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Akiva Weinberger
Akiva Weinberger
01:52
@TedShifrin So I'm planning on deferring a semester this fall
I'm not sure what to do in the meantime, but my plan is to email some professors and ask if I could intern with them or help with a research project
user1787812
user1787812
02:31
I need a clarification on the definition of multiplicative group as mentioned in this comment
Ted Shifrin
Ted Shifrin
02:53
@AkivaWeinberger Yeah, probably smart, although who knows how long this will go on. You're a little "young" for research, unless you have something in mind. What field(s) have you decided you're interested in pursuing for now?
@user1787812 You take the units in the ring (elements that have a multiplicative inverse). Those form a group under multiplication.
user1787812
user1787812
yes... but, why cannot it be {1,2,3,4,5,6,7,8,9,10,11}? Because this presentation proves that {1,2,3,4,5,6} is multiplicative group for Z7
Ted Shifrin
Ted Shifrin
Huh?
Yes, but $7$ is prime and $12$ is not!!
In $\Bbb Z_{12}$, for example, $\bar 2\cdot\bar 6 = \bar 0$, so neither $\bar 2$ nor $\bar 6$ has a multiplicative inverse.
You need to learn basics.
user1787812
user1787812
Got it...
 
2 hours later…
user76284
user76284
04:42
What’s the powerclass of the class of ordinals?
Thorgott
Thorgott
04:53
big
user21820
user21820
@Thorgott No, it doesn't even exist in NBG or MK.
Knight
Knight
@robjohn I think $x(0)= x_0$ and $x(t)$ is the value of $x$ at some $t$
robjohn
robjohn
05:47
@Knight if the equation is $\frac{\mathrm{d}x}{\mathrm{d}t}=f(x)$, then the equilibrium is stable when $f\!\left(x^\ast\right)=0\implies f'\!\left(x^\ast\right)\lt0$
There is a similar condition on the eigenvalues of the Jacobian in higher dimensions
the equilibrium is stable if all the eigenvalues have negative real part.
Threnody
Threnody
06:20
Would you guys classify Kuratwoski's Theorem for planar graphs as "Algebraic Graph Theory" or "Topological Graph Theory"?
Furthermore, would you classify Euler's Characteristic for Orientable Surfaces as "Algebraic Graph Theory" or "Topological Graph Theory"?
Thorgott
Thorgott
06:30
neither and neither
Kuratowski's theorem is pure graph theory and orientable surfaces aren't graph theory
Edward Evans
Edward Evans
waddup
Threnody
Threnody
So "Topological and Algebraic Graph Theory" makes no sense?
Thorgott
Thorgott
@Edward I slept for 2.5h and have an exam soon
how do I not die
Edward Evans
Edward Evans
You just have to think hard enough about not dying and you'll become immortal
what exam do you have?
(also just replace any liquid intake with red bull and coffee)
Thorgott
Thorgott
topology/manifolds
I don't drink coffee, but I'm on strong black tea
Edward Evans
Edward Evans
06:40
Oh nice, viel Erfolg!
I only have one exam this semester, which is just Algebra 2
 
2 hours later…
ronak jain
ronak jain
09:07
Hi everyone
Can someone tell me if in definite integral I have an upper limit of a point such that the integral is not defined there . Than what should I do ? It is given that the limit of the function at that point exist but the function is getting a undefined form at that point ?
maths student
maths student
09:23
f f:RtoR

and fn(x)=f(x/n)→0 uniformly on R

proof that f is a bounded function
uniformly continuous function is bounded since it is Lipschitz continuous.
Then how to tackle this problem.
Leaky Nun
Leaky Nun
@ronakjain then you split the integral at that point into two integrals
Thorgott
Thorgott
When the professor and assessor suddenly start discussing what the correct definition of a sheaf is during your oral examination
ronak jain
ronak jain
09:50
@LeakyNun can you explain that a bit more ?
Leaky Nun
Leaky Nun
can you provide an example?
ronak jain
ronak jain
Wait a minute....
user image
@LeakyNun look at the problem. If I put the lower limit to be 0 than f(x) gets undefined as f(x) is inversely proportional to x^2
Leaky Nun
Leaky Nun
that's not what you said
anyway
I don't understand what the question is
ronak jain
ronak jain
10:22
@LeakyNun when you will put the limit = 0 in right side . You will put x = 0 and the f(x) = 1/x^2 * integral. So the value after putting limit will become undefined.
ronak jain
ronak jain
11:05
@LeakyNun are you here ? Did you get the question now ?
Nisarg Jain
Nisarg Jain
11:32
anyone preparing for IITJAM?
 
