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12:26 AM
@copper.hat oh right, that was confusing me thanks
hm, but do I really need two variables? $e^{-Px}\int Q'(x)\frac{e^{Px}}{P} \ dx$ can be rewritten $g(x) = e^{-Px}F(x)$.. right?
where $P$ is a constant and $Q'(x) \to \infty$ as $x \to \infty$
and where I'm trying to determine whether $g(x)$ converges as $x \to \infty$
 
Bob
12:51 AM
@Avra is the author Stroud? or Booth?
I have 3 to 4 four pages of algebra
 
@shintuku you should distinguish between the $x$ outside and the $x$ as the integrand. presumably they are different. also, without limits it is hard to say what it going on.
 
Bob
I believe there is a mistake in it
what is the best way for me to find the mistake?
 
thanks a lot, I'm working on formulating the question properly so the comments on the ambiguity are super appreciated
I'm trying to prove there exists a limit and determine what it is for $g(x)$ as $x \to \infty$. Wouldn't in this case distinguishing between the interior and the exterior $x$ require writing $g$ as a two variable function?
in other words, I'm working on: prove there exists an $L \in \mathbb{R}$ such that $\lim \limits_{x \to \infty} g(x) = L$
 
 
1 hour later…
1:59 AM
whispering the secret is that $\int f \ dx = \int_a^x f \ dt$
oh! thank you so much shintuku!
you're welcome shintuku
 
2:42 AM
@shintuku Disagree. The LHS is an arbitrary antiderivative. If you mean the RHS, then write it explicitly.
 
hm. then, $\int f \ dx = \int_a^x f \ dt$ iff for any $C \in \mathbb{R}$ there exists an $a$ s.t. $F(x) + C = F(x) - F(a)$
wait it's not right give me a second
 
shin: LHS is indefinite integral and may often be regarded as a set that is $\int f dx=$ set of all functions which derivative is equal to f under appropriate situations. Can you equal a function to a set? If yes, you should define that equality. Professor Ted may correct me here if I am wrong.
 
Professor Ted is he wrong or correct?
 
$\int f \ dx$ is a set! that makes too much sense!
bless you Koro, that is an excellent idea and I will frame it in my bedroom
 
you’re welcome @shin
 
2:53 AM
$\int f(x) \ dx = \{F(x) + C:(F(x) + C)' = f(x) \}$
 
@TedShifrin @Bob. Yes!
 
or more exactly $\int f(x) \ dx = \{ F(x): F'(x) = f(x) \}$
 
Suppose that I have a vector space of 100-tuple that is $\mathbb R^{100} (\mathbb R)$ over field $\mathbb R$ and I am given two subspaces $V$ and $W$ such that $V=\{(x_1,x_2,...,x_{100}): 4|i\}$ and $W=\{(y_1,y_2,...,y_{100}): 5|i\}$, I want to find dimension of subspace $V\cap W$.
I say dim V=100-25=75 and dim W=100-20=80 so dim $(V\cap W)=100-45=55$
Is that correct?
@shintuku shin, I also said "under appropriate situations" that is, you should also state domain etc. also :)
 
3:08 AM
hm, I'll think of a formal statement for that
is it not sufficient for $F(x)$ to be an antiderivative, no matter the condition?
 
But for F' to make sense, F has to be differentiable, right? So it's better to state domain so that all symbols being used make sense :)
 
hm: if we write instead, not $=$, but $:=$, then if $F$ is not differentiable, it is not part of the set
right?
 
yeah I think. But suppose you give this definition (using :=) to someone, what will they make out of it? To make it more meaningful and complete, I think that you should state domain also.
 
makes sense, thanks!
 
something like this, $\int f dx:= \{F: A(\subset \mathbb R) \to \mathbb R: F'(x) =f(x) \text{for all $x\in A$}\}$ etc.
where $A$ is non empty subset of $\mathbb R$ , etc.
Any ideas here?
23 mins ago, by Koro
Suppose that I have a vector space of 100-tuple that is $\mathbb R^{100} (\mathbb R)$ over field $\mathbb R$ and I am given two subspaces $V$ and $W$ such that $V=\{(x_1,x_2,...,x_{100}): 4|i\}$ and $W=\{(y_1,y_2,...,y_{100}): 5|i\}$, I want to find dimension of subspace $V\cap W$.
 
