« first day (2789 days earlier)      last day (2233 days later) » 
02:00 - 05:0005:00 - 16:0016:00 - 20:0020:00 - 23:0023:00 - 00:00

Faust
Faust
5:00 AM
oh*
 
orbit-stabilizer
orbit-stabilizer
I'm still not comfortable with integrals for some reason..
 
Faust
Faust
atm im really intrested in most things to do with geometry hyperbolic especially and dif geo i also intrested in algerbra and topology
 
Zee
Zee
I have a set of notes that cover all you need from ODEs in 7 pages of rigoures math if anybody is interested
 
orbit-stabilizer
orbit-stabilizer
after 4 semesters of calculus and 2 real analysis courses lo
lol
 
Faust
Faust
i think derivatives are harder than integrals
integrals are sort of easy cause the hard part is easier
and contour integration is balls easy
 
orbit-stabilizer
orbit-stabilizer
5:02 AM
I don't mean in solving them. I mean philosophically. It just doesn't feel like it should work that way for some reason.
 
Faust
Faust
but i dont calculate many integrals anymore
ah
yeah i have tried exmaplian integrals to 1st years
explaining
im not veyr good at
gimmie abstract algerbra i can explain the pants off that
i dunno i hate analysis
so i avoid calculus cause it just turns into the same thing eventually
 
orbit-stabilizer
orbit-stabilizer
Man.. but sequences are so fun. Chapter 2,3,4,7 are great in Rudin
 
Zee
Zee
Calculus is not the same as analysis
 
Faust
Faust
if you like sequences do number theory
 
Zee
Zee
If fact most analysts don’t even do calculus , that’s more differential geometry
 
orbit-stabilizer
orbit-stabilizer
5:07 AM
I hate number theory
 
Faust
Faust
i see calculus at a highlevel as just a mislabel for analysis but what do i know im a shitty undergrad
1st year nt is boring
 
orbit-stabilizer
orbit-stabilizer
I doubt you're shitty
 
Faust
Faust
but its got alot of cool sequences in it
i still dont like it though but im taking my thrid class on it cause its an easy grade
 
Zee
Zee
Calculus is cool though...
 
Faust
Faust
it is
but i hate analysis
:p
 
Zee
Zee
5:08 AM
Why?
 
Faust
Faust
im autistic too many words
 
Zee
Zee
What pisses you off in particular?
 
Faust
Faust
its not that it pisses me off its just tireing to communicate in that language
i prefer topology where you make a general argument (not a proof) and that sufficeint
 
orbit-stabilizer
orbit-stabilizer
epsilons and deltas?
 
Faust
Faust
no its the level of detail required in the explantions
its very hard for me to communicate at a very strict level like that
i can do it but its very tiring to me
 
Zee
Zee
5:11 AM
Your kinda vague
 
Faust
Faust
actually scored a perfect 100% in my real analysis final
 
Zee
Zee
Don’t sound like you have a problem then...
 
orbit-stabilizer
orbit-stabilizer
wow. well that's good
 
Zee
Zee
Did you do measure theory ?
 
Faust
Faust
sorry its hard to explain my difficulty its associated with my disability its very hard for me to communicate with people in general and that type of math is very tiring to me so i try and avoid it.
not in that class
 
Zee
Zee
5:13 AM
well measure theory is closer to true analysis
Your always gonna do calculus unless you do pure analysis or pure algebra
 
orbit-stabilizer
orbit-stabilizer
But algebra is fine?
 
Faust
Faust
i really like topology though
algebra is fine
 
Zee
Zee
That has calculus in it ...
 