2 hours later…
Fargle
Fargle
13:05
@BalarkaSen you definitely should, Things Fall Apart is one of the best novels I've ever read
Balarka Sen
Balarka Sen
13:19
@Fargle my English literature teacher in high school recommended it to me once
I'll give it a go. Almost 1/3 through The Possessed by the way
How do I read a novel where all the characters are despicable scums?
Fargle
Fargle
eagerly
Balarka Sen
Balarka Sen
lmfao
hash man
hash man
14:05
Hi
I try to decipher this:
1
A: Why does the residue theorem fail in the calculation of $\int _\gamma\frac{zdz}{1+z+z^4}$?

ConradNote that the lowest absolute value for points on the rectangle is $\sqrt 2$, so the interior of the rectangle contains the disc of radius $\sqrt 2$ centered at the origin; but $(\sqrt 2)^4=4 > \sqrt 2 +1$ so all the roots of the equation $z^4+z+1=0$ are inside the rectangle. Since $\gamma$ conta...

can someone explain me why he was able to take infinite radius?
Thorgott
Thorgott
14:17
cause $\int_{\gamma}=\int_R$ for all large enough $R$, hence $\int_{\gamma}=\lim_{R\rightarrow\infty}\int_R$
hash man
hash man
14:47
why is this true @Thorgott
Thorgott
Thorgott
because the limit of a constant sequence is that constant
hash man
hash man
I can't understand why for large enough r the integral is 0
@Thorgott
I have like 4 poles.
Thorgott
Thorgott
that follows from the estimate in the post
hash man
hash man
I really can't see why when $1+z+z^4$=0 isn't a pole
Thorgott
Thorgott
I don't see that being claimed anywhere
those points definitely are poles
hash man
hash man
14:59
if so what will happen if I'll try to use the residue theorem inside the rectangle?
Thorgott
Thorgott
it would tell you what the integral evaluates to
hash man
hash man
what is the name of this result which I can take the external domain instead of the inner domain?
so you tell me if I'll take the inner domain the result will be 0?
Thorgott
Thorgott
it follows from Cauchy's integral theorem
the point is that there are no poles on the difference between a large enough circle and the rectangle
hash man
hash man
this is beautiful thanks
 
2 hours later…
Vio Ariton
Vio Ariton
16:45
I've to find the sum of all the 100 < $x^{\circ}$ < 200, $cos^{3}(5x) + cos^{3}(3x) = 8cos^{3}(4x)cos^{3}(x)$ - any hints? I got it into the form $cos(5x)cos(3x) = 0$ and I was thinking to find the first x intercept > 100 and the last < 200 of $cos[5(\frac{pi}{10} + \frac{2\pi}{10})]$ and $cos[3(\frac{pi}{6} + \frac{2\pi}{6})]$ and then maybe I could telescope the series somehow
 
