric linear operators?
 
yeah, not even there.
but i'm an idiot, so i should point that out.
 
haha. No, it's helpful to hear what you think about this
 
d/dx is unbounded on basically every domain. i don't see what you gain by emphasizing any quadratic form.
 
d/dx isn't self-adjoint, though
 
you can tweak parameters so that it is, but i see your point
 
o.9
4:20 AM
more illuminating to think of $A$ as a linear transformation instead of a symmetric matrix?
 
any step away from the matrix world is welcome.
 
o.9
You can think of $A$ as a flow network
Is that a good thing?
 
I'm not familiar with the flow network viewpoint. Would $A_{ij}$ tell you like the rate of flow from node $i$ to node $j$ in a graph?
 
o.9
yes I think an example of that viewpoint is like when you have a doubly stochastic matrix in a transition matrix for a finite markov chain or something like that.
 
Whoa
 
o.9
4:28 AM
but I think you wanted examples where it's better to just view something as a linear transformation right?
 
yeah, I'd like an example where a symmetric matrix $A$ arises and it's more illuminating to think of $A$ as representing a linear transformation than as representing a quadratic form
 
the Timestamps are not even showing up now.. .
 
My hypothesis is that it is always more illuminating to think of a symmetric matrix $A$ as representing a quadratic form. I could be wrong of course.
 
o.9
But why look at it simply as a linear transform when you can see it as at least a diagonal matrix with orthonormal basis.
 
Is pro stack exchange like this I might check it out agian
 
o.9
4:31 AM
Would spectral graph theory be an example of what you are looking for?
adjacency matrices are symmetric
and they work with the spectra, maybe that counts as looking at it as a symmetric matrix?
 
That could be a good example, but I don't know much about spectral graph theory
That is worth looking into though. I should probably read more about that.
Do you have any favorite books that introduce spectral graph theory? I know Strang at least touches on it in some of his books.
 
Is non spectral sequences in Strangs book
 
o.9
Sorry I don't I just know the stuff that my advisor needed me to learn
I know David Godsil has a book and he also answers stuff on this site, I think I used it once to check something and it worked for me.
 
I have found a trick.
 
Oh good to know, thanks
 
4:35 AM
In some university tests you are allowed to bring either a book or a readable piece of paper with you.
Print your own book, gg
but probably there are only some books allowed, so...
 
o.9
are you allowed to bring a non-readable piece of paper?
 
well, of course, but you can't bring a magnifying glass
wonder if you can bring glasses without a seeing-weakness
 
o.9
When I went to the ACM-ICPC finals there was a page limit for the written stuff you could take in.
But we took less pages cuz we were kind of nooby :'/
and didn't really know what else to put in there
 
print out the internet as a book
 
o.9
but there was a page limit
 
4:41 AM
@SAJW how can you print out the internet as a book
 
but not a font-limit
 
o.9
It was 25 pages max
 
i'm only joking ;)
 
o.9
This document may contain up to 25 pages of reference materials, single-sided, letter or A4 size, with pages numbered in the upper right-hand corner and your university name and team name printed in the upper left-hand corner. Text and illustrations must be readable by a person with correctable eyesight without magnification from a distance of 1/2 meter.
It may include hand-written comments and corrections on the fronts of pages only. The document must be in some type of notebook or folder with the name of your institution on the front.
apparently they took that shit kinda serious :/
 
you already made a joke rofl
 
o.9
4:45 AM
what is a person with correctable eyesight?
 
there mongrels then
 
I'm not sure, if I can read this chat from 1/2 meter
 
o.9
why did you delete your account fia? now everything renders weird :'(
 
sure text, but not $e^{e^x}$
 
Im probably banned for beginning topology here
I just care about learning about the cross cap so i bet koob face tracked my information to here
 
o.9
4:49 AM
It says you have an account on math.se with 20 rep but when I click it it says it doesn't exist.
 
thats odd
I blame koob face
 
cu later
 
o.9
that's a virus?
 
because i dont have another reason > _<
 
o.9
I just learned about koobface 30 seconds ago
cu later ali gater
 
4:51 AM
What is a virus - yes koob face is one
 
o.9
I don't know, malware
 
Okay pce.
Absolutely
It is a worm-virus
 
o.9
what makes it a worm?
 
its the way of the cross caps
 
o.9
Sorry I'm sleepy I'm going to sleep now
 
 
3 hours later…
7:29 AM
@robjohn Ok I see what you mean. (1+i, 1+i) is not in that vector space, so closure property won’t be satisfied. But what I cannot understand is how can (i,-i) be a basis vector in R^2. “i” itself isn’t a real number so the vector (i,-i) is itself not in R^2 space just like (1+i, 1+i) isn’t in R^2……What I can’t get is how did you get (i, -i) to be a basis vector because to me it isn’t in R^2 just like (1+i, 1+i) isn’t……Could you elaborate this a bit, please
 
 
2 hours later…
9:34 AM
Hello I was banned for like 7 years. Nice to come back.
I need to go back bye guys. I will be banned for like 3 more m****s years.
 