Faust
Faust
but its looser
 
Zee
Zee
Idk what to tell you , except to try measure theory before giving up on analysis , good luck
 
Faust
Faust
5:15 AM
i am planning on taking 435 at my uni im not giving up per se i just doubt ill do a masters in it
 
Joe Shmo
Joe Shmo
@TedShifrin I suspect that's the topological definition -- an m-manifold is locally homeomorphic to R^m at every point
 
Zee
Zee
IDK why you expect me to know what’s 435 but good luck
 
orbit-stabilizer
orbit-stabilizer
I think you can infer that it's measure theory
 
Joe Shmo
Joe Shmo
yes, it's pretentious, but it's so much clearer, and so much more succinct than any analytic definition. Explicit, implicit, parametric, blah blah
 
Faust
Faust
its the class that has measure theory in it
 
Zee
Zee
5:18 AM
Ya I can infer that but it’s kinda funny how you mentioned the specific number of it
 
Faust
Faust
i try my best to enjoy all branchs of mathematics who knows i could end up loving it i rather enjoyed graph theory despite not really liking combintorics
 
Zee
Zee
I suppose it’s a habit from talking to your school mates
 
Faust
Faust
that or im just odd ^^
 
Zee
Zee
Meh, your not that odd
 
Faust
Faust
you a math major?
 
anon
anon
5:20 AM
9 mins ago, by Zee
Your kinda vague
6 mins ago, by Zee
Your always gonna do calculus unless you do pure analysis or pure algebra
 
Daminark
Daminark
I found that my habits for referring to a class by number vs name are kind of all over the place
 
Zee
Zee
Ya , am done with my degree but I barley graduated :p
 
anon
anon
1 min ago, by Zee
Meh, your not that odd
 
Daminark
Daminark
(This is when I'm talking to friends)
 
orbit-stabilizer
orbit-stabilizer
what's with the quotes @anon
 
anon
anon
5:21 AM
it's you're
 
orbit-stabilizer
orbit-stabilizer
you're?
oh
 
anon
anon
lol
 
Zee
Zee
I sense a spammer
 
orbit-stabilizer
orbit-stabilizer
anon is cool
 
Faust
Faust
@Zee i have found that most people in mathematics have a skewed view of who is in fact "odd" but i'll take it anyway
 
orbit-stabilizer
orbit-stabilizer
5:22 AM
There are as many odd people as there are even :P
 
Zee
Zee
I think in math as long as you are producing result , your not odd until your starving yourself to death
Aka Godel level crazy
 
anon
anon
that was on purpose
 
Faust
Faust
show me the one to one correspondence :P
@Zee you trying to find a job in the industry or going to go back to grad school?
 
Zee
Zee
Am taking graduate courses now but am not enrolled in the school , I may have to sell my soul to industry in the near future but am fighting that
 
Faust
Faust
kool
not the soul part
but the other bit
 
orbit-stabilizer
orbit-stabilizer
5:26 AM
Hahahaha. Selling your soul to academia is much worse
 
Zee
Zee
I don’t think so
Academia sucks but atleast your doing something you like
 
Faust
Faust
i dunno im not particularly good at math but i have an eidetic memory and im autistic so its actually amusing to sit there and just do math all day so im hoping to go to grad school.
thought eithier way id like to keep my soul. O.o
 
Zee
Zee
Good memory is useful for algebra , maybe that’s why you like it
 
orbit-stabilizer
orbit-stabilizer
How is it having eidetic memory
Must make math a lot easier
 
Faust
Faust
i actually struggled really hard with algebra at first but now im rather good at it ^^ my first course almost killed me though.
 
Zee
Zee
5:29 AM
I bet
 
Faust
Faust
Surprising not as useful as youd think. yeah i can recall the relatvent theorem but if i can't understand what it says its not really all that useful to know it...
 
orbit-stabilizer
orbit-stabilizer
Can you recall the proofs of the theorems?
 