3:30 AM
the notation is slightly ambiguous, but for an index to be divisible by both 4 and 5 is for it to be divisible by 20. i think you can write down an explicit basis for this space.
 
hi leslie :)
why do you think that notation is ambiguous.
 
there is no mention of i except in 4 divides i and 5 divides i. i (the person) have to infer what i (the index) is.
 
Ahh, I see. My bad. I meant $1\le i\le 100$.
 
this was implicit, but not explicit. i think i understood but was not sure. i did understand.
 
I was thinking what lies in $V\cap W$. The elements whose coordinates with indices of the form $4k$ and $5m$ , where $1\le k\le 25$ and $1\le m\le 20$ are zero.
So we're talking 100 -25-20= 55 non zero elements in a vector. But I am not sure.
Also the correct answer seems to be 66
:'(
Since indices divisible by 20 are double counted, I should take that into account as well but even then I get 100-25-20+5=60.
Ahh I got it. I am right :)
dim $(V\cap W)=60$.
I misread 60 as 66.
:)
 
3:46 AM
@Koro here I think we'd have a complete description: For a continuous $f: I \subset A \to B$ and $F: I \to B$, we have $\int f := \{ F \ | \ F' = f \}$
 
yes, I think. It also tells (secretly) by FTC, that $F$ exists and is differentiable also :)
I think you should also say that $f$ is bounded. Otherwise good luck integrating 1/x on $(0, 1]$ for example. :)
or better yet, $I$ is a closed and bounded set (let's consider $\mathbb R$ only) so that I is compact hence continuity of f ensures boundedness of $f$ also.
 
i just thought of this, since we can say a statement like "$\int f$ does not exist", isn't it wrong to define it with these conditions?
otherwise the statement "$\int f$ does not exist" would necessarily be a contradiction
 
I think you mean the situation $\int f=\emptyset$. But, how is it interesting?
 
oh, right. if $f$ has no antiderivatives, then we can say the set $\int f$ is empty
 
4:12 AM
right, well, then we can say $\int f(x) \ dx = \int_a^x f(t) \ dt$ if and only if $F(a) = 0$ where $F'(x) = f(x)$
 
Hello. How do I search for a webpage with posts authored at least by persons ABC and XYZ? Neither "user:ABC,XYZ" nor "user"ABC user:XYZ" does this. Thanks!
correction: nor "user:ABC user:XYZ"
 
@shintuku shin, I am not sure what you did in this but I'll suggest to have a look at a chapter in Spivak that deals with this. I think that will help you understand these things more.
 
my goal is to contact (thank) a particular user who has previously answered my questions
 
hm you're right. I'll go read it hehe
 
I think it's dealt with in chapter on derivatives or on integrals, though I don't remember right now which chapter exactly :'(
@shin
 
4:21 AM
oh! wait: actually, the proper statement is: $\int_a^x f(t) \ dt \in \int f(x) \ dx$!
 
exactly!
 
woohooo
thanks a lot Koro!
 
you're welcome @shin
 
4:48 AM
i am looking away
 
why
don't tell me the fruit of my strenuous labour is a false statement
 
@RyanG what do you mean webpage?
 
5:13 AM
i mean MSE question-answer page containing posts by at least users ABC and XYZ . in this case i'm ABC, and i'd like to know which of my previous questions XYZ has answered.
 
5:26 AM
@shintuku i am not a fan of indefinite integrals.
 
i spent a few hours in a state of confusion about indefinite integrals, and am not a fan either
 
@RyanG You can check the queries which I posted on MathOverflow Meta: Is it possible to search for posts/questions of an user $X$ commented on/answered by the user $Y$?
If you use some of those queries, do not forget to switch the site to Mathematics.
 
Hello, if f,g are continuous functions onto the unit sphere, then why is $\|tf(x)+(1-t)g(x)\|=0$ only if t=1-t? Any help is appreciated!
Can anyone give a hint please? I'm super stuck!
 
5:44 AM
the norm being zero means the points add to zero. there may be some g--m-tr- there.
 