@Shashaank The vector space has dimension $2$ because it has $2$ basis vectors, but it is not a subspace of $\mathbb{R}^2$. It is a vector space over $\mathbb{R}$.
 
9:58 AM
This is just the same as Hermitian matrices being a vector space over $\mathbb{R}$. $2\times2$ Hermitian matrices have conjugate elements in the $(1,2)$ and $(2,1)$ positions; that is, if you really mean Hermitian.
Hermitian matrices have complex elements, yet they form a vector space over $\mathbb{R}$ of dimension $6$.
 
10:40 AM
Greetings
 
11:09 AM
@robjohn Ok, I see what you are saying. So is it totally correct to say that this vector space given by you, having a basis vector (i,-i) and (1,1) i
*is a perfectly good vector space over reals
 
cool
 
*of reals. The only thing is that this vector space is different from R^2….is that correct
I was thinking that when we say a vector space over reals, we mean that not only the coefficients, the scalars that multiplies with the vector has to be real but even the number that we write inside “(“ and “)” (like (1,0) or (0,1) ) have to be also real numbers. That is why I was thinking that how can (i,-i) can be a basis vector in the space over reals. But I see you are saying that it could be a vector space but it’s just that it’s not the same as R^2 ..
@robjohn I find it strange. I always thought that if we say, a vector space over reals we also mean that the elements of the tuple or the numbers which appear between “(“ and “)” have also got to be real number
 
11:33 AM
The complex numbers form a vector space over the reals.
 
 
2 hours later…
1:38 PM
Can we view the ordinary derivative (one variable case) a special case of total derivative?
 
2:27 PM
@robjohn. Similarily to my question I posted, growth rate of $n^{a}$ is $a\log{n}$
 
2:46 PM
2
A: Why can't I take the derivative of $x^x$ as $x(x^{x-1})$?

Michael HoppeTaking the basis as variable, the derivative is $x\cdot x^{x-1}$, taking the exponent as such the derivative is $x^x\cdot\ln(x)$. Adding both gives the derivative of $x^x$, namely $$\left(x^x\right)'=x^x+x^x\cdot\ln(x).$$ Astonishing at the first glance, isn't it?

Is this a valid argument?
 
you're right that you can do it via the multivariable chain rule.
d/dt F(t,t) = F_1(t,t) + F_2(t,t) where F_1 and F_2 are partial derivatives with respect to the first and second variable.
F_1 meaning hold y constant and F_2 meaning hold x constant. it feels like magic.
 
@leslietownes You mean $F(a,b) = a^b$?
 
@PeterJohn Yes, that is how $F$ would be defined.
 
yeah.
 
3:04 PM
@leslietownes Why do we say 'total' derivative not just derivative? Is there any difference except the dimension?
 
terminology varies a lot. i think 'total' is often used if partial derivatives are also in the picture. when context is clear i just say 'derivative.'
thankfully i don't teach anymore, so the context is always clear :)
 
@leslietownes So it's just a terminology. I can just say total derivative for one variable function right?
 
i wouldn't stop you from saying that. :)
in one variable there's basically only one thing any type of derivative could mean, so one tends not to use qualifiers.
but you can put them there.
 
if you said total derivative for one variable function, i'd think you made a typo because I only see it in a multivariable context
but from context i'd understand
 
if you were teaching a single variable calculus class people would wonder why you were saying 'total' and what non-total derivatives you weren't talking about.
but in the comfort and privacy of the internet i am fine with it.
 
3:29 PM
@PeterJohn I have added an answer that is a bit more explicit, but follows what that author is trying to say.
 
that's a good answer.
no offense to other respondents but a lot of junk in that chain of answers
 
4:12 PM
@leslietownes would I have written it otherwise? Thanks!
 
haha
all of your stuff is great, as you well know.
which brings me to the key point, which is: where's grandad?
 
hiding in the pool
 
123
Hi All..
If marks in Eng = 12/25 , Maths = 40/50 , Science = 15/25 then Marks obtained/Total maks = 67/100 = 100%
 
going back in a few days
 
123
If equivalent fractions are taken for all subjects to out of 100 then, Eng = 48/100 , Maths = 80/100 , Science = 60/100 , Marks obtained/ Total Marks = 188/300 = 62.66% .
Why??? They are unequal
Hello @copper.hat
 
4:16 PM
hi 123 :-)
 
@123 the weights are different in the first computation
 
123
How we take weight factor in calculation. I also thought about weight is the problem.
 
4:51 PM
In the first computation, the math is weighted twice as much as the English and Science.
in the second computation, the math is weighted equally with the other subjects
since your math score was the highest, the first computation gives 67% and while the second gives 62.66%
@123 it's as easy as 1-2- oh, nevermind.
 
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