Zee
Zee
ohhhh
 
Faust
Faust
if look at the proofs yes
 
orbit-stabilizer
orbit-stabilizer
oh wow
 
Zee
Zee
5:31 AM
Jesus that’s awesome
 
Faust
Faust
but that can be rathe rtime consuming to figure out ina test
 
orbit-stabilizer
orbit-stabilizer
that's nice
 
Faust
Faust
its better to learn how to prove it
then jsut memorize the proof
i know those may sound like the same thing but there not really
 
Zee
Zee
Proofs are useless , I think the only place for proofs in math is in research papers
 
Faust
Faust
but yeah if i am feeling really lazy and i dont want to figure out how to do something i can just look at a ton of questions and the solutions and get by pretty well even in math classes
though i literally have no idea what i am doing
so i try not to do that ^^
 
orbit-stabilizer
orbit-stabilizer
5:35 AM
fascinating ...
does it work well?
 
Zee
Zee
See above , (100 on analysis test)
 
Faust
Faust
i try and learn most of what im doing but at the very least combination of the two works really well i like textbook with lots of worked out examples or problems with solutions a prof can only ask a question in so many different ways so i memorize all the different ways its just a different value or letter but a concept i already know being asked in a different way
i did rather poorly in mathimatics at first for several reason ( i have alot of other learning disabitlies as well) but have a gpa of 9 for the last 3 semesters
 
Zee
Zee
I work in the polar opposite way but am not doing so well so maybe your way is the right path...
 
Faust
Faust
i think its better to learn how to solve the problem on your own
and certainly better for reasearch
if im mass memorizing stuff thats completely useless for research
 
Zee
Zee
My dream is to be known as the dumbest genius to ever live
 
Faust
Faust
5:40 AM
no point memorizing what everyone else has already done need to think critically and come up with your own ideas i think thats much more important
it took me till grade 6 to be able to read at a grade 1 level and was completely illiterate until grade 5
 
Zee
Zee
You can read ? I still haven’t learned to do so
 
Faust
Faust
language is really hard for me ^^
 
orbit-stabilizer
orbit-stabilizer
That's rough :/
 
Faust
Faust
(apparently english is particularly bad for my brain were other languages would be easier for me to understand)
 
Zee
Zee
you should try learning Chinese , don’t they need to memorize thousands of characters
 
Faust
Faust
5:44 AM
even at university we had a entrance exam for English placement i scored so badly that it said i had to register for english as a second language classes at university but because my transcript form highschool was from canada the universitys system wouldnt let me register for them
when i went to the office to get it sorted out the lady said thats not possible you couldn't of scored a 1 out of 6 you can speak english
sure enough i did, she had to manually overide it so i could register for normal people english
<-- somehow ESL without a first language lol
 
Zee
Zee
I remember when I enrolled in college , I didn’t wanna take English as second language since it’s harder than normal English , so I told the lady (in a thick accent) I only speak English , she stared at me then said “OK whatever floats your boat”
And gave me permission
Screw you guys , am going home
 
Faust
Faust
?
im not really sure who your talking to but didnt mean to offend you
well i am going sleep
was nice chatting with you guys enjoy your mathematics ^^
 
Balarka Sen
Balarka Sen
6:23 AM
Is this the future of power rangers?
Because if it is HELL YEAH I'm watchin' it.
 
orbit-stabilizer
orbit-stabilizer
shrugs
 
Zee
Zee
6:49 AM
You don’t get the things you dream of , you get the things you work hard for
 
Niing
Niing
7:30 AM
Hello
My friend asked me a question about modulo arithmetic: Is it possible that $a^k \mid b^k\ (\textrm{mod}\ m)$ implies $a \mid b\ (\textrm{mod}\ m)$?
So I'm thinking about the case $k=2$... Any idea is appreciated!
Give me a ping because I'm leaving
 
Secret
Secret
7:58 AM
$a^k \mid b^k\ (\textrm{mod}\ m) \implies \exists n\forall r [b^k = na^k + rm]$
$a \mid b\ (\textrm{mod}\ m) \implies \exists s\forall r [b = sa + rm]$
$b^k = na^k + rm \implies (sa + rm)^k = na^k + rm \implies s^ka^k + m (1 + (sa)^{k-1}r + (sa)^{k-2}r^2m + \cdots + r^km^{k-1}) = na^k + rm \implies s^ka^k + m\Bbb{Z} = na^k + m\Bbb{Z} \implies s^k = n$
So the implication is true only when given $b=sa$, $b^k = na^k$, that $s^k=n$
I am not sure about more generic finite fields, however
also typo, I forgot the binomial coefficients, but the conclusion remains unaffected
and also typo, there should be no 1 +
 