Does it imply that f(x) must be equal to -g(x)? @leslietownes
Since they must be antipodal to add to zero?
 
the scalar thing is throwing me off but i do think they need to point in antipodal directions or it won't work.
it's not clear to me that you wouldn't be able to even infer information about t.
i'm still thinking it over.
 
It is from this by the way :math.stackexchange.com/a/3143172/810585
 
6:10 AM
@pritchard you must be missing something, that is certainly not enough to conclude $t=1-t$.
 
@copper.hat I am trying to figure out the reasoning behind this question math.stackexchange.com/questions/3142878/…
Specifically how Connor Mallin shows why the assumption holds true
Geometrically, it seems to me that the points must be antipodal for norm to be zero; is that true?
 
6:40 AM
@MartinSleziak Thanks, thank worked.
*that
 
 
3 hours later…
9:17 AM
privyet
person told me that each and every branch of math is boring
 
 
1 hour later…
10:32 AM
the second derivative test is inconclusive
what do i do, i wasn't trained for this
 
10:47 AM
you use thought, shintuku, the power of thought! i summoned a directional derivative which was equal to zero, showing said point was not an extremum
wait that doesn't show anything
the multivariable definition of local extrema is either greater/lesser or equal
 
11:06 AM
but found an alternative way to deal with this sunglasses
 
 
4 hours later…
2:54 PM
@russianbot they sound boring
 
is modulus defined for negative numbers?
 
@SAJW you mean like $5\bmod3=2$?
and $-1\bmod3=2$
$-5\bmod3=1$
 
ah yes, and is that normal for python or any of the object oriented languages, because if then I found maybe a solution to my problem
 
It should be, but there is some wiggle room: it is valid to say that $-5\equiv-2\pmod3$
 
but my solution only works if the roboter can only drive parallel to some x-axis or y-axis
 
2:58 PM
so a mod function in some implementations may say $-5\bmod3=-2$
You'll have to check the docs for whatever language you're using
 
mmh, i have 4 directions and 0 north. -1 should then be the same as 3
so -1=3 mod 4 but yes I'll have to check
 
It looks here as if it works as I first specified
 
that looks like I want it, great
 
It's just a matter of a simple manipulation if it didn't handle it that way
So it wouldn't be game over
 
hehe
 
3:05 PM
BBL
 
so a turn against the clock at 90° is simply -1 to the direction
and the direction is at the end mod 4
i wonder if it may be simplier to say: the roboter changes which axis he drives parallel to if he turns 90°
but he can drive forward and backward and maybe that just outsources the problem to another places
 
3:40 PM
mmh, if direction mod 4 =0, if forward, then increase y-coordinate by distance.
distance is fixed
just an example how I gonna tackle it tomorrow
 
 
1 hour later…
4:57 PM
distance from the (0,0) is $\sqrt{x^2+y^2}$ right?
 
the euclidean distance from (x,y) to (0,0) is sqrt(x^2+y^2). :)
 
5:24 PM
exponents get added when for example $a^x*a^y$?
 
if it's the same base they do.
 
so $a*\sqrt{a}=a^{1+0.5}$?
 
yes. if a is a complex number there are issues with defining sqrt and the addition law may not hold anymore.
 
ah ok
and $\sqrt{3^{\frac{2}{3}}}=3^{\frac{1}{3}}$?
 
5:46 PM
mmh, I don't come any further: $b*(183-b^{1.5})^{\frac{1}{3}}+(183-b^{1.5})^{\frac{2}{3}}*\sqrt{b}=182$ what to do next?
to solve for b
 
6:12 PM
surprisingly some complex operations are complex.
 