1 hour later…
Vio Ariton
Vio Ariton
17:46
I didn't understand this one "An angle x is chosen at random from the interval $0 < x < \frac{\pi}{2}$. What is the probability that $sin^{2}(x)$, $cos^{2}(x)$ and $cos(x)sin(x)$ are not the lengths of the sides of a triangle." So what's the probability that choosing an x, the lengths of the right angle $a$, $b$ and $c$ aren't equal to sin, cos and sincos? or sin != a, cos != b, sincos != c?
user434058
Hi! Could anybody explain me the comments by zwim in this comment thread?
Ted Shifrin
Ted Shifrin
@FakeMod: The problem is that the indefinite integral is not a function, so it makes no sense to differentiate it with respect to the parameter. The constant of integration could very well be a function of the parameter as well. I think it's hopeless. You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative.
user434058
@TedShifrin Hmmm... I understand your second sentence, however, I don't see why the final expression of the indefinite integral won't be a function of the parameter we introduced in the integrand?
Ted Shifrin
Ted Shifrin
It's not a well-defined function. The indefinite integral is not a function. It's a symbol for "any antiderivative."
user434058
@TedShifrin Oh, yeah. So except the arbitrary integration constant, the rest is a well characterised function in the parameter that we introduced, and the integration variable, right?
Ted Shifrin
Ted Shifrin
18:00
But the integration constant is not actually a constant, as I already said. That's the fallacy. It may vary and be an arbitrary function of your parameter $b$.
user434058
@TedShifrin "You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative." But how do I get any closer to using Feynman's technique after doing that?
user434058
@TedShifrin Oh! I see.
Ted Shifrin
Ted Shifrin
I don't know why Feynman's name gets attached to this, incidentally. That's totally bizarre.
If it were a Feynman path integral, OK.
user434058
@TedShifrin Prolly, 'cause he thought of it first (I may be terribly wrong)
Ted Shifrin
Ted Shifrin
That's totally not true.
It was around in mathematics and in physics long before him.
user434058
18:04
@TedShifrin Then the only reason I can think of is that he popularized it.
Ted Shifrin
Ted Shifrin
Nah. I've never heard of this before.
user434058
@TedShifrin Anyways, I still don't know why you said ths \/
user434058
5 mins ago, by FakeMod
@TedShifrin "You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative." But how do I get any closer to using Feynman's technique after doing that?
Ted Shifrin
Ted Shifrin
If you consider $F(x,b) = \int_a^x g(t,b)\,dt$, then $\partial F/\partial b = \int_a^x \partial g/\partial b\,dt$ is valid.
user434058
@TedShifrin That's alright, but does it get me any closer to evaluating the integral this way?
Balarka Sen
Balarka Sen
18:07
I think the story is Feynman liked to say to people that no matter how complicated an integral a mathematician can give to him he can always do it very fast, and differentiation under the integral sign with respect to a different parameter thing was a trick upon his sleeve
Ted Shifrin
Ted Shifrin
I have no idea if it gets you closer to evaluating the integral. The differentiation under the integral trick works sometimes and doesn't work others.
user434058
@TedShifrin So any chances, whatsoever, of it working with indefinite integrals transformed to indefinite integrals?
Ted Shifrin
Ted Shifrin
No.
Charlie Shuffler
Charlie Shuffler
Hi all
Ted Shifrin
Ted Shifrin
I explained why it's nonsensical.
user434058
18:10
@TedShifrin Alright, thanks a lot! That makes sense :-)
Charlie Shuffler
Charlie Shuffler
is there an elegant name for the integration "constant" that comes up when taking the indefinite integral on the left and right hand side of an equation of the form $\frac{\partial f(x,y)}{\partial x} = c_1$, with $c_1$ a constant?
user434058
So, now what should I do to my question. Let it stay opened, even though it has been satifactorily answered?
Charlie Shuffler
Charlie Shuffler
the integration "constant" in such a case would actually be a function of y
Ted Shifrin
Ted Shifrin
It's not a constant, @CharlieShuffler. It's a function of $y$.
Aha.
No, there's no name for it.
Charlie Shuffler
Charlie Shuffler
Any clean explanation for why it appears?
I can't word it properly
Since its not a constant I struggle formulating an explanation
Ted Shifrin
Ted Shifrin
18:14
This is in every multivariable calculus book. You hold $y$ fixed when you do your "integration with respect to $x$"; therefore, the constant of integration depends on $y$.
This is very similar to the thing that @FakeMod was ignoring :P
Charlie Shuffler
Charlie Shuffler
Hmm fair enough, I think I can work with that explanation
Thanks !
user434058
:D
Ted Shifrin
Ted Shifrin
If you look at it this way, @CharlieShuffler, $\int_a^x g(t,y)\,dt$ is a function of $x$ AND $y$. (Here $g$ is your partial derivative.)
Charlie Shuffler
Charlie Shuffler
Interesting
Makes sense
 
4 hours later…
Artificial Stupidity
Artificial Stupidity
21:48
0
Q: Can Venn's diagram be used to represent the number of elements of sets?

Artificial StupidityConsider the following question: A school consists of 10 students. Five students love M (mathematics). Two students love both M and P (mathematics and physics). Three students love neither M nor P. I teach my students with the following diagram. Here the students in the school are denoted by a...

set-theory

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