Tanuj
Tanuj
8:14 AM
@Secret @Semiclassical Hey guys ! :)
Guys I'm trying to evaluate$$\int\limits_\frac{-\pi}{2}^\frac{\pi}{2}\left({x^2 +\ln\left|\frac{\pi+x}{\pi-x}\right|}\right)\cos(x) dx $$
I know $$\int\limits_\frac{-\pi}{2}^\frac{\pi}{2}\ln\left|\frac{\pi+x}{\pi-x}\right|\cos(x) dx$$
will be equal to $0$
So , I'm eventually left with $$\int\limits_\frac{-\pi}{2}^\frac{\pi}{2}x^2\cos(x)dx $$ , however , when I try to solve it using integration by parts , it basically never ends up integrating and I have to apply by parts numerous times. What am I doing wrong here ?
 
Secret
Secret
The integrals of the form:
$$\int x^n (\sin \text{or } \cos) x dx$$
does require a reduction formula, which is basically integrating by parts many times
 
Tanuj
Tanuj
8:31 AM
@Secret okay , sounds tedious.
 
Balarka Sen
Balarka Sen
@Tanuj Denote the integral by $I$. Consider the substitution $x \mapsto -x$
 
Secret
Secret
ah, I forgot symmetry considerations (I am super weak at that)
 
Tanuj
Tanuj
@BalarkaSen but why ? I see it has no effect on the integral , but how would I notice that ?
 
Balarka Sen
Balarka Sen
@Tanuj It has an effect on the integral though
The integrand changes from $(x^2 + \ln \text{garbage})\cos(x)$ to $(x^2 - \ln \text{garbage})\cos(x)$
 
Tanuj
Tanuj
@BalarkaSen yea , I mean it has no effect on the expression we are trying to integrate , is this the reason you're doing this ?
 
Balarka Sen
Balarka Sen
8:36 AM
It does have an effect on the expression you're trying to integrate.
 
Tanuj
Tanuj
@BalarkaSen how ? Oh you're doing it as a first step okay
@BalarkaSen okay
 
Balarka Sen
Balarka Sen
I think you get $\displaystyle I = \int_{-\pi/2}^{\pi/2} \left [ x^2 + \ln \left | \frac{\pi + x}{\pi - x} \right | \right ] \cos(x) dx = \int_{-\pi/2}^{\pi/2} \left [ x^2 - \ln \left | \frac{\pi + x}{\pi - x} \right | \right ] \cos(x) dx$
 
Tanuj
Tanuj
@BalarkaSen yea
 
Balarka Sen
Balarka Sen
So $\displaystyle 2I = \int_{-\pi/2}^{\pi/2} 2x^2 \cos(x) dx$
 
Tanuj
Tanuj
@BalarkaSen yea
brb .
 
Balarka Sen
Balarka Sen
8:40 AM
Oh I guess you already figured out that $\displaystyle I = \int_{-\pi/2}^{\pi/2} x^2 \cos(x) dx$ by splitting the integral up and using that the second half is an odd function.
That integral is easy to integrate using parts, if you use it correctly. Try $u = x^2$ and $dv = \cos(x) dx$.
You have to use it twice.
 