Hi, robjohn. How is it going? I recently took this test test.mensa.no and since I couldn't finish all within the given amount of time (25 minutes) I scored 135. With a bit of extra time I could figure out how to go further, and that would have probably led to about 140 or slightly above.
I point out that I consider these tests only out of curiosity, to see (if the case) how mathematics performed in the long term affects someone's power of pattern recognition, and not as a means of measuring intelligence (whatever that would mean).
Let me know if you had a chance to take the test. If not, and you are willing to take it, let me know at any time how things seemed to you.
 
intelligence is don't put on your shoes before your underwear :(
2
except if you are nudist haha
 
One more thing that these tests don't catch is when someone has ideas to attack a question, but more time is necessary, and the case when someone simply has no idea about what to do further.
 
such measurements make me uncomfortable
 
the best I laughed at as how pattern completion measures math intelligence (1,3,5,7,?)
together with written addition/substraction/multiplying/devision
if you only take that then I am average in math
 
6:21 PM
in The Nineteenth Byte, Jul 3 at 9:44, by ngn
IQ is the perfect metric of how good you are at solving IQ tests :)
 
:D
 
@pritchard i do not understand the full context of your question but it is possible for some curve that satisfies $f(x)+g(x) \neq 0$ to satisfy $tf(x)+(1-t)g(x) = 0$ for some $t \neq {1 \over 2}$.
 
@copper @leslie You like the OP's remark at the end of my answer here? :D
 
if event b is never gonna happen, is it correct to say a happended before b?
 
@TedShifrin i laughed. i appreciate your politeness and can understand the OP's bewilderment :-).
 
6:27 PM
it got a giggle out of me. it only seems like a basic relationship because of brouwer and others. i'd expect it was effort-intensive for him too.
 
Well, when one asks for deep connections, one has to be prepared to learn something :P
 
there's a principle of conservation of effort. it has to come somewhere. the bourbaki school has at times made it out like it's just a question of writing the right book, but in my view, a big nope.
but i'm an idiot so i should always mention that.
i should preface with it instead of following with it.
 
it is a bit frustrating when someone has the entitlement of simplicity.
 
7:11 PM
I tried the IQ test but did very poorly and now I'm sad :/
 
@Yorch. I think this test is not an indicator but for your preparation like anyother test!
@leslietownes. What do you think about IQ test!?\
 
Bob
if you did a Calculus problem that resulted in pages of algebra
and you believe there is an error in the algebra
does anybody have any tricks to find the error?
@Yorch I do not believe that IQ tests mean much
@Yorch I have a friend who has a very high IQ
@Yorch but he did not do well in college and his career did not go well
 
as a fellow countryman wrote "A cynic is a man who knows the price of everything, and the value of nothing"
all that can be measured...
 
Bob
@copper.hat I am still stuck on that problem. Would you do it over again from scratch?
 
@copper.hat. Deep.
 
Bob
7:23 PM
@copper.hat I have an answer but it is wrong.
 
what are some "never" predictions that didn't hold up? like a computer never needs more than x mb ram
(the wording might be wrong)
 
@Bob i answered it already? why would i do it again???
 
Bob
@copper.hat I do not have a correct answer. I posted my incorrect answer. I am thinking I should redo my work from scratch.
 
@SAJW ibm's watson predicted a world market of about 5 computers.
 
you mean 5 different companies selling them or literally 5 computers on earth?
 
7:28 PM
IQ tests measure something, more or less by tautology.
i do not think it measures what some think that it measures.
 
it measures the ability to prepare for such tests maybe?
i mean if you really train you at least will get 10% more points
like if it would be your life goal to score 150 in an IQ test, surely you could do it
but would that be intelligent?
 
i took one once and they let me skip a grade of elementary school, so i'm not too against them.
saved me some time. :)
 
school was the best time of my life
 
Bob
it was literal 5 to 10 computers in the world
 
i didn't like it very much.
i have a book somewhere from the 80s which has a physical map of the internet. there are maybe 15 sites.
 
7:35 PM
@robjohn , I assume $\frac{f(u,h)}{{(hu)}^m}$. Anyway, have you ever come across little-o with two variables before in other problems? This seems to be very rare.
 
i mean, i could agree to better times of the day for school, because I'm an owl haha
8 am here
 
@leslietownes. Wow. This is amazing to hear
 
@Bob i would not grind through the algebra as you are doing. it is not clear from your question, but you appear to be trying to compute a solution for some fixed $x_k$ and for arbitrary $y_0,...,y_3$. Since the solution is affine in $y$ you could solve the system I wrote for $y=0, (1,0,0,0),...,(0,0,0,1)$ and combine to get a general answer.
 