Secret
Secret
Btw, the reason of setting $u$ as the polynomial because we are exploiting the property that $\frac{d^n}{dx^n}$(deg(n) polynomial) = constant
 
Balarka Sen
Balarka Sen
$\displaystyle \int x^2 \cos(x) dx = x^2 \sin(x) - 2 \int x \sin(x) dx$, and now $\displaystyle \int x \sin(x) dx = - x \cos(x) + \int \cos(x) dx = -x \cos(x) + \sin(x)$ if I am not mistaken
 
Secret
Secret
looks fine
 
Balarka Sen
Balarka Sen
So $\int x^2 \cos(x) dx = x^2 \sin(x) - 2 \sin(x) + 2x \cos(x)$.
Or at least, that's the antiderivative, upto constant of integration
 
Secret
Secret
(unrelated) Sometimes I am wondering something more general:
Consider a generic integrable function $f$
$$\int f(x) dx$$
We are interested in finding the set of all substitutions $x \mapsto g(x)$ with the following property:
 
Balarka Sen
Balarka Sen
8:48 AM
@Tanuj So anyway, denote the above antiderivative by $F(x)$. Then your answer is $F(\pi/2) - F(-\pi/2)$ which turns out to be $\displaystyle\frac{\pi^2}{2} - 4$ if I am not wrong
 
Secret
Secret
$$\int f(x) dx = \lambda \int f(g(x))g'(x)dx$$
for some real number $\lambda$
or more generally the set of all symmetries such that:
$x \mapsto g(x)$ gives:
$$\int f(g(x)) g'(x)dx = h\left( \int f(x)dx\right)$$
where $h$ is some known and easy to manage function (e.g. linear, polynomial etc.) with the arguments of the original integral
 
Alessandro Codenotti
Alessandro Codenotti
@Balarka @Dami are you still interested in number theory?
 
Secret
Secret
I wonder if there is a way to find all symmetry transformations given any integrand...?
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti I am, but unfortunately I won't have time until 6th of next month :)
I will very actively be doing number theory after 6th though
Got those admission exams to prepare for
 
Alessandro Codenotti
Alessandro Codenotti
Would you be interesting in reading something about algebraic number theory together?
I'm taking an ANT course in uni this semester, but I'd like to see more than what we have in the syllabus
 
Secret
Secret
8:59 AM
$$f(g(x))g'(x) = h' (I) f(x)$$
grrrr, is there a way to free up the g
hmm...
$A(g)B(g)=h'(I)f$
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti I'll be doing more elementary-style stuff, but sure, occasional drifts to algebraic number theory will be interesting
I will likely not be doing anything fancy though
 
Tanuj
Tanuj
@Secret @BalarkaSen Thanks ! I totally messed up the integration by parts .
 
Alessandro Codenotti
Alessandro Codenotti
We did elementary stuff until now, but we'll start with the algebraic part of the course next week. I'm trying to prove that there are infinitely many primes of the form $8k-1$ right now actually
 
Balarka Sen
Balarka Sen
There's a concrete book I have by Pollard and Diamond on algebraic number theory which goes up to the proof of FLT for regular primes, starting ground-up very concretely. I might look into it
 
Secret
Secret
$((f \circ g) \cdot x) \cdot ((D \circ g) \cdot x) = ((D \circ h) \cdot I) \cdot f \cdot x$
ugh, how am I supposed to invert this...
 
Balarka Sen
Balarka Sen
9:11 AM
@Alessandro Fair, I doubt I'd be getting into the deep end though. So unlikely I'll spend time on any of the off-syllabus material you'd consider.
Mostly I don't want to theorize for the next month at all
 
Secret
Secret
$(Ag)(Bg) = (hI)f$
$(Bg) = (Ag)^{\leftarrow}(hI)f$
$g = B^{\leftarrow}((Ag)^{\leftarrow}(hI)f)$
.....
$(A \cdot B)*g = (Ag)*(Bg)$ hmm...
so if $A=f$, $B=\frac{d}{dx}$, what is $A \cdot B$...
 
Secret
Secret
9:28 AM
wait a sec... how about:
$$A(g)B(g) = h(I)f$$
$$g = (A^{\leftarrow}(\cdot)B^{\leftarrow}(\cdot))(h(I)f)$$
$$g = A^{\leftarrow}(h(I)f)B^{\leftarrow}(h(I)f)$$
Let $A, B$ as above. Then:
$$g = f^{\leftarrow}(h(I)f)\int h(I)f$$
and therefore:
$$g(x) = f^{\leftarrow}(h\left(\int f(x) dx\right)f(x))\int h\left(\int f(x) dx\right)f(x)dx$$
actually no, I cannot assume the preimage of $A()B()$ to be of the form $A^{\leftarrow}()B^{\leftarrow}()$, for example: $xe^x=y \implies x=W(y)$
hmm...
Try something simpler...
$$A(g)B(g)=X$$
Let $C$ be the preimage of $A()B()$ such that $C(A(g)B(g))=g$. Then $g = C(X)$
Now letting $A=f, B=\frac{d}{dx}$ we have the following functional equation:
$f(g)g'=X \implies F(g)=\int X \implies g = F^{\leftarrow} (\int X)$
So:
$$C = \left(\left(\int f () d()\right)^{\leftarrow}\circ \int \right)$$
And therefore:
$$g(x) = \left(\left(\int f () d()\right)^{\leftarrow}\circ \int \right) \left(h \left( \int f(x)dx \right)f(x)\right)$$
$$g(x) = F^{\leftarrow}\left( \int h F(x)f(x) dx\right)$$
but this is not useful because it is $F$ we want to calculate
hmmmmmmm...
wait a sec..
Given we want:
$$\int f(g(x)) g'(x)dx = h\left( \int f(x)dx\right)$$
we did found the relevant $g$ such that the above is true
Let $F = \int f$, thus we have:
$$F(x) = \int f(g(x))g'(x)dx = h(F) \implies (\text{id}-h)(F(x))=0$$
and thus:
$$F(x) = (\text{id}-h)^{-1}(0)$$
uh..., that does not make sense. Grr this is getting too long, going back to paper for now before writing the conclusion...
 
Niing
Niing
10:53 AM
@Secret Thank you
 
Secret
Secret
(Unrelated) Proposition: We are interested in finding the relation satisfied by all substitutions $x\mapsto g(x)$ such that the following holds:
 
Niing
Niing
A question about mathematical induction: in the induction step, if $P(k)$ then $P(k+1)$, when $k\ge b$. So when I want to prove this is true, should I assume both $k\ge b \land P(k)$?
 
Secret
Secret
Yes, you need to ensure $P(k)$ is true for all $k \geq b$ before you can use $P(k)\rightarrow P(k+1)$
 
Niing
Niing
I got confused when reading strong induction proof, so I check the idea of MI...
 
Secret
Secret
this ultimately means that $P(b)$ has to be proven true
So in an induction, we must prove that $P(b)$ is true, and then prove the inductive case (which will start from $b$, thus for some $k \leq b$, it will be true and thus the only step left is to prove the inductive case $k \to k+1$)
 
Secret
Secret
11:03 AM
(cont.)
$$\int f(x)dx \mapsto \int [f\circ g](x) g'(x)dx = h \left(\int f(x) dx\right) $$ for some known and nice functions $h$
Let $\int f(x) dx = F(x)$, and let $F^{\leftarrow}$ be the preimage of $F$, thus we have:
$$g(x) = [F^{\leftarrow} \circ h \circ F](x)$$
Differentiate both sides and using $a^{\leftarrow '}(x) = \frac{1}{[a' \circ a^{\leftarrow}](x)}$ to get:
$$g'(x) = \frac{[h' \circ F](x)f(x)}{[f \circ F^{\leftarrow} \circ h \circ F](x)}$$
and therefore:
$$\int [f\circ F^{\leftarrow} \circ h \circ F](x) \frac{[h' \circ F](x)f(x)}{[f\circ F^{\leftarrow} \circ h \circ F](x)}dx = [h \circ F](x)$$
 
user342146
11:30 AM
can someone give me more opinion about this?
 
user342146
4
Q: Can we write it as $A$ is closed under the action of $T$?