@schn just to see how another problem deals with it, @robjohn
 
do i need to understand quantum mechanics to be able to program in the distant future?
*quantum math
 
7:40 PM
i need to find that book. i bet that image could go viral on twitter. which is the only thing that is important to me.
 
@SAJW. Program what!!
 
software for clients
 
@SAJW. Not at all! You just need discrete math as far as you can go for all what you see on the web. For image processing, you might need more advanced math like wavelets and differential equations. Same as before goes with machine and deep learning. Anything else can be covered basically in discrete math.
 
i think the short answer to your question is no.
i was composing my useless answer while avra provided a more useful one. :)
 
my current perspective is that quantum computing is job creation.
 
7:46 PM
@leslietownes. You go first, sorry
 
the quantum computing literature is largely a junkyard populated by people who only read and cite papers with 'quantum' in the title.
which is not how you do research.
or shouldn't be, anyway. end rant
 
@leslietownes. Exactly! Some might be fooled that quantum computing is about quantum mechanics!
 
there are some very interesting things happening in that field and talented people working in it, but a lot of it is people riding the wave.
pardon my french.
i'm just mad because i didn't get to ride a wave.
 
8:02 PM
@schn You are trying to generalize where different cases need to be handled separately. Instead of trying to build up a calculus of little-o, it might be more informative to look at places where it is used, and see how it works there.
 
8:31 PM
@leslietownes How about this one?
 
-1 for not including berkeley.
that's what i thought was so cool about the map in the book, the basement of evans hall was on it. i spent a lot of time there.
they should put it on the tour.
 
are longitude and altitude the coordinate system on a sphere?
*sphere's surface
 
they provide one of many possible coordinate systems on the sphere.
wait, altitude?
what is that
different points at the same longitude with the same altitude would mess up that being a coordinate system.
 
sorry
latitude
 
8:46 PM
just curious, what did you think the coordinates are for?
 
I only saw those lines on a globe, never really put thought into it.
 
9:08 PM
It's just spherical coordinates you learn in multivariable calculus, by the way.
 
finding the longitude was a problem for which there was given a royalty by the king of england. you can get the latitude somehow from astrological measurements but not the longitude.
it was done with a reliable clock.
 
The Foucault pendulum recovers the latitude for you. See my diff geo text.
astrological measurements?
 
astronomical measurements.
optics, stars, telescopes and s--t.
i misspoke. but that is what a double scorpio would do.
 
Not to mention a triple.
 
i am a sagittarius rising.
you strike me as a triple scorpio although you probably aren't.
it's all hogwash.
 
9:16 PM
Nope. Allegedly triple Aquarius.
 
i think astrology is BS except everyone who knows anything about it can always guess my signs. and my wife and best friend are double scorpios too.
we're just a hive of scorpions, lashing out at those who have wronged us. or may wrong us.
or the unfortunate toads who try to help us.
 
 
1 hour later…
10:31 PM
dava sobel has a readable book about harrison.
my wife wanted to go on a cruise at some point (not my fave activity), i brought along my student sextant so i could take readings in between sips. managed to meet the ship's navigator, but that stuff is relegated to a 101 navigation course and quickly transplanted by gps, loran and all that modern junk.
 
11:16 PM
@leslietownes SRI (now MSRI) is associated with Berkeley, is it not?
 
Yes.
 
So take back that -1!
:-)
 
I’m losted.
 
@TedShifrin Referring to the comment leslie left on my comment
3 hours ago, by robjohn
@leslietownes How about this one?
 
Hmm, is that SRI connected to MSRI? MSRI came along in the early 80s.
SRI in Menlo Park?
Anyhow, nothing to do with MSRI, which is associated to Berkeley.
 
11:37 PM
was my score bad?
or not so bad
 
Cycling is a very mathematical discipline
the wheels are shaped like circles
 
Unless you go to MoMath, where the wheels are square. :) youtu.be/Q3e0gQdRDW0
 
So then are auto-racing and skiing.
 
that's pretty cool :)
 
11:47 PM
Yeah, it’s a demo of the theorem that given any shape wheel, there is a surface for which the wheel works. But for kids!
In theory, the more complicated the shape, the more the mathematician has an advantage. Everybody gets circles. But the air resistance of cycling is another matter.
 

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