Baby ElephantI have a doubt in writing a math proof: Let $T: X \to X$ is continuous map and $A\subseteq X$. A set $A$ is $T$ invariant. Can we write it as $A$ is closed under the action of $T$? I am asking it because sometime we write in linear algebra that a subspace $A$ is closed under scalar multiplicatio...

proof-writing
 
Secret
Secret
Grrr, of course it does not work: Indefinite integrals HAVE NO GLOBAL SYMMETRY UNDER AN ARBITRARY SUBSTITUTION in general!
So the actual proposition that needs to be investigated is the following:
$$\int_a^b f(x)dx \mapsto \int_{g(a)}^{g(b)} [f\circ g](x) g'(x)dx = h \left(\int_a^b f(x) dx\right) + k(x)\mid_{a}^b$$
For those who are confused on what I am doing, I am trying to generalise the following problem solving method of the following integral:
$$\int_0^{\pi} \frac{x \sin x}{1+ \cos^2 x}dx$$
which one method of solving it is to use the subsitution $x \mapsto \pi - u$ which causes the integral to become something of the form:
2*Original integral = some easy integral
Statement: Let $f,g,h,k$ be integrable functions, with $K(x) = \int_a^b k(x) dx$ and $h$ given, with $h$ being some "nice" function. Then find $g$ such that when the substitution $x \mapsto g(x)$ is made on the integral $F(x) = \int f(x) dx$, the following holds:
$$\int_a^b f(x) dx = \int_{g(a)}^{g(b)} [f\circ g](x) g'(x)dx = h \left(\int_a^b f(x) dx\right)+ K(x)\mid_a^b$$
To approach this, we first consider the more general indefinite integral for some variable $u$:
$$\int_a^u f(x) dx = \int_{g(a)}^{g(u)} [f\circ g](x) g'(x)dx = h \left(\int_a^u f(x) dx\right)+ K(x)\mid_a^u$$
$$F(u) - F(a) = [F \circ g](u) - [F \circ g](a) = [h \circ F](u) + K(u) - K(a)$$
 
Silent
Silent
12:02 PM
Suppose $a,b,c$ integers and $c\ne 0$, and $c$ divides $ab$. If $p$ is a prime that does not divide $a$ nor $b$, then how to show that $p$ does not divide $\dfrac{ab}c$?
 
Secret
Secret
$p \not\mid a \land p \not\mid b \implies p \not\mid ab \implies p \not\mid nc \implies p \not\mid n=\frac{ab}{c}$
 
Secret
Secret
12:33 PM
typo
$$\int_a^u f(x) dx = \int_{g(a)}^{g(u)} [f\circ g](x) g'(x)dx = h \left(\int_a^u f(x) dx\right)+ K(x)\mid_{g(a)}^{g(u)}$$
After rearranging, the required $g$ is then obtained as follows:
$$g(u) = F^{\leftarrow}([h \circ F](u)+K(u)+[F\circ g](a) - [K\circ g](a))$$
and thus we have:
$$[\text{id} - h] \circ F(x) = [K\circ F^{\leftarrow}]([h \circ F](u)+K(u)+[F\circ g](a) - [K\circ g](a)) - [K \circ F^{\leftarrow}]([h \circ F](a)+K(a)+[F\circ g](a) - [K\circ g](a))$$
 
Secret
Secret
12:54 PM
Now$F^{\leftarrow}$ is found by considering the following:
$$F(g(u)) = y \implies \int f(g(u))d(g(u)) = y \implies f(g(u)) = \frac{dy}{d(g(u))} \implies g(u) = f^{\leftarrow} (\frac{dy}{d(g(u))})$$
Thus:
$$F^{\leftarrow} = f^{\leftarrow} \circ \frac{dy}{d(g(u))}$$
and $\frac{dy}{d(g(u))} = y' \frac{1}{[g'\circ g](u)}$
 
ÍgjøgnumMeg
ÍgjøgnumMeg
If $M$ is a torsion-free $R$-module (with $R$ a pid) then does the structure theorem automatically imply that $M$ is a free $R$-module? That is, by the structure theorem, $M \cong R^f$ for some $f$ so $M$ is free.... is there no subtlety to this that I'm missing?
 
Secret
Secret
1:27 PM
and thus:
$$[f \circ g](u) [g' \circ g^{\leftarrow}](u) = [h' \circ F](u)f(u)+[k \circ g](u)g'(u)$$
$$[f \circ g](u) [g' \circ g^{\leftarrow}](u)-[k \circ g](u)g'(u) = [h' \circ F](u)f(u)$$
 
Rolling Stones
Rolling Stones
I have heard that proof by contradiction isn't acceptable by some logicians and mathematicians but I would ask why such an issue is challenging?
 
Secret
Secret

♦  Logic

This room is meant for discussion about logic, including found...
1
 
Rolling Stones
Rolling Stones
@Secret Thanks!
 
Secret
Secret
(cont.)
 
Lozansky
Lozansky
$\int_{-1}^{1} (x^2-1)^n dx =...?$
 
Secret
Secret
1:42 PM
try $x^2-1=u$ as substitution and then you should get some arcsin integrals?
 
Secret
Secret
2:11 PM
---
grrr, I don't understand why I cannot isolate the g without reference to $F$
Let's abstract from this:
$\int_0^{\pi} \frac{x \sin x}{1+ \cos^2 x}dx$
$x=\pi - u$
$\int_0^{\pi} \frac{x \sin x}{1+ \cos^2 x}dx = -\int_{\pi}^{0} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du = \int_{0}^{\pi} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du$
$= \int_{0}^{\pi} \frac{- u \sin (-u)}{1+ \cos^2 (- u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du = -\int_{0}^{\pi} \frac{u \sin (u)}{1+ \cos^2 (u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du$
 
Secret
Secret
2:32 PM
hmm...
maybe I should solve for $k$ instead of $g$...
 
Secret
Secret
2:50 PM
grrrr... why I cannot seemed to generalise:
$$\int_0^{\pi} \frac{x \sin x}{1+ \cos^2 x}dx = -\int_{\pi}^{0} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du = \int_{0}^{\pi} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du = \int_{0}^{\pi} \frac{- u \sin (-u)}{1+ \cos^2 (- u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du = -\int_{0}^{\pi} \frac{u \sin (u)}{1+ \cos^2 (u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du$$
with the more general case:
$$\int_a^u f(x) dx = \int_{g(a)}^{g(u)} [f\circ g](x) g'(x)dx = h \left(\int_a^u f(x) dx\right)+ K(x)\mid_{g(a)}^{g(u)}$$
I am not even assuming the integral $F(x)$ to have a closed form, why is it not working
Fine, if generic substitutions $g$ is causing problem, I am going linear!
Attempt 9:
Let $x=au+b$. Then $dx=adu$
$$\int_c^y f(x) dx = \int_{\frac{c-b}{a}}^{\frac{y-b}{a}} f(au+b) adu = h \left(\int_c^y f(x) dx\right)+ K(x)\mid_{\frac{c-b}{a}}^{\frac{y-b}{a}}$$
ok, so $K$ is a relation of $F, f, g$ thus there is no escape
I need to think of a good way to write a generic integrand $f$. Clearly, it does not stop at $f(x)$
A generic integrand has arbitrary number and permutation of composition, pointwise multiplication and addition
 
Secret
Secret
3:57 PM
$$f(x) = \left(\mathop{Op}_{i \in S} g_i\right)(x)$$
where $S$ is a well ordered set consists of $\circ, \cdot,h_{i}()$. Thus $\mathop{Op}_{i \in S}$ expresses the set of all functions formed by finite instance of composition and pointwise multiplication
 
02:00 - 05:0005:00 - 16:0016:00 - 20:0020:00 - 23:0023:00 - 00:00

« first day (2789 days earlier)      last day (2233 days later) »